3.31 \(\int \frac{(a+a \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x))}{(c+d \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=451 \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m \left (c d (3 A+4 B m-B+3 C)+d^2 (-4 A m+A-3 B+3 C)-2 c^2 (2 C m+C)\right ) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{3}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{3 d f (2 m+1) (c-d)^2 (c+d) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \left (-d^2 (-2 A m+A+3 C)+B c d (1-2 m)+2 c^2 C (m+1)\right ) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{3}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{3 a d f (2 m+3) (c-d)^2 (c+d) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{2 \cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}} \]

[Out]

(2*(c^2*C - B*c*d + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(3*d*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2))
+ (Sqrt[2]*(d^2*(A - 3*B + 3*C - 4*A*m) + c*d*(3*A - B + 3*C + 4*B*m) - 2*c^2*(C + 2*C*m))*AppellF1[1/2 + m,
1/2, 3/2, 3/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^
m*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(3*(c - d)^2*d*(c + d)*f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin
[e + f*x]]) + (Sqrt[2]*(B*c*d*(1 - 2*m) + 2*c^2*C*(1 + m) - d^2*(A + 3*C - 2*A*m))*AppellF1[3/2 + m, 1/2, 3/2,
5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*S
qrt[(c + d*Sin[e + f*x])/(c - d)])/(3*a*(c - d)^2*d*(c + d)*f*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin[
e + f*x]])

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Rubi [A]  time = 1.16287, antiderivative size = 451, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3043, 2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m \left (c d (3 A+4 B m-B+3 C)+d^2 (-4 A m+A-3 B+3 C)-2 c^2 (2 C m+C)\right ) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{1}{2};\frac{1}{2},\frac{3}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{3 d f (2 m+1) (c-d)^2 (c+d) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \left (-d^2 (-2 A m+A+3 C)+B c d (1-2 m)+2 c^2 C (m+1)\right ) \sqrt{\frac{c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac{3}{2};\frac{1}{2},\frac{3}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{3 a d f (2 m+3) (c-d)^2 (c+d) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{2 \cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(2*(c^2*C - B*c*d + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(3*d*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2))
+ (Sqrt[2]*(d^2*(A - 3*B + 3*C - 4*A*m) + c*d*(3*A - B + 3*C + 4*B*m) - 2*c^2*(C + 2*C*m))*AppellF1[1/2 + m,
1/2, 3/2, 3/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^
m*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(3*(c - d)^2*d*(c + d)*f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin
[e + f*x]]) + (Sqrt[2]*(B*c*d*(1 - 2*m) + 2*c^2*C*(1 + m) - d^2*(A + 3*C - 2*A*m))*AppellF1[3/2 + m, 1/2, 3/2,
5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*S
qrt[(c + d*Sin[e + f*x])/(c - d)])/(3*a*(c - d)^2*d*(c + d)*f*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[c + d*Sin[
e + f*x]])

Rule 3043

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

Rule 2987

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x
], x] + Dist[B/b, Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f,
A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]

Rule 2788

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

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac{2 \int \frac{(a+a \sin (e+f x))^m \left (-\frac{1}{2} a \left (2 (c C-B d) \left (\frac{3 d}{2}-c m\right )+2 A d \left (\frac{3 c}{2}-d m\right )\right )+\frac{1}{2} a \left (3 C d^2-d (B c-A d) (1-2 m)-2 c^2 C (1+m)\right ) \sin (e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 a d \left (c^2-d^2\right )}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac{\left (B c d (1-2 m)+2 c^2 C (1+m)-d^2 (A+3 C-2 A m)\right ) \int \frac{(a+a \sin (e+f x))^{1+m}}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 a d \left (c^2-d^2\right )}+\frac{\left (d^2 (A-3 B+3 C-4 A m)+c d (3 A-B+3 C+4 B m)-2 c^2 (C+2 C m)\right ) \int \frac{(a+a \sin (e+f x))^m}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d \left (c^2-d^2\right )}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac{\left (a \left (B c d (1-2 m)+2 c^2 C (1+m)-d^2 (A+3 C-2 A m)\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{a-a x} (c+d x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 d \left (c^2-d^2\right ) f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 \left (d^2 (A-3 B+3 C-4 A m)+c d (3 A-B+3 C+4 B m)-2 c^2 (C+2 C m)\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{a-a x} (c+d x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 d \left (c^2-d^2\right ) f \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac{\left (a \left (B c d (1-2 m)+2 c^2 C (1+m)-d^2 (A+3 C-2 A m)\right ) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} (c+d x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 \sqrt{2} d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 \left (d^2 (A-3 B+3 C-4 A m)+c d (3 A-B+3 C+4 B m)-2 c^2 (C+2 C m)\right ) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} (c+d x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 \sqrt{2} d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac{\left (a^2 \left (B c d (1-2 m)+2 c^2 C (1+m)-d^2 (A+3 C-2 A m)\right ) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 \sqrt{2} d (a c-a d) \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\left (a^3 \left (d^2 (A-3 B+3 C-4 A m)+c d (3 A-B+3 C+4 B m)-2 c^2 (C+2 C m)\right ) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m}}{\sqrt{\frac{1}{2}-\frac{x}{2}} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 \sqrt{2} d (a c-a d) \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac{\sqrt{2} \left (d^2 (A-3 B+3 C-4 A m)+c d (3 A-B+3 C+4 B m)-2 c^2 (C+2 C m)\right ) F_1\left (\frac{1}{2}+m;\frac{1}{2},\frac{3}{2};\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}{3 (c-d)^2 d (c+d) f (1+2 m) \sqrt{1-\sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{2} \left (B c d (1-2 m)+2 c^2 C (1+m)-d^2 (A+3 C-2 A m)\right ) F_1\left (\frac{3}{2}+m;\frac{1}{2},\frac{3}{2};\frac{5}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt{1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}{3 (c-d)^2 d (c+d) f (3+2 m) (a-a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 33.6076, size = 20723, normalized size = 45.95 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/(c + d*Sin[e + f*x])^(5/2),x]

[Out]

Result too large to show

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Maple [F]  time = 0.73, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(5/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c+d*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 (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(d*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{{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )} \sqrt{d \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*sqrt(d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(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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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((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)+C*sin(f*x+e)**2)/(c+d*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 (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(d*sin(f*x + e) + c)^(5/2), x)