\(\int (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n \, dx\) [185]

Optimal result

Integrand size = 23, antiderivative size = 89 \[ \int (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n \, dx=\frac {2 \cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\sin ^2(e+f x)\right ) (a \sin (e+f x))^{3/2} (b \tan (e+f x))^{1+n}}{b f (5+2 n)} \]

[Out]

2*(cos(f*x+e)^2)^(1/2+1/2*n)*hypergeom([1/2+1/2*n, 5/4+1/2*n],[9/4+1/2*n],sin(f*x+e)^2)*(a*sin(f*x+e))^(3/2)*(
b*tan(f*x+e))^(1+n)/b/f/(5+2*n)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2682, 2657} \[ \int (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n \, dx=\frac {2 (a \sin (e+f x))^{3/2} \cos ^2(e+f x)^{\frac {n+1}{2}} (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{4} (2 n+5),\frac {1}{4} (2 n+9),\sin ^2(e+f x)\right )}{b f (2 n+5)} \]

[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 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*} \text {integral}& = \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}} \operatorname {Hypergeometric2F1}\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] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 32.84 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.34 \[ \int (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n \, dx=\frac {8 (9+2 n) \operatorname {AppellF1}\left (\frac {5}{4}+\frac {n}{2},n,\frac {5}{2},\frac {9}{4}+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sin \left (\frac {1}{2} (e+f x)\right ) (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n}{f (5+2 n) \left (2 (9+2 n) \operatorname {AppellF1}\left (\frac {5}{4}+\frac {n}{2},n,\frac {5}{2},\frac {9}{4}+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (5 \operatorname {AppellF1}\left (\frac {9}{4}+\frac {n}{2},n,\frac {7}{2},\frac {13}{4}+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {9}{4}+\frac {n}{2},1+n,\frac {5}{2},\frac {13}{4}+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-1+\cos (e+f x))\right )} \]

[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])))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[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)

Sympy [F(-1)]

Timed out. \[ \int (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int (a \sin (e+f x))^{3/2} (b \tan (e+f x))^n \, dx=\int {\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]

[In]

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

[Out]

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