\(\int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx\) [174]

Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{1+n}}{b f (1+m+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2682, 2657} \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {n+1}{2}} (a \sin (e+f x))^m (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{2} (m+n+1),\frac {1}{2} (m+n+3),\sin ^2(e+f x)\right )}{b f (m+n+1)} \]

[In]

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

[Out]

((Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (1 + m + n)/2, (3 + m + n)/2, Sin[e + f*x]^2]*(a*Si
n[e + f*x])^m*(b*Tan[e + f*x])^(1 + n))/(b*f*(1 + m + 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))^{m+n} \, dx}{b} \\ & = \frac {\cos ^2(e+f x)^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sin ^2(e+f x)\right ) (a \sin (e+f x))^m (b \tan (e+f x))^{1+n}}{b f (1+m+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 = 1.82 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.13 \[ \int (a \sin (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {(3+m+n) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),n,1+m,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x) (a \sin (e+f x))^m (b \tan (e+f x))^n}{f (1+m+n) \left ((3+m+n) \operatorname {AppellF1}\left (\frac {1}{2} (1+m+n),n,1+m,\frac {1}{2} (3+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((1+m) \operatorname {AppellF1}\left (\frac {1}{2} (3+m+n),n,2+m,\frac {1}{2} (5+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (\frac {1}{2} (3+m+n),1+n,1+m,\frac {1}{2} (5+m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

((3 + m + n)*AppellF1[(1 + m + n)/2, n, 1 + m, (3 + m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[e +
f*x]*(a*Sin[e + f*x])^m*(b*Tan[e + f*x])^n)/(f*(1 + m + n)*((3 + m + n)*AppellF1[(1 + m + n)/2, n, 1 + m, (3
+ m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((1 + m)*AppellF1[(3 + m + n)/2, n, 2 + m, (5 + m + n
)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[(3 + m + n)/2, 1 + n, 1 + m, (5 + m + n)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((a*sin(f*x + e))^m*(b*tan(f*x + e))^n, x)

Sympy [F]

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

[In]

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

[Out]

Integral((a*sin(e + f*x))**m*(b*tan(e + f*x))**n, x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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