3.742 \(\int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=391 \[ -\frac {2 \left (-a^2 d^2+2 a b c d+b^2 \left (8 c^2-9 d^2\right )\right ) (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (4 a^3 c d^3-3 a^2 b d^2 \left (c^2+3 d^2\right )-6 a b^2 c d \left (c^2-3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f \left (c^2-d^2\right )^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 \left (a c d+b \left (c^2-2 d^2\right )\right ) (b c-a d)^2 \cos (e+f x)}{3 d^2 f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}+\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}} \]

[Out]

2/3*(-a*d+b*c)^2*cos(f*x+e)*(a+b*sin(f*x+e))/d/(c^2-d^2)/f/(c+d*sin(f*x+e))^(3/2)+8/3*(-a*d+b*c)^2*(a*c*d+b*(c
^2-2*d^2))*cos(f*x+e)/d^2/(c^2-d^2)^2/f/(c+d*sin(f*x+e))^(1/2)-2/3*(4*a^3*c*d^3-6*a*b^2*c*d*(c^2-3*d^2)-3*a^2*
b*d^2*(c^2+3*d^2)+b^3*(8*c^4-15*c^2*d^2+3*d^4))*(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(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))^(1/2)/d^3/(c^2-d^2)^2/f/((c+d*si
n(f*x+e))/(c+d))^(1/2)+2/3*(-a*d+b*c)*(2*a*b*c*d-a^2*d^2+b^2*(8*c^2-9*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^2-d^2)/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.75, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2792, 3021, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (-a^2 d^2+2 a b c d+b^2 \left (8 c^2-9 d^2\right )\right ) (b c-a d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (-3 a^2 b d^2 \left (c^2+3 d^2\right )+4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )+b^3 \left (-15 c^2 d^2+8 c^4+3 d^4\right )\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^3 f \left (c^2-d^2\right )^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {8 \left (a c d+b \left (c^2-2 d^2\right )\right ) (b c-a d)^2 \cos (e+f x)}{3 d^2 f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}+\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(2*(b*c - a*d)^2*Cos[e + f*x]*(a + b*Sin[e + f*x]))/(3*d*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2)) + (8*(b*c -
a*d)^2*(a*c*d + b*(c^2 - 2*d^2))*Cos[e + f*x])/(3*d^2*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) + (2*(4*a^3*c
*d^3 - 6*a*b^2*c*d*(c^2 - 3*d^2) - 3*a^2*b*d^2*(c^2 + 3*d^2) + b^3*(8*c^4 - 15*c^2*d^2 + 3*d^4))*EllipticE[(e
- Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*d^3*(c^2 - d^2)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)]) - (2*(b*c - a*d)*(2*a*b*c*d - a^2*d^2 + b^2*(8*c^2 - 9*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d
)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d^3*(c^2 - 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 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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+b \sin (e+f x))^3}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (2 b (b c-a d)^2-3 a d \left (\left (a^2+b^2\right ) c-2 a b d\right )\right )-\frac {1}{2} \left (5 a^2 b c d+3 b^3 c d-a^3 d^2+a b^2 \left (2 c^2-9 d^2\right )\right ) \sin (e+f x)+\frac {1}{2} b \left (2 a b c d-a^2 d^2-b^2 \left (4 c^2-3 d^2\right )\right ) \sin ^2(e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d \left (c^2-d^2\right )}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac {8 (b c-a d)^2 \left (a c d+b \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}+\frac {4 \int \frac {-\frac {1}{4} d \left (12 a^2 b c d^2-a^3 d \left (3 c^2+d^2\right )-3 a b^2 d \left (c^2+3 d^2\right )-2 b^3 \left (c^3-3 c d^2\right )\right )+\frac {1}{4} \left (4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )-3 a^2 b d^2 \left (c^2+3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d^2 \left (c^2-d^2\right )^2}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac {8 (b c-a d)^2 \left (a c d+b \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )-3 a^2 b d^2 \left (c^2+3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{3 d^3 \left (c^2-d^2\right )^2}--\frac {\left (4 \left (-\frac {1}{4} d^2 \left (12 a^2 b c d^2-a^3 d \left (3 c^2+d^2\right )-3 a b^2 d \left (c^2+3 d^2\right )-2 b^3 \left (c^3-3 c d^2\right )\right )-\frac {1}{4} c \left (4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )-3 a^2 b d^2 \left (c^2+3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right )\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d^3 \left (c^2-d^2\right )^2}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac {8 (b c-a d)^2 \left (a c d+b \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (\left (4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )-3 a^2 b d^2 \left (c^2+3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{3 d^3 \left (c^2-d^2\right )^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}--\frac {\left (4 \left (-\frac {1}{4} d^2 \left (12 a^2 b c d^2-a^3 d \left (3 c^2+d^2\right )-3 a b^2 d \left (c^2+3 d^2\right )-2 b^3 \left (c^3-3 c d^2\right )\right )-\frac {1}{4} c \left (4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )-3 a^2 b d^2 \left (c^2+3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right )\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}{3 d^3 \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 (b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac {8 (b c-a d)^2 \left (a c d+b \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (4 a^3 c d^3-6 a b^2 c d \left (c^2-3 d^2\right )-3 a^2 b d^2 \left (c^2+3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\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)}}{3 d^3 \left (c^2-d^2\right )^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (b c-a d) \left (8 b^2 c^2+2 a b c d-a^2 d^2-9 b^2 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}}}{3 d^3 \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 3.73, size = 357, normalized size = 0.91 \[ \frac {2 \left (\frac {(-c-d \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \left (\left (4 a^3 c d^3-3 a^2 b d^2 \left (c^2+3 d^2\right )-6 a b^2 c d \left (c^2-3 d^2\right )+b^3 \left (8 c^4-15 c^2 d^2+3 d^4\right )\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )-c F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )+d^2 \left (a^3 d \left (3 c^2+d^2\right )-12 a^2 b c d^2+3 a b^2 d \left (c^2+3 d^2\right )+2 b^3 \left (c^3-3 c d^2\right )\right ) F\left (\frac {1}{4} (-2 e-2 f x+\pi )|\frac {2 d}{c+d}\right )\right )}{(c-d)^2 (c+d)^2}-\frac {d (b c-a d)^2 \cos (e+f x) \left (d \left (-4 a c d-5 b c^2+9 b d^2\right ) \sin (e+f x)-5 a c^2 d+a d^3-4 b c^3+8 b c d^2\right )}{\left (c^2-d^2\right )^2}\right )}{3 d^3 f (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^3/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(2*(((d^2*(-12*a^2*b*c*d^2 + a^3*d*(3*c^2 + d^2) + 3*a*b^2*d*(c^2 + 3*d^2) + 2*b^3*(c^3 - 3*c*d^2))*EllipticF[
(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] + (4*a^3*c*d^3 - 6*a*b^2*c*d*(c^2 - 3*d^2) - 3*a^2*b*d^2*(c^2 + 3*d^2) +
b^3*(8*c^4 - 15*c^2*d^2 + 3*d^4))*((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])*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/((c - d)^2*(c +
d)^2) - (d*(b*c - a*d)^2*Cos[e + f*x]*(-4*b*c^3 - 5*a*c^2*d + 8*b*c*d^2 + a*d^3 + d*(-5*b*c^2 - 4*a*c*d + 9*b*
d^2)*Sin[e + f*x]))/(c^2 - d^2)^2))/(3*d^3*f*(c + d*Sin[e + f*x])^(3/2))

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, a b^{2} \cos \left (f x + e\right )^{2} - a^{3} - 3 \, a b^{2} + {\left (b^{3} \cos \left (f x + e\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} + {\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)^3/(d*sin(f*x + e) + c)^(5/2), x)

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maple [B]  time = 6.79, size = 1379, normalized size = 3.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^3/(c+d*sin(f*x+e))^(5/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(b^2/d^3*(2*b*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)
))+6*d*a*(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))-4*c*b*(
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)))+1/d^3*(a^3*d^3-3
*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*(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))))+3*b/d^3*(a^2*d^2-2*a*b*c*d+b^2*c^2)*(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)*Ellip
ticF(((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)))))/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 \[ \int \frac {{\left (b \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^3/(d*sin(f*x + e) + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^3/(c + d*sin(e + f*x))^(5/2),x)

[Out]

int((a + b*sin(e + f*x))^3/(c + d*sin(e + f*x))^(5/2), 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+b*sin(f*x+e))**3/(c+d*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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