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

Optimal. Leaf size=153 \[ \frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.45, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac {\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {(b B-a C) \tan (c+d x)}{b^2 d}+\frac {C \tan (c+d x) \sec (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 2.61, size = 472, normalized size = 3.08 \[ \frac {\cos (c+d x) (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 a (\sin (c)+i \cos (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {4 b \sin \left (\frac {d x}{2}\right ) (b B-a C)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 b \sin \left (\frac {d x}{2}\right ) (b B-a C)}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^2 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^2 C}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{2 b^3 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-2*(2*A*b^2 - 2*a*b*B + 2*a^2*C +
b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(2*A*b^2 - 2*a*b*B + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]] + (8*a*(A*b^2 + a*(-(b*B) + a*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])]*(I*Cos[c] + Sin[c]))/(Sqrt[a^2 - b^2]*Sqrt[(Cos
[c] - I*Sin[c])^2]) + (b^2*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b*(b*B - a*C)*Sin[(d*x)/2])/((Cos[c
/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (b^2*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*
b*(b*B - a*C)*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(2*b^3*d*(A + 2*C
+ 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x]))

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fricas [B]  time = 14.95, size = 657, normalized size = 4.29 \[ \left [\frac {2 \, {\left (C a^{3} - B a^{2} b + A a b^{2}\right )} \sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )^{2} \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 ) + {\left (2 \, C a^{4} - 2 \, B a^{3} b + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, C a^{4} - 2 \, B a^{3} b + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{2} b^{2} - C b^{4} - 2 \, {\left (C a^{3} b - B a^{2} b^{2} - C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, {\left (C a^{3} - B a^{2} b + A a 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 )^{2} - {\left (2 \, C a^{4} - 2 \, B a^{3} b + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{4} - 2 \, B a^{3} b + {\left (2 \, A - C\right )} a^{2} b^{2} + 2 \, B a b^{3} - {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C a^{2} b^{2} - C b^{4} - 2 \, {\left (C a^{3} b - B a^{2} b^{2} - C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(C*a^3 - B*a^2*b + A*a*b^2)*sqrt(a^2 - b^2)*cos(d*x + c)^2*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*c
os(d*x + c) + b^2)) + (2*C*a^4 - 2*B*a^3*b + (2*A - C)*a^2*b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log
(sin(d*x + c) + 1) - (2*C*a^4 - 2*B*a^3*b + (2*A - C)*a^2*b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(
-sin(d*x + c) + 1) + 2*(C*a^2*b^2 - C*b^4 - 2*(C*a^3*b - B*a^2*b^2 - C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x +
c))/((a^2*b^3 - b^5)*d*cos(d*x + c)^2), -1/4*(4*(C*a^3 - B*a^2*b + A*a*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)^2 - (2*C*a^4 - 2*B*a^3*b + (2*A - C)*a^2
*b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*C*a^4 - 2*B*a^3*b + (2*A - C)*a^2*
b^2 + 2*B*a*b^3 - (2*A + C)*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(C*a^2*b^2 - C*b^4 - 2*(C*a^3*b - B
*a^2*b^2 - C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5)*d*cos(d*x + c)^2)]

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giac [B]  time = 0.31, size = 287, normalized size = 1.88 \[ \frac {\frac {{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} - \frac {{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {4 \, {\left (C a^{3} - B a^{2} b + A a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{3}} + \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*C*a^2 - 2*B*a*b + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 - (2*C*a^2 - 2*B*a*b + 2*A*b
^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - 4*(C*a^3 - B*a^2*b + A*a*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^3) + 2*(2*C*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*b*tan(1/2*d*x + 1/2*c)^3 + C*b*tan(1/2*d*x + 1/2*c)^3 -
2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2
- 1)^2*b^2))/d

