Integrand size = 32, antiderivative size = 55 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {3}{4 \sqrt {1+2 \sin (x)}}-\frac {4}{\sqrt [4]{1+2 \sin (x)}}-\frac {1}{2} \sqrt {1+2 \sin (x)}+\frac {1}{12} (1+2 \sin (x))^{3/2} \] Output:
-4/(1+2*sin(x))^(1/4)+1/12*(1+2*sin(x))^(3/2)+3/4/(1+2*sin(x))^(1/2)-1/2*( 1+2*sin(x))^(1/2)
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=-\frac {-3+\cos (2 x)+4 \sin (x)+24 \sqrt [4]{1+2 \sin (x)}}{6 \sqrt {1+2 \sin (x)}} \] Input:
Integrate[(Cos[x]*(-Cos[x]^2 + 2*(1 + 2*Sin[x])^(1/4)))/(1 + 2*Sin[x])^(3/ 2),x]
Output:
-1/6*(-3 + Cos[2*x] + 4*Sin[x] + 24*(1 + 2*Sin[x])^(1/4))/Sqrt[1 + 2*Sin[x ]]
Time = 0.40 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4856, 25, 7267, 25, 2010, 2009}
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.
\(\displaystyle \int \frac {\cos (x) \left (2 \sqrt [4]{2 \sin (x)+1}-\cos ^2(x)\right )}{(2 \sin (x)+1)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x) \left (2 \sqrt [4]{2 \sin (x)+1}-\cos (x)^2\right )}{(2 \sin (x)+1)^{3/2}}dx\) |
\(\Big \downarrow \) 4856 |
\(\displaystyle \int -\frac {-\sin ^2(x)-2 \sqrt [4]{2 \sin (x)+1}+1}{(2 \sin (x)+1)^{3/2}}d\sin (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {-\sin ^2(x)-2 \sqrt [4]{2 \sin (x)+1}+1}{(2 \sin (x)+1)^{3/2}}d\sin (x)\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {1}{2} \int -\csc ^3(x) \left (-(2 \sin (x)+1)^2+2 (2 \sin (x)+1)-8 \sqrt [4]{2 \sin (x)+1}+3\right )d\sqrt [4]{2 \sin (x)+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \csc ^3(x) \left (-(2 \sin (x)+1)^2+2 (2 \sin (x)+1)-8 \sqrt [4]{2 \sin (x)+1}+3\right )d\sqrt [4]{2 \sin (x)+1}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{2} \int \left (3 \csc ^3(x)-8 \csc ^2(x)-(2 \sin (x)+1)^{5/4}+2 \sqrt [4]{2 \sin (x)+1}\right )d\sqrt [4]{2 \sin (x)+1}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{6} (2 \sin (x)+1)^{3/2}-\sqrt {2 \sin (x)+1}+\frac {3 \csc ^2(x)}{2}-8 \csc (x)\right )\) |
Input:
Int[(Cos[x]*(-Cos[x]^2 + 2*(1 + 2*Sin[x])^(1/4)))/(1 + 2*Sin[x])^(3/2),x]
Output:
(-8*Csc[x] + (3*Csc[x]^2)/2 - Sqrt[1 + 2*Sin[x]] + (1 + 2*Sin[x])^(3/2)/6) /2
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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])
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.66 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(-\frac {4}{\left (1+2 \sin \left (x \right )\right )^{\frac {1}{4}}}+\frac {\left (1+2 \sin \left (x \right )\right )^{\frac {3}{2}}}{12}+\frac {3}{4 \sqrt {1+2 \sin \left (x \right )}}-\frac {\sqrt {1+2 \sin \left (x \right )}}{2}\) | \(42\) |
default | \(-\frac {4}{\left (1+2 \sin \left (x \right )\right )^{\frac {1}{4}}}+\frac {\left (1+2 \sin \left (x \right )\right )^{\frac {3}{2}}}{12}+\frac {3}{4 \sqrt {1+2 \sin \left (x \right )}}-\frac {\sqrt {1+2 \sin \left (x \right )}}{2}\) | \(42\) |
parts | \(-\frac {4}{\left (1+2 \sin \left (x \right )\right )^{\frac {1}{4}}}+\frac {\left (1+2 \sin \left (x \right )\right )^{\frac {3}{2}}}{12}+\frac {3}{4 \sqrt {1+2 \sin \left (x \right )}}-\frac {\sqrt {1+2 \sin \left (x \right )}}{2}\) | \(42\) |
Input:
int(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x,method=_R ETURNVERBOSE)
Output:
-4/(1+2*sin(x))^(1/4)+1/12*(1+2*sin(x))^(3/2)+3/4/(1+2*sin(x))^(1/2)-1/2*( 1+2*sin(x))^(1/2)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=-\frac {{\left (\cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 2\right )} \sqrt {2 \, \sin \left (x\right ) + 1} + 12 \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {3}{4}}}{3 \, {\left (2 \, \sin \left (x\right ) + 1\right )}} \] Input:
integrate(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x, al gorithm="fricas")
Output:
-1/3*((cos(x)^2 + 2*sin(x) - 2)*sqrt(2*sin(x) + 1) + 12*(2*sin(x) + 1)^(3/ 4))/(2*sin(x) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (48) = 96\).
