3.164 \(\int \frac{(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=114 \[ \frac{g^4 2^{m+\frac{9}{4}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac{5}{4},-m-\frac{1}{4};-\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{5 a c^2 f \sqrt{c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}} \]
[Out]
(2^(9/4 + m)*g^4*Cos[e + f*x]*Hypergeometric2F1[-5/4, -1/4 - m, -1/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])
^(-1/4 - m)*(a + a*Sin[e + f*x])^(1 + m))/(5*a*c^2*f*(g*Cos[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])
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Rubi [A] time = 0.357219, antiderivative size = 114, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used =
{2853, 2689, 70, 69} \[ \frac{g^4 2^{m+\frac{9}{4}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{4}} (a \sin (e+f x)+a)^{m+1} \, _2F_1\left (-\frac{5}{4},-m-\frac{1}{4};-\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right )}{5 a c^2 f \sqrt{c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[((g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m)/(c - c*Sin[e + f*x])^(5/2),x]
[Out]
(2^(9/4 + m)*g^4*Cos[e + f*x]*Hypergeometric2F1[-5/4, -1/4 - m, -1/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])
^(-1/4 - m)*(a + a*Sin[e + f*x])^(1 + m))/(5*a*c^2*f*(g*Cos[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])
Rule 2853
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e
+ f*x])^FracPart[m])/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m])), Int[(g*Cos[e + f*x])^(2*m + p)*(c +
d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 -
b^2, 0] && (FractionQ[m] || !FractionQ[n])
Rule 2689
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a^2*
(g*Cos[e + f*x])^(p + 1))/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Rule 70
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] &&
(RationalQ[m] || !SimplerQ[n + 1, m + 1])
Rule 69
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))
Rubi steps
\begin{align*} \int \frac{(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{5/2}} \, dx &=\frac{\left (g^5 \cos (e+f x)\right ) \int \frac{(a+a \sin (e+f x))^{\frac{5}{2}+m}}{(g \cos (e+f x))^{7/2}} \, dx}{a^2 c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{\left (g^4 \cos (e+f x) (a-a \sin (e+f x))^{5/4} (a+a \sin (e+f x))^{3/4}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{4}+m}}{(a-a x)^{9/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{\left (2^{\frac{1}{4}+m} g^4 \cos (e+f x) (a-a \sin (e+f x))^{5/4} (a+a \sin (e+f x))^{1+m} \left (\frac{a+a \sin (e+f x)}{a}\right )^{-\frac{1}{4}-m}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{4}+m}}{(a-a x)^{9/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{2^{\frac{9}{4}+m} g^4 \cos (e+f x) \, _2F_1\left (-\frac{5}{4},-\frac{1}{4}-m;-\frac{1}{4};\frac{1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac{1}{4}-m} (a+a \sin (e+f x))^{1+m}}{5 a c^2 f (g \cos (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 52.3, size = 3845, normalized size = 33.73 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((g*Cos[e + f*x])^(3/2)*(a + a*Sin[e + f*x])^m)/(c - c*Sin[e + f*x])^(5/2),x]
[Out]
-(((g*Cos[e + f*x])^(3/2)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a + a*Sin[e + f*x])^m*((-5*Cos[(-e + Pi/2 -
f*x)/2]^(-1 + 2*m)*Sqrt[Cos[e + f*x]]*(-1 + Cot[(-e + Pi/2 - f*x)/4]^2)*(-3*AppellF1[1/4, -1/2 - 2*m, 2*m, 5/
4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[-3/4, -1/2 - 2*m, 2*m, 1/4, Tan[(-e + P
i/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2)*Csc[(-e + Pi/2 - f*x)/4]^2)/(12*Sqrt
[2]*(8*m*AppellF1[5/4, -1/2 - 2*m, 1 + 2*m, 9/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (2
+ 8*m)*AppellF1[5/4, 1/2 - 2*m, 2*m, 9/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 5*Cot[(-
e + Pi/2 - f*x)/4]^2*(AppellF1[1/4, -1/2 - 2*m, 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4
]^2] - 8*m*AppellF1[1/4, -1/2 - 2*m, 1 + 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] -
2*AppellF1[1/4, 1/2 - 2*m, 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 8*m*AppellF1[1
/4, 1/2 - 2*m, 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + AppellF1[-3/4, -1/2 - 2*m,
2*m, 1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2))) - (10*Sqrt[2
]*Cos[(-e + Pi/2 - f*x)/4]^2*Cos[(-e + Pi/2 - f*x)/2]^(-2 + 2*m)*Cos[e + f*x]^(3/2)*(-1 + Cot[(-e + Pi/2 - f*x
)/4]^2)*(-3*AppellF1[1/4, -1/2 - 2*m, 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + App
ellF1[-3/4, -1/2 - 2*m, 2*m, 1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*
x)/4]^2)*Csc[(-e + Pi/2 - f*x)/2]^3*Sin[(-e + Pi/2 - f*x)/4]^4)/(3*(40*AppellF1[-3/4, -1/2 - 2*m, 2*m, 1/4, Ta
n[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cos[(-e + Pi/2 - f*x)/4]^4 + 4*Sin[(-e + Pi/2 - f*x)/4]
^2*(5*AppellF1[1/4, -1/2 - 2*m, 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(1 + Cos[(-
e + Pi/2 - f*x)/2]) - 2*(10*(4*m*AppellF1[1/4, -1/2 - 2*m, 1 + 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e
