3.2.87 \(\int \frac {(b \tan (e+f x))^n}{\sqrt {a \sin (e+f x)}} \, dx\) [187]

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

[Out]

2*(cos(f*x+e)^2)^(1/2+1/2*n)*hypergeom([1/4+1/2*n, 1/2+1/2*n],[5/4+1/2*n],sin(f*x+e)^2)*(b*tan(f*x+e))^(1+n)/b
/f/(1+2*n)/(a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 = {2682, 2657} \begin {gather*} \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)}} \end {gather*}

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 2657

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)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2682

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]

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*}

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Mathematica [A]
time = 11.42, size = 89, normalized size = 1.00 \begin {gather*} \frac {\cos ^2(e+f x)^{\frac {1}{2} (-1+n)} \, _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 ) \sin (2 (e+f x)) (b \tan (e+f x))^n}{(f+2 f n) \sqrt {a \sin (e+f x)}} \end {gather*}

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 = 0.20, size = 0, normalized size = 0.00 \[\int \frac {\left (b \tan \left (f x +e \right )\right )^{n}}{\sqrt {a \sin \left (f x +e \right )}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \tan {\left (e + f x \right )}\right )^{n}}{\sqrt {a \sin {\left (e + f x \right )}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{\sqrt {a\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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