Time = 22.27 (sec) , antiderivative size = 230, normalized size of antiderivative = 4.18 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {4 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}} \sin ^{2}{\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {2 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}} \sin {\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} + \frac {3 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}} \cos ^{2}{\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {2 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {24 \sin {\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {12}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} \] Input:
integrate(cos(x)*(-cos(x)**2+2*(1+2*sin(x))**(1/4))/(1+2*sin(x))**(3/2),x)
Output:
4*(2*sin(x) + 1)**(3/4)*sin(x)**2/(6*(2*sin(x) + 1)**(1/4)*sin(x) + 3*(2*s in(x) + 1)**(1/4)) - 2*(2*sin(x) + 1)**(3/4)*sin(x)/(6*(2*sin(x) + 1)**(1/ 4)*sin(x) + 3*(2*sin(x) + 1)**(1/4)) + 3*(2*sin(x) + 1)**(3/4)*cos(x)**2/( 6*(2*sin(x) + 1)**(1/4)*sin(x) + 3*(2*sin(x) + 1)**(1/4)) - 2*(2*sin(x) + 1)**(3/4)/(6*(2*sin(x) + 1)**(1/4)*sin(x) + 3*(2*sin(x) + 1)**(1/4)) - 24* sin(x)/(6*(2*sin(x) + 1)**(1/4)*sin(x) + 3*(2*sin(x) + 1)**(1/4)) - 12/(6* (2*sin(x) + 1)**(1/4)*sin(x) + 3*(2*sin(x) + 1)**(1/4))
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {1}{12} \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {3}{2}} - \frac {16 \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {1}{4}} - 3}{4 \, \sqrt {2 \, \sin \left (x\right ) + 1}} - \frac {1}{2} \, \sqrt {2 \, \sin \left (x\right ) + 1} \] Input:
integrate(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x, al gorithm="maxima")
Output:
1/12*(2*sin(x) + 1)^(3/2) - 1/4*(16*(2*sin(x) + 1)^(1/4) - 3)/sqrt(2*sin(x ) + 1) - 1/2*sqrt(2*sin(x) + 1)
Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {1}{12} \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {3}{2}} - \frac {16 \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {1}{4}} - 3}{4 \, \sqrt {2 \, \sin \left (x\right ) + 1}} - \frac {1}{2} \, \sqrt {2 \, \sin \left (x\right ) + 1} \] Input:
integrate(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x, al gorithm="giac")
Output:
1/12*(2*sin(x) + 1)^(3/2) - 1/4*(16*(2*sin(x) + 1)^(1/4) - 3)/sqrt(2*sin(x ) + 1) - 1/2*sqrt(2*sin(x) + 1)
Timed out. \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=-\int -\frac {\cos \left (x\right )\,\left (2\,{\left (2\,\sin \left (x\right )+1\right )}^{1/4}-{\cos \left (x\right )}^2\right )}{{\left (2\,\sin \left (x\right )+1\right )}^{3/2}} \,d x \] Input:
int((cos(x)*(2*(2*sin(x) + 1)^(1/4) - cos(x)^2))/(2*sin(x) + 1)^(3/2),x)
Output:
-int(-(cos(x)*(2*(2*sin(x) + 1)^(1/4) - cos(x)^2))/(2*sin(x) + 1)^(3/2), x )
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {3 \sqrt {2 \sin \left (x \right )+1}\, \cos \left (x \right )^{2}-12 \left (2 \sin \left (x \right )+1\right )^{\frac {3}{4}}+4 \sqrt {2 \sin \left (x \right )+1}\, \sin \left (x \right )^{2}-2 \sqrt {2 \sin \left (x \right )+1}\, \sin \left (x \right )-2 \sqrt {2 \sin \left (x \right )+1}}{6 \sin \left (x \right )+3} \] Input:
int(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x)
Output:
(3*sqrt(2*sin(x) + 1)*cos(x)**2 - 12*(2*sin(x) + 1)**(3/4) + 4*sqrt(2*sin( x) + 1)*sin(x)**2 - 2*sqrt(2*sin(x) + 1)*sin(x) - 2*sqrt(2*sin(x) + 1))/(3 *(2*sin(x) + 1))