3.676 \(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=186 \[ -\frac {a \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {2 a^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (3 a^2 C+b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 b^3 d}-\frac {a C \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {C \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.65, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4103, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac {\left (3 a^2 C+b^2 (3 A+2 C)\right ) \tan (c+d x)}{3 b^3 d}-\frac {a \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}+\frac {2 a^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}-\frac {a C \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac {C \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

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

Rule 4103

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

Rubi steps

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

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Mathematica [C]  time = 3.97, size = 657, normalized size = 3.53 \[ \frac {\cos (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 b \sin \left (\frac {d x}{2}\right ) \left (3 a^2 C+3 A b^2+2 b^2 C\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 b \sin \left (\frac {d x}{2}\right ) \left (3 a^2 C+3 A b^2+2 b^2 C\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+6 a \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 a \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-\frac {24 i a^2 (\cos (c)-i \sin (c)) \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b^2 C \left ((3 a+b) \sin \left (\frac {c}{2}\right )+(b-3 a) \cos \left (\frac {c}{2}\right )\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {b^2 C \left ((3 a+b) \sin \left (\frac {c}{2}\right )+(3 a-b) \cos \left (\frac {c}{2}\right )\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \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 C \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\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 C \sin \left (\frac {d x}{2}\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}\right )}{6 b^4 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]

[Out]

(Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*(6*a*(2*A*b^2 + (2*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2
] - Sin[(c + d*x)/2]] - 6*a*(2*A*b^2 + (2*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((24*I)*a^2
*(A*b^2 + a^2*C)*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[
(Cos[c] - I*Sin[c])^2])]*(Cos[c] - I*Sin[c]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2]) + (2*b^3*C*Sin[(d*
x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (b^2*C*((-3*a + b)*Cos[c/2] + (3*a +
b)*Sin[c/2]))/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*b*(3*A*b^2 + 3*a^2*C + 2*b^
2*C)*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (2*b^3*C*Sin[(d*x)/2])/((Co
s[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + (b^2*C*((3*a - b)*Cos[c/2] + (3*a + b)*Sin[c/2])
)/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*b*(3*A*b^2 + 3*a^2*C + 2*b^2*C)*Sin[(d*
x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(6*b^4*d*(A + 2*C + A*Cos[2*(c + d*x)])
*(a + b*Sec[c + d*x]))

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fricas [A]  time = 3.08, size = 663, normalized size = 3.56 \[ \left [\frac {6 \, {\left (C a^{4} + A a^{2} b^{2}\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 \, C a^{5} + {\left (2 \, A - C\right )} a^{3} b^{2} - {\left (2 \, A + C\right )} a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, C a^{5} + {\left (2 \, A - C\right )} a^{3} b^{2} - {\left (2 \, A + C\right )} a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{2} b^{3} - 2 \, C b^{5} + 2 \, {\left (3 \, C a^{4} b + {\left (3 \, A - C\right )} a^{2} b^{3} - {\left (3 \, A + 2 \, C\right )} b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - C a b^{4}\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 (C a^{4} + A a^{2} b^{2}\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 \, C a^{5} + {\left (2 \, A - C\right )} a^{3} b^{2} - {\left (2 \, A + C\right )} a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, C a^{5} + {\left (2 \, A - C\right )} a^{3} b^{2} - {\left (2 \, A + C\right )} a b^{4}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{2} b^{3} - 2 \, C b^{5} + 2 \, {\left (3 \, C a^{4} b + {\left (3 \, A - C\right )} a^{2} b^{3} - {\left (3 \, A + 2 \, C\right )} b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - C a b^{4}\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)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/12*(6*(C*a^4 + A*a^2*b^2)*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*C*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*C
*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*C*a^2*b^3 - 2*C*b^5 +
2*(3*C*a^4*b + (3*A - C)*a^2*b^3 - (3*A + 2*C)*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - C*a*b^4)*cos(d*x + c))*si
n(d*x + c))/((a^2*b^4 - b^6)*d*cos(d*x + c)^3), 1/12*(12*(C*a^4 + A*a^2*b^2)*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*C*a^5 + (2*A - C)*a^3*b^2 - (2
*A + C)*a*b^4)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3*(2*C*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*cos(d*
x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*C*a^2*b^3 - 2*C*b^5 + 2*(3*C*a^4*b + (3*A - C)*a^2*b^3 - (3*A + 2*C)*b^
5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - C*a*b^4)*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 = 8.94, size = 372, normalized size = 2.00 \[ -\frac {\frac {3 \, {\left (2 \, C a^{3} + 2 \, A a b^{2} + C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {3 \, {\left (2 \, C a^{3} + 2 \, A a b^{2} + C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {12 \, {\left (C a^{4} + A a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{4}} + \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(2*C*a^3 + 2*A*a*b^2 + C*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - 3*(2*C*a^3 + 2*A*a*b^2 + C*a*
b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 12*(C*a^4 + A*a^2*b^2)*(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*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b
^2*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^2*tan(1/
2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c)
+ 6*C*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 = 0.51, size = 554, normalized size = 2.98 \[ \frac {2 a^{2} \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^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\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 ) C}{d \,b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {C}{3 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {A}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {a^{2} C}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {C a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {C}{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}{d \,b^{2}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,b^{4}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{2 d \,b^{2}}-\frac {C}{3 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {a^{2} C}{d \,b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {C}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {C a}{2 d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {C}{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}{d \,b^{2}}-\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,b^{4}}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{2 d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)

