3.263 \(\int \frac{(a+a \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=163 \[ \frac{2 a^2 (4 A+5 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{5 d \sqrt{\sec (c+d x)}} \]

[Out]

(2*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(4*A + 5*C)*Sqrt[Sec[c + d*x
]]*Sin[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sqrt[Sec[
c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

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Rubi [A]  time = 0.464401, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {4087, 4017, 4015, 3801, 215} \[ \frac{2 a^2 (4 A+5 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}+\frac{2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{5 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(2*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*a^2*(4*A + 5*C)*Sqrt[Sec[c + d*x
]]*Sin[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sqrt[Sec[
c + d*x]]) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

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 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [B]  time = 6.29136, size = 428, normalized size = 2.63 \[ \frac{4 C \sin (c+d x) \cos ^3(c+d x) \sqrt{\sec ^2(c+d x)-1} (a (\sec (c+d x)+1))^{3/2} \left (\log \left (\sec ^{\frac{3}{2}}(c+d x)+\sqrt{\sec (c+d x)+1} \sqrt{\sec ^2(c+d x)-1}+\sqrt{\sec (c+d x)}\right )-\log (\sec (c+d x)+1)\right ) \left (A+C \sec ^2(c+d x)\right )}{d \left (1-\cos ^2(c+d x)\right ) (\sec (c+d x)+1)^{3/2} (A \cos (2 c+2 d x)+A+2 C)}+\frac{(a (\sec (c+d x)+1))^{3/2} \sqrt{(\cos (c+d x)+1) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac{(17 A+20 C) \sin (c) \cos (d x)}{5 d}+\frac{(17 A+20 C) \cos (c) \sin (d x)}{5 d}-\frac{4 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (4 A \sin \left (\frac{d x}{2}\right )+5 C \sin \left (\frac{d x}{2}\right )\right )}{5 d}-\frac{4 (4 A+5 C) \tan \left (\frac{c}{2}\right )}{5 d}+\frac{4 A \sin (2 c) \cos (2 d x)}{5 d}+\frac{A \sin (3 c) \cos (3 d x)}{5 d}+\frac{4 A \cos (2 c) \sin (2 d x)}{5 d}+\frac{A \cos (3 c) \sin (3 d x)}{5 d}\right )}{\sec ^{\frac{3}{2}}(c+d x) (\sec (c+d x)+1)^{3/2} (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(4*C*Cos[c + d*x]^3*(-Log[1 + Sec[c + d*x]] + Log[Sqrt[Sec[c + d*x]] + Sec[c + d*x]^(3/2) + Sqrt[1 + Sec[c + d
*x]]*Sqrt[-1 + Sec[c + d*x]^2]])*(a*(1 + Sec[c + d*x]))^(3/2)*Sqrt[-1 + Sec[c + d*x]^2]*(A + C*Sec[c + d*x]^2)
*Sin[c + d*x])/(d*(1 - Cos[c + d*x]^2)*(A + 2*C + A*Cos[2*c + 2*d*x])*(1 + Sec[c + d*x])^(3/2)) + (Sqrt[(1 + C
os[c + d*x])*Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2)*(A + C*Sec[c + d*x]^2)*(((17*A + 20*C)*Cos[d*x]*Sin[c]
)/(5*d) + (4*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (A*Cos[3*d*x]*Sin[3*c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(4*
A*Sin[(d*x)/2] + 5*C*Sin[(d*x)/2]))/(5*d) + ((17*A + 20*C)*Cos[c]*Sin[d*x])/(5*d) + (4*A*Cos[2*c]*Sin[2*d*x])/
(5*d) + (A*Cos[3*c]*Sin[3*d*x])/(5*d) - (4*(4*A + 5*C)*Tan[c/2])/(5*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c
+ d*x]^(3/2)*(1 + Sec[c + d*x])^(3/2))

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Maple [A]  time = 0.592, size = 222, normalized size = 1.4 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{10\,d\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 5\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) \sin \left ( dx+c \right ) -5\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) \sin \left ( dx+c \right ) +4\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+12\,A\cos \left ( dx+c \right ) +20\,C\cos \left ( dx+c \right ) -24\,A-20\,C \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(5*C*(-2/(cos(d*x+c)+1))^(1/2)*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(c
os(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*sin(d*x+c)-5*C*(-2/(cos(d*x+c)+1))^(1/2)*2^(1/2)*arctan(1/4*2^(
1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*sin(d*x+c)+4*A*cos(d*x+c)^3+8*A*cos(d*x+c)^2+12*A*co
s(d*x+c)+20*C*cos(d*x+c)-24*A-20*C)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)

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Maxima [B]  time = 2.1075, size = 655, normalized size = 4.02 \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))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/20*(sqrt(2)*(20*a*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) + 5*a*co
s(2/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 20*a*cos(5/2*d*x + 5/2*c)*si
n(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) - 5*a*cos(5/2*d*x + 5/2*c)*sin(2/5*arctan2(sin(5/2*
d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 2*a*sin(5/2*d*x + 5/2*c) + 5*a*sin(3/5*arctan2(sin(5/2*d*x + 5/2*c), co
s(5/2*d*x + 5/2*c))) + 20*a*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*A*sqrt(a) + 5*sqrt(2
)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt
(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*
cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/
2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(
1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c
) + 2) + 8*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a))/d

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Fricas [A]  time = 0.60433, size = 1034, normalized size = 6.34 \begin{align*} \left [\frac{5 \,{\left (C a \cos \left (d x + c\right ) + C a\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac{4 \,{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac{4 \,{\left (A a \cos \left (d x + c\right )^{3} + 3 \, A a \cos \left (d x + c\right )^{2} +{\left (6 \, A + 5 \, C\right )} a \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 )}{\sqrt{\cos \left (d x + c\right )}}}{10 \,{\left (d \cos \left (d x + c\right ) + d\right )}}, \frac{5 \,{\left (C a \cos \left (d x + c\right ) + C a\right )} \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac{2 \,{\left (A a \cos \left (d x + c\right )^{3} + 3 \, A a \cos \left (d x + c\right )^{2} +{\left (6 \, A + 5 \, C\right )} a \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 )}{\sqrt{\cos \left (d x + c\right )}}}{5 \,{\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/10*(5*(C*a*cos(d*x + c) + C*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*c
os(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x +
c)^3 + cos(d*x + c)^2)) + 4*(A*a*cos(d*x + c)^3 + 3*A*a*cos(d*x + c)^2 + (6*A + 5*C)*a*cos(d*x + c))*sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/5*(5*(C*a*cos(d*x + c
) + C*a)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a
*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) + 2*(A*a*cos(d*x + c)^3 + 3*A*a*cos(d*x + c)^2 + (6*A + 5*C)*a*cos(d*
x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)]

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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))**(3/2)*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2),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{3}{2}}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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