3.2.88 \(\int \frac {(b \tan (e+f x))^n}{(a \sin (e+f x))^{3/2}} \, dx\) [188]

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

[Out]

-2*(cos(f*x+e)^2)^(1/2+1/2*n)*hypergeom([-1/4+1/2*n, 1/2+1/2*n],[3/4+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))^(3/2)

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Rubi [A]
time = 0.08, 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+3);\sin ^2(e+f x)\right )}{b f (1-2 n) (a \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^n/(a*Sin[e + f*x])^(3/2),x]

[Out]

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

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

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Mathematica [A]
time = 12.13, size = 90, normalized size = 1.01 \begin {gather*} \frac {2 b \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} (3+2 n);\sin ^2(e+f x)\right ) \sqrt {a \sin (e+f x)} (b \tan (e+f x))^{-1+n}}{a^2 f (-1+2 n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^n/(a*Sin[e + f*x])^(3/2),x]

[Out]

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

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Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \frac {\left (b \tan \left (f x +e \right )\right )^{n}}{\left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^n/(a*sin(f*x+e))^(3/2),x)

[Out]

int((b*tan(f*x+e))^n/(a*sin(f*x+e))^(3/2),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))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^n/(a*sin(f*x + e))^(3/2), 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))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(a*sin(f*x + e))*(b*tan(f*x + e))^n/(a^2*cos(f*x + e)^2 - a^2), 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}}{\left (a \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, 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))**(3/2),x)

[Out]

Integral((b*tan(e + f*x))**n/(a*sin(e + f*x))**(3/2), 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))^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^n/(a*sin(f*x + e))^(3/2), 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}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,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))^(3/2),x)

[Out]

int((b*tan(e + f*x))^n/(a*sin(e + f*x))^(3/2), x)

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