3.209 \(\int \frac{(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=134 \[ \frac{(A (1-2 m)-B (2 m+3)) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{2 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(2*f*(c - c*Sin[e + f*x])^(3/2)) + ((A*(1 - 2*m) - B*(3 + 2*m))*
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)/(4*c*f*(1 +
2*m)*Sqrt[c - c*Sin[e + f*x]])

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Rubi [A]  time = 0.29977, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2972, 2745, 2667, 68} \[ \frac{(A (1-2 m)-B (2 m+3)) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{2 f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(2*f*(c - c*Sin[e + f*x])^(3/2)) + ((A*(1 - 2*m) - B*(3 + 2*m))*
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)/(4*c*f*(1 +
2*m)*Sqrt[c - c*Sin[e + f*x]])

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 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m}{2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (B c \left (-\frac{3}{2}-m\right )-A c \left (-\frac{1}{2}+m\right )\right ) \int \frac{(a+a \sin (e+f x))^m}{\sqrt{c-c \sin (e+f x)}} \, dx}{2 c^2}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m}{2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (\left (B c \left (-\frac{3}{2}-m\right )-A c \left (-\frac{1}{2}+m\right )\right ) \cos (e+f x)\right ) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac{1}{2}+m} \, dx}{2 c^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m}{2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (a \left (B c \left (-\frac{3}{2}-m\right )-A c \left (-\frac{1}{2}+m\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 )}{2 c^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{(A+B) \cos (e+f x) (a+a \sin (e+f x))^m}{2 f (c-c \sin (e+f x))^{3/2}}+\frac{(A (1-2 m)-B (3+2 m)) \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}{4 c f (1+2 m) \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 10.1848, size = 369, normalized size = 2.75 \[ -\frac{\sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{2 m} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 (a \sin (e+f x)+a)^m \left (\frac{4^{-m} (A-3 B) \, _2F_1\left (2 m,2 m;2 m+1;\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )}{m}+\frac{4^{-m} (A+B) \left (\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )-1\right ) \, _2F_1\left (2 m,2 m+1;2 (m+1);\frac{1}{2} \left (1-\tan ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )\right )}{2 m+1}-\frac{(A-3 B) \sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{-2 m} \, _2F_1\left (1,2 m;2 m+1;\cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )}{m}-\frac{2 (A+B) \cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right ) \sec ^2\left (\frac{1}{4} \left (-e-f x+\frac{\pi }{2}\right )\right )^{-2 m} \, _2F_1\left (2,2 m+1;2 m+2;\cos \left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )\right )\right )}{2 m+1}\right )}{8 \sqrt{2} f (c-c \sin (e+f x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-((Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a + a*Sin[e + f*x])^m*(((A - 3*B
)*Hypergeometric2F1[2*m, 2*m, 1 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2])/(4^m*m) - ((A - 3*B)*Hypergeometri
c2F1[1, 2*m, 1 + 2*m, Cos[(-e + Pi/2 - f*x)/2]])/(m*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)) - (2*(A + B)*Cos[(-e +
Pi/2 - f*x)/2]*Hypergeometric2F1[2, 1 + 2*m, 2 + 2*m, Cos[(-e + Pi/2 - f*x)/2]])/((1 + 2*m)*(Sec[(-e + Pi/2 -
f*x)/4]^2)^(2*m)) + ((A + B)*Hypergeometric2F1[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))/(4^m*(1 + 2*m))))/(8*Sqrt[2]*f*(c - c*Sin[e + f*x])^(3/2))

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Maple [F]  time = 0.284, 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 ) \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{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-c*sin(f*x+e))^(3/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (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{3}{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-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, 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-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c^2*cos(f*x + e)^2 + 2*c^2*si
n(f*x + e) - 2*c^2), 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-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (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{3}{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-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(3/2), x)