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maple [B]  time = 0.67, size = 499, normalized size = 3.26 \[ -\frac {2 a \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 \sqrt {\left (a -b \right ) \left (a +b \right )}}+\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 ) B}{d \,b^{2} \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 ) C}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {C}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {B}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {C a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {C}{2 d b \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}{d b}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B a}{d \,b^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} C}{d \,b^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{2 d b}-\frac {C}{2 d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {B}{d b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {C a}{d \,b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {C}{2 d b \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}{d b}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B a}{d \,b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} C}{d \,b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{2 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*a/b/((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^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))*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))*C+1/2/d/b/(tan(1/2*d*x+1/2*c)-1)^2*C-1/d/b/(tan(1/2*d*x+1/2*c)-1)*B+1/d
/b^2/(tan(1/2*d*x+1/2*c)-1)*C*a+1/2/d/b/(tan(1/2*d*x+1/2*c)-1)*C-1/d/b*ln(tan(1/2*d*x+1/2*c)-1)*A+1/d/b^2*ln(t
an(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^2*C-1/2/d/b*ln(tan(1/2*d*x+1/2*c)-1)*C-1/2/d/b/(ta
n(1/2*d*x+1/2*c)+1)^2*C-1/d/b/(tan(1/2*d*x+1/2*c)+1)*B+1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*C*a+1/2/d/b/(tan(1/2*d*x
+1/2*c)+1)*C+1/d/b*ln(tan(1/2*d*x+1/2*c)+1)*A-1/d/b^2*ln(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^2*C+1/2/d/b*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)^2*(A+B*sec(d*x+c)+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 = 12.10, size = 5489, normalized size = 35.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(2*B*b - 2*C*a + C*b))/b^2 + (tan(c/2 + (d*x)/2)^3*(2*C*a - 2*B*b + C*b))/b^2)/(d*(tan(c/
2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (atan(((((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^
7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^
4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*
b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b
^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*
B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 + (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C
*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2
+ b^2*(A + C/2) - B*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2
+ b^2*(A + C/2) - B*a*b)*1i)/b^3 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 -
3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3
- 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 -
12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5
- 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*
C*a^5*b^2))/b^4 - (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*
b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B
*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) -
B*a*b)*1i)/b^3)/((16*(4*C^3*a^8 - 4*A^3*a*b^7 - 6*C^3*a^7*b + 4*A^3*a^2*b^6 + 4*B^3*a^4*b^4 - 4*B^3*a^5*b^3 -
C^3*a^3*b^5 + 2*C^3*a^4*b^4 - 5*C^3*a^5*b^3 + 6*C^3*a^6*b^2 - A*C^2*a*b^7 - 4*A^2*C*a*b^7 - 12*B*C^2*a^7*b - 1
2*A*B^2*a^3*b^5 + 12*A*B^2*a^4*b^4 + 12*A^2*B*a^2*b^6 - 12*A^2*B*a^3*b^5 + 2*A*C^2*a^2*b^6 - 9*A*C^2*a^3*b^5 +
12*A*C^2*a^4*b^4 - 16*A*C^2*a^5*b^3 + 12*A*C^2*a^6*b^2 + 6*A^2*C*a^2*b^6 - 14*A^2*C*a^3*b^5 + 12*A^2*C*a^4*b^
4 + B*C^2*a^2*b^6 - 2*B*C^2*a^3*b^5 + 9*B*C^2*a^4*b^4 - 12*B*C^2*a^5*b^3 + 16*B*C^2*a^6*b^2 - 4*B^2*C*a^3*b^5
+ 6*B^2*C*a^4*b^4 - 14*B^2*C*a^5*b^3 + 12*B^2*C*a^6*b^2 + 8*A*B*C*a^2*b^6 - 12*A*B*C*a^3*b^5 + 28*A*B*C*a^4*b^
4 - 24*A*B*C*a^5*b^3))/b^6 - (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a
*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2
*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C
*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A
*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b
^2))/b^4 + (((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4
*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B*a*b)*(
8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) - B*a*b))
