3.311 \(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=187 \[ -\frac {2 a^3 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {\left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {B \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

[Out]

1/2*(2*a^2+b^2)*(A*b-B*a)*arctanh(sin(d*x+c))/b^4/d-2*a^3*(A*b-B*a)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+
b)^(1/2))/b^4/d/(a-b)^(1/2)/(a+b)^(1/2)-1/3*(3*A*a*b-3*B*a^2-2*B*b^2)*tan(d*x+c)/b^3/d+1/2*(A*b-B*a)*sec(d*x+c
)*tan(d*x+c)/b^2/d+1/3*B*sec(d*x+c)^2*tan(d*x+c)/b/d

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Rubi [A]  time = 0.68, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {4033, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac {\left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {2 a^3 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {(A b-a B) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {B \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]

[Out]

((2*a^2 + b^2)*(A*b - a*B)*ArcTanh[Sin[c + d*x]])/(2*b^4*d) - (2*a^3*(A*b - a*B)*ArcTanh[(Sqrt[a - b]*Tan[(c +
d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^4*Sqrt[a + b]*d) - ((3*a*A*b - 3*a^2*B - 2*b^2*B)*Tan[c + d*x])/(3*b^3*
d) + ((A*b - a*B)*Sec[c + d*x]*Tan[c + d*x])/(2*b^2*d) + (B*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*d)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 4033

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2))/(
b*f*(m + n)), x] + Dist[d^2/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2)
+ B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] &&  !IGtQ[m, 1]

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 4092

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2)
+ A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m
}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx &=\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a B+2 b B \sec (c+d x)+3 (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b}\\ &=\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}+\frac {\int \frac {\sec (c+d x) \left (3 a (A b-a B)+b (3 A b+a B) \sec (c+d x)-2 \left (3 a A b-3 a^2 B-2 b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^2}\\ &=-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}+\frac {\int \frac {\sec (c+d x) \left (3 a b (A b-a B)+3 \left (2 a^2+b^2\right ) (A b-a B) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^3}\\ &=-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}-\frac {\left (a^3 (A b-a B)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^4}+\frac {\left (\left (2 a^2+b^2\right ) (A b-a B)\right ) \int \sec (c+d x) \, dx}{2 b^4}\\ &=\frac {\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}-\frac {\left (a^3 (A b-a B)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{b^5}\\ &=\frac {\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}-\frac {\left (2 a^3 (A b-a B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=\frac {\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac {2 a^3 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}-\frac {\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac {B \sec ^2(c+d x) \tan (c+d x)}{3 b d}\\ \end {align*}

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Mathematica [B]  time = 3.01, size = 422, normalized size = 2.26 \[ \frac {\frac {4 b \left (3 a^2 B-3 a A b+2 b^2 B\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b \left (3 a^2 B-3 a A b+2 b^2 B\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+6 \left (2 a^2+b^2\right ) (a B-A b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \left (2 a^2+b^2\right ) (a B-A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {24 a^3 (A b-a B) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {b^2 (B (b-3 a)+3 A b)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^2 (B (b-3 a)+3 A b)}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]

[Out]

((24*a^3*(A*b - a*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + 6*(2*a^2 + b^2)*(
-(A*b) + a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*(2*a^2 + b^2)*(-(A*b) + a*B)*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] + (b^2*(3*A*b + (-3*a + b)*B))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^3*B*Sin[(c +
d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*b*(-3*a*A*b + 3*a^2*B + 2*b^2*B)*Sin[(c + d*x)/2])/(Cos[
(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*b^3*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (b^2*(
3*A*b + (-3*a + b)*B))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b*(-3*a*A*b + 3*a^2*B + 2*b^2*B)*Sin[(c +
d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(12*b^4*d)

