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

Optimal. Leaf size=89 \[ \frac{2 (a \sin (e+f x))^{3/2} \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+5);\frac{1}{4} (2 n+9);\sin ^2(e+f x)\right )}{b f (2 n+5)} \]

[Out]

(2*(Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (5 + 2*n)/4, (9 + 2*n)/4, Sin[e + f*x]^2]*(a*Sin[
e + f*x])^(3/2)*(b*Tan[e + f*x])^(1 + n))/(b*f*(5 + 2*n))

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Rubi [A]  time = 0.120763, 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} \[ \frac{2 (a \sin (e+f x))^{3/2} \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{1}{4} (2 n+5);\frac{1}{4} (2 n+9);\sin ^2(e+f x)\right )}{b f (2 n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (5 + 2*n)/4, (9 + 2*n)/4, Sin[e + f*x]^2]*(a*Sin[
e + f*x])^(3/2)*(b*Tan[e + f*x])^(1 + n))/(b*f*(5 + 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 (a \sin (e+f x))^{3/2} (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{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} (5+2 n);\frac{1}{4} (9+2 n);\sin ^2(e+f x)\right ) (a \sin (e+f x))^{3/2} (b \tan (e+f x))^{1+n}}{b f (5+2 n)}\\ \end{align*}

Mathematica [C]  time = 2.44902, size = 297, normalized size = 3.34 \[ \frac{8 (2 n+9) \sin \left (\frac{1}{2} (e+f x)\right ) \cos ^3\left (\frac{1}{2} (e+f x)\right ) (a \sin (e+f x))^{3/2} F_1\left (\frac{n}{2}+\frac{5}{4};n,\frac{5}{2};\frac{n}{2}+\frac{9}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) (b \tan (e+f x))^n}{f (2 n+5) \left (2 (2 n+9) \cos ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{n}{2}+\frac{5}{4};n,\frac{5}{2};\frac{n}{2}+\frac{9}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 (\cos (e+f x)-1) \left (5 F_1\left (\frac{n}{2}+\frac{9}{4};n,\frac{7}{2};\frac{n}{2}+\frac{13}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 n F_1\left (\frac{n}{2}+\frac{9}{4};n+1,\frac{5}{2};\frac{n}{2}+\frac{13}{4};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*(9 + 2*n)*AppellF1[5/4 + n/2, n, 5/2, 9/4 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^
3*Sin[(e + f*x)/2]*(a*Sin[e + f*x])^(3/2)*(b*Tan[e + f*x])^n)/(f*(5 + 2*n)*(2*(9 + 2*n)*AppellF1[5/4 + n/2, n,
5/2, 9/4 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(5*AppellF1[9/4 + n/2, n, 7/2
, 13/4 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*n*AppellF1[9/4 + n/2, 1 + n, 5/2, 13/4 + n/2, Tan[(
e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x])))

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Maple [F]  time = 0.144, size = 0, normalized size = 0. \begin{align*} \int \left ( a\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \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))^(3/2)*(b*tan(f*x+e))^n,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}} \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))^(3/2)*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sin(f*x+e))^(3/2)*(b*tan(f*x+e))^n,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(-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))**(3/2)*(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 \left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}} \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))^(3/2)*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

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