Integrand size = 15, antiderivative size = 73 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {5 \text {arcsinh}\left (\sqrt {2} \cos (x)\right )}{16 \sqrt {2}}-\frac {5}{16} \cos (x) \sqrt {1+2 \cos ^2(x)}-\frac {5}{24} \cos (x) \left (1+2 \cos ^2(x)\right )^{3/2}-\frac {1}{6} \cos (x) \left (1+2 \cos ^2(x)\right )^{5/2} \] Output:
-5/24*cos(x)*(1+2*cos(x)^2)^(3/2)-1/6*cos(x)*(1+2*cos(x)^2)^(5/2)-5/32*arc sinh(cos(x)*2^(1/2))*2^(1/2)-5/16*cos(x)*(1+2*cos(x)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=\frac {1}{96} \left (-2 \sqrt {2+\cos (2 x)} (92 \cos (x)+23 \cos (3 x)+2 \cos (5 x))-15 \sqrt {2} \log \left (\sqrt {2} \cos (x)+\sqrt {2+\cos (2 x)}\right )\right ) \] Input:
Integrate[(1 + 2*Cos[x]^2)^(5/2)*Sin[x],x]
Output:
(-2*Sqrt[2 + Cos[2*x]]*(92*Cos[x] + 23*Cos[3*x] + 2*Cos[5*x]) - 15*Sqrt[2] *Log[Sqrt[2]*Cos[x] + Sqrt[2 + Cos[2*x]]])/96
Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 25, 3669, 211, 211, 211, 222}
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.
\(\displaystyle \int \sin (x) \left (2 \cos ^2(x)+1\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (2 \sin \left (x+\frac {\pi }{2}\right )^2+1\right )^{5/2} \left (-\cos \left (x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cos \left (x+\frac {\pi }{2}\right ) \left (2 \sin \left (x+\frac {\pi }{2}\right )^2+1\right )^{5/2}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle -\int \left (2 \cos ^2(x)+1\right )^{5/2}d\cos (x)\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {5}{6} \int \left (2 \cos ^2(x)+1\right )^{3/2}d\cos (x)-\frac {1}{6} \cos (x) \left (2 \cos ^2(x)+1\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {5}{6} \left (\frac {3}{4} \int \sqrt {2 \cos ^2(x)+1}d\cos (x)+\frac {1}{4} \cos (x) \left (2 \cos ^2(x)+1\right )^{3/2}\right )-\frac {1}{6} \cos (x) \left (2 \cos ^2(x)+1\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {2 \cos ^2(x)+1}}d\cos (x)+\frac {1}{2} \sqrt {2 \cos ^2(x)+1} \cos (x)\right )+\frac {1}{4} \cos (x) \left (2 \cos ^2(x)+1\right )^{3/2}\right )-\frac {1}{6} \cos (x) \left (2 \cos ^2(x)+1\right )^{5/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {5}{6} \left (\frac {3}{4} \left (\frac {\text {arcsinh}\left (\sqrt {2} \cos (x)\right )}{2 \sqrt {2}}+\frac {1}{2} \cos (x) \sqrt {2 \cos ^2(x)+1}\right )+\frac {1}{4} \cos (x) \left (2 \cos ^2(x)+1\right )^{3/2}\right )-\frac {1}{6} \cos (x) \left (2 \cos ^2(x)+1\right )^{5/2}\) |
Input:
Int[(1 + 2*Cos[x]^2)^(5/2)*Sin[x],x]
Output:
-1/6*(Cos[x]*(1 + 2*Cos[x]^2)^(5/2)) - (5*((Cos[x]*(1 + 2*Cos[x]^2)^(3/2)) /4 + (3*(ArcSinh[Sqrt[2]*Cos[x]]/(2*Sqrt[2]) + (Cos[x]*Sqrt[1 + 2*Cos[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[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
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]
Time = 1.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(-\frac {5 \cos \left (x \right ) \left (1+2 \cos \left (x \right )^{2}\right )^{\frac {3}{2}}}{24}-\frac {\cos \left (x \right ) \left (1+2 \cos \left (x \right )^{2}\right )^{\frac {5}{2}}}{6}-\frac {5 \,\operatorname {arcsinh}\left (\cos \left (x \right ) \sqrt {2}\right ) \sqrt {2}}{32}-\frac {5 \cos \left (x \right ) \sqrt {1+2 \cos \left (x \right )^{2}}}{16}\) | \(56\) |
default | \(-\frac {5 \cos \left (x \right ) \left (1+2 \cos \left (x \right )^{2}\right )^{\frac {3}{2}}}{24}-\frac {\cos \left (x \right ) \left (1+2 \cos \left (x \right )^{2}\right )^{\frac {5}{2}}}{6}-\frac {5 \,\operatorname {arcsinh}\left (\cos \left (x \right ) \sqrt {2}\right ) \sqrt {2}}{32}-\frac {5 \cos \left (x \right ) \sqrt {1+2 \cos \left (x \right )^{2}}}{16}\) | \(56\) |
Input:
int((1+2*cos(x)^2)^(5/2)*sin(x),x,method=_RETURNVERBOSE)
Output:
