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

Optimal. Leaf size=229 \[ \frac{2 (A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}-\frac{(A+7 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{2 (A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a^2 d}-\frac{(A+7 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{(A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

((A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + (2*(A + 5*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A + 7*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(a^2*d) + (2*(A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d) - ((A + 7*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x]
)/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.379498, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4085, 4019, 3787, 3768, 3771, 2639, 2641} \[ -\frac{(A+7 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{2 (A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a^2 d}-\frac{(A+7 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{2 (A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + (2*(A + 5*C)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - ((A + 7*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/
(a^2*d) + (2*(A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d) - ((A + 7*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x]
)/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac{(A+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (-\frac{1}{2} a (A-5 C)-\frac{3}{2} a (A+3 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(A+7 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \sec ^{\frac{3}{2}}(c+d x) \left (\frac{3}{2} a^2 (A+7 C)-3 a^2 (A+5 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(A+7 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(A+5 C) \int \sec ^{\frac{5}{2}}(c+d x) \, dx}{a^2}-\frac{(A+7 C) \int \sec ^{\frac{3}{2}}(c+d x) \, dx}{2 a^2}\\ &=-\frac{(A+7 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac{2 (A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac{(A+7 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{(A+5 C) \int \sqrt{\sec (c+d x)} \, dx}{3 a^2}+\frac{(A+7 C) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+7 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac{2 (A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac{(A+7 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\left ((A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2}+\frac{\left ((A+7 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=\frac{(A+7 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{2 (A+5 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}-\frac{(A+7 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac{2 (A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac{(A+7 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 7.4472, size = 884, normalized size = 3.86 \[ -\frac{2 \sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \left (C \sec ^2(c+d x)+A\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}-\frac{14 \sqrt{2} C e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \sec \left (\frac{c}{2}\right ) \left (C \sec ^2(c+d x)+A\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{8 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{40 C \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2}+\frac{\sqrt{\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \left (\frac{4 \sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+C \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{4 (A+C) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{16 \sec \left (\frac{c}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )+4 C \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{4 (A+7 C) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{d}+\frac{16 C \sec (c) \sec (c+d x) \sin (d x)}{3 d}+\frac{16 (5 \cos (c) C+C+A \cos (c)) \sec (c) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (2 c+2 d x) A+A+2 C) (\sec (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]
^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a +
a*Sec[c + d*x])^2) - (14*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x
))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hyperge
ometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + A*
Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (8*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c
+ d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(
a + a*Sec[c + d*x])^2) + (40*C*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[
c/2]*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x]
)^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((-4*(A + 7*C)*Cos[d*x]*Csc[c/2]*Sec[c/
2])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (16*Sec[c/2]*Sec[c/2 + (d*
x)/2]*(A*Sin[(d*x)/2] + 4*C*Sin[(d*x)/2]))/(3*d) + (16*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/(3*d) + (16*(C + A*Cos[
c] + 5*C*Cos[c])*Sec[c]*Tan[c/2])/(3*d) + (4*(A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/((A + 2*C + A*Cos[
2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)

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Maple [B]  time = 7.02, size = 738, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(1/3*(A+C)*(2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(
2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2
*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(
1/2)/cos(1/2*d*x+1/2*c)/(sin(1/2*d*x+1/2*c)^2-1)+4*C*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)
^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+4*C*(cos(1/
2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-8*C*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(-2*sin(1
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*s
in(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{2}\right )} \sqrt{\sec \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*sqrt(sec(d*x + c))/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a
^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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