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fricas [B]  time = 1.18, size = 743, normalized size = 3.97 \[ \left [-\frac {6 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 3 \, {\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, B a^{2} b^{3} - 2 \, B b^{5} + 2 \, {\left (3 \, B a^{4} b - 3 \, A a^{3} b^{2} - B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac {12 \, {\left (B a^{4} - A a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B a^{2} b^{3} - 2 \, B b^{5} + 2 \, {\left (3 \, B a^{4} b - 3 \, A a^{3} b^{2} - B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[-1/12*(6*(B*a^4 - A*a^3*b)*sqrt(a^2 - b^2)*cos(d*x + c)^3*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c
)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x +
c) + b^2)) + 3*(2*B*a^5 - 2*A*a^4*b - B*a^3*b^2 + A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d*x + c)^3*log(sin(d*x + c
) + 1) - 3*(2*B*a^5 - 2*A*a^4*b - B*a^3*b^2 + A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) +
1) - 2*(2*B*a^2*b^3 - 2*B*b^5 + 2*(3*B*a^4*b - 3*A*a^3*b^2 - B*a^2*b^3 + 3*A*a*b^4 - 2*B*b^5)*cos(d*x + c)^2 -
3*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c)^3), 1
/12*(12*(B*a^4 - A*a^3*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x
+ c)))*cos(d*x + c)^3 - 3*(2*B*a^5 - 2*A*a^4*b - B*a^3*b^2 + A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d*x + c)^3*log(s
in(d*x + c) + 1) + 3*(2*B*a^5 - 2*A*a^4*b - B*a^3*b^2 + A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d*x + c)^3*log(-sin(d
*x + c) + 1) + 2*(2*B*a^2*b^3 - 2*B*b^5 + 2*(3*B*a^4*b - 3*A*a^3*b^2 - B*a^2*b^3 + 3*A*a*b^4 - 2*B*b^5)*cos(d*
x + c)^2 - 3*(B*a^3*b^2 - A*a^2*b^3 - B*a*b^4 + A*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x
+ c)^3)]

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giac [B]  time = 0.28, size = 412, normalized size = 2.20 \[ -\frac {\frac {3 \, {\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {3 \, {\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {12 \, {\left (B a^{4} - A a^{3} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{4}} + \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(2*B*a^3 - 2*A*a^2*b + B*a*b^2 - A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 3*(2*B*a^3 - 2*A*a^2*
b + B*a*b^2 - A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 12*(B*a^4 - A*a^3*b)*(pi*floor(1/2*(d*x + c)/pi
+ 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a
^2 + b^2)*b^4) + 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*b*tan(1/2*d*x + 1/
2*c)^5 - 3*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 1
2*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 4*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*tan(1/2*d*x + 1/2*c) - 6*A*a*b*tan(1
/2*d*x + 1/2*c) - 3*B*a*b*tan(1/2*d*x + 1/2*c) + 3*A*b^2*tan(1/2*d*x + 1/2*c) + 6*B*b^2*tan(1/2*d*x + 1/2*c))/
((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^3))/d

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maple [B]  time = 0.58, size = 688, normalized size = 3.68 \[ \frac {2 a^{4} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3} B}{d \,b^{4}}-\frac {a B}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {B}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A}{2 d b}+\frac {A}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{3 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A}{2 d b}+\frac {A a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {B}{3 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {A}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {B}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {A}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) A \,a^{2}}{d \,b^{3}}+\frac {a B}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {a^{2} B}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) A \,a^{2}}{d \,b^{3}}-\frac {a^{2} B}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B a}{2 d \,b^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B a}{2 d \,b^{2}}+\frac {A a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {B a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3} B}{d \,b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)

[Out]

2/d*a^4/b^4/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-2/d*a^3/b^3/((a-b)*(a+
b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A-1/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*a^3*B-1/2/d
/b^2/(tan(1/2*d*x+1/2*c)-1)^2*a*B+1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^2*B+1/2/d/b*ln(tan(1/2*d*x+1/2*c)+1)*A+1/2/d/
b/(tan(1/2*d*x+1/2*c)+1)*A-1/d/b/(tan(1/2*d*x+1/2*c)+1)*B+1/2/d/b/(tan(1/2*d*x+1/2*c)-1)*A-1/d/b/(tan(1/2*d*x+
1/2*c)-1)*B-1/3/d*B/b/(tan(1/2*d*x+1/2*c)+1)^3-1/2/d/b*ln(tan(1/2*d*x+1/2*c)-1)*A+1/d/b^2/(tan(1/2*d*x+1/2*c)-
1)*A*a-1/3/d*B/b/(tan(1/2*d*x+1/2*c)-1)^3+1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2*A-1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2*
B-1/2/d/b/(tan(1/2*d*x+1/2*c)+1)^2*A+1/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*A*a^2+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2
*a*B-1/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a^2*B-1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)*B*a-1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1
)*A*a^2-1/d/b^3/(tan(1/2*d*x+1/2*c)-1)*a^2*B+1/2/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*B*a-1/2/d/b^2*ln(tan(1/2*d*x+1
/2*c)+1)*B*a+1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*A*a-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)*B*a+1/d/b^4*ln(tan(1/2*d*x+1/
2*c)-1)*a^3*B

