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

Optimal. Leaf size=89 \[ \frac{2 \sqrt{a \sin (e+f x)} \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+3);\frac{1}{4} (2 n+7);\sin ^2(e+f x)\right )}{b f (2 n+3)} \]

[Out]

(2*(Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (3 + 2*n)/4, (7 + 2*n)/4, Sin[e + f*x]^2]*Sqrt[a*
Sin[e + f*x]]*(b*Tan[e + f*x])^(1 + n))/(b*f*(3 + 2*n))

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Rubi [A]  time = 0.106581, 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} \[ \frac{2 \sqrt{a \sin (e+f x)} \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+3);\frac{1}{4} (2 n+7);\sin ^2(e+f x)\right )}{b f (2 n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (3 + 2*n)/4, (7 + 2*n)/4, Sin[e + f*x]^2]*Sqrt[a*
Sin[e + f*x]]*(b*Tan[e + f*x])^(1 + n))/(b*f*(3 + 2*n))

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 \sqrt{a \sin (e+f x)} (b \tan (e+f x))^n \, 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} (3+2 n);\frac{1}{4} (7+2 n);\sin ^2(e+f x)\right ) \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{1+n}}{b f (3+2 n)}\\ \end{align*}

Mathematica [A]  time = 1.57254, size = 91, normalized size = 1.02 \[ \frac{\sin (2 (e+f x)) \sqrt{a \sin (e+f x)} \cos ^2(e+f x)^{\frac{n-1}{2}} (b \tan (e+f x))^n \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+3);\frac{1}{4} (2 n+7);\sin ^2(e+f x)\right )}{f (2 n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^((-1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (3 + 2*n)/4, (7 + 2*n)/4, Sin[e + f*x]^2]*Sqrt[a*S
in[e + f*x]]*Sin[2*(e + f*x)]*(b*Tan[e + f*x])^n)/(f*(3 + 2*n))

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Maple [F]  time = 0.136, size = 0, normalized size = 0. \begin{align*} \int \sqrt{a\sin \left ( fx+e \right ) } \left ( b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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