[Out]

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

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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)^3*(A+C*sec(d*x+c)^2)/(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 = 8.04, size = 3914, normalized size = 21.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))),x)

[Out]

(a^2*atan(((a^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A
^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 1
3*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5
*b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6 + (a^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*(4*A*a^3*b^10
- 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 - (8*
a^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6
- a^2*b^4))))/(b^6 - a^2*b^4))*1i)/(b^6 - a^2*b^4) + (a^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*tan(c/2
+ (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a
^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12
*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6 - (a^2*((a + b)*(a - b)
)^(1/2)*(A*b^2 + C*a^2)*((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5
*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 + (8*a^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8*a*
b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4))))/(b^6 - a^2*b^4))*1i)/(b^6 - a^2*b^4))/((16*(4*C^3*a^11
- 6*C^3*a^10*b - 4*A^3*a^4*b^7 + 4*A^3*a^5*b^6 - C^3*a^6*b^5 + 2*C^3*a^7*b^4 - 5*C^3*a^8*b^3 + 6*C^3*a^9*b^2
- A*C^2*a^4*b^7 + 2*A*C^2*a^5*b^6 - 9*A*C^2*a^6*b^5 + 12*A*C^2*a^7*b^4 - 16*A*C^2*a^8*b^3 + 12*A*C^2*a^9*b^2 -
4*A^2*C*a^4*b^7 + 6*A^2*C*a^5*b^6 - 14*A^2*C*a^6*b^5 + 12*A^2*C*a^7*b^4))/b^9 - (a^2*((a + b)*(a - b))^(1/2)*
(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4
*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*
a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))
/b^6 + (a^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*
b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 - (8*a^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))
^(1/2)*(A*b^2 + C*a^2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4))))/(b^6 - a^2*b^4)))/(b^6 - a
^2*b^4) + (a^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^
2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13
*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*
b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6 - (a^2*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*((8*(4*A*a^3*b^10 -
8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 + (8*a
^2*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 -
a^2*b^4))))/(b^6 - a^2*b^4)))/(b^6 - a^2*b^4)))*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2)*2i)/(d*(b^6 - a^2*b^4
)) - (atan((((C*a^3 + b^2*(A*a + (C*a)/2))*((((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 -
