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

Optimal. Leaf size=83 \[ \frac{\cos ^2(e+f x)^{\frac{n+1}{2}} (a \sin (e+f x))^m (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\sin ^2(e+f x)\right )}{b f (m+n+1)} \]

[Out]

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

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Rubi [A]  time = 0.102836, antiderivative size = 83, normalized size of antiderivative = 1., 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 = {2602, 2577} \[ \frac{\cos ^2(e+f x)^{\frac{n+1}{2}} (a \sin (e+f x))^m (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{2} (m+n+1);\frac{1}{2} (m+n+3);\sin ^2(e+f x)\right )}{b f (m+n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

Mathematica [C]  time = 1.90652, size = 260, normalized size = 3.13 \[ \frac{(m+n+3) \sin (e+f x) (a \sin (e+f x))^m (b \tan (e+f x))^n F_1\left (\frac{1}{2} (m+n+1);n,m+1;\frac{1}{2} (m+n+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{f (m+n+1) \left ((m+n+3) F_1\left (\frac{1}{2} (m+n+1);n,m+1;\frac{1}{2} (m+n+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left ((m+1) F_1\left (\frac{1}{2} (m+n+3);n,m+2;\frac{1}{2} (m+n+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-n F_1\left (\frac{1}{2} (m+n+3);n+1,m+1;\frac{1}{2} (m+n+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a*Sin[e + f*x])^m*(b*Tan[e + f*x])^n,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]*Sin[e +
f*x]*(a*Sin[e + f*x])^m*(b*Tan[e + f*x])^n)/(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*((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] - 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])*Tan[(e + f*x)/2]^2))

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Maple [F]  time = 0.915, size = 0, normalized size = 0. \begin{align*} \int \left ( a\sin \left ( fx+e \right ) \right ) ^{m} \left ( b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*sin(f*x+e))^m*(b*tan(f*x+e))^n,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 )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(f*x + e))^m*(b*tan(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 )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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