3.5.21 \(\int \cos (x) (-\cos ^2(x)-5 \sin ^2(x))^{3/2} \, dx\) [421]

Optimal. Leaf size=58 \[ \frac {3}{16} \tan ^{-1}\left (\frac {2 \sin (x)}{\sqrt {-1-4 \sin ^2(x)}}\right )-\frac {3}{8} \sin (x) \sqrt {-1-4 \sin ^2(x)}+\frac {1}{4} \sin (x) \left (-1-4 \sin ^2(x)\right )^{3/2} \]

[Out]

3/16*arctan(2*sin(x)/(-1-4*sin(x)^2)^(1/2))+1/4*sin(x)*(-1-4*sin(x)^2)^(3/2)-3/8*sin(x)*(-1-4*sin(x)^2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4441, 201, 223, 209} \begin {gather*} \frac {3}{16} \text {ArcTan}\left (\frac {2 \sin (x)}{\sqrt {-4 \sin ^2(x)-1}}\right )+\frac {1}{4} \sin (x) \left (-4 \sin ^2(x)-1\right )^{3/2}-\frac {3}{8} \sin (x) \sqrt {-4 \sin ^2(x)-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]*(-Cos[x]^2 - 5*Sin[x]^2)^(3/2),x]

[Out]

(3*ArcTan[(2*Sin[x])/Sqrt[-1 - 4*Sin[x]^2]])/16 - (3*Sin[x]*Sqrt[-1 - 4*Sin[x]^2])/8 + (Sin[x]*(-1 - 4*Sin[x]^
2)^(3/2))/4

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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*} \int \cos (x) \left (-\cos ^2(x)-5 \sin ^2(x)\right )^{3/2} \, dx &=\text {Subst}\left (\int \left (-1-4 x^2\right )^{3/2} \, dx,x,\sin (x)\right )\\ &=\frac {1}{4} \sin (x) \left (-1-4 \sin ^2(x)\right )^{3/2}-\frac {3}{4} \text {Subst}\left (\int \sqrt {-1-4 x^2} \, dx,x,\sin (x)\right )\\ &=-\frac {3}{8} \sin (x) \sqrt {-1-4 \sin ^2(x)}+\frac {1}{4} \sin (x) \left (-1-4 \sin ^2(x)\right )^{3/2}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{\sqrt {-1-4 x^2}} \, dx,x,\sin (x)\right )\\ &=-\frac {3}{8} \sin (x) \sqrt {-1-4 \sin ^2(x)}+\frac {1}{4} \sin (x) \left (-1-4 \sin ^2(x)\right )^{3/2}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1+4 x^2} \, dx,x,\frac {\sin (x)}{\sqrt {-1-4 \sin ^2(x)}}\right )\\ &=\frac {3}{16} \tan ^{-1}\left (\frac {2 \sin (x)}{\sqrt {-1-4 \sin ^2(x)}}\right )-\frac {3}{8} \sin (x) \sqrt {-1-4 \sin ^2(x)}+\frac {1}{4} \sin (x) \left (-1-4 \sin ^2(x)\right )^{3/2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 61, normalized size = 1.05 \begin {gather*} \frac {\sqrt {-3+2 \cos (2 x)} \left (-3 \sinh ^{-1}(2 \sin (x))+2 \sqrt {3-2 \cos (2 x)} (-11 \sin (x)+2 \sin (3 x))\right )}{16 \sqrt {1+4 \sin ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*(-Cos[x]^2 - 5*Sin[x]^2)^(3/2),x]

[Out]

(Sqrt[-3 + 2*Cos[2*x]]*(-3*ArcSinh[2*Sin[x]] + 2*Sqrt[3 - 2*Cos[2*x]]*(-11*Sin[x] + 2*Sin[3*x])))/(16*Sqrt[1 +
4*Sin[x]^2])

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Maple [A]
time = 0.18, size = 82, normalized size = 1.41

method	result	size
default	\(\frac {\sqrt {\left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (\sin ^{2}\left (x \right )\right )}\, \left (-32 \sqrt {-4 \left (\sin ^{4}\left (x \right )\right )-\left (\sin ^{2}\left (x \right )\right )}\, \left (\sin ^{2}\left (x \right )\right )-20 \sqrt {-4 \left (\sin ^{4}\left (x \right )\right )-\left (\sin ^{2}\left (x \right )\right )}+3 \arcsin \left (8 \left (\sin ^{2}\left (x \right )\right )+1\right )\right )}{32 \sin \left (x \right ) \sqrt {4 \left (\cos ^{2}\left (x \right )\right )-5}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*(-cos(x)^2-5*sin(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/32*((4*cos(x)^2-5)*sin(x)^2)^(1/2)*(-32*(-4*sin(x)^4-sin(x)^2)^(1/2)*sin(x)^2-20*(-4*sin(x)^4-sin(x)^2)^(1/2
)+3*arcsin(8*sin(x)^2+1))/sin(x)/(4*cos(x)^2-5)^(1/2)

