3.4.37 \(\int \frac {1}{(a+a \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x)} \, dx\) [337]

Optimal. Leaf size=195 \[ \frac {49 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {13 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-1/5*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^3-8/15*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^2-
13/6*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a^3+a^3*sec(d*x+c))+49/10*(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^3/d-13/6*(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^3/d

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Rubi [A]
time = 0.23, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3317, 3902, 4105, 3872, 3856, 2719, 2720} \begin {gather*} -\frac {13 \sin (c+d x) \sqrt {\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {13 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {49 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {8 \sin (c+d x) \sqrt {\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Cos[c + d*x])^3*Sec[c + d*x]^(7/2)),x]

[Out]

(49*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - (13*Sqrt[Cos[c + d*x]]*Ellip
ticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x
])^3) - (8*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (13*Sqrt[Sec[c + d*x]]*Sin[c + d
*x])/(6*d*(a^3 + a^3*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 3317

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

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 3902

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

Rule 4105

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

Rubi steps

\begin {align*} \int \frac {1}{(a+a \cos (c+d x))^3 \sec ^{\frac {7}{2}}(c+d x)} \, dx &=\int \frac {1}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-\frac {11 a}{2}+\frac {5}{2} a \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {-\frac {41 a^2}{2}+12 a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \frac {-\frac {147 a^3}{4}+\frac {65}{4} a^3 \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {13 \int \sqrt {\sec (c+d x)} \, dx}{12 a^3}+\frac {49 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\left (13 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}+\frac {\left (49 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac {49 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {13 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {8 \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {13 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 2.19, size = 378, normalized size = 1.94 \begin {gather*} \frac {2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (147 \left (1+e^{2 i (c+d x)}\right )+147 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+65 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}-\frac {1}{32} \left (1134 \cos \left (\frac {1}{2} (c-d x)\right )+1071 \cos \left (\frac {1}{2} (3 c+d x)\right )+923 \cos \left (\frac {1}{2} (c+3 d x)\right )+694 \cos \left (\frac {1}{2} (5 c+3 d x)\right )+470 \cos \left (\frac {1}{2} (3 c+5 d x)\right )+265 \cos \left (\frac {1}{2} (7 c+5 d x)\right )+117 \cos \left (\frac {1}{2} (5 c+7 d x)\right )+30 \cos \left (\frac {1}{2} (9 c+7 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}\right )}{15 a^3 d (1+\cos (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Cos[c + d*x])^3*Sec[c + d*x]^(7/2)),x]

[Out]

(2*Cos[(c + d*x)/2]^6*(((2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(147*(1 + E^((2*I)*(c +
d*x))) + 147*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c +
d*x))] + 65*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4,
-E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) - ((1134*Cos[(c - d*x)/2] + 1071*Cos[(3*c + d*x)
/2] + 923*Cos[(c + 3*d*x)/2] + 694*Cos[(5*c + 3*d*x)/2] + 470*Cos[(3*c + 5*d*x)/2] + 265*Cos[(7*c + 5*d*x)/2]
+ 117*Cos[(5*c + 7*d*x)/2] + 30*Cos[(9*c + 7*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^5*Sqrt[Sec[c + d*x]])
/32))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]
time = 0.22, size = 270, normalized size = 1.38

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (348 \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+130 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+294 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-578 \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+264 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-37 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3\right )}{60 a^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*cos(d*x+c))^3/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*cos(1/2*d*x+1/2*c)^8+130*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+294*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),
2^(1/2))-578*cos(1/2*d*x+1/2*c)^6+264*cos(1/2*d*x+1/2*c)^4-37*cos(1/2*d*x+1/2*c)^2+3)/a^3/(-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)^5/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(1/(a+a*cos(d*x+c))^3/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate(1/((a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.15, size = 353, normalized size = 1.81 \begin {gather*} -\frac {65 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 65 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 147 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 3 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\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 ) + 147 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 3 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\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 (87 \, \cos \left (d x + c\right )^{3} + 146 \, \cos \left (d x + c\right )^{2} + 65 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^3/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/60*(65*(-I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weie
rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 65*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2)*cos(d*x + c
)^2 + 3*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 147*(-
I*sqrt(2)*cos(d*x + c)^3 - 3*I*sqrt(2)*cos(d*x + c)^2 - 3*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(
-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 147*(I*sqrt(2)*cos(d*x + c)^3 + 3*I*sqrt(2
)*cos(d*x + c)^2 + 3*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos
(d*x + c) - I*sin(d*x + c))) + 2*(87*cos(d*x + c)^3 + 146*cos(d*x + c)^2 + 65*cos(d*x + c))*sin(d*x + c)/sqrt(
cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))**3/sec(d*x+c)**(7/2),x)

[Out]

Timed out

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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(1/(a+a*cos(d*x+c))^3/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^3*sec(d*x + c)^(7/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^3),x)

[Out]

int(1/((1/cos(c + d*x))^(7/2)*(a + a*cos(c + d*x))^3), x)

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