3.453 \(\int \frac {(5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} (2+\sqrt {-1+5 \sin ^2(x)})} \, dx\)

Optimal. Leaf size=101 \[ 2 \sqrt [4]{5 \sin ^2(x)-1}-\frac {\sqrt [4]{5 \sin ^2(x)-1}}{2 \left (\sqrt {5 \sin ^2(x)-1}+2\right )}-\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{5 \sin ^2(x)-1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{5 \sin ^2(x)-1}}{\sqrt {2}}\right )}{2 \sqrt {2}} \]

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Rubi [A]  time = 1.41, antiderivative size = 126, normalized size of antiderivative = 1.25, number of steps used = 14, number of rules used = 10, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4361, 6742, 6697, 341, 50, 63, 203, 470, 522, 207} \[ 2 \sqrt [4]{4-5 \cos ^2(x)}-\frac {\sqrt [4]{4-5 \cos ^2(x)}}{2 \left (\sqrt {4-5 \cos ^2(x)}+2\right )}-2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[((5*Cos[x]^2 - Sqrt[-1 + 5*Sin[x]^2])*Tan[x])/((-1 + 5*Sin[x]^2)^(1/4)*(2 + Sqrt[-1 + 5*Sin[x]^2])),x]

[Out]

ArcTan[(4 - 5*Cos[x]^2)^(1/4)/Sqrt[2]]/Sqrt[2] - 2*Sqrt[2]*ArcTan[(4 - 5*Cos[x]^2)^(1/4)/Sqrt[2]] - ArcTanh[(4
- 5*Cos[x]^2)^(1/4)/Sqrt[2]]/(2*Sqrt[2]) + 2*(4 - 5*Cos[x]^2)^(1/4) - (4 - 5*Cos[x]^2)^(1/4)/(2*(2 + Sqrt[4 -
5*Cos[x]^2]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

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

Rule 207

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

Rule 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 4361

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[(b*
c)^(-1), Subst[Int[SubstFor[1/x, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[
c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])

Rule 6697

Int[(u_.)*(v_)^(m_.)*((a_.) + (b_.)*(y_)^(n_))^(p_.), x_Symbol] :> Module[{q, r}, Dist[q*r, Subst[Int[x^m*(a +
b*x^n)^p, x], x, y], x] /;  !FalseQ[r = Divides[y^m, v^m, x]] &&  !FalseQ[q = DerivativeDivides[y, u, x]]] /;
FreeQ[{a, b, m, n, p}, x]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {\left (5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}\right ) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \, dx &=-\operatorname {Subst}\left (\int \frac {5 x^2-\sqrt {4-5 x^2}}{\sqrt [4]{4-5 x^2} \left (2 x+x \sqrt {4-5 x^2}\right )} \, dx,x,\cos (x)\right )\\ &=-\operatorname {Subst}\left (\int \left (\frac {5 x}{\sqrt [4]{4-5 x^2} \left (2+\sqrt {4-5 x^2}\right )}-\frac {\sqrt [4]{4-5 x^2}}{x \left (2+\sqrt {4-5 x^2}\right )}\right ) \, dx,x,\cos (x)\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{4-5 x^2} \left (2+\sqrt {4-5 x^2}\right )} \, dx,x,\cos (x)\right )\right )+\operatorname {Subst}\left (\int \frac {\sqrt [4]{4-5 x^2}}{x \left (2+\sqrt {4-5 x^2}\right )} \, dx,x,\cos (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [4]{4-5 x}}{\left (2+\sqrt {4-5 x}\right ) x} \, dx,x,\cos ^2(x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (2+\sqrt {x}\right ) \sqrt [4]{x}} \, dx,x,4-5 \cos ^2(x)\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^4}{\left (-2+x^2\right ) \left (2+x^2\right )^2} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )+\operatorname {Subst}\left (\int \frac {\sqrt {x}}{2+x} \, dx,x,\sqrt {4-5 \cos ^2(x)}\right )\\ &=2 \sqrt [4]{4-5 \cos ^2(x)}-\frac {\sqrt [4]{4-5 \cos ^2(x)}}{2 \left (2+\sqrt {4-5 \cos ^2(x)}\right )}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {-4+6 x^2}{\left (-2+x^2\right ) \left (2+x^2\right )} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )-2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (2+x)} \, dx,x,\sqrt {4-5 \cos ^2(x)}\right )\\ &=2 \sqrt [4]{4-5 \cos ^2(x)}-\frac {\sqrt [4]{4-5 \cos ^2(x)}}{2 \left (2+\sqrt {4-5 \cos ^2(x)}\right )}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-2+x^2} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )-4 \operatorname {Subst}\left (\int \frac {1}{2+x^2} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )+\operatorname {Subst}\left (\int \frac {1}{2+x^2} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{\sqrt {2}}-2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}+2 \sqrt [4]{4-5 \cos ^2(x)}-\frac {\sqrt [4]{4-5 \cos ^2(x)}}{2 \left (2+\sqrt {4-5 \cos ^2(x)}\right )}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 89, normalized size = 0.88 \[ \frac {1}{4} \left (-2 \sqrt [4]{4-5 \cos ^2(x)} \left (\frac {1}{\sqrt {4-5 \cos ^2(x)}+2}-4\right )-6 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt [4]{3-5 \cos (2 x)}}{2^{3/4}}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt [4]{3-5 \cos (2 x)}}{2^{3/4}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((5*Cos[x]^2 - Sqrt[-1 + 5*Sin[x]^2])*Tan[x])/((-1 + 5*Sin[x]^2)^(1/4)*(2 + Sqrt[-1 + 5*Sin[x]^2])),
x]

[Out]

(-6*Sqrt[2]*ArcTan[(3 - 5*Cos[2*x])^(1/4)/2^(3/4)] - Sqrt[2]*ArcTanh[(3 - 5*Cos[2*x])^(1/4)/2^(3/4)] - 2*(4 -
5*Cos[x]^2)^(1/4)*(-4 + (2 + Sqrt[4 - 5*Cos[x]^2])^(-1)))/4

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}\right ) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[((5*Cos[x]^2 - Sqrt[-1 + 5*Sin[x]^2])*Tan[x])/((-1 + 5*Sin[x]^2)^(1/4)*(2 + Sqrt[-1 + 5*Sin
[x]^2])),x]

[Out]

Could not integrate

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fricas [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((5*cos(x)^2-(-1+5*sin(x)^2)^(1/2))*tan(x)/(-1+5*sin(x)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2)),x, algorit
hm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, \cos \relax (x)^{2} - \sqrt {5 \, \sin \relax (x)^{2} - 1}\right )} \tan \relax (x)}{{\left (5 \, \sin \relax (x)^{2} - 1\right )}^{\frac {1}{4}} {\left (\sqrt {5 \, \sin \relax (x)^{2} - 1} + 2\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*cos(x)^2-(-1+5*sin(x)^2)^(1/2))*tan(x)/(-1+5*sin(x)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2)),x, algorit
hm="giac")

[Out]

integrate((5*cos(x)^2 - sqrt(5*sin(x)^2 - 1))*tan(x)/((5*sin(x)^2 - 1)^(1/4)*(sqrt(5*sin(x)^2 - 1) + 2)), x)

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maple [F]  time = 1.08, size = 0, normalized size = 0.00 \[\int \frac {\left (5 \left (\cos ^{2}\relax (x )\right )-\sqrt {-1+5 \left (\sin ^{2}\relax (x )\right )}\right ) \tan \relax (x )}{\left (-1+5 \left (\sin ^{2}\relax (x )\right )\right )^{\frac {1}{4}} \left (2+\sqrt {-1+5 \left (\sin ^{2}\relax (x )\right )}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*cos(x)^2-(-1+5*sin(x)^2)^(1/2))*tan(x)/(-1+5*sin(x)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2)),x)

[Out]

int((5*cos(x)^2-(-1+5*sin(x)^2)^(1/2))*tan(x)/(-1+5*sin(x)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2)),x)

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maxima [A]  time = 0.99, size = 100, normalized size = 0.99 \[ -\frac {3}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (5 \, \sin \relax (x)^{2} - 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - {\left (5 \, \sin \relax (x)^{2} - 1\right )}^{\frac {1}{4}}}{\sqrt {2} + {\left (5 \, \sin \relax (x)^{2} - 1\right )}^{\frac {1}{4}}}\right ) + 2 \, {\left (5 \, \sin \relax (x)^{2} - 1\right )}^{\frac {1}{4}} - \frac {{\left (5 \, \sin \relax (x)^{2} - 1\right )}^{\frac {1}{4}}}{2 \, {\left (\sqrt {5 \, \sin \relax (x)^{2} - 1} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*cos(x)^2-(-1+5*sin(x)^2)^(1/2))*tan(x)/(-1+5*sin(x)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2)),x, algorit
hm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tan}\relax (x)\,\left (5\,{\cos \relax (x)}^2-\sqrt {5\,{\sin \relax (x)}^2-1}\right )}{{\left (5\,{\sin \relax (x)}^2-1\right )}^{1/4}\,\left (\sqrt {5\,{\sin \relax (x)}^2-1}+2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x)*(5*cos(x)^2 - (5*sin(x)^2 - 1)^(1/2)))/((5*sin(x)^2 - 1)^(1/4)*((5*sin(x)^2 - 1)^(1/2) + 2)),x)

[Out]

int((tan(x)*(5*cos(x)^2 - (5*sin(x)^2 - 1)^(1/2)))/((5*sin(x)^2 - 1)^(1/4)*((5*sin(x)^2 - 1)^(1/2) + 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \sqrt {5 \sin ^{2}{\relax (x )} - 1} + 5 \cos ^{2}{\relax (x )}\right ) \tan {\relax (x )}}{\left (\sqrt {5 \sin ^{2}{\relax (x )} - 1} + 2\right ) \sqrt [4]{5 \sin ^{2}{\relax (x )} - 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*cos(x)**2-(-1+5*sin(x)**2)**(1/2))*tan(x)/(-1+5*sin(x)**2)**(1/4)/(2+(-1+5*sin(x)**2)**(1/2)),x)

[Out]

Integral((-sqrt(5*sin(x)**2 - 1) + 5*cos(x)**2)*tan(x)/((sqrt(5*sin(x)**2 - 1) + 2)*(5*sin(x)**2 - 1)**(1/4)),
x)

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