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

Optimal. Leaf size=79 \[ \frac{2 \sqrt [4]{\cos ^2(e+f x)} \sqrt{b \tan (e+f x)} (a \sin (e+f x))^m \, _2F_1\left (\frac{1}{4},\frac{1}{4} (2 m+1);\frac{1}{4} (2 m+5);\sin ^2(e+f x)\right )}{b f (2 m+1)} \]

[Out]

(2*(Cos[e + f*x]^2)^(1/4)*Hypergeometric2F1[1/4, (1 + 2*m)/4, (5 + 2*m)/4, Sin[e + f*x]^2]*(a*Sin[e + f*x])^m*
Sqrt[b*Tan[e + f*x]])/(b*f*(1 + 2*m))

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Rubi [A]  time = 0.102909, 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} \[ \frac{2 \sqrt [4]{\cos ^2(e+f x)} \sqrt{b \tan (e+f x)} (a \sin (e+f x))^m \, _2F_1\left (\frac{1}{4},\frac{1}{4} (2 m+1);\frac{1}{4} (2 m+5);\sin ^2(e+f x)\right )}{b f (2 m+1)} \]

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)^(1/4)*Hypergeometric2F1[1/4, (1 + 2*m)/4, (5 + 2*m)/4, Sin[e + f*x]^2]*(a*Sin[e + f*x])^m*
Sqrt[b*Tan[e + f*x]])/(b*f*(1 + 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 \frac{(a \sin (e+f x))^m}{\sqrt{b \tan (e+f x)}} \, dx &=\frac{\left (a \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}\right ) \int \sqrt{\cos (e+f x)} (a \sin (e+f x))^{-\frac{1}{2}+m} \, dx}{b \sqrt{a \sin (e+f x)}}\\ &=\frac{2 \sqrt [4]{\cos ^2(e+f x)} \, _2F_1\left (\frac{1}{4},\frac{1}{4} (1+2 m);\frac{1}{4} (5+2 m);\sin ^2(e+f x)\right ) (a \sin (e+f x))^m \sqrt{b \tan (e+f x)}}{b f (1+2 m)}\\ \end{align*}

Mathematica [A]  time = 2.81801, size = 87, normalized size = 1.1 \[ \frac{2 \sqrt{b \tan (e+f x)} \sec ^2(e+f x)^{m/2} (a \sin (e+f x))^m \, _2F_1\left (\frac{m+2}{2},\frac{1}{4} (2 m+1);\frac{1}{4} (2 m+5);-\tan ^2(e+f x)\right )}{b f (2 m+1)} \]

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, (1 + 2*m)/4, (5 + 2*m)/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(m/2)*(a*Sin[e + f
*x])^m*Sqrt[b*Tan[e + f*x]])/(b*f*(1 + 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}{\frac{1}{\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 \frac{\left (a \sin \left (f x + e\right )\right )^{m}}{\sqrt{b \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/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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

[Out]

Integral((a*sin(e + f*x))**m/sqrt(b*tan(e + f*x)), x)

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

[Out]

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