3.3.32 \(\int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [232]

Optimal. Leaf size=190 \[ \frac {(A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(3 A+5 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]

[Out]

1/3*(3*A+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d-(A+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))-(A+3*C)*sin
(d*x+c)*sec(d*x+c)^(1/2)/a/d+(A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2
*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d+1/3*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*
c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

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Rubi [A]
time = 0.15, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4170, 3872, 3853, 3856, 2719, 2720} \begin {gather*} -\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(3 A+5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}-\frac {(A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}+\frac {(3 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 d}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

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

Rule 2720

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

Rule 3853

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 3856

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 3872

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 4170

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)]

Rubi steps

\begin {align*} \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {\int \sec ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a (A+3 C)-\frac {1}{2} a (3 A+5 C) \sec (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(A+3 C) \int \sec ^{\frac {3}{2}}(c+d x) \, dx}{2 a}+\frac {(3 A+5 C) \int \sec ^{\frac {5}{2}}(c+d x) \, dx}{2 a}\\ &=-\frac {(A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {(A+3 C) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{2 a}+\frac {(3 A+5 C) \int \sqrt {\sec (c+d x)} \, dx}{6 a}\\ &=-\frac {(A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\left ((A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a}+\frac {\left ((3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a}\\ &=\frac {(A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(3 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 d}-\frac {(A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}+\frac {(3 A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 4.66, size = 324, normalized size = 1.71 \begin {gather*} \frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (-i (A+3 C) e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}+e^{2 i (c+d x)}+e^{3 i (c+d x)}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+2 (3 A+5 C) \sqrt {\cos (c+d x)} \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+i \sin \left (\frac {1}{2} (c+d x)\right )\right ) (\cos (c+d x)-i \sin (c+d x))+2 i (3 A+5 C+6 C \cos (c+d x)+(3 A+7 C) \cos (2 (c+d x))-2 i C \sin (c+d x)+2 i C \sin (2 (c+d x)))\right ) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right )}{6 a d (1+\sec (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(((-I)*(A + 3*C)*Sqrt[1 + E^((2*I)*(c + d*x))]*(1 + E^(I*(c + d*x)) + E^(
(2*I)*(c + d*x)) + E^((3*I)*(c + d*x)))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)
) + 2*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2])*EllipticF[(c + d*x)/2, 2]*(Cos[
(c + d*x)/2] + I*Sin[(c + d*x)/2])*(Cos[c + d*x] - I*Sin[c + d*x]) + (2*I)*(3*A + 5*C + 6*C*Cos[c + d*x] + (3*
A + 7*C)*Cos[2*(c + d*x)] - (2*I)*C*Sin[c + d*x] + (2*I)*C*Sin[2*(c + d*x)]))*(Cos[(c + 3*d*x)/2] + I*Sin[(c +
3*d*x)/2]))/(6*a*d*E^(I*d*x)*(1 + Sec[c + d*x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(458\) vs. \(2(224)=448\).
time = 6.93, size = 459, normalized size = 2.42

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\frac {\left (A +C \right ) \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}+2 C \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}{6 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}}\right )-\frac {2 C \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(459\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((A+C)*(cos(1/2*d*x+1/2*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)*(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)+2*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)))-2*C/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2
*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.00, size = 302, normalized size = 1.59 \begin {gather*} \frac {{\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, C \cos \left (d x + c\right ) - 2 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(3*I*A + 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(3*I*A + 5*I*C)*cos(d*x
+ c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c)^2 +
sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*
x + c))) - 3*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A + 3*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0
, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(A + 3*C)*cos(d*x + c)^2 + 4*C*cos(d*x + c
) - 2*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^2 + a*d*cos(d*x + c))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x)),x)

[Out]

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x)), x)

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