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

Optimal. Leaf size=137 \[ \frac{\left (2 a^2 C+b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}-\frac{a C \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.380863, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4093, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (2 a^2 C+b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}-\frac{a C \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4093

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + 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*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[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 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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 1.96736, size = 428, normalized size = 3.12 \[ \frac{\cos (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (-2 \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 a (\sin (c)+i \cos (c)) \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}-\frac{4 a b C \sin \left (\frac{d x}{2}\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{4 a b C \sin \left (\frac{d x}{2}\right )}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{b^2 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{2 b^3 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.076, size = 362, normalized size = 2.6 \begin{align*} -2\,{\frac{Aa}{db\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{{a}^{3}C}{d{b}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{A}{db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{{a}^{2}C}{d{b}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{C}{2\,db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{aC}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{A}{db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{{a}^{2}C}{d{b}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{C}{2\,db}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{C}{2\,db} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{aC}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.02013, size = 1254, normalized size = 9.15 \begin{align*} \left [\frac{2 \,{\left (C a^{3} + 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} +{\left (2 \, A - C\right )} a^{2} b^{2} -{\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} +{\left (2 \, A - C\right )} a^{2} b^{2} -{\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 - C a b^{3}\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} + 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} +{\left (2 \, A - C\right )} a^{2} b^{2} -{\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} +{\left (2 \, A - C\right )} a^{2} b^{2} -{\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 - C a b^{3}\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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + 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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]  time = 1.3369, size = 327, normalized size = 2.39 \begin{align*} \frac{\frac{{\left (2 \, C a^{2} + 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 \, 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} + 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} + 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 ) + 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*C*a^2 + 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*A*b^2 + C*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1))/b^3 - 4*(C*a^3 + A*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arcta
n(-(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 + C*b*tan(1/2*d*x + 1/2*c)^3 - 2*C*a*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