Integrand size = 25, antiderivative size = 115 \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \]
d*AppellF1(1/2,-1/2*m+1/2,-p,3/2,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)*(d*cos(f* x+e))^(-1+m)*(cos(f*x+e)^2)^(-1/2*m+1/2)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^p/f /((1+b*sin(f*x+e)^2/a)^p)
Time = 1.43 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.98 \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \tan (e+f x)}{f \left (3 a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},1-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )-a (-1+m) \operatorname {AppellF1}\left (\frac {3}{2},\frac {3-m}{2},-p,\frac {5}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )\right ) \sin ^2(e+f x)\right )} \]
(3*a*AppellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2 )/a)]*(d*Cos[e + f*x])^m*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(f*(3*a*Ap pellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] + (2*b*p*AppellF1[3/2, (1 - m)/2, 1 - p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)] - a*(-1 + m)*AppellF1[3/2, (3 - m)/2, -p, 5/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)])*Sin[e + f*x]^2))
Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3672, 334, 333}
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.
\(\displaystyle \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \cos (e+f x))^m \left (a+b \sin (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 3672 |
\(\displaystyle \frac {d \cos ^2(e+f x)^{\frac {1-m}{2}} (d \cos (e+f x))^{m-1} \int \left (1-\sin ^2(e+f x)\right )^{\frac {m-1}{2}} \left (b \sin ^2(e+f x)+a\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {d \cos ^2(e+f x)^{\frac {1-m}{2}} (d \cos (e+f x))^{m-1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \left (1-\sin ^2(e+f x)\right )^{\frac {m-1}{2}} \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {d \sin (e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} (d \cos (e+f x))^{m-1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{f}\) |
(d*AppellF1[1/2, (1 - m)/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/ a)]*(d*Cos[e + f*x])^(-1 + m)*(Cos[e + f*x]^2)^((1 - m)/2)*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^p)/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)
3.4.72.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[f f*d^(2*IntPart[(m - 1)/2] + 1)*((d*Cos[e + f*x])^(2*FracPart[(m - 1)/2])/(f *(Cos[e + f*x]^2)^FracPart[(m - 1)/2])) Subst[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, d, e, f, m, p}, x] && !IntegerQ[m]
\[\int \left (\cos \left (f x +e \right ) d \right )^{m} {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
\[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (d \cos \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \cos (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\left (d\,\cos \left (e+f\,x\right )\right )}^m\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]