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

Optimal. Leaf size=135 \[ \frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4081, 3998, 3770, 3831, 2659, 208} \[ \frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {C \tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4081

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

Rubi steps

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

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Mathematica [C]  time = 2.39, size = 331, normalized size = 2.45 \[ \frac {2 (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {b \left (a^2 C+A b^2\right ) (b \sin (c)-a \sin (d x))}{a (a-b) (a+b) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )}+\frac {2 a (\sin (c)+i \cos (c)) \left (C \left (a^2-2 b^2\right )-A b^2\right ) (a \cos (c+d x)+b) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^{3/2} \sqrt {(\cos (c)-i \sin (c))^2}}-C (a \cos (c+d x)+b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+C (a \cos (c+d x)+b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{b^2 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 2.51, size = 676, normalized size = 5.01 \[ \left [\frac {{\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3} + {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{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 (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C a^{4} b + {\left (A - C\right )} a^{2} b^{3} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}, -\frac {2 \, {\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3} + {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2}\right )} \cos \left (d x + c\right )\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 ) - {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{4} b + {\left (A - C\right )} a^{2} b^{3} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((C*a^3*b - (A + 2*C)*a*b^3 + (C*a^4 - (A + 2*C)*a^2*b^2)*cos(d*x + c))*sqrt(a^2 - b^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*cos(d*x + c) + b^2)) + (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^
4)*cos(d*x + c))*log(sin(d*x + c) + 1) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(
d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 +
a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b^7)*d), -1/2*(2*(C*a^3*b - (A + 2*C)*a*b^3 + (C*a^4 - (A + 2*
C)*a^2*b^2)*cos(d*x + c))*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))) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1)
+ (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C
*a^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d*x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a
^2*b^5 + b^7)*d)]

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giac [A]  time = 0.30, size = 231, normalized size = 1.71 \[ \frac {\frac {2 \, {\left (C a^{3} - A a b^{2} - 2 \, C a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.65, size = 350, normalized size = 2.59 \[ \frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} C}{d b \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 a \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \,b^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 a \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) C}{d \,b^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) C}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*A+2/d/b/(a^2-b^2)*tan(1
/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*a^2*C+2/d*a/(a-b)/(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^3/b^2/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(ta
n(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C+4/d*a/(a-b)/(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))*C-1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*C+1/d/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)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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 = 11.22, size = 3838, normalized size = 28.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (C*atan(((C*((C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*
b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^
7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 - (32*tan(c/2 + (d*x)/2)
*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^
2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 - (C*((C*((32*(A*a^4*b^5 - A*a^
2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4
- a^3*b^3) + (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/
(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 + (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*
C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b
^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2)/((C*((C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 -
3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(
2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))
)/b^2 - (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4
+ 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 - (6
4*(C^3*a^5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C^3*a^3*b^2 + A*C^2*a*b^4 - A*C^2*a^4*b + 3*A*C^2*a^2
*b^3 - A*C^2*a^3*b^2 + A^2*C*a^2*b^3))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (C*((C*((32*(A*a^4*b^5 - A*a^2*b^7
- A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3
*b^3) + (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(
a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 + (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^
5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b
^5 - a^2*b^3 - a^3*b^2)))/b^2))*2i)/(b^2*d) - (2*tan(c/2 + (d*x)/2)*(A*b^2 + C*a^2))/(d*(a + b)*(a*b - b^2)*(a
+ b - tan(c/2 + (d*x)/2)^2*(a - b))) - (a*atan(((a*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(2*C^2
*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A
*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (a*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 -
C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*a*
tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*
a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))
*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(A*b^2 - C*a^
2 + 2*C*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) + (a*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)
/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4
*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (a*((32*(A*a^4*b^5 - A*a^2*b^7 - A*
a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3
) - (32*a*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^
3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 -
a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(A*
b^2 - C*a^2 + 2*C*b^2)*1i)/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))/((64*(C^3*a^5 + 2*C^3*a*b^4 - C^3*a^4*b +
2*C^3*a^2*b^3 - 3*C^3*a^3*b^2 + A*C^2*a*b^4 - A*C^2*a^4*b + 3*A*C^2*a^2*b^3 - A*C^2*a^3*b^2 + A^2*C*a^2*b^3))/
(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (a*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^
6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 -
2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (a*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2
*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*a*tan(c/2 + (d*
x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a
^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(
a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(A*b^2 - C*a^2 + 2*C*b^2))
/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) + (a*((a + b)^3*(a - b)^3)^(1/2)*((32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 +
C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2
*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (a*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9
+ C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*a*tan(c/
2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^
6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((a +
b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(A*b^2 - C*a^2 + 2*
C*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(A*b^2 - C*a^2 + 2*C*b^2)*2i)/(d
*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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