Integrand size = 23, antiderivative size = 214 \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {(3 a+b (7+2 p)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}-\frac {\cos ^2(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2+2 a b (5+2 p)+b^2 \left (15+16 p+4 p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)} \]
-(3*a+b*(7+2*p))*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(p+1)/b^2/f/(4*p^2+16*p+15) -cos(f*x+e)^2*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(p+1)/b/f/(5+2*p)+(3*a^2+2*a*b *(5+2*p)+b^2*(4*p^2+16*p+15))*hypergeom([1/2, -p],[3/2],-b*sin(f*x+e)^2/a) *sin(f*x+e)*(a+b*sin(f*x+e)^2)^p/b^2/f/(4*p^2+16*p+15)/((1+b*sin(f*x+e)^2/ a)^p)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.44 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},-2,-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \cos ^4(e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p}{f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},-2,-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+2 \left (b p \operatorname {AppellF1}\left (\frac {3}{2},-2,1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-2 a \operatorname {AppellF1}\left (\frac {3}{2},-1,-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \]
(3*a*AppellF1[1/2, -2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*C os[e + f*x]^4*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(3*a*AppellF1[1/2, -2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] + 2*(b*p*AppellF1[3 /2, -2, 1 - p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] - 2*a*AppellF 1[3/2, -1, -p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)])*Sin[e + f*x] ^2))
Time = 0.38 (sec) , antiderivative size = 211, 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, 3669, 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.
\(\displaystyle \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^5 \left (a+b \sin (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \left (1-\sin ^2(e+f x)\right )^2 \left (b \sin ^2(e+f x)+a\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {\frac {\int \left (b \sin ^2(e+f x)+a\right )^p \left (-\left ((3 a+b (2 p+7)) \sin ^2(e+f x)\right )+a+b (2 p+5)\right )d\sin (e+f x)}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {\frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \int \left (b \sin ^2(e+f x)+a\right )^pd\sin (e+f x)}{b (2 p+3)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+3)}}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {\frac {\frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^pd\sin (e+f x)}{b (2 p+3)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+3)}}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {\frac {\frac {\left (3 a^2+2 a b (2 p+5)+b^2 \left (4 p^2+16 p+15\right )\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right )}{b (2 p+3)}-\frac {(3 a+b (2 p+7)) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+3)}}{b (2 p+5)}-\frac {\sin (e+f x) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{p+1}}{b (2 p+5)}}{f}\) |
(-((Sin[e + f*x]*(1 - Sin[e + f*x]^2)*(a + b*Sin[e + f*x]^2)^(1 + p))/(b*( 5 + 2*p))) + (-(((3*a + b*(7 + 2*p))*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^( 1 + p))/(b*(3 + 2*p))) + ((3*a^2 + 2*a*b*(5 + 2*p) + b^2*(15 + 16*p + 4*p^ 2))*Hypergeometric2F1[1/2, -p, 3/2, -((b*Sin[e + f*x]^2)/a)]*Sin[e + f*x]* (a + b*Sin[e + f*x]^2)^p)/(b*(3 + 2*p)*(1 + (b*Sin[e + f*x]^2)/a)^p))/(b*( 5 + 2*p)))/f
3.4.73.3.1 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[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
\[\int \left (\cos ^{5}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
\[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\cos \left (e+f\,x\right )}^5\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]