\(\int \frac {\cos (x) (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)})}{(1+2 \sin (x))^{3/2}} \, dx\) [396]

Optimal result

Integrand size = 32, antiderivative size = 55 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {3}{4 \sqrt {1+2 \sin (x)}}-\frac {4}{\sqrt [4]{1+2 \sin (x)}}-\frac {1}{2} \sqrt {1+2 \sin (x)}+\frac {1}{12} (1+2 \sin (x))^{3/2} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4441, 14} \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {1}{12} (2 \sin (x)+1)^{3/2}-\frac {1}{2} \sqrt {2 \sin (x)+1}-\frac {4}{\sqrt [4]{2 \sin (x)+1}}+\frac {3}{4 \sqrt {2 \sin (x)+1}} \]

[In]

Int[(Cos[x]*(-Cos[x]^2 + 2*(1 + 2*Sin[x])^(1/4)))/(1 + 2*Sin[x])^(3/2),x]

[Out]

3/(4*Sqrt[1 + 2*Sin[x]]) - 4/(1 + 2*Sin[x])^(1/4) - Sqrt[1 + 2*Sin[x]]/2 + (1 + 2*Sin[x])^(3/2)/12

Rule 14

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]]

Rule 4441

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[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])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {-1+x^2+2 \sqrt [4]{1+2 x}}{(1+2 x)^{3/2}} \, dx,x,\sin (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {-3+8 x-2 x^4+x^8}{x^3} \, dx,x,\sqrt [4]{1+2 \sin (x)}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {3}{x^3}+\frac {8}{x^2}-2 x+x^5\right ) \, dx,x,\sqrt [4]{1+2 \sin (x)}\right ) \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=-\frac {-3+\cos (2 x)+4 \sin (x)+24 \sqrt [4]{1+2 \sin (x)}}{6 \sqrt {1+2 \sin (x)}} \]

[In]

Integrate[(Cos[x]*(-Cos[x]^2 + 2*(1 + 2*Sin[x])^(1/4)))/(1 + 2*Sin[x])^(3/2),x]

[Out]

-1/6*(-3 + Cos[2*x] + 4*Sin[x] + 24*(1 + 2*Sin[x])^(1/4))/Sqrt[1 + 2*Sin[x]]

Maple [A] (verified)

Time = 0.49 (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	\(-\frac {4}{\left (1+2 \sin \left (x \right )\right )^{\frac {1}{4}}}+\frac {\left (1+2 \sin \left (x \right )\right )^{\frac {3}{2}}}{12}+\frac {3}{4 \sqrt {1+2 \sin \left (x \right )}}-\frac {\sqrt {1+2 \sin \left (x \right )}}{2}\)	\(42\)

[In]

int(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=-\frac {{\left (\cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 2\right )} \sqrt {2 \, \sin \left (x\right ) + 1} + 12 \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {3}{4}}}{3 \, {\left (2 \, \sin \left (x\right ) + 1\right )}} \]

[In]

integrate(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x, algorithm="fricas")

[Out]

-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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (48) = 96\).

Time = 37.51 (sec) , antiderivative size = 230, normalized size of antiderivative = 4.18 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {4 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}} \sin ^{2}{\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {2 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}} \sin {\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} + \frac {3 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}} \cos ^{2}{\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {2 \left (2 \sin {\left (x \right )} + 1\right )^{\frac {3}{4}}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {24 \sin {\left (x \right )}}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} - \frac {12}{6 \sqrt [4]{2 \sin {\left (x \right )} + 1} \sin {\left (x \right )} + 3 \sqrt [4]{2 \sin {\left (x \right )} + 1}} \]

[In]

integrate(cos(x)*(-cos(x)**2+2*(1+2*sin(x))**(1/4))/(1+2*sin(x))**(3/2),x)

[Out]

4*(2*sin(x) + 1)**(3/4)*sin(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)*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))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {1}{12} \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {3}{2}} - \frac {16 \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {1}{4}} - 3}{4 \, \sqrt {2 \, \sin \left (x\right ) + 1}} - \frac {1}{2} \, \sqrt {2 \, \sin \left (x\right ) + 1} \]

[In]

integrate(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=\frac {1}{12} \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {3}{2}} - \frac {16 \, {\left (2 \, \sin \left (x\right ) + 1\right )}^{\frac {1}{4}} - 3}{4 \, \sqrt {2 \, \sin \left (x\right ) + 1}} - \frac {1}{2} \, \sqrt {2 \, \sin \left (x\right ) + 1} \]

[In]

integrate(cos(x)*(-cos(x)^2+2*(1+2*sin(x))^(1/4))/(1+2*sin(x))^(3/2),x, algorithm="giac")

[Out]

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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx=-\int -\frac {\cos \left (x\right )\,\left (2\,{\left (2\,\sin \left (x\right )+1\right )}^{1/4}-{\cos \left (x\right )}^2\right )}{{\left (2\,\sin \left (x\right )+1\right )}^{3/2}} \,d x \]

[In]

int((cos(x)*(2*(2*sin(x) + 1)^(1/4) - cos(x)^2))/(2*sin(x) + 1)^(3/2),x)

[Out]

-int(-(cos(x)*(2*(2*sin(x) + 1)^(1/4) - cos(x)^2))/(2*sin(x) + 1)^(3/2), x)