3.427 \(\int \frac {\cos (3 x)}{-\sqrt {-1+8 \cos ^2(x)}+\sqrt {3 \cos ^2(x)-\sin ^2(x)}} \, dx\)

Optimal. Leaf size=112 \[ \frac {5 \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )}{4 \sqrt {2}}+\frac {3}{4} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {3}}\right )-\frac {1}{2} \sin (x) \sqrt {4 \cos ^2(x)-1}-\frac {1}{2} \sin (x) \sqrt {8 \cos ^2(x)-1}-\frac {3}{4} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {4 \cos ^2(x)-1}}\right )-\frac {3}{4} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {8 \cos ^2(x)-1}}\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6742, 402, 216, 377, 204, 195} \[ \frac {5 \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )}{4 \sqrt {2}}+\frac {3}{4} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {3}}\right )-\frac {1}{2} \sin (x) \sqrt {7-8 \sin ^2(x)}-\frac {1}{2} \sin (x) \sqrt {3-4 \sin ^2(x)}-\frac {3}{4} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {7-8 \sin ^2(x)}}\right )-\frac {3}{4} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {3-4 \sin ^2(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cos[3*x]/(-Sqrt[-1 + 8*Cos[x]^2] + Sqrt[3*Cos[x]^2 - Sin[x]^2]),x]

[Out]

(5*ArcSin[2*Sqrt[2/7]*Sin[x]])/(4*Sqrt[2]) + (3*ArcSin[(2*Sin[x])/Sqrt[3]])/4 - (3*ArcTan[Sin[x]/Sqrt[7 - 8*Si
n[x]^2]])/4 - (3*ArcTan[Sin[x]/Sqrt[3 - 4*Sin[x]^2]])/4 - (Sin[x]*Sqrt[7 - 8*Sin[x]^2])/2 - (Sin[x]*Sqrt[3 - 4
*Sin[x]^2])/2

