Integrand size = 21, antiderivative size = 67 \[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f} \] Output:
hypergeom([1/4, -p],[5/4],-b*sin(f*x+e)^4/a)*sin(f*x+e)*(a+b*sin(f*x+e)^4) ^p/f/((1+b*sin(f*x+e)^4/a)^p)
Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right ) \sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (1+\frac {b \sin ^4(e+f x)}{a}\right )^{-p}}{f} \] Input:
Integrate[Cos[e + f*x]*(a + b*Sin[e + f*x]^4)^p,x]
Output:
(Hypergeometric2F1[1/4, -p, 5/4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p)/(f*(1 + (b*Sin[e + f*x]^4)/a)^p)
Time = 0.24 (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, 3702, 779, 778}
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.
\(\displaystyle \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x) \left (a+b \sin (e+f x)^4\right )^pdx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \frac {\int \left (b \sin ^4(e+f x)+a\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {\left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \int \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^pd\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {\sin (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \left (\frac {b \sin ^4(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b \sin ^4(e+f x)}{a}\right )}{f}\) |
Input:
Int[Cos[e + f*x]*(a + b*Sin[e + f*x]^4)^p,x]
Output:
(Hypergeometric2F1[1/4, -p, 5/4, -((b*Sin[e + f*x]^4)/a)]*Sin[e + f*x]*(a + b*Sin[e + f*x]^4)^p)/(f*(1 + (b*Sin[e + f*x]^4)/a)^p)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x ^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p, x], x ] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p, 0] && !IntegerQ[1/n] && !ILtQ[Si mplify[1/n + p], 0] && !(IntegerQ[p] || GtQ[a, 0])
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])
\[\int \cos \left (f x +e \right ) \left (a +b \sin \left (f x +e \right )^{4}\right )^{p}d x\]
Input:
int(cos(f*x+e)*(a+b*sin(f*x+e)^4)^p,x)
Output:
int(cos(f*x+e)*(a+b*sin(f*x+e)^4)^p,x)
\[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right ) \,d x } \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)^4)^p,x, algorithm="fricas")
Output:
integral((b*cos(f*x + e)^4 - 2*b*cos(f*x + e)^2 + a + b)^p*cos(f*x + e), x )
Timed out. \[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)**4)**p,x)
Output:
Timed out
\[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right ) \,d x } \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)^4)^p,x, algorithm="maxima")
Output:
integrate((b*sin(f*x + e)^4 + a)^p*cos(f*x + e), x)
\[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{4} + a\right )}^{p} \cos \left (f x + e\right ) \,d x } \] Input:
integrate(cos(f*x+e)*(a+b*sin(f*x+e)^4)^p,x, algorithm="giac")
Output:
integrate((b*sin(f*x + e)^4 + a)^p*cos(f*x + e), x)
Time = 35.81 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^4+a\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},-p;\ \frac {5}{4};\ -\frac {b\,{\sin \left (e+f\,x\right )}^4}{a}\right )}{f\,{\left (\frac {b\,{\sin \left (e+f\,x\right )}^4}{a}+1\right )}^p} \] Input:
int(cos(e + f*x)*(a + b*sin(e + f*x)^4)^p,x)
Output:
(sin(e + f*x)*(a + b*sin(e + f*x)^4)^p*hypergeom([1/4, -p], 5/4, -(b*sin(e + f*x)^4)/a))/(f*((b*sin(e + f*x)^4)/a + 1)^p)
\[ \int \cos (e+f x) \left (a+b \sin ^4(e+f x)\right )^p \, dx=\frac {\left (\sin \left (f x +e \right )^{4} b +a \right )^{p} \sin \left (f x +e \right )+16 \left (\int \frac {\left (\sin \left (f x +e \right )^{4} b +a \right )^{p} \cos \left (f x +e \right )}{4 \sin \left (f x +e \right )^{4} b p +\sin \left (f x +e \right )^{4} b +4 a p +a}d x \right ) a f \,p^{2}+4 \left (\int \frac {\left (\sin \left (f x +e \right )^{4} b +a \right )^{p} \cos \left (f x +e \right )}{4 \sin \left (f x +e \right )^{4} b p +\sin \left (f x +e \right )^{4} b +4 a p +a}d x \right ) a f p}{f \left (4 p +1\right )} \] Input:
int(cos(f*x+e)*(a+b*sin(f*x+e)^4)^p,x)
Output:
((sin(e + f*x)**4*b + a)**p*sin(e + f*x) + 16*int(((sin(e + f*x)**4*b + a) **p*cos(e + f*x))/(4*sin(e + f*x)**4*b*p + sin(e + f*x)**4*b + 4*a*p + a), x)*a*f*p**2 + 4*int(((sin(e + f*x)**4*b + a)**p*cos(e + f*x))/(4*sin(e + f *x)**4*b*p + sin(e + f*x)**4*b + 4*a*p + a),x)*a*f*p)/(f*(4*p + 1))