Integrand size = 14, antiderivative size = 90 \[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{f} \] Output:
AppellF1(1/2,1/2,-p,3/2,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)*(cos(f*x+e)^2)^(1/ 2)*(a+b*sin(f*x+e)^2)^p*tan(f*x+e)/f/((1+b*sin(f*x+e)^2/a)^p)
Time = 0.64 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.61 \[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {2^{-1-p} \operatorname {AppellF1}\left (1+p,\frac {1}{2},\frac {1}{2},2+p,\frac {2 a+b-b \cos (2 (e+f x))}{2 (a+b)},\frac {2 a+b-b \cos (2 (e+f x))}{2 a}\right ) \sqrt {\frac {b \cos ^2(e+f x)}{a+b}} (2 a+b-b \cos (2 (e+f x)))^{1+p} \csc (2 (e+f x)) \sqrt {-\frac {b \sin ^2(e+f x)}{a}}}{b f (1+p)} \] Input:
Integrate[(a + b*Sin[e + f*x]^2)^p,x]
Output:
(2^(-1 - p)*AppellF1[1 + p, 1/2, 1/2, 2 + p, (2*a + b - b*Cos[2*(e + f*x)] )/(2*(a + b)), (2*a + b - b*Cos[2*(e + f*x)])/(2*a)]*Sqrt[(b*Cos[e + f*x]^ 2)/(a + b)]*(2*a + b - b*Cos[2*(e + f*x)])^(1 + p)*Csc[2*(e + f*x)]*Sqrt[- ((b*Sin[e + f*x]^2)/a)])/(b*f*(1 + p))
Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3664, 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.
\(\displaystyle \int \left (a+b \sin ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \sin (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 3664 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\left (b \sin ^2(e+f x)+a\right )^p}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\left (\frac {b \sin ^2(e+f x)}{a}+1\right )^p}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-p,\frac {3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{f}\) |
Input:
Int[(a + b*Sin[e + f*x]^2)^p,x]
Output:
(AppellF1[1/2, 1/2, -p, 3/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*Sqrt [Cos[e + f*x]^2]*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f* x])) Subst[Int[(a + b*ff^2*x^2)^p/Sqrt[1 - ff^2*x^2], x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && !IntegerQ[p]
\[\int \left (a +b \sin \left (f x +e \right )^{2}\right )^{p}d x\]
Input:
int((a+b*sin(f*x+e)^2)^p,x)
Output:
int((a+b*sin(f*x+e)^2)^p,x)
\[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \,d x } \] Input:
integrate((a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")
Output:
integral((-b*cos(f*x + e)^2 + a + b)^p, x)
Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:
integrate((a+b*sin(f*x+e)**2)**p,x)
Output:
Timed out
\[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \,d x } \] Input:
integrate((a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")
Output:
integrate((b*sin(f*x + e)^2 + a)^p, x)
\[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \,d x } \] Input:
integrate((a+b*sin(f*x+e)^2)^p,x, algorithm="giac")
Output:
integrate((b*sin(f*x + e)^2 + a)^p, x)
Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \] Input:
int((a + b*sin(e + f*x)^2)^p,x)
Output:
int((a + b*sin(e + f*x)^2)^p, x)
\[ \int \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int \left (\sin \left (f x +e \right )^{2} b +a \right )^{p}d x \] Input:
int((a+b*sin(f*x+e)^2)^p,x)
Output:
int((sin(e + f*x)**2*b + a)**p,x)