3.137 \(\int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=85 \[ \frac{2^{m+\frac{1}{2}} \cos (e+f x) (1-\sin (e+f x))^{-m-\frac{1}{2}} (a-a \sin (e+f x))^m F_1\left (\frac{1}{2};-n,\frac{1}{2}-m;\frac{3}{2};\sin (e+f x)+1,\frac{1}{2} (\sin (e+f x)+1)\right )}{f} \]

[Out]

(2^(1/2 + m)*AppellF1[1/2, -n, 1/2 - m, 3/2, 1 + Sin[e + f*x], (1 + Sin[e + f*x])/2]*Cos[e + f*x]*(1 - Sin[e +
f*x])^(-1/2 - m)*(a - a*Sin[e + f*x])^m)/f

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Rubi [A]  time = 0.109815, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2787, 2785, 133} \[ \frac{2^{m+\frac{1}{2}} \cos (e+f x) (1-\sin (e+f x))^{-m-\frac{1}{2}} (a-a \sin (e+f x))^m F_1\left (\frac{1}{2};-n,\frac{1}{2}-m;\frac{3}{2};\sin (e+f x)+1,\frac{1}{2} (\sin (e+f x)+1)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(-Sin[e + f*x])^n*(a - a*Sin[e + f*x])^m,x]

[Out]

(2^(1/2 + m)*AppellF1[1/2, -n, 1/2 - m, 3/2, 1 + Sin[e + f*x], (1 + Sin[e + f*x])/2]*Cos[e + f*x]*(1 - Sin[e +
f*x])^(-1/2 - m)*(a - a*Sin[e + f*x])^m)/f

Rule 2787

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
m*(d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 2785

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

Rule 133

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

Rubi steps

\begin{align*} \int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx &=\left ((1-\sin (e+f x))^{-m} (a-a \sin (e+f x))^m\right ) \int (1-\sin (e+f x))^m (-\sin (e+f x))^n \, dx\\ &=\frac{\left (\cos (e+f x) (1-\sin (e+f x))^{-\frac{1}{2}-m} (a-a \sin (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{(1-x)^n (2-x)^{-\frac{1}{2}+m}}{\sqrt{x}} \, dx,x,1+\sin (e+f x)\right )}{f \sqrt{1+\sin (e+f x)}}\\ &=\frac{2^{\frac{1}{2}+m} F_1\left (\frac{1}{2};-n,\frac{1}{2}-m;\frac{3}{2};1+\sin (e+f x),\frac{1}{2} (1+\sin (e+f x))\right ) \cos (e+f x) (1-\sin (e+f x))^{-\frac{1}{2}-m} (a-a \sin (e+f x))^m}{f}\\ \end{align*}

Mathematica [B]  time = 0.604021, size = 301, normalized size = 3.54 \[ -\frac{(2 m+3) \cos (e+f x) (-\sin (e+f x))^n (a-a \sin (e+f x))^m F_1\left (m+\frac{1}{2};-n,m+n+1;m+\frac{3}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )}{f (2 m+1) \left ((2 m+3) F_1\left (m+\frac{1}{2};-n,m+n+1;m+\frac{3}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-2 \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \left (n F_1\left (m+\frac{3}{2};1-n,m+n+1;m+\frac{5}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+(m+n+1) F_1\left (m+\frac{3}{2};-n,m+n+2;m+\frac{5}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(-Sin[e + f*x])^n*(a - a*Sin[e + f*x])^m,x]

[Out]

-(((3 + 2*m)*AppellF1[1/2 + m, -n, 1 + m + n, 3/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]
^2]*Cos[e + f*x]*(-Sin[e + f*x])^n*(a - a*Sin[e + f*x])^m)/(f*(1 + 2*m)*((3 + 2*m)*AppellF1[1/2 + m, -n, 1 + m
+ n, 3/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2] - 2*(n*AppellF1[3/2 + m, 1 - n, 1 +
m + n, 5/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2] + (1 + m + n)*AppellF1[3/2 + m, -n,
2 + m + n, 5/2 + m, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2])*Tan[(2*e - Pi + 2*f*x)/4]^2))
)

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Maple [F]  time = 0.675, size = 0, normalized size = 0. \begin{align*} \int \left ( -\sin \left ( fx+e \right ) \right ) ^{n} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-sin(f*x+e))^n*(a-a*sin(f*x+e))^m,x)

[Out]

int((-sin(f*x+e))^n*(a-a*sin(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \left (-\sin \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sin(f*x+e))^n*(a-a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((-a*sin(f*x + e) + a)^m*(-sin(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \left (-\sin \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sin(f*x+e))^n*(a-a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((-a*sin(f*x + e) + a)^m*(-sin(f*x + e))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- \sin{\left (e + f x \right )}\right )^{n} \left (- a \left (\sin{\left (e + f x \right )} - 1\right )\right )^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sin(f*x+e))**n*(a-a*sin(f*x+e))**m,x)

[Out]

Integral((-sin(e + f*x))**n*(-a*(sin(e + f*x) - 1))**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a \sin \left (f x + e\right ) + a\right )}^{m} \left (-\sin \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-sin(f*x+e))^n*(a-a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((-a*sin(f*x + e) + a)^m*(-sin(f*x + e))^n, x)