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Maxima [C] Result contains complex when optimal does not.
time = 3.28, size = 36, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, {\left (-4 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {3}{2}} \sin \left (x\right ) - \frac {3}{8} \, \sqrt {-4 \, \sin \left (x\right )^{2} - 1} \sin \left (x\right ) - \frac {3}{16} i \, \operatorname {arsinh}\left (2 \, \sin \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(-cos(x)^2-5*sin(x)^2)^(3/2),x, algorithm="maxima")

[Out]

1/4*(-4*sin(x)^2 - 1)^(3/2)*sin(x) - 3/8*sqrt(-4*sin(x)^2 - 1)*sin(x) - 3/16*I*arcsinh(2*sin(x))

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Fricas [C] Result contains complex when optimal does not.
time = 1.23, size = 123, normalized size = 2.12 \begin {gather*} \frac {1}{128} \, {\left (12 i \, e^{\left (4 i \, x\right )} \log \left (-\frac {1}{2} \, \sqrt {e^{\left (4 i \, x\right )} - 3 \, e^{\left (2 i \, x\right )} + 1} {\left (4 \, e^{\left (2 i \, x\right )} - 5\right )} + 2 \, e^{\left (4 i \, x\right )} - \frac {11}{2} \, e^{\left (2 i \, x\right )} + \frac {5}{2}\right ) - 12 i \, e^{\left (4 i \, x\right )} \log \left (\sqrt {e^{\left (4 i \, x\right )} - 3 \, e^{\left (2 i \, x\right )} + 1} - e^{\left (2 i \, x\right )} - 1\right ) - 8 \, {\left (2 i \, e^{\left (6 i \, x\right )} - 11 i \, e^{\left (4 i \, x\right )} + 11 i \, e^{\left (2 i \, x\right )} - 2 i\right )} \sqrt {e^{\left (4 i \, x\right )} - 3 \, e^{\left (2 i \, x\right )} + 1} - 145 i \, e^{\left (4 i \, x\right )}\right )} e^{\left (-4 i \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(-cos(x)^2-5*sin(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/128*(12*I*e^(4*I*x)*log(-1/2*sqrt(e^(4*I*x) - 3*e^(2*I*x) + 1)*(4*e^(2*I*x) - 5) + 2*e^(4*I*x) - 11/2*e^(2*I
*x) + 5/2) - 12*I*e^(4*I*x)*log(sqrt(e^(4*I*x) - 3*e^(2*I*x) + 1) - e^(2*I*x) - 1) - 8*(2*I*e^(6*I*x) - 11*I*e
^(4*I*x) + 11*I*e^(2*I*x) - 2*I)*sqrt(e^(4*I*x) - 3*e^(2*I*x) + 1) - 145*I*e^(4*I*x))*e^(-4*I*x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(-cos(x)**2-5*sin(x)**2)**(3/2),x)

[Out]

Timed out

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Giac [C] Result contains complex when optimal does not.
time = 1.14, size = 41, normalized size = 0.71 \begin {gather*} -\frac {1}{8} i \, {\left (8 \, \sin \left (x\right )^{2} + 5\right )} \sqrt {4 \, \sin \left (x\right )^{2} + 1} \sin \left (x\right ) + \frac {3}{16} i \, \log \left (\sqrt {4 \, \sin \left (x\right )^{2} + 1} - 2 \, \sin \left (x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*(-cos(x)^2-5*sin(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/8*I*(8*sin(x)^2 + 5)*sqrt(4*sin(x)^2 + 1)*sin(x) + 3/16*I*log(sqrt(4*sin(x)^2 + 1) - 2*sin(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \cos \left (x\right )\,{\left (-{\cos \left (x\right )}^2-5\,{\sin \left (x\right )}^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*(- cos(x)^2 - 5*sin(x)^2)^(3/2),x)

[Out]

int(cos(x)*(- cos(x)^2 - 5*sin(x)^2)^(3/2), x)

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