Integrand size = 23, antiderivative size = 220 \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {(3 a-2 b (2+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p} \sin ^2(e+f x)}{b f (5+2 p)} \]
(3*a-2*b*(2+p))*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(p+1)/b^2/f/(4*p^2+16*p+15 )-(3*a^2-4*a*b*(p+1)+4*b^2*(p^2+3*p+2))*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^p* hypergeom([1/2, -p],[3/2],b*cos(f*x+e)^2/(a+b))/b^2/f/(4*p^2+16*p+15)/((1- b*cos(f*x+e)^2/(a+b))^p)-cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(p+1)*sin(f*x+e)^ 2/b/f/(5+2*p)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.58 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.45 \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (3,\frac {1}{2},-p,4,\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {a+b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{6 f} \]
(AppellF1[3, 1/2, -p, 4, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt[Cos [e + f*x]^2]*Sin[e + f*x]^5*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(6*f*(( a + b*Sin[e + f*x]^2)/a)^p)
Time = 0.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3665, 318, 299, 238, 237}
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.
| \(\displaystyle \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^5 \left (a+b \sin (e+f x)^2\right )^pdx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \left (1-\cos ^2(e+f x)\right )^2 \left (-b \cos ^2(e+f x)+a+b\right )^pd\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (1-\cos ^2(e+f x)\right ) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\int \left (-b \cos ^2(e+f x)+a+b\right )^p \left (-\left ((3 a-2 b (p+2)) \cos ^2(e+f x)\right )+a-2 b (p+2)\right )d\cos (e+f x)}{b (2 p+5)}}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (1-\cos ^2(e+f x)\right ) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\frac {(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \int \left (-b \cos ^2(e+f x)+a+b\right )^pd\cos (e+f x)}{b (2 p+3)}}{b (2 p+5)}}{f}\) |
| \(\Big \downarrow \) 238 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (1-\cos ^2(e+f x)\right ) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\frac {(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \int \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^pd\cos (e+f x)}{b (2 p+3)}}{b (2 p+5)}}{f}\) |
| \(\Big \downarrow \) 237 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (1-\cos ^2(e+f x)\right ) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+5)}-\frac {\frac {(3 a-2 b (p+2)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b (2 p+3)}-\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b (2 p+3)}}{b (2 p+5)}}{f}\) |
-(((Cos[e + f*x]*(1 - Cos[e + f*x]^2)*(a + b - b*Cos[e + f*x]^2)^(1 + p))/ (b*(5 + 2*p)) - (((3*a - 2*b*(2 + p))*Cos[e + f*x]*(a + b - b*Cos[e + f*x] ^2)^(1 + p))/(b*(3 + 2*p)) - ((3*a^2 - 4*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^ 2))*Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^p*Hypergeometric2F1[1/2, -p, 3 /2, (b*Cos[e + f*x]^2)/(a + b)])/(b*(3 + 2*p)*(1 - (b*Cos[e + f*x]^2)/(a + b))^p))/(b*(5 + 2*p)))/f)
3.2.72.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
\[\int \left (\sin ^{5}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
\[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(-b*cos(f*x + e)^2 + a + b)^p*sin(f*x + e), x)
Timed out. \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
\[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \sin ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]