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

Optimal. Leaf size=89 \[ \frac{2 \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\sin ^2(e+f x)\right )}{b f (2 n+1) \sqrt{a \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.106694, antiderivative size = 89, 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} \[ \frac{2 \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\sin ^2(e+f x)\right )}{b f (2 n+1) \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{(b \tan (e+f x))^n}{\sqrt{a \sin (e+f x)}} \, 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))^{-\frac{1}{2}+n} \, dx}{b}\\ &=\frac{2 \cos ^2(e+f x)^{\frac{1+n}{2}} \, _2F_1\left (\frac{1+n}{2},\frac{1}{4} (1+2 n);\frac{1}{4} (5+2 n);\sin ^2(e+f x)\right ) (b \tan (e+f x))^{1+n}}{b f (1+2 n) \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.25083, size = 89, normalized size = 1. \[ \frac{\sin (2 (e+f x)) \cos ^2(e+f x)^{\frac{n-1}{2}} (b \tan (e+f x))^n \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\sin ^2(e+f x)\right )}{(2 f n+f) \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.135, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\tan \left ( fx+e \right ) \right ) ^{n}{\frac{1}{\sqrt{a\sin \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^n/(a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \sin \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{n}}{a \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^n/(a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))**n/(a*sin(f*x+e))**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^n/(a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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