3.167 \(\int (a \sin (e+f x))^m \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=68 \[ \frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) (a \sin (e+f x))^{m+3} \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(e+f x)\right )}{a^3 f (m+3)} \]

[Out]

(Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[e + f*x]^2]*Sec[e + f*x]*(a*Sin[e + f*x
])^(3 + m))/(a^3*f*(3 + m))

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Rubi [A]  time = 0.0888997, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2600, 2577} \[ \frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) (a \sin (e+f x))^{m+3} \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(e+f x)\right )}{a^3 f (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sin[e + f*x])^m*Tan[e + f*x]^2,x]

[Out]

(Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[e + f*x]^2]*Sec[e + f*x]*(a*Sin[e + f*x
])^(3 + m))/(a^3*f*(3 + m))

Rule 2600

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

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 \tan ^2(e+f x) \, dx &=\frac{\int \sec ^2(e+f x) (a \sin (e+f x))^{2+m} \, dx}{a^2}\\ &=\frac{\sqrt{\cos ^2(e+f x)} \, _2F_1\left (\frac{3}{2},\frac{3+m}{2};\frac{5+m}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (a \sin (e+f x))^{3+m}}{a^3 f (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.0844246, size = 71, normalized size = 1.04 \[ \frac{\sin ^2(e+f x) \sqrt{\cos ^2(e+f x)} \tan (e+f x) (a \sin (e+f x))^m \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(e+f x)\right )}{f (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cos[e + f*x]^2]*Hypergeometric2F1[3/2, (3 + m)/2, (5 + m)/2, Sin[e + f*x]^2]*Sin[e + f*x]^2*(a*Sin[e + f
*x])^m*Tan[e + f*x])/(f*(3 + m))

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int \left ( a\sin \left ( fx+e \right ) \right ) ^{m} \left ( \tan \left ( fx+e \right ) \right ) ^{2}\, 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)^2,x)

[Out]

int((a*sin(f*x+e))^m*tan(f*x+e)^2,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 )^{2}\,{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)^2,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e))^m*tan(f*x + e)^2, 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 )^{2}, 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)^2,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e))^m*tan(f*x + e)^2, 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 ^{2}{\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)**2,x)

[Out]

Integral((a*sin(e + f*x))**m*tan(e + f*x)**2, 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 )^{2}\,{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)^2,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^m*tan(f*x + e)^2, x)