Rule 195

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 204

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

Rule 216

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\cos (3 x)}{-\sqrt {-1+8 \cos ^2(x)}+\sqrt {3 \cos ^2(x)-\sin ^2(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {-1+4 x^2}{\sqrt {7-8 x^2}-\sqrt {3-4 x^2}} \, dx,x,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \left (-\frac {1}{\sqrt {7-8 x^2}-\sqrt {3-4 x^2}}+\frac {4 x^2}{\sqrt {7-8 x^2}-\sqrt {3-4 x^2}}\right ) \, dx,x,\sin (x)\right )\\ &=4 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {7-8 x^2}-\sqrt {3-4 x^2}} \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \frac {1}{\sqrt {7-8 x^2}-\sqrt {3-4 x^2}} \, dx,x,\sin (x)\right )\\ &=4 \operatorname {Subst}\left (\int \left (-\frac {1}{4} \sqrt {7-8 x^2}-\frac {1}{4} \sqrt {3-4 x^2}-\frac {\sqrt {7-8 x^2}}{4 \left (-1+x^2\right )}-\frac {\sqrt {3-4 x^2}}{4 \left (-1+x^2\right )}\right ) \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \left (-\frac {\sqrt {7-8 x^2}}{4 \left (-1+x^2\right )}-\frac {\sqrt {3-4 x^2}}{4 \left (-1+x^2\right )}\right ) \, dx,x,\sin (x)\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt {7-8 x^2}}{-1+x^2} \, dx,x,\sin (x)\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {\sqrt {3-4 x^2}}{-1+x^2} \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \sqrt {7-8 x^2} \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \sqrt {3-4 x^2} \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \frac {\sqrt {7-8 x^2}}{-1+x^2} \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \frac {\sqrt {3-4 x^2}}{-1+x^2} \, dx,x,\sin (x)\right )\\ &=-\frac {1}{2} \sin (x) \sqrt {7-8 \sin ^2(x)}-\frac {1}{2} \sin (x) \sqrt {3-4 \sin ^2(x)}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {7-8 x^2} \left (-1+x^2\right )} \, dx,x,\sin (x)\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-4 x^2} \left (-1+x^2\right )} \, dx,x,\sin (x)\right )-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-4 x^2}} \, dx,x,\sin (x)\right )-2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {7-8 x^2}} \, dx,x,\sin (x)\right )-\frac {7}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {7-8 x^2}} \, dx,x,\sin (x)\right )+4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-4 x^2}} \, dx,x,\sin (x)\right )+8 \operatorname {Subst}\left (\int \frac {1}{\sqrt {7-8 x^2}} \, dx,x,\sin (x)\right )-\operatorname {Subst}\left (\int \frac {1}{\sqrt {3-4 x^2}} \, dx,x,\sin (x)\right )+\operatorname {Subst}\left (\int \frac {1}{\sqrt {7-8 x^2} \left (-1+x^2\right )} \, dx,x,\sin (x)\right )+\operatorname {Subst}\left (\int \frac {1}{\sqrt {3-4 x^2} \left (-1+x^2\right )} \, dx,x,\sin (x)\right )\\ &=-\frac {11 \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )}{4 \sqrt {2}}+2 \sqrt {2} \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )+\frac {3}{4} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {3}}\right )-\frac {1}{2} \sin (x) \sqrt {7-8 \sin ^2(x)}-\frac {1}{2} \sin (x) \sqrt {3-4 \sin ^2(x)}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\frac {\sin (x)}{\sqrt {7-8 \sin ^2(x)}}\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\frac {\sin (x)}{\sqrt {3-4 \sin ^2(x)}}\right )+\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\frac {\sin (x)}{\sqrt {7-8 \sin ^2(x)}}\right )+\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\frac {\sin (x)}{\sqrt {3-4 \sin ^2(x)}}\right )\\ &=-\frac {11 \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )}{4 \sqrt {2}}+2 \sqrt {2} \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )+\frac {3}{4} \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {3}}\right )-\frac {3}{4} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {7-8 \sin ^2(x)}}\right )-\frac {3}{4} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {3-4 \sin ^2(x)}}\right )-\frac {1}{2} \sin (x) \sqrt {7-8 \sin ^2(x)}-\frac {1}{2} \sin (x) \sqrt {3-4 \sin ^2(x)}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 156, normalized size = 1.39 \[ \frac {1}{8} \left (5 \sqrt {2} \sin ^{-1}\left (2 \sqrt {\frac {2}{7}} \sin (x)\right )+6 \sin ^{-1}\left (\frac {2 \sin (x)}{\sqrt {3}}\right )-4 \sin (x) \sqrt {2 \cos (2 x)+1}-4 \sin (x) \sqrt {4 \cos (2 x)+3}+3 \tan ^{-1}\left (\frac {7-8 \sin (x)}{\sqrt {4 \cos (2 x)+3}}\right )+3 \tan ^{-1}\left (\frac {3-4 \sin (x)}{\sqrt {2 \cos (2 x)+1}}\right )-3 \tan ^{-1}\left (\frac {4 \sin (x)+3}{\sqrt {2 \cos (2 x)+1}}\right )-3 \tan ^{-1}\left (\frac {8 \sin (x)+7}{\sqrt {4 \cos (2 x)+3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[3*x]/(-Sqrt[-1 + 8*Cos[x]^2] + Sqrt[3*Cos[x]^2 - Sin[x]^2]),x]

[Out]

(5*Sqrt[2]*ArcSin[2*Sqrt[2/7]*Sin[x]] + 6*ArcSin[(2*Sin[x])/Sqrt[3]] + 3*ArcTan[(7 - 8*Sin[x])/Sqrt[3 + 4*Cos[
2*x]]] + 3*ArcTan[(3 - 4*Sin[x])/Sqrt[1 + 2*Cos[2*x]]] - 3*ArcTan[(3 + 4*Sin[x])/Sqrt[1 + 2*Cos[2*x]]] - 3*Arc
Tan[(7 + 8*Sin[x])/Sqrt[3 + 4*Cos[2*x]]] - 4*Sqrt[1 + 2*Cos[2*x]]*Sin[x] - 4*Sqrt[3 + 4*Cos[2*x]]*Sin[x])/8

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos (3 x)}{-\sqrt {-1+8 \cos ^2(x)}+\sqrt {3 \cos ^2(x)-\sin ^2(x)}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Cos[3*x]/(-Sqrt[-1 + 8*Cos[x]^2] + Sqrt[3*Cos[x]^2 - Sin[x]^2]),x]

[Out]

