3.5.53 \(\int \frac {(5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} (2+\sqrt {-1+5 \sin ^2(x)})} \, dx\) [453]

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

[Out]

2*(-1+5*sin(x)^2)^(1/4)-3/2*arctan(1/2*(-1+5*sin(x)^2)^(1/4)*2^(1/2))*2^(1/2)-1/4*arctanh(1/2*(-1+5*sin(x)^2)^
(1/4)*2^(1/2))*2^(1/2)-1/2*(-1+5*sin(x)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2))

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Rubi [A]
time = 0.92, 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 = {4446, 6874, 6829, 348, 52, 65, 209, 481, 536, 213} \begin {gather*} -2 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )+\frac {\text {ArcTan}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{\sqrt {2}}+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 )}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{2 \sqrt {2}} \end {gather*}

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 52

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 65

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 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 213

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

Rule 348

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 481

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 536

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 4446

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 6829

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 6874

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 &=-\text {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 )\\ &=-\text {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 \text {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 )+\text {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} \text {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} \text {Subst}\left (\int \frac {1}{\left (2+\sqrt {x}\right ) \sqrt [4]{x}} \, dx,x,4-5 \cos ^2(x)\right )\\ &=2 \text {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 )+\text {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} \text {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 \text {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} \text {Subst}\left (\int \frac {1}{-2+x^2} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )-4 \text {Subst}\left (\int \frac {1}{2+x^2} \, dx,x,\sqrt [4]{4-5 \cos ^2(x)}\right )+\text {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 = 0.29, size = 89, normalized size = 0.88 \begin {gather*} \frac {1}{4} \left (-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 )-2 \sqrt [4]{4-5 \cos ^2(x)} \left (-4+\frac {1}{2+\sqrt {4-5 \cos ^2(x)}}\right )\right ) \end {gather*}

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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Maple [F]
time = 0.70, size = 0, normalized size = 0.00 \[\int \frac {\left (5 \left (\cos ^{2}\left (x \right )\right )-\sqrt {-1+5 \left (\sin ^{2}\left (x \right )\right )}\right ) \tan \left (x \right )}{\left (-1+5 \left (\sin ^{2}\left (x \right )\right )\right )^{\frac {1}{4}} \left (2+\sqrt {-1+5 \left (\sin ^{2}\left (x \right )\right )}\right )}\, dx\]

