3.107 \(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=352 \[ -\frac {14 a^3 c^2 (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {14 a^3 c^2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{15 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 c^2 \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{15 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{11 f g} \]

[Out]

-2/33*a*c^2*(g*cos(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(3/2)/f/g/(c-c*sin(f*x+e))^(1/2)+14/99*c^2*(g*cos(f*x+e))^(5
/2)*(a+a*sin(f*x+e))^(5/2)/f/g/(c-c*sin(f*x+e))^(1/2)-14/45*a^3*c^2*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^
(1/2)/(c-c*sin(f*x+e))^(1/2)+14/15*a^3*c^2*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2
*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-2/1
5*a^2*c^2*(g*cos(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/f/g/(c-c*sin(f*x+e))^(1/2)+2/11*c*(g*cos(f*x+e))^(5/2)*(
a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2)/f/g

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Rubi [A]  time = 1.76, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2851, 2842, 2640, 2639} \[ -\frac {14 a^3 c^2 (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 c^2 \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{15 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 a^3 c^2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{15 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{11 f g} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(-14*a^3*c^2*(g*Cos[e + f*x])^(5/2))/(45*f*g*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (14*a^3*c^2*
g*Sqrt[Cos[e + f*x]]*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(15*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c
*Sin[e + f*x]]) - (2*a^2*c^2*(g*Cos[e + f*x])^(5/2)*Sqrt[a + a*Sin[e + f*x]])/(15*f*g*Sqrt[c - c*Sin[e + f*x]]
) - (2*a*c^2*(g*Cos[e + f*x])^(5/2)*(a + a*Sin[e + f*x])^(3/2))/(33*f*g*Sqrt[c - c*Sin[e + f*x]]) + (14*c^2*(g
*Cos[e + f*x])^(5/2)*(a + a*Sin[e + f*x])^(5/2))/(99*f*g*Sqrt[c - c*Sin[e + f*x]]) + (2*c*(g*Cos[e + f*x])^(5/
2)*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])/(11*f*g)

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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2842

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

Rule 2851

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e +
f*x])^n)/(f*g*(m + n + p)), x] + Dist[(a*(2*m + p - 1))/(m + n + p), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && Eq
Q[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + n + p, 0] &&  !LtQ[0, n, m] && IntegersQ[2*m, 2*n, 2*p]

Rubi steps

\begin {align*} \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2} \, dx &=\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {1}{11} (7 c) \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx\\ &=\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {1}{33} \left (7 c^2\right ) \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {1}{3} \left (a c^2\right ) \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a^2 c^2 (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{15 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {1}{15} \left (7 a^2 c^2\right ) \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {14 a^3 c^2 (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 c^2 (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{15 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {1}{15} \left (7 a^3 c^2\right ) \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {14 a^3 c^2 (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 c^2 (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{15 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {\left (7 a^3 c^2 g \cos (e+f x)\right ) \int \sqrt {g \cos (e+f x)} \, dx}{15 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {14 a^3 c^2 (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 c^2 (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{15 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}+\frac {\left (7 a^3 c^2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{15 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {14 a^3 c^2 (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {14 a^3 c^2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 a^2 c^2 (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{15 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{33 f g \sqrt {c-c \sin (e+f x)}}+\frac {14 c^2 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{99 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{11 f g}\\ \end {align*}

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Mathematica [A]  time = 1.28, size = 189, normalized size = 0.54 \[ \frac {c (\sin (e+f x)-1) (a (\sin (e+f x)+1))^{5/2} \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{3/2} \left (\sqrt {\cos (e+f x)} (-836 \sin (2 (e+f x))-110 \sin (4 (e+f x))+450 \cos (e+f x)+225 \cos (3 (e+f x))+45 \cos (5 (e+f x)))-3696 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )\right )}{3960 f \cos ^{\frac {3}{2}}(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(c*(g*Cos[e + f*x])^(3/2)*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]]*(-3696*Ell
ipticE[(e + f*x)/2, 2] + Sqrt[Cos[e + f*x]]*(450*Cos[e + f*x] + 225*Cos[3*(e + f*x)] + 45*Cos[5*(e + f*x)] - 8
36*Sin[2*(e + f*x)] - 110*Sin[4*(e + f*x)])))/(3960*f*Cos[e + f*x]^(3/2)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])
^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c g \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{2} c g \cos \left (f x + e\right )^{3}\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c*g*cos(f*x + e)^3*sin(f*x + e) + a^2*c*g*cos(f*x + e)^3)*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x +
e) + a)*sqrt(-c*sin(f*x + e) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(3/2), x)

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maple [C]  time = 0.68, size = 382, normalized size = 1.09 \[ -\frac {2 \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} \left (45 \left (\cos ^{6}\left (f x +e \right )\right ) \sin \left (f x +e \right )+231 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )-231 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+55 \left (\cos ^{6}\left (f x +e \right )\right )+231 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )-231 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+22 \left (\cos ^{4}\left (f x +e \right )\right )+154 \left (\cos ^{2}\left (f x +e \right )\right )-231 \cos \left (f x +e \right )\right ) \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{495 f \left (1+\sin \left (f x +e \right )\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(3/2),x)

[Out]

-2/495/f*(-c*(sin(f*x+e)-1))^(3/2)*(45*cos(f*x+e)^6*sin(f*x+e)+231*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos
(f*x+e)+1))^(1/2)*sin(f*x+e)*cos(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)-231*I*(1/(cos(f*x+e)+1))^(1/
2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*cos(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)+55*cos(f*
x+e)^6+231*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))
/sin(f*x+e),I)-231*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*EllipticF(I*(-1+cos
(f*x+e))/sin(f*x+e),I)+22*cos(f*x+e)^4+154*cos(f*x+e)^2-231*cos(f*x+e))*(g*cos(f*x+e))^(3/2)*(a*(1+sin(f*x+e))
)^(5/2)/(1+sin(f*x+e))/sin(f*x+e)/cos(f*x+e)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2),x)

[Out]

int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2), 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((g*cos(f*x+e))**(3/2)*(a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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