\(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx\) [108]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 42, antiderivative size = 290 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {22 a^3 c (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {22 a^3 c g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {22 a^2 c (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{9 f g \sqrt {c-c \sin (e+f x)}} \]

[Out]

-2/21*a*c*(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)+2/9*c*(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)-22/45*a^3*c*(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)+22/15*a^3*c*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)-22/105*a^2*c*
(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)

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2930, 2921, 2721, 2719} \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {22 a^3 c (g \cos (e+f x))^{5/2}}{45 f g \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {22 a^3 c g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{15 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {22 a^2 c \sqrt {a \sin (e+f x)+a} (g \cos (e+f x))^{5/2}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (a \sin (e+f x)+a)^{3/2} (g \cos (e+f x))^{5/2}}{21 f g \sqrt {c-c \sin (e+f x)}}+\frac {2 c (a \sin (e+f x)+a)^{5/2} (g \cos (e+f x))^{5/2}}{9 f g \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(-22*a^3*c*(g*Cos[e + f*x])^(5/2))/(45*f*g*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (22*a^3*c*g*Sq
rt[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]]) - (22*a^2*c*(g*Cos[e + f*x])^(5/2)*Sqrt[a + a*Sin[e + f*x]])/(105*f*g*Sqrt[c - c*Sin[e + f*x]]) -
(2*a*c*(g*Cos[e + f*x])^(5/2)*(a + a*Sin[e + f*x])^(3/2))/(21*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))/(9*f*g*Sqrt[c - c*Sin[e + f*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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2921

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 2930

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] &&
EqQ[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*} \text {integral}& = \frac {2 c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}{9 f g \sqrt {c-c \sin (e+f x)}}+\frac {1}{3} c \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 (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 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}}{9 f g \sqrt {c-c \sin (e+f x)}}+\frac {1}{21} (11 a c) \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 {22 a^2 c (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 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}}{9 f g \sqrt {c-c \sin (e+f x)}}+\frac {1}{15} \left (11 a^2 c\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 {22 a^3 c (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 {22 a^2 c (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 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}}{9 f g \sqrt {c-c \sin (e+f x)}}+\frac {1}{15} \left (11 a^3 c\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 {22 a^3 c (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 {22 a^2 c (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 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}}{9 f g \sqrt {c-c \sin (e+f x)}}+\frac {\left (11 a^3 c 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 {22 a^3 c (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 {22 a^2 c (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 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}}{9 f g \sqrt {c-c \sin (e+f x)}}+\frac {\left (11 a^3 c 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 {22 a^3 c (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 {22 a^3 c 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 {22 a^2 c (g \cos (e+f x))^{5/2} \sqrt {a+a \sin (e+f x)}}{105 f g \sqrt {c-c \sin (e+f x)}}-\frac {2 a c (g \cos (e+f x))^{5/2} (a+a \sin (e+f x))^{3/2}}{21 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}}{9 f g \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.09 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.97 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {a^3 e^{-4 i (e+f x)} \left (i+e^{i (e+f x)}\right ) g \sqrt {g \cos (e+f x)} \left (\sqrt {1+e^{2 i (e+f x)}} \left (-35+180 i e^{i (e+f x)}+238 e^{2 i (e+f x)}+540 i e^{3 i (e+f x)}+3696 e^{4 i (e+f x)}+540 i e^{5 i (e+f x)}-238 e^{6 i (e+f x)}+180 i e^{7 i (e+f x)}+35 e^{8 i (e+f x)}\right )-2464 e^{6 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (e+f x)}\right )\right ) \sqrt {c-c \sin (e+f x)}}{2520 \left (-i+e^{i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}} f \sqrt {a (1+\sin (e+f x))}} \]

[In]

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

[Out]

(a^3*(I + E^(I*(e + f*x)))*g*Sqrt[g*Cos[e + f*x]]*(Sqrt[1 + E^((2*I)*(e + f*x))]*(-35 + (180*I)*E^(I*(e + f*x)
) + 238*E^((2*I)*(e + f*x)) + (540*I)*E^((3*I)*(e + f*x)) + 3696*E^((4*I)*(e + f*x)) + (540*I)*E^((5*I)*(e + f
*x)) - 238*E^((6*I)*(e + f*x)) + (180*I)*E^((7*I)*(e + f*x)) + 35*E^((8*I)*(e + f*x))) - 2464*E^((6*I)*(e + f*
x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(e + f*x))])*Sqrt[c - c*Sin[e + f*x]])/(2520*E^((4*I)*(e + f*x)
)*(-I + E^(I*(e + f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))]*f*Sqrt[a*(1 + Sin[e + f*x])])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.42 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.71

method result size
default \(\frac {2 \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {g \cos \left (f x +e \right )}\, a^{2} g \left (231 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-231 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-35 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+462 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sec \left (f x +e \right )-462 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sec \left (f x +e \right )-90 \left (\cos ^{3}\left (f x +e \right )\right )-35 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+231 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (\sec ^{2}\left (f x +e \right )\right )-231 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (\sec ^{2}\left (f x +e \right )\right )-90 \left (\cos ^{2}\left (f x +e \right )\right )+77 \cos \left (f x +e \right ) \sin \left (f x +e \right )+77 \sin \left (f x +e \right )+231 \tan \left (f x +e \right )\right )}{315 f \left (1+\cos \left (f x +e \right )\right )}\) \(496\)

[In]

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/315/f*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)*(g*cos(f*x+e))^(1/2)*a^2*g/(1+cos(f*x+e))*(231*I*(c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)-231*I*(cos(f*x
+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)-35*cos(f*x+e)^3*sin(
f*x+e)+462*I*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)
*sec(f*x+e)-462*I*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e
)),I)*sec(f*x+e)-90*cos(f*x+e)^3-35*cos(f*x+e)^2*sin(f*x+e)+231*I*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(
f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*sec(f*x+e)^2-231*I*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/
(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*sec(f*x+e)^2-90*cos(f*x+e)^2+77*cos(f*x+e)*sin(f*
x+e)+77*sin(f*x+e)+231*tan(f*x+e))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.55 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {-231 i \, \sqrt {2} \sqrt {a c g} a^{2} g {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 231 i \, \sqrt {2} \sqrt {a c g} a^{2} g {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (90 \, a^{2} g \cos \left (f x + e\right )^{2} + 7 \, {\left (5 \, a^{2} g \cos \left (f x + e\right )^{2} - 11 \, a^{2} g\right )} \sin \left (f x + e\right )\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, f} \]

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/315*(-231*I*sqrt(2)*sqrt(a*c*g)*a^2*g*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin
(f*x + e))) + 231*I*sqrt(2)*sqrt(a*c*g)*a^2*g*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) -
 I*sin(f*x + e))) - 2*(90*a^2*g*cos(f*x + e)^2 + 7*(5*a^2*g*cos(f*x + e)^2 - 11*a^2*g)*sin(f*x + e))*sqrt(g*co
s(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/f

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate((g*cos(f*x+e))**(3/2)*(a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {c-c\,\sin \left (e+f\,x\right )} \,d x \]

[In]

int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(1/2),x)

[Out]

int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(1/2), x)