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

Optimal. Leaf size=145 \[ \frac{(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac{3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{3 C \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

(-3*C*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((2*A + 27*C)*Tan[c + d*x])/(15*a^3*d) - ((A + C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((A - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2)
+ (3*C*Tan[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.427468, antiderivative size = 145, 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 = {4085, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac{(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac{3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{3 C \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac{(A-9 C) \tan (c+d x) \sec ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*C*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((2*A + 27*C)*Tan[c + d*x])/(15*a^3*d) - ((A + C)*Sec[c + d*x]^3*Tan[c
+ d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((A - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2)
+ (3*C*Tan[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

Rule 4085

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

Rule 4019

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

Rule 4008

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^3(c+d x) (-a (2 A-3 C)-a (A+6 C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\sec ^2(c+d x) \left (-2 a^2 (A-9 C)-a^2 (2 A+27 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sec (c+d x) \left (-45 a^3 C+a^3 (2 A+27 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(3 C) \int \sec (c+d x) \, dx}{a^3}+\frac{(2 A+27 C) \int \sec ^2(c+d x) \, dx}{15 a^3}\\ &=-\frac{3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(2 A+27 C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=-\frac{3 C \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac{(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 3.01238, size = 457, normalized size = 3.15 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\sec \left (\frac{c}{2}\right ) \sec (c) \sec (c+d x) \left (-10 A \sin \left (c-\frac{d x}{2}\right )+10 A \sin \left (c+\frac{d x}{2}\right )-20 A \sin \left (2 c+\frac{d x}{2}\right )+22 A \sin \left (2 c+\frac{3 d x}{2}\right )+10 A \sin \left (c+\frac{5 d x}{2}\right )+10 A \sin \left (3 c+\frac{5 d x}{2}\right )+2 A \sin \left (2 c+\frac{7 d x}{2}\right )+2 A \sin \left (4 c+\frac{7 d x}{2}\right )-5 (4 A+51 C) \sin \left (\frac{d x}{2}\right )+(22 A+567 C) \sin \left (\frac{3 d x}{2}\right )-600 C \sin \left (c-\frac{d x}{2}\right )+375 C \sin \left (c+\frac{d x}{2}\right )-480 C \sin \left (2 c+\frac{d x}{2}\right )-60 C \sin \left (c+\frac{3 d x}{2}\right )+402 C \sin \left (2 c+\frac{3 d x}{2}\right )-225 C \sin \left (3 c+\frac{3 d x}{2}\right )+315 C \sin \left (c+\frac{5 d x}{2}\right )+30 C \sin \left (2 c+\frac{5 d x}{2}\right )+240 C \sin \left (3 c+\frac{5 d x}{2}\right )-45 C \sin \left (4 c+\frac{5 d x}{2}\right )+72 C \sin \left (2 c+\frac{7 d x}{2}\right )+15 C \sin \left (3 c+\frac{7 d x}{2}\right )+57 C \sin \left (4 c+\frac{7 d x}{2}\right )\right )+2880 C \cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{60 a^3 d (\sec (c+d x)+1)^3 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(2880*C*Cos[(c + d*x)/2]^5*(Log[Cos[(c + d*x)/2] - Sin[(
c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]*(-5*(4*A + 51*C)*Sin[(
d*x)/2] + (22*A + 567*C)*Sin[(3*d*x)/2] - 10*A*Sin[c - (d*x)/2] - 600*C*Sin[c - (d*x)/2] + 10*A*Sin[c + (d*x)/
2] + 375*C*Sin[c + (d*x)/2] - 20*A*Sin[2*c + (d*x)/2] - 480*C*Sin[2*c + (d*x)/2] - 60*C*Sin[c + (3*d*x)/2] + 2
2*A*Sin[2*c + (3*d*x)/2] + 402*C*Sin[2*c + (3*d*x)/2] - 225*C*Sin[3*c + (3*d*x)/2] + 10*A*Sin[c + (5*d*x)/2] +
315*C*Sin[c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] + 10*A*Sin[3*c + (5*d*x)/2] + 240*C*Sin[3*c + (5*d*x)/2]
- 45*C*Sin[4*c + (5*d*x)/2] + 2*A*Sin[2*c + (7*d*x)/2] + 72*C*Sin[2*c + (7*d*x)/2] + 15*C*Sin[3*c + (7*d*x)/2
] + 2*A*Sin[4*c + (7*d*x)/2] + 57*C*Sin[4*c + (7*d*x)/2])))/(60*a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[
c + d*x])^3)

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Maple [A]  time = 0.062, size = 204, normalized size = 1.4 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{3}}}-{\frac{C}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{3}}} \end{align*}

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+a*sec(d*x+c))^3,x)

[Out]

1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*A+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+1/6/d/a^3*tan(1/2*d*x+1/2*c)^3*A+1/2/d/a^3
*C*tan(1/2*d*x+1/2*c)^3+1/4/d/a^3*A*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)-1/d/a^3*C/(tan(1/2*d*x+
1/2*c)+1)-3/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)*C-1/d/a^3*C/(tan(1/2*d*x+1/2*c)-1)+3/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)
*C

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Maxima [A]  time = 0.958932, size = 315, normalized size = 2.17 \begin{align*} \frac{3 \, C{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac{A{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \end{align*}

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+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*C*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x
+ c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 -
60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) + A*(15*sin
(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)
/a^3)/d

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Fricas [A]  time = 0.513473, size = 576, normalized size = 3.97 \begin{align*} -\frac{45 \,{\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \,{\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \,{\left (A + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (7 \, A + 117 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}

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+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/30*(45*(C*cos(d*x + c)^4 + 3*C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(sin(d*x + c) + 1)
- 45*(C*cos(d*x + c)^4 + 3*C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*
(2*(A + 36*C)*cos(d*x + c)^3 + 3*(2*A + 57*C)*cos(d*x + c)^2 + (7*A + 117*C)*cos(d*x + c) + 15*C)*sin(d*x + c)
)/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

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+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c
+ d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

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Giac [A]  time = 1.24574, size = 240, normalized size = 1.66 \begin{align*} -\frac{\frac{180 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{180 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{120 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

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+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(180*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 180*C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 120*C*tan
(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x
+ 1/2*c)^5 + 10*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^12*tan(1/2*d*x + 1/
2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d