+ Pi/2 - f*x)/4]^2] + (1 + 4*m)*AppellF1[1/4, 1/2 - 2*m, 2*m, 5/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2
- f*x)/4]^2])*Cos[(-e + Pi/2 - f*x)/4]^2 + 8*m*AppellF1[5/4, -1/2 - 2*m, 1 + 2*m, 9/4, Tan[(-e + Pi/2 - f*x)/
4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sin[(-e + Pi/2 - f*x)/4]^2 + 2*(1 + 4*m)*AppellF1[5/4, 1/2 - 2*m, 2*m, 9/4,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sin[(-e + Pi/2 - f*x)/4]^2)))) - ((Cos[(-e + Pi/2 -
f*x)/4]^2)^(2*m)*Cos[(-e + Pi/2 - f*x)/2]^(1 + 2*m)*Cos[e + f*x]*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-
e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 -
f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Tan
[(-e + Pi/2 - f*x)/4])/(60*Sqrt[2]*(Cos[Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^4*Sqrt[2
- 2*Tan[(-e + Pi/2 - f*x)/4]^2]*(-((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[e + f*x]]*(5*AppellF1[3/4, -1/2
- 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m,
-1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 -
f*x)/4]^2)^(2 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]^2)/(240*(2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2)^(3/2)) - ((Cos[(-e +
Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[e + f*x]]*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2,
-Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e +
Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2 + 2*m))/(480*Sqrt[2 - 2*Tan[(-e
+ Pi/2 - f*x)/4]^2]) + (m*(Cos[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m)*Sqrt[Cos[e + f*x]]*(5*AppellF1[3/4, -1/2 - 2
*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/
4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)
/4]^2)^(1 + 2*m)*Sin[(-e + Pi/2 - f*x)/4]^2)/(120*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2]) - ((Cos[(-e + Pi/2 -
f*x)/4]^2)^(2*m)*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]
^2] + 3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e
+ Pi/2 - f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Sin[e + f*x]*Tan[(-e + Pi/2 - f*x)/4])/(240*Sqrt[C
os[e + f*x]]*Sqrt[2 - 2*Tan[(-e + Pi/2 - f*x)/4]^2]) - ((1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[
e + f*x]]*(5*AppellF1[3/4, -1/2 - 2*m, 2*m, 7/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 3*
AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2
- f*x)/4]^4)*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Tan[(-e + Pi/2 - f*x)/4]^2)/(240*Sqrt[2 - 2*Tan[(-e + Pi/2
- f*x)/4]^2]) - ((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Sqrt[Cos[e + f*x]]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)
*Tan[(-e + Pi/2 - f*x)/4]*(-3*AppellF1[-5/4, -1/2 - 2*m, 2*m, -1/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/
2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*Csc[(-e + Pi/2 - f*x)/4]^2 + 3*Cot[(-e + Pi/2 - f*x)/4]^4*(-5*m*Appe
llF1[-1/4, -1/2 - 2*m, 1 + 2*m, 3/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 -
f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4] + (5*(-1/2 - 2*m)*AppellF1[-1/4, 1/2 - 2*m, 2*m, 3/4, Tan[(-e + Pi/2 - f*x
)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/2) + 5*((-3*m*Appell
F1[7/4, -1/2 - 2*m, 1 + 2*m, 11/4, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f
*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/7 + (3*(-1/2 - 2*m)*AppellF1[7/4, 1/2 - 2*m, 2*m, 11/4, Tan[(-e + Pi/2 - f*
x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/14)))/(120*Sqrt[2 -
2*Tan[(-e + Pi/2 - f*x)/4]^2])))))/(f*Cos[(-e + Pi/2 - f*x)/2]^(2*m)*Cos[e + f*x]^(3/2)*(c - c*Sin[e + f*x])^
(5/2)))
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Maple [F] time = 0.23, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( g\cos \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}
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))^m/(c-c*sin(f*x+e))^(5/2),x)
[Out]
int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(5/2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
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))^m/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{g \cos \left (f x + e\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m} g \cos \left (f x + e\right )}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} -{\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}
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))^m/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
integral(-sqrt(g*cos(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m*g*cos(f*x + e)/(3*c^3*cos(f*x
+ e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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))**m/(c-c*sin(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
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))^m/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")
[Out]
integrate((g*cos(f*x + e))^(3/2)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(5/2), x)