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 6.93, size = 4667, normalized size = 24.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + b/cos(c + d*x))),x)

[Out]

- ((tan(c/2 + (d*x)/2)*(A*b^2 + 2*B*a^2 + 2*B*b^2 - 2*A*a*b - B*a*b))/b^3 + (tan(c/2 + (d*x)/2)^5*(2*B*a^2 - A
*b^2 + 2*B*b^2 - 2*A*a*b + B*a*b))/b^3 - (4*tan(c/2 + (d*x)/2)^3*(3*B*a^2 + B*b^2 - 3*A*a*b))/(3*b^3))/(d*(3*t
an(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1)) - (atan(((((8*tan(c/2 + (d*x)/2)*(A^
2*b^9 - 8*B^2*a^9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*
b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a
^6*b^3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*
B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 - (((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9
+ 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 - (4*tan(c/2 + (d*x
)/2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/b^10)*(A*b^3 - 2*B*a^3 + 2*A
*a^2*b - B*a*b^2))/(2*b^4))*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2)*1i)/(2*b^4) + (((8*tan(c/2 + (d*x)/2)*(A^2
*b^9 - 8*B^2*a^9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b
^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^
6*b^3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B
*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 + (((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9
+ 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 + (4*tan(c/2 + (d*x)
/2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/b^10)*(A*b^3 - 2*B*a^3 + 2*A*
a^2*b - B*a*b^2))/(2*b^4))*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2)*1i)/(2*b^4))/((16*(4*B^3*a^11 - 6*B^3*a^10*
b + A^3*a^3*b^8 - 2*A^3*a^4*b^7 + 5*A^3*a^5*b^6 - 6*A^3*a^6*b^5 + 6*A^3*a^7*b^4 - 4*A^3*a^8*b^3 - B^3*a^6*b^5
+ 2*B^3*a^7*b^4 - 5*B^3*a^8*b^3 + 6*B^3*a^9*b^2 - 12*A*B^2*a^10*b + 3*A*B^2*a^5*b^6 - 6*A*B^2*a^6*b^5 + 15*A*B
^2*a^7*b^4 - 18*A*B^2*a^8*b^3 + 18*A*B^2*a^9*b^2 - 3*A^2*B*a^4*b^7 + 6*A^2*B*a^5*b^6 - 15*A^2*B*a^6*b^5 + 18*A
^2*B*a^7*b^4 - 18*A^2*B*a^8*b^3 + 12*A^2*B*a^9*b^2))/b^9 - (((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^9 - 3*A^
2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8
*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^7*b^2
- 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B*a^5*b^4 + 32*A*B*a^6*b
^3 - 32*A*B*a^7*b^2))/b^6 - (((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^2*b^11 - 2*B*a^
3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 - (4*tan(c/2 + (d*x)/2)*(8*a*b^10 - 16*a^2*
b^9 + 8*a^3*b^8)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/b^10)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/(2*b^
4))*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/(2*b^4) + (((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^9 - 3*A^2*a*
b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2
*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^7*b^2 - 2*
A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 -
32*A*B*a^7*b^2))/b^6 + (((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^2*b^11 - 2*B*a^3*b^
10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 + (4*tan(c/2 + (d*x)/2)*(8*a*b^10 - 16*a^2*b^9
+ 8*a^3*b^8)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/b^10)*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/(2*b^4))*