Could not integrate

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fricas [B]  time = 1.51, size = 195, normalized size = 1.74 \[ -\frac {5}{32} \, \sqrt {2} \arctan \left (\frac {{\left (512 \, \sqrt {2} \cos \relax (x)^{4} - 576 \, \sqrt {2} \cos \relax (x)^{2} + 113 \, \sqrt {2}\right )} \sqrt {8 \, \cos \relax (x)^{2} - 1}}{16 \, {\left (128 \, \cos \relax (x)^{4} - 88 \, \cos \relax (x)^{2} + 9\right )} \sin \relax (x)}\right ) - \frac {1}{2} \, \sqrt {8 \, \cos \relax (x)^{2} - 1} \sin \relax (x) - \frac {1}{2} \, \sqrt {4 \, \cos \relax (x)^{2} - 1} \sin \relax (x) + \frac {3}{8} \, \arctan \left (\frac {4 \, {\left (8 \, \cos \relax (x)^{2} - 5\right )} \sqrt {4 \, \cos \relax (x)^{2} - 1} \sin \relax (x) - 9 \, \cos \relax (x) \sin \relax (x)}{64 \, \cos \relax (x)^{4} - 71 \, \cos \relax (x)^{2} + 16}\right ) + \frac {3}{8} \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x)}\right ) + \frac {3}{8} \, \arctan \left (\frac {9 \, \cos \relax (x)^{2} - 2}{2 \, \sqrt {8 \, \cos \relax (x)^{2} - 1} \sin \relax (x)}\right ) + \frac {3}{4} \, \arctan \left (\frac {\sqrt {4 \, \cos \relax (x)^{2} - 1}}{\sin \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(3*x)/(-(-1+8*cos(x)^2)^(1/2)+(3*cos(x)^2-sin(x)^2)^(1/2)),x, algorithm="fricas")

[Out]

-5/32*sqrt(2)*arctan(1/16*(512*sqrt(2)*cos(x)^4 - 576*sqrt(2)*cos(x)^2 + 113*sqrt(2))*sqrt(8*cos(x)^2 - 1)/((1
28*cos(x)^4 - 88*cos(x)^2 + 9)*sin(x))) - 1/2*sqrt(8*cos(x)^2 - 1)*sin(x) - 1/2*sqrt(4*cos(x)^2 - 1)*sin(x) +
3/8*arctan((4*(8*cos(x)^2 - 5)*sqrt(4*cos(x)^2 - 1)*sin(x) - 9*cos(x)*sin(x))/(64*cos(x)^4 - 71*cos(x)^2 + 16)
) + 3/8*arctan(sin(x)/cos(x)) + 3/8*arctan(1/2*(9*cos(x)^2 - 2)/(sqrt(8*cos(x)^2 - 1)*sin(x))) + 3/4*arctan(sq
rt(4*cos(x)^2 - 1)/sin(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\cos \left (3 \, x\right )}{\sqrt {8 \, \cos \relax (x)^{2} - 1} - \sqrt {3 \, \cos \relax (x)^{2} - \sin \relax (x)^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(3*x)/(-(-1+8*cos(x)^2)^(1/2)+(3*cos(x)^2-sin(x)^2)^(1/2)),x, algorithm="giac")

[Out]

integrate(-cos(3*x)/(sqrt(8*cos(x)^2 - 1) - sqrt(3*cos(x)^2 - sin(x)^2)), x)

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\cos \left (3 x \right )}{-\sqrt {-1+8 \left (\cos ^{2}\relax (x )\right )}+\sqrt {3 \left (\cos ^{2}\relax (x )\right )-\left (\sin ^{2}\relax (x )\right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(3*x)/(-(-1+8*cos(x)^2)^(1/2)+(3*cos(x)^2-sin(x)^2)^(1/2)),x)

[Out]

int(cos(3*x)/(-(-1+8*cos(x)^2)^(1/2)+(3*cos(x)^2-sin(x)^2)^(1/2)),x)

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maxima [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(3*x)/(-(-1+8*cos(x)^2)^(1/2)+(3*cos(x)^2-sin(x)^2)^(1/2)),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int -\frac {\cos \left (3\,x\right )}{\sqrt {3\,{\cos \relax (x)}^2-{\sin \relax (x)}^2}-\sqrt {8\,{\cos \relax (x)}^2-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(3*x)/((3*cos(x)^2 - sin(x)^2)^(1/2) - (8*cos(x)^2 - 1)^(1/2)),x)

[Out]

-int(-cos(3*x)/((3*cos(x)^2 - sin(x)^2)^(1/2) - (8*cos(x)^2 - 1)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (3 x \right )}}{\sqrt {- \sin ^{2}{\relax (x )} + 3 \cos ^{2}{\relax (x )}} - \sqrt {8 \cos ^{2}{\relax (x )} - 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(3*x)/(-(-1+8*cos(x)**2)**(1/2)+(3*cos(x)**2-sin(x)**2)**(1/2)),x)

[Out]

Integral(cos(3*x)/(sqrt(-sin(x)**2 + 3*cos(x)**2) - sqrt(8*cos(x)**2 - 1)), x)

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