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

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

[Out]

-2*(cos(f*x+e)^2)^(1/2+1/2*n)*hypergeom([1/2+1/2*n, -1/4+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 = 0.12, 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 = {2602, 2577} \[ -\frac {2 \cos ^2(e+f x)^{\frac {n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac {n+1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\sin ^2(e+f x)\right )}{b f (1-2 n) (a \sin (e+f x))^{3/2}} \]

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 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]

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]

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 = 1.96, size = 90, normalized size = 1.01 \[ \frac {2 b \sqrt {a \sin (e+f x)} \cos ^2(e+f x)^{\frac {n-1}{2}} (b \tan (e+f x))^{n-1} \, _2F_1\left (\frac {n+1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\sin ^2(e+f x)\right )}{a^2 f (2 n-1)} \]

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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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sin \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{n}}{a^{2} \cos \left (f x + e\right )^{2} - a^{2}}, x\right ) \]

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

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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maple [F]  time = 0.40, 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 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{n}}{\left (a \sin \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

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

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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