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

Optimal. Leaf size=230 \[ -\frac{\left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{32 c^2 f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}} \]

[Out]

((A + B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(8*a*f*(c - c*Sin[e + f*x])^(5/2)) + ((A*(5 - 2*m) - B
*(3 + 2*m) - C*(11 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(16*c*f*(c - c*Sin[e + f*x])^(3/2)) - ((B*(5 -
8*m - 4*m^2) - A*(3 - 8*m + 4*m^2) - C*(19 + 24*m + 4*m^2))*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 +
m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^m)/(32*c^2*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.69715, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.104, Rules used = {3035, 2972, 2745, 2667, 68} \[ -\frac{\left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{32 c^2 f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{16 c f (c-c \sin (e+f x))^{3/2}}+\frac{(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}} \]

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 - c*Sin[e + f*x])^(5/2),x]

[Out]

((A + B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(8*a*f*(c - c*Sin[e + f*x])^(5/2)) + ((A*(5 - 2*m) - B
*(3 + 2*m) - C*(11 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(16*c*f*(c - c*Sin[e + f*x])^(3/2)) - ((B*(5 -
8*m - 4*m^2) - A*(3 - 8*m + 4*m^2) - C*(19 + 24*m + 4*m^2))*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 +
m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^m)/(32*c^2*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

Rule 3035

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[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(2*b*c*f*(2*m + 1)), x] - Dist[1/(2*b*c*d*(2*m + 1)), Int[
(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n -
1) - C*(c^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (LtQ[m, -2^(-1)] || (EqQ[m +
n + 2, 0] && NeQ[2*m + 1, 0]))

Rule 2972

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

Rule 2745

Int[((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])/Cos[e + f*x]^(2
*FracPart[m]), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

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-c \sin (e+f x))^{5/2}} \, dx &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}-\frac{\int \frac{(a+a \sin (e+f x))^m \left (-\frac{1}{2} a^2 (A (9-2 m)-(B+C) (7+2 m))-\frac{1}{2} a^2 ((A+B) (1-2 m)-C (15+2 m)) \sin (e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 a^2 c}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac{\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c-c \sin (e+f x)}} \, dx}{32 c^2}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac{\left (\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x)\right ) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac{1}{2}+m} \, dx}{32 c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac{\left (a \left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-\frac{1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{32 c^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac{(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac{\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x) \, _2F_1\left (1,\frac{1}{2}+m;\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{32 c^2 f (1+2 m) \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 16.3837, size = 829, normalized size = 3.6 \[ -\frac{\cos ^{-2 m}\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5 (\sin (e+f x) a+a)^m \left (\frac{1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )}{\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1}\right )^{2 m} \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1\right )^{2 m} \left (\frac{16 (A-B-3 C) \, _2F_1\left (2,2 m+1;2 (m+1);\frac{1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )}{\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1}\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1\right )^{-2 m-1}}{2 m+1}-\frac{8 (A+B+C) m \, _2F_1\left (2,2 m+1;2 (m+1);\frac{1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )}{\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1}\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1\right )^{-2 m-1}}{2 m+1}+(A+B+C) \cot ^4\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1\right )^{1-2 m}-\frac{2 (3 A-5 B+19 C) \, _2F_1\left (1,2 m;2 m+1;\frac{1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )}{\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1}\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )+1\right )^{-2 m}}{m}+\frac{2^{1-2 m} (3 A-5 B+19 C) \, _2F_1\left (2 m,2 m;2 m+1;\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )}{m}+\frac{2^{1-2 m} (3 A-5 B-13 C) \, _2F_1\left (2 m,2 m+1;2 (m+1);\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right )}{2 m+1}+\frac{4^{1-m} (A+B+C) \, _2F_1\left (2 m-1,2 m+1;2 (m+1);\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right ) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right )}{2 m+1}\right )}{128 \sqrt{2} f (c-c \sin (e+f x))^{5/2}} \]

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 - c*Sin[e + f*x])^(5/2),x]

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(a + a*Sin[e + f*x])^m*((1 - Tan[(-e + Pi/2 - f*x)/4]^2)/(1 + Tan[(-
e + Pi/2 - f*x)/4]^2))^(2*m)*(1 + Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)*((2^(1 - 2*m)*(3*A - 5*B + 19*C)*Hypergeom
etric2F1[2*m, 2*m, 1 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2])/m + (2^(1 - 2*m)*(3*A - 5*B - 13*C)*Hypergeom
etric2F1[2*m, 1 + 2*m, 2*(1 + m), (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2))/(1 +
2*m) + (4^(1 - m)*(A + B + C)*Hypergeometric2F1[-1 + 2*m, 1 + 2*m, 2*(1 + m), (1 - Tan[(-e + Pi/2 - f*x)/4]^2)
/2]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2))/(1 + 2*m) + (16*(A - B - 3*C)*Hypergeometric2F1[2, 1 + 2*m, 2*(1 + m),
(1 - Tan[(-e + Pi/2 - f*x)/4]^2)/(1 + Tan[(-e + Pi/2 - f*x)/4]^2)]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 + Tan[
(-e + Pi/2 - f*x)/4]^2)^(-1 - 2*m))/(1 + 2*m) - (8*(A + B + C)*m*Hypergeometric2F1[2, 1 + 2*m, 2*(1 + m), (1 -
Tan[(-e + Pi/2 - f*x)/4]^2)/(1 + Tan[(-e + Pi/2 - f*x)/4]^2)]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 + Tan[(-e
+ Pi/2 - f*x)/4]^2)^(-1 - 2*m))/(1 + 2*m) + (A + B + C)*Cot[(-e + Pi/2 - f*x)/4]^4*(-1 + Tan[(-e + Pi/2 - f*x)
/4]^2)*(1 + Tan[(-e + Pi/2 - f*x)/4]^2)^(1 - 2*m) - (2*(3*A - 5*B + 19*C)*Hypergeometric2F1[1, 2*m, 1 + 2*m, (
1 - Tan[(-e + Pi/2 - f*x)/4]^2)/(1 + Tan[(-e + Pi/2 - f*x)/4]^2)])/(m*(1 + Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m)))
)/(128*Sqrt[2]*f*Cos[(-e + Pi/2 - f*x)/2]^(2*m)*(c - c*Sin[e + f*x])^(5/2))

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Maple [F]  time = 0.739, 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-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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*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-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 (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 (-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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*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/(-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{{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*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(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(3*c^3*c
os(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((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)+C*sin(f*x+e)**2)/(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 (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 (-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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*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/(-c*sin(f*x + e) + c)^(5/2), x)