3.1.31 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x))}{(c+d \sin (e+f x))^{5/2}} \, dx\) [31]

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

[Out]

2/3*(A*d^2-B*c*d+C*c^2)*cos(f*x+e)*(a+a*sin(f*x+e))^m/d/(c^2-d^2)/f/(c+d*sin(f*x+e))^(3/2)+1/3*(d^2*(-4*A*m+A-
3*B+3*C)+c*d*(4*B*m+3*A-B+3*C)-2*c^2*(2*C*m+C))*AppellF1(1/2+m,3/2,1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*s
in(f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^m*2^(1/2)*((c+d*sin(f*x+e))/(c-d))^(1/2)/(c-d)^2/d/(c+d)/f/(1+2*m)/(1-s
in(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)+1/3*(B*c*d*(1-2*m)+2*c^2*C*(1+m)-d^2*(-2*A*m+A+3*C))*AppellF1(3/2+m,3/
2,1/2,5/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*2^(1/2)*((c+d*sin(f*
x+e))/(c-d))^(1/2)/a/(c-d)^2/d/(c+d)/f/(3+2*m)/(1-sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.78, antiderivative size = 451, normalized size of antiderivative = 1.00, 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 = {3122, 3066, 2867, 145, 144, 143} \begin {gather*} \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}} \end {gather*}

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 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 144

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 145

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 2867

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 3066

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 3122

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])

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(20654\) vs. \(2(451)=902\).
time = 64.50, size = 20654, normalized size = 45.80 \begin {gather*} \text {Result too large to show} \end {gather*}

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 = 1.41, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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)

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