6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 - (8*tan(c/2 + (d*x)/2)*(C*a^3 + b^2*(A*a + (C*a)/2)
)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/b^10)*(C*a^3 + b^2*(A*a + (C*a)/2)))/b^4 + (8*tan(c/2 + (d*x)/2)*(8*C^2
*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a^
3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20
*A*C*a^4*b^5 + 28*A*C*a^5*b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6)*1i)/b^4 - ((C*a^3 + b^2*(A*a + (C*a)/2)
)*((((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 +
2*C*a*b^12))/b^9 + (8*tan(c/2 + (d*x)/2)*(C*a^3 + b^2*(A*a + (C*a)/2))*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/b^
10)*(C*a^3 + b^2*(A*a + (C*a)/2)))/b^4 - (8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*
A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 -
16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*b^4 - 32*A*C*a^
6*b^3 + 16*A*C*a^7*b^2))/b^6)*1i)/b^4)/(((C*a^3 + b^2*(A*a + (C*a)/2))*((((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2*
C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*A*a*b^12 + 2*C*a*b^12))/b^9 - (8*tan(c/2 + (d*x)/2)*
(C*a^3 + b^2*(A*a + (C*a)/2))*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/b^10)*(C*a^3 + b^2*(A*a + (C*a)/2)))/b^4 +
(8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*
b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a
^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*b^4 - 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6))/b^4 - (16*
(4*C^3*a^11 - 6*C^3*a^10*b - 4*A^3*a^4*b^7 + 4*A^3*a^5*b^6 - C^3*a^6*b^5 + 2*C^3*a^7*b^4 - 5*C^3*a^8*b^3 + 6*C
^3*a^9*b^2 - A*C^2*a^4*b^7 + 2*A*C^2*a^5*b^6 - 9*A*C^2*a^6*b^5 + 12*A*C^2*a^7*b^4 - 16*A*C^2*a^8*b^3 + 12*A*C^
2*a^9*b^2 - 4*A^2*C*a^4*b^7 + 6*A^2*C*a^5*b^6 - 14*A^2*C*a^6*b^5 + 12*A^2*C*a^7*b^4))/b^9 + ((C*a^3 + b^2*(A*a
+ (C*a)/2))*((((8*(4*A*a^3*b^10 - 8*A*a^2*b^11 - 2*C*a^2*b^11 + 2*C*a^3*b^10 - 6*C*a^4*b^9 + 4*C*a^5*b^8 + 4*
A*a*b^12 + 2*C*a*b^12))/b^9 + (8*tan(c/2 + (d*x)/2)*(C*a^3 + b^2*(A*a + (C*a)/2))*(8*a*b^10 - 16*a^2*b^9 + 8*a
^3*b^8))/b^10)*(C*a^3 + b^2*(A*a + (C*a)/2)))/b^4 - (8*tan(c/2 + (d*x)/2)*(8*C^2*a^9 - 16*C^2*a^8*b - 4*A^2*a^
2*b^7 + 12*A^2*a^3*b^6 - 16*A^2*a^4*b^5 + 8*A^2*a^5*b^4 - C^2*a^2*b^7 + 3*C^2*a^3*b^6 - 7*C^2*a^4*b^5 + 13*C^2
*a^5*b^4 - 16*C^2*a^6*b^3 + 16*C^2*a^7*b^2 - 4*A*C*a^2*b^7 + 12*A*C*a^3*b^6 - 20*A*C*a^4*b^5 + 28*A*C*a^5*b^4
- 32*A*C*a^6*b^3 + 16*A*C*a^7*b^2))/b^6))/b^4))*(C*a^3 + b^2*(A*a + (C*a)/2))*2i)/(b^4*d) - ((tan(c/2 + (d*x)/
2)^5*(2*A*b^2 + 2*C*a^2 + 2*C*b^2 + C*a*b))/b^3 - (4*tan(c/2 + (d*x)/2)^3*(3*A*b^2 + 3*C*a^2 + C*b^2))/(3*b^3)
+ (tan(c/2 + (d*x)/2)*(2*A*b^2 + 2*C*a^2 + 2*C*b^2 - C*a*b))/b^3)/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d
*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x)), x)

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