-5/24*cos(x)*(1+2*cos(x)^2)^(3/2)-1/6*cos(x)*(1+2*cos(x)^2)^(5/2)-5/32*arc sinh(cos(x)*2^(1/2))*2^(1/2)-5/16*cos(x)*(1+2*cos(x)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {1}{48} \, {\left (32 \, \cos \left (x\right )^{5} + 52 \, \cos \left (x\right )^{3} + 33 \, \cos \left (x\right )\right )} \sqrt {2 \, \cos \left (x\right )^{2} + 1} + \frac {5}{256} \, \sqrt {2} \log \left (2048 \, \cos \left (x\right )^{8} + 2048 \, \cos \left (x\right )^{6} + 640 \, \cos \left (x\right )^{4} + 64 \, \cos \left (x\right )^{2} - 8 \, {\left (128 \, \sqrt {2} \cos \left (x\right )^{7} + 96 \, \sqrt {2} \cos \left (x\right )^{5} + 20 \, \sqrt {2} \cos \left (x\right )^{3} + \sqrt {2} \cos \left (x\right )\right )} \sqrt {2 \, \cos \left (x\right )^{2} + 1} + 1\right ) \] Input:
integrate((1+2*cos(x)^2)^(5/2)*sin(x),x, algorithm="fricas")
Output:
-1/48*(32*cos(x)^5 + 52*cos(x)^3 + 33*cos(x))*sqrt(2*cos(x)^2 + 1) + 5/256 *sqrt(2)*log(2048*cos(x)^8 + 2048*cos(x)^6 + 640*cos(x)^4 + 64*cos(x)^2 - 8*(128*sqrt(2)*cos(x)^7 + 96*sqrt(2)*cos(x)^5 + 20*sqrt(2)*cos(x)^3 + sqrt (2)*cos(x))*sqrt(2*cos(x)^2 + 1) + 1)
Timed out. \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=\text {Timed out} \] Input:
integrate((1+2*cos(x)**2)**(5/2)*sin(x),x)
Output:
Timed out
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {1}{6} \, {\left (2 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {5}{2}} \cos \left (x\right ) - \frac {5}{24} \, {\left (2 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \cos \left (x\right ) - \frac {5}{32} \, \sqrt {2} \operatorname {arsinh}\left (\sqrt {2} \cos \left (x\right )\right ) - \frac {5}{16} \, \sqrt {2 \, \cos \left (x\right )^{2} + 1} \cos \left (x\right ) \] Input:
integrate((1+2*cos(x)^2)^(5/2)*sin(x),x, algorithm="maxima")
Output:
-1/6*(2*cos(x)^2 + 1)^(5/2)*cos(x) - 5/24*(2*cos(x)^2 + 1)^(3/2)*cos(x) - 5/32*sqrt(2)*arcsinh(sqrt(2)*cos(x)) - 5/16*sqrt(2*cos(x)^2 + 1)*cos(x)
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {1}{48} \, {\left (4 \, {\left (8 \, \cos \left (x\right )^{2} + 13\right )} \cos \left (x\right )^{2} + 33\right )} \sqrt {2 \, \cos \left (x\right )^{2} + 1} \cos \left (x\right ) + \frac {5}{32} \, \sqrt {2} \log \left (-\sqrt {2} \cos \left (x\right ) + \sqrt {2 \, \cos \left (x\right )^{2} + 1}\right ) \] Input:
integrate((1+2*cos(x)^2)^(5/2)*sin(x),x, algorithm="giac")
Output:
-1/48*(4*(8*cos(x)^2 + 13)*cos(x)^2 + 33)*sqrt(2*cos(x)^2 + 1)*cos(x) + 5/ 32*sqrt(2)*log(-sqrt(2)*cos(x) + sqrt(2*cos(x)^2 + 1))
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.59 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {5\,\sqrt {2}\,\mathrm {asinh}\left (\sqrt {2}\,\cos \left (x\right )\right )}{32}-\frac {\sqrt {2}\,\sqrt {{\cos \left (x\right )}^2+\frac {1}{2}}\,\left (\frac {4\,{\cos \left (x\right )}^5}{3}+\frac {13\,{\cos \left (x\right )}^3}{6}+\frac {11\,\cos \left (x\right )}{8}\right )}{2} \] Input:
int(sin(x)*(2*cos(x)^2 + 1)^(5/2),x)
Output:
- (5*2^(1/2)*asinh(2^(1/2)*cos(x)))/32 - (2^(1/2)*(cos(x)^2 + 1/2)^(1/2)*( (11*cos(x))/8 + (13*cos(x)^3)/6 + (4*cos(x)^5)/3))/2
\[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {2 \sqrt {2 \cos \left (x \right )^{2}+1}\, \cos \left (x \right )^{5}}{3}-\frac {13 \sqrt {2 \cos \left (x \right )^{2}+1}\, \cos \left (x \right )^{3}}{12}-\frac {11 \sqrt {2 \cos \left (x \right )^{2}+1}\, \cos \left (x \right )}{16}+\frac {5 \left (\int \frac {\sqrt {2 \cos \left (x \right )^{2}+1}\, \sin \left (x \right )}{2 \cos \left (x \right )^{2}+1}d x \right )}{16} \] Input:
int((1+2*cos(x)^2)^(5/2)*sin(x),x)
Output:
( - 32*sqrt(2*cos(x)**2 + 1)*cos(x)**5 - 52*sqrt(2*cos(x)**2 + 1)*cos(x)** 3 - 33*sqrt(2*cos(x)**2 + 1)*cos(x) + 15*int((sqrt(2*cos(x)**2 + 1)*sin(x) )/(2*cos(x)**2 + 1),x))/48