Integrand size = 21, antiderivative size = 67 \[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \] Output:
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/f/((1+b*sin(f*x+e)^2/a)^p)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{f} \] Input:
Integrate[Cos[e + f*x]*(a + b*Sin[e + f*x]^2)^p,x]
Output:
(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)/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3669, 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.
\(\displaystyle \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x) \left (a+b \sin (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \frac {\int \left (b \sin ^2(e+f x)+a\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {\left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \int \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \sin ^2(e+f x)}{a}\right )}{f}\) |
Input:
Int[Cos[e + f*x]*(a + b*Sin[e + f*x]^2)^p,x]
Output:
(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)/(f*(1 + (b*Sin[e + f*x]^2)/a)^p)
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[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]
\[\int \cos \left (f x +e \right ) \left (a +b \sin \left (f x +e \right )^{2}\right )^{p}d x\]
Input:
int(cos(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)
Output:
int(cos(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)
\[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right ) \,d x } \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="fricas")
Output:
integral((-b*cos(f*x + e)^2 + a + b)^p*cos(f*x + e), x)
Timed out. \[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)**2)**p,x)
Output:
Timed out
\[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right ) \,d x } \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="maxima")
Output:
integrate((b*sin(f*x + e)^2 + a)^p*cos(f*x + e), x)
\[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \cos \left (f x + e\right ) \,d x } \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)^2)^p,x, algorithm="giac")
Output:
integrate((b*sin(f*x + e)^2 + a)^p*cos(f*x + e), x)
Time = 36.52 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ -\frac {b\,{\sin \left (e+f\,x\right )}^2}{a}\right )}{f\,{\left (\frac {b\,{\sin \left (e+f\,x\right )}^2}{a}+1\right )}^p} \] Input:
int(cos(e + f*x)*(a + b*sin(e + f*x)^2)^p,x)
Output:
(sin(e + f*x)*(a + b*sin(e + f*x)^2)^p*hypergeom([1/2, -p], 3/2, -(b*sin(e + f*x)^2)/a))/(f*((b*sin(e + f*x)^2)/a + 1)^p)
\[ \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\left (\sin \left (f x +e \right )^{2} b +a \right )^{p} \sin \left (f x +e \right )+4 \left (\int \frac {\left (\sin \left (f x +e \right )^{2} b +a \right )^{p} \cos \left (f x +e \right )}{2 \sin \left (f x +e \right )^{2} b p +\sin \left (f x +e \right )^{2} b +2 a p +a}d x \right ) a f \,p^{2}+2 \left (\int \frac {\left (\sin \left (f x +e \right )^{2} b +a \right )^{p} \cos \left (f x +e \right )}{2 \sin \left (f x +e \right )^{2} b p +\sin \left (f x +e \right )^{2} b +2 a p +a}d x \right ) a f p}{f \left (2 p +1\right )} \] Input:
int(cos(f*x+e)*(a+b*sin(f*x+e)^2)^p,x)
Output:
((sin(e + f*x)**2*b + a)**p*sin(e + f*x) + 4*int(((sin(e + f*x)**2*b + a)* *p*cos(e + f*x))/(2*sin(e + f*x)**2*b*p + sin(e + f*x)**2*b + 2*a*p + a),x )*a*f*p**2 + 2*int(((sin(e + f*x)**2*b + a)**p*cos(e + f*x))/(2*sin(e + f* x)**2*b*p + sin(e + f*x)**2*b + 2*a*p + a),x)*a*f*p)/(f*(2*p + 1))