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 = 3.13, size = 100, normalized size = 0.99 \begin {gather*} -\frac {3}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}}{\sqrt {2} + {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}}\right ) + 2 \, {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}} - \frac {{\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}}{2 \, {\left (\sqrt {5 \, \sin \left (x\right )^{2} - 1} + 2\right )}} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (81) = 162\).
time = 51.85, size = 461, normalized size = 4.56 \begin {gather*} \frac {70 \, {\left (5 \, \sqrt {2} \cos \left (x\right )^{4} - 4 \, \sqrt {2} \cos \left (x\right )^{2}\right )} \arctan \left (-\frac {2 \, {\left ({\left (5 \, \sqrt {2} \cos \left (x\right )^{2} - 4 \, \sqrt {2}\right )} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {3}{4}} - 2 \, \sqrt {2} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {5}{4}}\right )}}{5 \, {\left (5 \, \cos \left (x\right )^{4} - 4 \, \cos \left (x\right )^{2}\right )}}\right ) - 50 \, {\left (5 \, \sqrt {2} \cos \left (x\right )^{4} - 4 \, \sqrt {2} \cos \left (x\right )^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {2} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {3}{4}} + 2 \, \sqrt {2} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {1}{4}}\right )}}{5 \, \cos \left (x\right )^{2}}\right ) + 35 \, {\left (5 \, \sqrt {2} \cos \left (x\right )^{4} - 4 \, \sqrt {2} \cos \left (x\right )^{2}\right )} \log \left (-\frac {125 \, \cos \left (x\right )^{6} - 1700 \, \cos \left (x\right )^{4} - 8 \, {\left (15 \, \sqrt {2} \cos \left (x\right )^{2} - 16 \, \sqrt {2}\right )} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {5}{4}} + 2560 \, \cos \left (x\right )^{2} + 4 \, {\left (25 \, \sqrt {2} \cos \left (x\right )^{4} - 100 \, \sqrt {2} \cos \left (x\right )^{2} + 64 \, \sqrt {2}\right )} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {3}{4}} - 16 \, {\left (25 \, \cos \left (x\right )^{4} - 60 \, \cos \left (x\right )^{2} + 32\right )} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 1024}{5 \, \cos \left (x\right )^{6} - 4 \, \cos \left (x\right )^{4}}\right ) + 25 \, {\left (5 \, \sqrt {2} \cos \left (x\right )^{4} - 4 \, \sqrt {2} \cos \left (x\right )^{2}\right )} \log \left (-\frac {25 \, \cos \left (x\right )^{4} - 320 \, \cos \left (x\right )^{2} - 4 \, {\left (5 \, \sqrt {2} \cos \left (x\right )^{2} - 16 \, \sqrt {2}\right )} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {3}{4}} - 16 \, {\left (5 \, \cos \left (x\right )^{2} - 8\right )} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 8 \, {\left (15 \, \sqrt {2} \cos \left (x\right )^{2} - 16 \, \sqrt {2}\right )} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {1}{4}} + 256}{\cos \left (x\right )^{4}}\right ) + 16 \, {\left (5 \, \cos \left (x\right )^{2} - 2 \, {\left (10 \, \cos \left (x\right )^{2} - 1\right )} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 4\right )} {\left (-5 \, \cos \left (x\right )^{2} + 4\right )}^{\frac {3}{4}}}{160 \, {\left (5 \, \cos \left (x\right )^{4} - 4 \, \cos \left (x\right )^{2}\right )}} \end {gather*}

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]

1/160*(70*(5*sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos(x)^2)*arctan(-2/5*((5*sqrt(2)*cos(x)^2 - 4*sqrt(2))*(-5*cos(x)^2
+ 4)^(3/4) - 2*sqrt(2)*(-5*cos(x)^2 + 4)^(5/4))/(5*cos(x)^4 - 4*cos(x)^2)) - 50*(5*sqrt(2)*cos(x)^4 - 4*sqrt(
2)*cos(x)^2)*arctan(2/5*(sqrt(2)*(-5*cos(x)^2 + 4)^(3/4) + 2*sqrt(2)*(-5*cos(x)^2 + 4)^(1/4))/cos(x)^2) + 35*(
5*sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos(x)^2)*log(-(125*cos(x)^6 - 1700*cos(x)^4 - 8*(15*sqrt(2)*cos(x)^2 - 16*sqrt
(2))*(-5*cos(x)^2 + 4)^(5/4) + 2560*cos(x)^2 + 4*(25*sqrt(2)*cos(x)^4 - 100*sqrt(2)*cos(x)^2 + 64*sqrt(2))*(-5
*cos(x)^2 + 4)^(3/4) - 16*(25*cos(x)^4 - 60*cos(x)^2 + 32)*sqrt(-5*cos(x)^2 + 4) - 1024)/(5*cos(x)^6 - 4*cos(x
)^4)) + 25*(5*sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos(x)^2)*log(-(25*cos(x)^4 - 320*cos(x)^2 - 4*(5*sqrt(2)*cos(x)^2
- 16*sqrt(2))*(-5*cos(x)^2 + 4)^(3/4) - 16*(5*cos(x)^2 - 8)*sqrt(-5*cos(x)^2 + 4) - 8*(15*sqrt(2)*cos(x)^2 - 1
6*sqrt(2))*(-5*cos(x)^2 + 4)^(1/4) + 256)/cos(x)^4) + 16*(5*cos(x)^2 - 2*(10*cos(x)^2 - 1)*sqrt(-5*cos(x)^2 +
4) - 4)*(-5*cos(x)^2 + 4)^(3/4))/(5*cos(x)^4 - 4*cos(x)^2)

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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