3.39 \(\int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=87 \[ -\frac {(a \sin (e+f x))^m (b \cot (e+f x))^{n+1} \sin ^2(e+f x)^{\frac {1}{2} (-m+n+1)} \, _2F_1\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1);\frac {n+3}{2};\cos ^2(e+f x)\right )}{b f (n+1)} \]

[Out]

-(b*cot(f*x+e))^(1+n)*hypergeom([1/2+1/2*n, 1/2-1/2*m+1/2*n],[3/2+1/2*n],cos(f*x+e)^2)*(a*sin(f*x+e))^m*(sin(f
*x+e)^2)^(1/2-1/2*m+1/2*n)/b/f/(1+n)

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Rubi [A]  time = 0.10, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2603, 2617} \[ -\frac {(a \sin (e+f x))^m (b \cot (e+f x))^{n+1} \sin ^2(e+f x)^{\frac {1}{2} (-m+n+1)} \, _2F_1\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1);\frac {n+3}{2};\cos ^2(e+f x)\right )}{b f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*Cot[e + f*x])^(1 + n)*Hypergeometric2F1[(1 + n)/2, (1 - m + n)/2, (3 + n)/2, Cos[e + f*x]^2]*(a*Sin[e +
f*x])^m*(Sin[e + f*x]^2)^((1 - m + n)/2))/(b*f*(1 + n)))

Rule 2603

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f*
x])^FracPart[m]*(Sec[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sec[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin {align*} \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx &=\left (\left (\frac {\csc (e+f x)}{a}\right )^m (a \sin (e+f x))^m\right ) \int (b \cot (e+f x))^n \left (\frac {\csc (e+f x)}{a}\right )^{-m} \, dx\\ &=-\frac {(b \cot (e+f x))^{1+n} \, _2F_1\left (\frac {1+n}{2},\frac {1}{2} (1-m+n);\frac {3+n}{2};\cos ^2(e+f x)\right ) (a \sin (e+f x))^m \sin ^2(e+f x)^{\frac {1}{2} (1-m+n)}}{b f (1+n)}\\ \end {align*}

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Mathematica [C]  time = 1.80, size = 289, normalized size = 3.32 \[ \frac {(m-n+3) \sin (e+f x) (a \sin (e+f x))^m (b \cot (e+f x))^n F_1\left (\frac {1}{2} (m-n+1);-n,m+1;\frac {1}{2} (m-n+3);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f (m-n+1) \left ((m-n+3) F_1\left (\frac {1}{2} (m-n+1);-n,m+1;\frac {1}{2} (m-n+3);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac {1}{2} (e+f x)\right ) \left (n F_1\left (\frac {1}{2} (m-n+3);1-n,m+1;\frac {1}{2} (m-n+5);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(m+1) F_1\left (\frac {1}{2} (m-n+3);-n,m+2;\frac {1}{2} (m-n+5);\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \cot \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.96, size = 0, normalized size = 0.00 \[ \int \left (b \cot \left (f x +e \right )\right )^{n} \left (a \sin \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\mathrm {cot}\left (e+f\,x\right )\right )}^n\,{\left (a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin {\left (e + f x \right )}\right )^{m} \left (b \cot {\left (e + f x \right )}\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sin(e + f*x))**m*(b*cot(e + f*x))**n, x)

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