/b^3 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b +
16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b
^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 +
16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^
4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 - (((8*(4*A*b
^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9
- 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2 + b^2*(A + C/2) - B*a*b)*(8*a*b^8 - 16*a^2*b^7 +
8*a^3*b^6))/b^7)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3)*(C*a^2 + b^2*(A + C/2) - B*a*b))/b^3))*(C*a^2 + b^2*(A
+ C/2) - B*a*b)*2i)/(b^3*d) + (a*atan(((a*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a
^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12
*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b
^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 1
6*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^
3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 + (a*((a + b)*(a - b))^(1/2)*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*
b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9)
)/b^6 + (8*a*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2 - B*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*
b^6))/(b^4*(b^5 - a^2*b^3)))*(A*b^2 + C*a^2 - B*a*b))/(b^5 - a^2*b^3))*(A*b^2 + C*a^2 - B*a*b)*1i)/(b^5 - a^2*
b^3) + (a*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C
^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8
*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12
*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 -
28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a
^5*b^2))/b^4 - (a*((a + b)*(a - b))^(1/2)*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 +
2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*a*tan(c/2 + (d*x)/2)*(
(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2 - B*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(A*b
^2 + C*a^2 - B*a*b))/(b^5 - a^2*b^3))*(A*b^2 + C*a^2 - B*a*b)*1i)/(b^5 - a^2*b^3))/((16*(4*C^3*a^8 - 4*A^3*a*b
^7 - 6*C^3*a^7*b + 4*A^3*a^2*b^6 + 4*B^3*a^4*b^4 - 4*B^3*a^5*b^3 - C^3*a^3*b^5 + 2*C^3*a^4*b^4 - 5*C^3*a^5*b^3
+ 6*C^3*a^6*b^2 - A*C^2*a*b^7 - 4*A^2*C*a*b^7 - 12*B*C^2*a^7*b - 12*A*B^2*a^3*b^5 + 12*A*B^2*a^4*b^4 + 12*A^2
*B*a^2*b^6 - 12*A^2*B*a^3*b^5 + 2*A*C^2*a^2*b^6 - 9*A*C^2*a^3*b^5 + 12*A*C^2*a^4*b^4 - 16*A*C^2*a^5*b^3 + 12*A
*C^2*a^6*b^2 + 6*A^2*C*a^2*b^6 - 14*A^2*C*a^3*b^5 + 12*A^2*C*a^4*b^4 + B*C^2*a^2*b^6 - 2*B*C^2*a^3*b^5 + 9*B*C
^2*a^4*b^4 - 12*B*C^2*a^5*b^3 + 16*B*C^2*a^6*b^2 - 4*B^2*C*a^3*b^5 + 6*B^2*C*a^4*b^4 - 14*B^2*C*a^5*b^3 + 12*B
^2*C*a^6*b^2 + 8*A*B*C*a^2*b^6 - 12*A*B*C*a^3*b^5 + 28*A*B*C*a^4*b^4 - 24*A*B*C*a^5*b^3))/b^6 - (a*((a + b)*(a
- b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6
*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*
a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*
b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5 - 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A
*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 + (a*((a
+ b)*(a - b))^(1/2)*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^
3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*a*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2
)*(A*b^2 + C*a^2 - B*a*b)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(A*b^2 + C*a^2 - B*a*b))/
(b^5 - a^2*b^3))*(A*b^2 + C*a^2 - B*a*b))/(b^5 - a^2*b^3) + (a*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*
(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4
+ 4*B^2*a^2*b^5 - 12*B^2*a^3*b^4 + 16*B^2*a^4*b^3 - 8*B^2*a^5*b^2 + 7*C^2*a^2*b^5 - 13*C^2*a^3*b^4 + 16*C^2*a^
4*b^3 - 16*C^2*a^5*b^2 + 4*A*C*b^7 - 8*A*B*a*b^6 - 12*A*C*a*b^6 - 4*B*C*a*b^6 + 16*B*C*a^6*b + 24*A*B*a^2*b^5
- 32*A*B*a^3*b^4 + 16*A*B*a^4*b^3 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2 + 12*B*C
*a^2*b^5 - 20*B*C*a^3*b^4 + 28*B*C*a^4*b^3 - 32*B*C*a^5*b^2))/b^4 - (a*((a + b)*(a - b))^(1/2)*((8*(4*A*b^10 +
2*C*b^10 + 4*A*a^2*b^8 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 4*
B*a*b^9 - 2*C*a*b^9))/b^6 - (8*a*tan(c/2 + (d*x)/2)*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2 - B*a*b)*(8*a*b^8 -
16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(A*b^2 + C*a^2 - B*a*b))/(b^5 - a^2*b^3))*(A*b^2 + C*a^2 - B*
a*b))/(b^5 - a^2*b^3)))*((a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2 - B*a*b)*2i)/(d*(b^5 - a^2*b^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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