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

Optimal. Leaf size=55 \[ \frac {1}{12} (2 \sin (x)+1)^{3/2}-\frac {1}{2} \sqrt {2 \sin (x)+1}-\frac {4}{\sqrt [4]{2 \sin (x)+1}}+\frac {3}{4 \sqrt {2 \sin (x)+1}} \]

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Rubi [A]  time = 0.15, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4356, 14} \[ \frac {1}{12} (2 \sin (x)+1)^{3/2}-\frac {1}{2} \sqrt {2 \sin (x)+1}-\frac {4}{\sqrt [4]{2 \sin (x)+1}}+\frac {3}{4 \sqrt {2 \sin (x)+1}} \]

Antiderivative was successfully verified.

[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 4356

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*} \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx &=\operatorname {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} \operatorname {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} \operatorname {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*}

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Mathematica [A]  time = 0.08, size = 36, normalized size = 0.65 \[ -\frac {4 \sin (x)+24 \sqrt [4]{2 \sin (x)+1}+\cos (2 x)-3}{6 \sqrt {2 \sin (x)+1}} \]

Antiderivative was successfully verified.

[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]]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos (x) \left (-\cos ^2(x)+2 \sqrt [4]{1+2 \sin (x)}\right )}{(1+2 \sin (x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 0.80, size = 40, normalized size = 0.73 \[ -\frac {{\left (\cos \relax (x)^{2} + 2 \, \sin \relax (x) - 2\right )} \sqrt {2 \, \sin \relax (x) + 1} + 12 \, {\left (2 \, \sin \relax (x) + 1\right )}^{\frac {3}{4}}}{3 \, {\left (2 \, \sin \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.65, size = 43, normalized size = 0.78 \[ \frac {1}{12} \, {\left (2 \, \sin \relax (x) + 1\right )}^{\frac {3}{2}} - \frac {16 \, {\left (2 \, \sin \relax (x) + 1\right )}^{\frac {1}{4}} - 3}{4 \, \sqrt {2 \, \sin \relax (x) + 1}} - \frac {1}{2} \, \sqrt {2 \, \sin \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.48, size = 31, normalized size = 0.56 \[\frac {\sin ^{2}\relax (x )-2 \sin \relax (x )-12 \left (1+2 \sin \relax (x )\right )^{\frac {1}{4}}+1}{3 \sqrt {1+2 \sin \relax (x )}}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/(1+2*sin(x))^(1/2)*(sin(x)^2-2*sin(x)-12*(1+2*sin(x))^(1/4)+1)

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maxima [A]  time = 0.43, size = 43, normalized size = 0.78 \[ \frac {1}{12} \, {\left (2 \, \sin \relax (x) + 1\right )}^{\frac {3}{2}} - \frac {16 \, {\left (2 \, \sin \relax (x) + 1\right )}^{\frac {1}{4}} - 3}{4 \, \sqrt {2 \, \sin \relax (x) + 1}} - \frac {1}{2} \, \sqrt {2 \, \sin \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ -\int -\frac {\cos \relax (x)\,\left (2\,{\left (2\,\sin \relax (x)+1\right )}^{1/4}-{\cos \relax (x)}^2\right )}{{\left (2\,\sin \relax (x)+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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