Integrand size = 18, antiderivative size = 69 \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\frac {625}{32} \arcsin \left (\frac {2 \sin (x)}{\sqrt {5}}\right )+\frac {125}{16} \sin (x) \sqrt {5-4 \sin ^2(x)}+\frac {25}{24} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2} \] Output:
625/32*arcsin(2/5*sin(x)*5^(1/2))+25/24*sin(x)*(5-4*sin(x)^2)^(3/2)+1/6*si n(x)*(5-4*sin(x)^2)^(5/2)+125/16*sin(x)*(5-4*sin(x)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\frac {1}{96} \left (1875 \arcsin \left (\frac {2 \sin (x)}{\sqrt {5}}\right )+2 \sqrt {3+2 \cos (2 x)} (515 \sin (x)+90 \sin (3 x)+8 \sin (5 x))\right ) \] Input:
Integrate[Cos[x]*(5*Cos[x]^2 + Sin[x]^2)^(5/2),x]
Output:
(1875*ArcSin[(2*Sin[x])/Sqrt[5]] + 2*Sqrt[3 + 2*Cos[2*x]]*(515*Sin[x] + 90 *Sin[3*x] + 8*Sin[5*x]))/96
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4856, 211, 211, 211, 223}
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.
\(\displaystyle \int \cos (x) \left (\sin ^2(x)+5 \cos ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (x) \left (\sin (x)^2+5 \cos (x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4856 |
\(\displaystyle \int \left (5-4 \sin ^2(x)\right )^{5/2}d\sin (x)\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {25}{6} \int \left (5-4 \sin ^2(x)\right )^{3/2}d\sin (x)+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {25}{6} \left (\frac {15}{4} \int \sqrt {5-4 \sin ^2(x)}d\sin (x)+\frac {1}{4} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}\right )+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {25}{6} \left (\frac {15}{4} \left (\frac {5}{2} \int \frac {1}{\sqrt {5-4 \sin ^2(x)}}d\sin (x)+\frac {1}{2} \sqrt {5-4 \sin ^2(x)} \sin (x)\right )+\frac {1}{4} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}\right )+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {25}{6} \left (\frac {15}{4} \left (\frac {5}{4} \arcsin \left (\frac {2 \sin (x)}{\sqrt {5}}\right )+\frac {1}{2} \sin (x) \sqrt {5-4 \sin ^2(x)}\right )+\frac {1}{4} \sin (x) \left (5-4 \sin ^2(x)\right )^{3/2}\right )+\frac {1}{6} \sin (x) \left (5-4 \sin ^2(x)\right )^{5/2}\) |
Input:
Int[Cos[x]*(5*Cos[x]^2 + Sin[x]^2)^(5/2),x]
Output:
(Sin[x]*(5 - 4*Sin[x]^2)^(5/2))/6 + (25*((Sin[x]*(5 - 4*Sin[x]^2)^(3/2))/4 + (15*((5*ArcSin[(2*Sin[x])/Sqrt[5]])/4 + (Sin[x]*Sqrt[5 - 4*Sin[x]^2])/2 ))/4))/6
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Sin[c*(a + b*x)], x]}, Simp[d/(b*c) Subst[Int[SubstFor[1, Sin[c*(a + b *x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a + b*x )]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])
Time = 2.58 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.49
method | result | size |
default | \(-\frac {\sqrt {\left (4 \cos \left (x \right )^{2}+1\right ) \sin \left (x \right )^{2}}\, \left (-512 \sqrt {-4 \sin \left (x \right )^{4}+5 \sin \left (x \right )^{2}}\, \sin \left (x \right )^{4}+2080 \sqrt {-4 \sin \left (x \right )^{4}+5 \sin \left (x \right )^{2}}\, \sin \left (x \right )^{2}-1875 \arcsin \left (-1+\frac {8 \sin \left (x \right )^{2}}{5}\right )-3300 \sqrt {-4 \sin \left (x \right )^{4}+5 \sin \left (x \right )^{2}}\right )}{192 \sin \left (x \right ) \sqrt {4 \cos \left (x \right )^{2}+1}}\) | \(103\) |
Input:
int(cos(x)*(5*cos(x)^2+sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/192*((4*cos(x)^2+1)*sin(x)^2)^(1/2)*(-512*(-4*sin(x)^4+5*sin(x)^2)^(1/2 )*sin(x)^4+2080*(-4*sin(x)^4+5*sin(x)^2)^(1/2)*sin(x)^2-1875*arcsin(-1+8/5 *sin(x)^2)-3300*(-4*sin(x)^4+5*sin(x)^2)^(1/2))/sin(x)/(4*cos(x)^2+1)^(1/2 )
Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28 \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\frac {1}{48} \, {\left (128 \, \cos \left (x\right )^{4} + 264 \, \cos \left (x\right )^{2} + 433\right )} \sqrt {4 \, \cos \left (x\right )^{2} + 1} \sin \left (x\right ) + \frac {625}{64} \, \arctan \left (\frac {4 \, {\left (8 \, \cos \left (x\right )^{2} - 3\right )} \sqrt {4 \, \cos \left (x\right )^{2} + 1} \sin \left (x\right ) - 25 \, \cos \left (x\right ) \sin \left (x\right )}{64 \, \cos \left (x\right )^{4} - 23 \, \cos \left (x\right )^{2} - 16}\right ) + \frac {625}{64} \, \arctan \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right ) \] Input:
integrate(cos(x)*(5*cos(x)^2+sin(x)^2)^(5/2),x, algorithm="fricas")
Output:
1/48*(128*cos(x)^4 + 264*cos(x)^2 + 433)*sqrt(4*cos(x)^2 + 1)*sin(x) + 625 /64*arctan((4*(8*cos(x)^2 - 3)*sqrt(4*cos(x)^2 + 1)*sin(x) - 25*cos(x)*sin (x))/(64*cos(x)^4 - 23*cos(x)^2 - 16)) + 625/64*arctan(sin(x)/cos(x))
Timed out. \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cos(x)*(5*cos(x)**2+sin(x)**2)**(5/2),x)
Output:
Timed out
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\frac {1}{6} \, {\left (-4 \, \sin \left (x\right )^{2} + 5\right )}^{\frac {5}{2}} \sin \left (x\right ) + \frac {25}{24} \, {\left (-4 \, \sin \left (x\right )^{2} + 5\right )}^{\frac {3}{2}} \sin \left (x\right ) + \frac {125}{16} \, \sqrt {-4 \, \sin \left (x\right )^{2} + 5} \sin \left (x\right ) + \frac {625}{32} \, \arcsin \left (\frac {2}{5} \, \sqrt {5} \sin \left (x\right )\right ) \] Input:
integrate(cos(x)*(5*cos(x)^2+sin(x)^2)^(5/2),x, algorithm="maxima")
Output:
1/6*(-4*sin(x)^2 + 5)^(5/2)*sin(x) + 25/24*(-4*sin(x)^2 + 5)^(3/2)*sin(x) + 125/16*sqrt(-4*sin(x)^2 + 5)*sin(x) + 625/32*arcsin(2/5*sqrt(5)*sin(x))
Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.59 \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\frac {1}{48} \, {\left (8 \, {\left (16 \, \sin \left (x\right )^{2} - 65\right )} \sin \left (x\right )^{2} + 825\right )} \sqrt {-4 \, \sin \left (x\right )^{2} + 5} \sin \left (x\right ) + \frac {625}{32} \, \arcsin \left (\frac {2}{5} \, \sqrt {5} \sin \left (x\right )\right ) \] Input:
integrate(cos(x)*(5*cos(x)^2+sin(x)^2)^(5/2),x, algorithm="giac")
Output:
1/48*(8*(16*sin(x)^2 - 65)*sin(x)^2 + 825)*sqrt(-4*sin(x)^2 + 5)*sin(x) + 625/32*arcsin(2/5*sqrt(5)*sin(x))
Timed out. \[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\int \cos \left (x\right )\,{\left (5\,{\cos \left (x\right )}^2+{\sin \left (x\right )}^2\right )}^{5/2} \,d x \] Input:
int(cos(x)*(5*cos(x)^2 + sin(x)^2)^(5/2),x)
Output:
int(cos(x)*(5*cos(x)^2 + sin(x)^2)^(5/2), x)
\[ \int \cos (x) \left (5 \cos ^2(x)+\sin ^2(x)\right )^{5/2} \, dx=\int \sqrt {5 \cos \left (x \right )^{2}+\sin \left (x \right )^{2}}\, \cos \left (x \right ) \sin \left (x \right )^{4}d x +25 \left (\int \sqrt {5 \cos \left (x \right )^{2}+\sin \left (x \right )^{2}}\, \cos \left (x \right )^{5}d x \right )+10 \left (\int \sqrt {5 \cos \left (x \right )^{2}+\sin \left (x \right )^{2}}\, \cos \left (x \right )^{3} \sin \left (x \right )^{2}d x \right ) \] Input:
int(cos(x)*(5*cos(x)^2+sin(x)^2)^(5/2),x)
Output:
int(sqrt(5*cos(x)**2 + sin(x)**2)*cos(x)*sin(x)**4,x) + 25*int(sqrt(5*cos( x)**2 + sin(x)**2)*cos(x)**5,x) + 10*int(sqrt(5*cos(x)**2 + sin(x)**2)*cos (x)**3*sin(x)**2,x)