(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2))/(2*b^4)))*(A*b^3 - 2*B*a^3 + 2*A*a^2*b - B*a*b^2)*1i)/(b^4*d) - (a^3*
atan(((a^3*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^9 - 3*A^2*a*b^8 + 16*
B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2
+ B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8
+ 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a
^7*b^2))/b^6 + (a^3*((a + b)*(a - b))^(1/2)*((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^
2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 + (8*a^3*tan(c/2 + (d*x)/2)*
((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4)))*(A*b - B*a))/(
b^6 - a^2*b^4))*1i)/(b^6 - a^2*b^4) + (a^3*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2)*(A^2*b^9
- 8*B^2*a^9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b^4 +
16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^
3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B*a^5
*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 - (a^3*((a + b)*(a - b))^(1/2)*((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A
*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/
b^9 - (8*a^3*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*
(b^6 - a^2*b^4)))*(A*b - B*a))/(b^6 - a^2*b^4))*1i)/(b^6 - a^2*b^4))/((16*(4*B^3*a^11 - 6*B^3*a^10*b + A^3*a^3
*b^8 - 2*A^3*a^4*b^7 + 5*A^3*a^5*b^6 - 6*A^3*a^6*b^5 + 6*A^3*a^7*b^4 - 4*A^3*a^8*b^3 - B^3*a^6*b^5 + 2*B^3*a^7
*b^4 - 5*B^3*a^8*b^3 + 6*B^3*a^9*b^2 - 12*A*B^2*a^10*b + 3*A*B^2*a^5*b^6 - 6*A*B^2*a^6*b^5 + 15*A*B^2*a^7*b^4
- 18*A*B^2*a^8*b^3 + 18*A*B^2*a^9*b^2 - 3*A^2*B*a^4*b^7 + 6*A^2*B*a^5*b^6 - 15*A^2*B*a^6*b^5 + 18*A^2*B*a^7*b^
4 - 18*A^2*B*a^8*b^3 + 12*A^2*B*a^9*b^2))/b^9 + (a^3*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*tan(c/2 + (d*x)/2
)*(A^2*b^9 - 8*B^2*a^9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*b^6 + 16*A^2*a^4*b^5 - 16*A^2
*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a^4*b^5 - 13*B^2*a^5*b^4 + 16*
B^2*a^6*b^3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*B*a^3*b^6 + 26*A*B*a^4*b^5 -
32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 + (a^3*((a + b)*(a - b))^(1/2)*((8*(2*A*b^13 + 2*A*a^2*
b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4*B*a^5*b^8 - 2*A*a*b^12 - 2*B
*a*b^12))/b^9 + (8*a^3*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b
^8))/(b^6*(b^6 - a^2*b^4)))*(A*b - B*a))/(b^6 - a^2*b^4)))/(b^6 - a^2*b^4) - (a^3*((a + b)*(a - b))^(1/2)*(A*b
- B*a)*((8*tan(c/2 + (d*x)/2)*(A^2*b^9 - 8*B^2*a^9 - 3*A^2*a*b^8 + 16*B^2*a^8*b + 7*A^2*a^2*b^7 - 13*A^2*a^3*
b^6 + 16*A^2*a^4*b^5 - 16*A^2*a^5*b^4 + 16*A^2*a^6*b^3 - 8*A^2*a^7*b^2 + B^2*a^2*b^7 - 3*B^2*a^3*b^6 + 7*B^2*a
^4*b^5 - 13*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 16*B^2*a^7*b^2 - 2*A*B*a*b^8 + 16*A*B*a^8*b + 6*A*B*a^2*b^7 - 14*A*
B*a^3*b^6 + 26*A*B*a^4*b^5 - 32*A*B*a^5*b^4 + 32*A*B*a^6*b^3 - 32*A*B*a^7*b^2))/b^6 - (a^3*((a + b)*(a - b))^(
1/2)*((8*(2*A*b^13 + 2*A*a^2*b^11 - 6*A*a^3*b^10 + 4*A*a^4*b^9 + 2*B*a^2*b^11 - 2*B*a^3*b^10 + 6*B*a^4*b^9 - 4
*B*a^5*b^8 - 2*A*a*b^12 - 2*B*a*b^12))/b^9 - (8*a^3*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*
a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4)))*(A*b - B*a))/(b^6 - a^2*b^4)))/(b^6 - a^2*b^4)))*((a
+ b)*(a - b))^(1/2)*(A*b - B*a)*2i)/(d*(b^6 - a^2*b^4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)**4/(a + b*sec(c + d*x)), x)

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