Integrand size = 23, antiderivative size = 93 \[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+n p)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+n p),\frac {1}{2} (4+n p),\frac {1}{2} (6+n p),\sin ^2(e+f x)\right ) \sin ^3(e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (4+n p)} \]
(cos(f*x+e)^2)^(1/2*n*p+1/2)*hypergeom([1/2*n*p+2, 1/2*n*p+1/2],[1/2*n*p+3 ],sin(f*x+e)^2)*sin(f*x+e)^3*tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p/f/(n*p+4)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 3.58 (sec) , antiderivative size = 506, normalized size of antiderivative = 5.44 \[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {4 (4+n p) \left (\operatorname {AppellF1}\left (1+\frac {n p}{2},n p,3,2+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-\operatorname {AppellF1}\left (1+\frac {n p}{2},n p,4,2+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p) \left (2 (4+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},n p,3,2+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )-2 (4+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},n p,4,2+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (3 \operatorname {AppellF1}\left (2+\frac {n p}{2},n p,4,3+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-4 \operatorname {AppellF1}\left (2+\frac {n p}{2},n p,5,3+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n p \left (-\operatorname {AppellF1}\left (2+\frac {n p}{2},1+n p,3,3+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\operatorname {AppellF1}\left (2+\frac {n p}{2},1+n p,4,3+\frac {n p}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) (-1+\cos (e+f x))\right )} \]
(4*(4 + n*p)*(AppellF1[1 + (n*p)/2, n*p, 3, 2 + (n*p)/2, Tan[(e + f*x)/2]^ 2, -Tan[(e + f*x)/2]^2] - AppellF1[1 + (n*p)/2, n*p, 4, 2 + (n*p)/2, Tan[( e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cos[(e + f*x)/2]^3*Sin[(e + f*x)/2]*S in[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p)/(f*(2 + n*p)*(2*(4 + n*p)*AppellF1 [1 + (n*p)/2, n*p, 3, 2 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2 ]*Cos[(e + f*x)/2]^2 - 2*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 4, 2 + (n*p) /2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(3*App ellF1[2 + (n*p)/2, n*p, 4, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x) /2]^2] - 4*AppellF1[2 + (n*p)/2, n*p, 5, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*p*(-AppellF1[2 + (n*p)/2, 1 + n*p, 3, 3 + (n*p)/2 , Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[2 + (n*p)/2, 1 + n*p , 4, 3 + (n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]))*(-1 + Cos[e + f*x])))
Time = 0.44 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4142, 3042, 3082, 3042, 3057}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^3 \left (b (c \tan (e+f x))^n\right )^pdx\) |
| \(\Big \downarrow \) 4142 |
| \(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \sin ^3(e+f x) (c \tan (e+f x))^{n p}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int \sin (e+f x)^3 (c \tan (e+f x))^{n p}dx\) |
| \(\Big \downarrow \) 3082 |
| \(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos ^{-n p}(e+f x) \sin ^{n p+3}(e+f x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sin ^{-n p}(e+f x) \cos ^{n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p \int \cos (e+f x)^{-n p} \sin (e+f x)^{n p+3}dx\) |
| \(\Big \downarrow \) 3057 |
| \(\displaystyle \frac {\sin ^3(e+f x) \tan (e+f x) \cos ^2(e+f x)^{\frac {1}{2} (n p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (n p+1),\frac {1}{2} (n p+4),\frac {1}{2} (n p+6),\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+4)}\) |
((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (4 + n*p)/2 , (6 + n*p)/2, Sin[e + f*x]^2]*Sin[e + f*x]^3*Tan[e + f*x]*(b*(c*Tan[e + f *x])^n)^p)/(f*(4 + n*p))
3.2.70.3.1 Defintions of rubi rules used
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*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(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]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[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]
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr acPart[p])) Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ b, c, e, f, n, p}, x] && !IntegerQ[p] && !IntegerQ[n] && (EqQ[u, 1] || Ma tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
\[\int \sin \left (f x +e \right )^{3} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]
\[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \sin \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\text {Timed out} \]
\[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \sin \left (f x + e\right )^{3} \,d x } \]
\[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \sin \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \sin ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \]