3.171 \(\int (a \sin (e+f x))^m \sqrt{b \tan (e+f x)} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \cos ^2(e+f x)^{3/4} (b \tan (e+f x))^{3/2} (a \sin (e+f x))^m \, _2F_1\left (\frac{3}{4},\frac{1}{4} (2 m+3);\frac{1}{4} (2 m+7);\sin ^2(e+f x)\right )}{b f (2 m+3)} \]

[Out]

(2*(Cos[e + f*x]^2)^(3/4)*Hypergeometric2F1[3/4, (3 + 2*m)/4, (7 + 2*m)/4, Sin[e + f*x]^2]*(a*Sin[e + f*x])^m*
(b*Tan[e + f*x])^(3/2))/(b*f*(3 + 2*m))

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Rubi [A]  time = 0.103751, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2602, 2577} \[ \frac{2 \cos ^2(e+f x)^{3/4} (b \tan (e+f x))^{3/2} (a \sin (e+f x))^m \, _2F_1\left (\frac{3}{4},\frac{1}{4} (2 m+3);\frac{1}{4} (2 m+7);\sin ^2(e+f x)\right )}{b f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^(3/4)*Hypergeometric2F1[3/4, (3 + 2*m)/4, (7 + 2*m)/4, Sin[e + f*x]^2]*(a*Sin[e + f*x])^m*
(b*Tan[e + f*x])^(3/2))/(b*f*(3 + 2*m))

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 \sqrt{b \tan (e+f x)} \, dx &=\frac{\left (a \cos ^{\frac{3}{2}}(e+f x) (b \tan (e+f x))^{3/2}\right ) \int \frac{(a \sin (e+f x))^{\frac{1}{2}+m}}{\sqrt{\cos (e+f x)}} \, dx}{b (a \sin (e+f x))^{3/2}}\\ &=\frac{2 \cos ^2(e+f x)^{3/4} \, _2F_1\left (\frac{3}{4},\frac{1}{4} (3+2 m);\frac{1}{4} (7+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{3/2}}{b f (3+2 m)}\\ \end{align*}

Mathematica [A]  time = 3.17363, size = 87, normalized size = 1.1 \[ \frac{2 (b \tan (e+f x))^{3/2} \sec ^2(e+f x)^{m/2} (a \sin (e+f x))^m \, _2F_1\left (\frac{m+2}{2},\frac{1}{4} (2 m+3);\frac{1}{4} (2 m+7);-\tan ^2(e+f x)\right )}{b f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[(2 + m)/2, (3 + 2*m)/4, (7 + 2*m)/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(m/2)*(a*Sin[e + f
*x])^m*(b*Tan[e + f*x])^(3/2))/(b*f*(3 + 2*m))

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Maple [F]  time = 0.132, size = 0, normalized size = 0. \begin{align*} \int \left ( a\sin \left ( fx+e \right ) \right ) ^{m}\sqrt{b\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*(b*tan(f*x+e))^(1/2),x)

[Out]

int((a*sin(f*x+e))^m*(b*tan(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{m}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \tan \left (f x + e\right )} \left (a \sin \left (f x + e\right )\right )^{m}, 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))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*tan(f*x + e))*(a*sin(f*x + e))^m, 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))**(1/2),x)

[Out]

Timed out

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Giac [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))^(1/2),x, algorithm="giac")

[Out]

Timed out