Integrand size = 23, antiderivative size = 226 \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{f}-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{3 f}+\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{5 f} \]
hypergeom([-p, 1/n],[1+1/n],-b*sin(f*x+e)^n/a)*sin(f*x+e)*(a+b*sin(f*x+e)^ n)^p/f/((1+b*sin(f*x+e)^n/a)^p)-2/3*hypergeom([-p, 3/n],[(3+n)/n],-b*sin(f *x+e)^n/a)*sin(f*x+e)^3*(a+b*sin(f*x+e)^n)^p/f/((1+b*sin(f*x+e)^n/a)^p)+1/ 5*hypergeom([-p, 5/n],[(5+n)/n],-b*sin(f*x+e)^n/a)*sin(f*x+e)^5*(a+b*sin(f *x+e)^n)^p/f/((1+b*sin(f*x+e)^n/a)^p)
Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.69 \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\frac {\sin (e+f x) \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right )-10 \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {3+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {5+n}{n},-\frac {b \sin ^n(e+f x)}{a}\right ) \sin ^4(e+f x)\right ) \left (a+b \sin ^n(e+f x)\right )^p \left (1+\frac {b \sin ^n(e+f x)}{a}\right )^{-p}}{15 f} \]
(Sin[e + f*x]*(15*Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*Sin[e + f *x]^n)/a)] - 10*Hypergeometric2F1[3/n, -p, (3 + n)/n, -((b*Sin[e + f*x]^n) /a)]*Sin[e + f*x]^2 + 3*Hypergeometric2F1[5/n, -p, (5 + n)/n, -((b*Sin[e + f*x]^n)/a)]*Sin[e + f*x]^4)*(a + b*Sin[e + f*x]^n)^p)/(15*f*(1 + (b*Sin[e + f*x]^n)/a)^p)
Time = 0.41 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3702, 2432, 2009}
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.
\(\displaystyle \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^5 \left (a+b \sin (e+f x)^n\right )^pdx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \left (1-\sin ^2(e+f x)\right )^2 \left (b \sin ^n(e+f x)+a\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 2432 |
\(\displaystyle \frac {\int \left (\sin ^4(e+f x) \left (b \sin ^n(e+f x)+a\right )^p-2 \sin ^2(e+f x) \left (b \sin ^n(e+f x)+a\right )^p+\left (b \sin ^n(e+f x)+a\right )^p\right )d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sin (e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{n},-p,1+\frac {1}{n},-\frac {b \sin ^n(e+f x)}{a}\right )+\frac {1}{5} \sin ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{n},-p,\frac {n+5}{n},-\frac {b \sin ^n(e+f x)}{a}\right )-\frac {2}{3} \sin ^3(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \left (\frac {b \sin ^n(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{n},-p,\frac {n+3}{n},-\frac {b \sin ^n(e+f x)}{a}\right )}{f}\) |
((Hypergeometric2F1[n^(-1), -p, 1 + n^(-1), -((b*Sin[e + f*x]^n)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^n)^p)/(1 + (b*Sin[e + f*x]^n)/a)^p - (2*Hyperg eometric2F1[3/n, -p, (3 + n)/n, -((b*Sin[e + f*x]^n)/a)]*Sin[e + f*x]^3*(a + b*Sin[e + f*x]^n)^p)/(3*(1 + (b*Sin[e + f*x]^n)/a)^p) + (Hypergeometric 2F1[5/n, -p, (5 + n)/n, -((b*Sin[e + f*x]^n)/a)]*Sin[e + f*x]^5*(a + b*Sin [e + f*x]^n)^p)/(5*(1 + (b*Sin[e + f*x]^n)/a)^p))/f
3.5.31.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly Q[Pq, x^n])
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
\[\int \left (\cos ^{5}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{n}\left (f x +e \right )\right )\right )}^{p}d x\]
\[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
\[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{n} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^n(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^5\,{\left (a+b\,{\sin \left (e+f\,x\right )}^n\right )}^p \,d x \]