3.275 \(\int \frac{(a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=266 \[ \frac{2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (32 A+33 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (568 A+759 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{10 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

[Out]

(2*a^3*(232*A + 297*C)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(568*A + 759
*C)*Sin[c + d*x])/(693*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^3*(568*A + 759*C)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(32*A + 33*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x
])/(231*d*Sec[c + d*x]^(5/2)) + (10*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)) + (
2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

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Rubi [A]  time = 0.797018, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4087, 4017, 4015, 3805, 3804} \[ \frac{2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{2 a^2 (32 A+33 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^3 (568 A+759 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{10 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{11 d \sec ^{\frac{9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(2*a^3*(232*A + 297*C)*Sin[c + d*x])/(693*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) + (2*a^3*(568*A + 759
*C)*Sin[c + d*x])/(693*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a^3*(568*A + 759*C)*Sqrt[Sec[c + d*
x]]*Sin[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a^2*(32*A + 33*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x
])/(231*d*Sec[c + d*x]^(5/2)) + (10*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(99*d*Sec[c + d*x]^(7/2)) + (
2*A*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(11*d*Sec[c + d*x]^(9/2))

Rule 4087

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rule 4017

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

Rule 4015

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

Rule 3805

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

Rule 3804

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Simp[(-2*a*Co
t[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^
2, 0]

Rubi steps

\begin{align*} \int \frac{(a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac{11}{2}}(c+d x)} \, dx &=\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{2 \int \frac{(a+a \sec (c+d x))^{5/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (4 A+11 C) \sec (c+d x)\right )}{\sec ^{\frac{9}{2}}(c+d x)} \, dx}{11 a}\\ &=\frac{10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{4 \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3}{4} a^2 (32 A+33 C)+\frac{1}{4} a^2 (56 A+99 C) \sec (c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx}{99 a}\\ &=\frac{2 a^2 (32 A+33 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{5}{8} a^3 (232 A+297 C)+\frac{1}{8} a^3 (776 A+1089 C) \sec (c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx}{693 a}\\ &=\frac{2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (32 A+33 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{1}{231} \left (a^2 (568 A+759 C)\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (32 A+33 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}+\frac{1}{693} \left (2 a^2 (568 A+759 C)\right ) \int \frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^3 (232 A+297 C) \sin (c+d x)}{693 d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{2 a^3 (568 A+759 C) \sin (c+d x)}{693 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{4 a^3 (568 A+759 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a^2 (32 A+33 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{231 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{10 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{99 d \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 A (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{11 d \sec ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 2.07714, size = 127, normalized size = 0.48 \[ \frac{a^2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (2 (6989 A+6666 C) \cos (c+d x)+16 (325 A+198 C) \cos (2 (c+d x))+1735 A \cos (3 (c+d x))+448 A \cos (4 (c+d x))+63 A \cos (5 (c+d x))+22928 A+396 C \cos (3 (c+d x))+27456 C)}{5544 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(11/2),x]

[Out]

(a^2*(22928*A + 27456*C + 2*(6989*A + 6666*C)*Cos[c + d*x] + 16*(325*A + 198*C)*Cos[2*(c + d*x)] + 1735*A*Cos[
3*(c + d*x)] + 396*C*Cos[3*(c + d*x)] + 448*A*Cos[4*(c + d*x)] + 63*A*Cos[5*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*
x])]*Tan[(c + d*x)/2])/(5544*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 0.384, size = 154, normalized size = 0.6 \begin{align*} -{\frac{2\,{a}^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 63\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+224\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+355\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+99\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+426\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+396\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+568\,A\cos \left ( dx+c \right ) +759\,C\cos \left ( dx+c \right ) +1136\,A+1518\,C \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{693\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x)

[Out]

-2/693/d*a^2*(-1+cos(d*x+c))*(63*A*cos(d*x+c)^5+224*A*cos(d*x+c)^4+355*A*cos(d*x+c)^3+99*C*cos(d*x+c)^3+426*A*
cos(d*x+c)^2+396*C*cos(d*x+c)^2+568*A*cos(d*x+c)+759*C*cos(d*x+c)+1136*A+1518*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))
^(1/2)*cos(d*x+c)^6*(1/cos(d*x+c))^(11/2)/sin(d*x+c)

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Maxima [B]  time = 2.15987, size = 1141, normalized size = 4.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

1/22176*(sqrt(2)*(31878*a^2*cos(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x +
11/2*c) + 8778*a^2*cos(8/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) +
3465*a^2*cos(6/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 1287*a^2*c
os(4/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) + 385*a^2*cos(2/11*arc
tan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c)))*sin(11/2*d*x + 11/2*c) - 31878*a^2*cos(11/2*d*x + 11/2*c
)*sin(10/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 8778*a^2*cos(11/2*d*x + 11/2*c)*sin(8/1
1*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 3465*a^2*cos(11/2*d*x + 11/2*c)*sin(6/11*arctan2(
sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 1287*a^2*cos(11/2*d*x + 11/2*c)*sin(4/11*arctan2(sin(11/2*d
*x + 11/2*c), cos(11/2*d*x + 11/2*c))) - 385*a^2*cos(11/2*d*x + 11/2*c)*sin(2/11*arctan2(sin(11/2*d*x + 11/2*c
), cos(11/2*d*x + 11/2*c))) + 126*a^2*sin(11/2*d*x + 11/2*c) + 385*a^2*sin(9/11*arctan2(sin(11/2*d*x + 11/2*c)
, cos(11/2*d*x + 11/2*c))) + 1287*a^2*sin(7/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 3465
*a^2*sin(5/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 8778*a^2*sin(3/11*arctan2(sin(11/2*d*
x + 11/2*c), cos(11/2*d*x + 11/2*c))) + 31878*a^2*sin(1/11*arctan2(sin(11/2*d*x + 11/2*c), cos(11/2*d*x + 11/2
*c))))*A*sqrt(a) + 132*sqrt(2)*(315*a^2*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d
*x + 7/2*c) + 77*a^2*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*a^
2*cos(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/
2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan
2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 21*a^2*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2
*c), cos(7/2*d*x + 7/2*c))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/
2*d*x + 7/2*c))) + 77*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a^2*sin(1/7*arcta
n2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*C*sqrt(a))/d

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Fricas [A]  time = 0.505933, size = 408, normalized size = 1.53 \begin{align*} \frac{2 \,{\left (63 \, A a^{2} \cos \left (d x + c\right )^{6} + 224 \, A a^{2} \cos \left (d x + c\right )^{5} +{\left (355 \, A + 99 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 6 \,{\left (71 \, A + 66 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} +{\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (568 \, A + 759 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{693 \,{\left (d \cos \left (d x + c\right ) + d\right )} \sqrt{\cos \left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

2/693*(63*A*a^2*cos(d*x + c)^6 + 224*A*a^2*cos(d*x + c)^5 + (355*A + 99*C)*a^2*cos(d*x + c)^4 + 6*(71*A + 66*C
)*a^2*cos(d*x + c)^3 + (568*A + 759*C)*a^2*cos(d*x + c)^2 + 2*(568*A + 759*C)*a^2*cos(d*x + c))*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sin(d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(11/2), x)