\(\int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx\) [180]

Optimal result

Integrand size = 19, antiderivative size = 78 \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(e+f x)\right ) \sin ^3(e+f x) (b \tan (e+f x))^{1+n}}{b f (4+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2682, 2657} \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {\sin ^3(e+f x) \cos ^2(e+f x)^{\frac {n+1}{2}} (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {n+4}{2},\frac {n+6}{2},\sin ^2(e+f x)\right )}{b f (n+4)} \]

[In]

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

[Out]

((Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (4 + n)/2, (6 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x]^
3*(b*Tan[e + f*x])^(1 + n))/(b*f*(4 + 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 (\cos ^{1+n}(e+f x) \sin ^{-1-n}(e+f x) (b \tan (e+f x))^{1+n}\right ) \int \cos ^{-n}(e+f x) \sin ^{3+n}(e+f x) \, dx}{b} \\ & = \frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(e+f x)\right ) \sin ^3(e+f x) (b \tan (e+f x))^{1+n}}{b f (4+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 = 2.30 (sec) , antiderivative size = 456, normalized size of antiderivative = 5.85 \[ \int \sin ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {4 (4+n) \left (\operatorname {AppellF1}\left (1+\frac {n}{2},n,3,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-\operatorname {AppellF1}\left (1+\frac {n}{2},n,4,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^3(e+f x) (b \tan (e+f x))^n}{f (2+n) \left (-2 (4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,4,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (3 \operatorname {AppellF1}\left (2+\frac {n}{2},n,4,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 \operatorname {AppellF1}\left (2+\frac {n}{2},n,5,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n \left (-\operatorname {AppellF1}\left (2+\frac {n}{2},1+n,3,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\operatorname {AppellF1}\left (2+\frac {n}{2},1+n,4,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) (-1+\cos (e+f x))+(4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,3,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))\right )} \]

[In]

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

[Out]

(4*(4 + n)*(AppellF1[1 + n/2, n, 3, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - AppellF1[1 + n/2, n, 4
, 2 + 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]*Sin[e + f*x]^3*(b*Tan
[e + f*x])^n)/(f*(2 + n)*(-2*(4 + n)*AppellF1[1 + n/2, n, 4, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
*Cos[(e + f*x)/2]^2 + 2*(3*AppellF1[2 + n/2, n, 4, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 4*Appel
lF1[2 + n/2, n, 5, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*(-AppellF1[2 + n/2, 1 + n, 3, 3 + n/2
, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[2 + n/2, 1 + n, 4, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2]))*(-1 + Cos[e + f*x]) + (4 + n)*AppellF1[1 + n/2, n, 3, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2]*(1 + Cos[e + f*x])))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-(cos(f*x + e)^2 - 1)*(b*tan(f*x + e))^n*sin(f*x + e), x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(sin(f*x+e)^3*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sin(f*x+e)^3*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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