3.502 \(\int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=336 \[ -\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 f (c+d)^3 \sqrt {c+d \sin (e+f x)}}+\frac {4 a^3 \left (4 c^2+11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^3 f (c+d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^3 f (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac {2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

[Out]

2/5*(c-d)*cos(f*x+e)*(a^3+a^3*sin(f*x+e))/d/(c+d)/f/(c+d*sin(f*x+e))^(5/2)+8/15*a^3*(c-d)*(c+3*d)*cos(f*x+e)/d
^2/(c+d)^2/f/(c+d*sin(f*x+e))^(3/2)-4/15*a^3*(4*c^2+15*c*d+27*d^2)*cos(f*x+e)/d^2/(c+d)^3/f/(c+d*sin(f*x+e))^(
1/2)+4/15*a^3*(4*c^2+15*c*d+27*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(co
s(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/d^3/(c+d)^3/f/((c+d*sin(f*x+e))/(c+d))
^(1/2)-4/15*a^3*(4*c^2+11*c*d+15*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(
cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/d^3/(c+d)^2/f/(c+d*sin(f*x+e
))^(1/2)

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Rubi [A]  time = 0.72, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2762, 2968, 3021, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 f (c+d)^3 \sqrt {c+d \sin (e+f x)}}+\frac {4 a^3 \left (4 c^2+11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^3 f (c+d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^3 f (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac {2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*(c - d)*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(5*d*(c + d)*f*(c + d*Sin[e + f*x])^(5/2)) + (8*a^3*(c - d)*
(c + 3*d)*Cos[e + f*x])/(15*d^2*(c + d)^2*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^3*(4*c^2 + 15*c*d + 27*d^2)*Cos
[e + f*x])/(15*d^2*(c + d)^3*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^3*(4*c^2 + 15*c*d + 27*d^2)*EllipticE[(e - Pi/
2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(15*d^3*(c + d)^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) +
(4*a^3*(4*c^2 + 11*c*d + 15*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c +
d)])/(15*d^3*(c + d)^2*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{7/2}} \, dx &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {(2 a) \int \frac {(a+a \sin (e+f x)) (a (c-6 d)-a (2 c+3 d) \sin (e+f x))}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {(2 a) \int \frac {a^2 (c-6 d)+\left (a^2 (c-6 d)-a^2 (2 c+3 d)\right ) \sin (e+f x)-a^2 (2 c+3 d) \sin ^2(e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}+\frac {(4 a) \int \frac {\frac {3}{2} a^2 (c-d) d (c+9 d)+\frac {1}{2} a^2 (c-d) \left (4 c^2+11 c d+15 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 (c-d) d^2 (c+d)^2}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {(8 a) \int \frac {\frac {1}{4} a^2 (c-15 d) (c-d)^2 d+\frac {1}{4} a^2 (c-d)^2 \left (4 c^2+15 c d+27 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 (c-d)^2 d^2 (c+d)^3}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (2 a^3 \left (4 c^2+11 c d+15 d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^3 (c+d)^2}-\frac {\left (2 a^3 \left (4 c^2+15 c d+27 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 d^3 (c+d)^3}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 a^3 \left (4 c^2+15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{15 d^3 (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^3 \left (4 c^2+11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{15 d^3 (c+d)^2 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac {8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^3 \left (4 c^2+15 c d+27 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d^3 (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^3 \left (4 c^2+11 c d+15 d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d^3 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.12, size = 298, normalized size = 0.89 \[ -\frac {2 a^3 (\sin (e+f x)+1)^3 \left (d \cos (e+f x) \left (4 c^4+15 c^3 d+2 d^2 \left (4 c^2+15 c d+27 d^2\right ) \sin ^2(e+f x)+55 c^2 d^2+d \left (9 c^3+45 c^2 d+115 c d^2+15 d^3\right ) \sin (e+f x)+15 c d^3+3 d^4\right )-2 (c+d \sin (e+f x))^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left (\left (4 c^2+15 c d+27 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )+d^2 (c-15 d) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )\right )}{15 d^3 f (c+d)^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6 (c+d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(-2*a^3*(1 + Sin[e + f*x])^3*(-2*((c - 15*d)*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] + (4*c^2 + 15
*c*d + 27*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (
2*d)/(c + d)]))*(c + d*Sin[e + f*x])^2*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + d*Cos[e + f*x]*(4*c^4 + 15*c^3*d +
55*c^2*d^2 + 15*c*d^3 + 3*d^4 + d*(9*c^3 + 45*c^2*d + 115*c*d^2 + 15*d^3)*Sin[e + f*x] + 2*d^2*(4*c^2 + 15*c*
d + 27*d^2)*Sin[e + f*x]^2)))/(15*d^3*(c + d)^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c + d*Sin[e + f*x])
^(5/2))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{d^{4} \cos \left (f x + e\right )^{4} + c^{4} + 6 \, c^{2} d^{2} + d^{4} - 2 \, {\left (3 \, c^{2} d^{2} + d^{4}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (c d^{3} \cos \left (f x + e\right )^{2} - c^{3} d - c d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

integral(-(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c)/
(d^4*cos(f*x + e)^4 + c^4 + 6*c^2*d^2 + d^4 - 2*(3*c^2*d^2 + d^4)*cos(f*x + e)^2 - 4*(c*d^3*cos(f*x + e)^2 - c
^3*d - c*d^3)*sin(f*x + e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^3/(d*sin(f*x + e) + c)^(7/2), x)

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maple [B]  time = 6.60, size = 1589, normalized size = 4.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(7/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*a^3*(2/d^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c
+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))
/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+3*(-c+d)/d^3*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^
(1/2)+2*c/(c^2-d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(
c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/
2))+2/(c^2-d^2)*d*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-
d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c
+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+3/d^3*(c^2-2*c*d+d^2)*(2/3/(c^2-d^
2)/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x
+e)-c)*cos(f*x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-
sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c
+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+8/3*c*d/(c^2-d^2)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d
*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)
*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)
/(c+d))^(1/2))))+1/d^3*(-c^3+3*c^2*d-3*c*d^2+d^3)*(2/5/(c^2-d^2)/d^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(
sin(f*x+e)+c/d)^3+16/15*c/(c^2-d^2)^2/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+2/15*d*cos(
f*x+e)^2/(c^2-d^2)^3*(23*c^2+9*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(15*c^3+17*c*d^2)/(15*c^6-45*c^4
*d^2+45*c^2*d^4-15*d^6)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)
*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))
^(1/2))+2/15*d*(23*c^2+9*d^2)/(c^2-d^2)^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2
)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e)
)/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))))/cos(f*x+e
)/(c+d*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(d*sin(f*x + e) + c)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3/(c + d*sin(e + f*x))^(7/2),x)

[Out]

int((a + a*sin(e + f*x))^3/(c + d*sin(e + f*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**(7/2),x)

[Out]

Timed out

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