3.162 \(\int (a \sin (e+f x))^m \tan (e+f x) \, dx\)

Optimal. Leaf size=48 \[ \frac{(a \sin (e+f x))^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )}{a^2 f (m+2)} \]

[Out]

(Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2]*(a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m))

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Rubi [A]  time = 0.0348985, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2592, 364} \[ \frac{(a \sin (e+f x))^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(e+f x)\right )}{a^2 f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, Sin[e + f*x]^2]*(a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m))

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (a \sin (e+f x))^m \tan (e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{1+m}}{a^2-x^2} \, dx,x,a \sin (e+f x)\right )}{f}\\ &=\frac{\, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};\sin ^2(e+f x)\right ) (a \sin (e+f x))^{2+m}}{a^2 f (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0325041, size = 53, normalized size = 1.1 \[ \frac{\sin ^2(e+f x) (a \sin (e+f x))^m \, _2F_1\left (1,\frac{m+2}{2};\frac{m+2}{2}+1;\sin ^2(e+f x)\right )}{f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, (2 + m)/2, 1 + (2 + m)/2, Sin[e + f*x]^2]*Sin[e + f*x]^2*(a*Sin[e + f*x])^m)/(f*(2 + m))

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Maple [F]  time = 0.546, size = 0, normalized size = 0. \begin{align*} \int \left ( a\sin \left ( fx+e \right ) \right ) ^{m}\tan \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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