\(\int \frac {(3+\sin ^2(x)) \tan ^3(x)}{(-2+\cos ^2(x)) (5-4 \sec ^2(x))^{3/2}} \, dx\) [438]

Optimal result

Integrand size = 31, antiderivative size = 73 \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {5}}\right )}{5 \sqrt {5}}-\frac {2}{15 \sqrt {5-4 \sec ^2(x)}} \]

[Out]

-1/18*arctanh(1/3*(5-4*sec(x)^2)^(1/2)*3^(1/2))*3^(1/2)-1/25*arctanh(1/5*(5-4*sec(x)^2)^(1/2)*5^(1/2))*5^(1/2)
-2/15/(5-4*sec(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {4458, 6857, 267, 272, 53, 65, 212, 528, 457, 87, 162, 213} \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {5}}\right )}{5 \sqrt {5}}-\frac {2}{15 \sqrt {5-4 \sec ^2(x)}} \]

[In]

Int[((3 + Sin[x]^2)*Tan[x]^3)/((-2 + Cos[x]^2)*(5 - 4*Sec[x]^2)^(3/2)),x]

[Out]

-1/6*ArcTanh[Sqrt[5 - 4*Sec[x]^2]/Sqrt[3]]/Sqrt[3] - ArcTanh[Sqrt[5 - 4*Sec[x]^2]/Sqrt[5]]/(5*Sqrt[5]) - 2/(15
*Sqrt[5 - 4*Sec[x]^2])

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 87

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

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 272

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

Rule 457

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

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 4458

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, Dist
[-(b*c*d^(n - 1))^(-1), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2)/x^n, Cos[c*(a + b*x)]/d, u, x], x], x, Co
s[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2]
&& NonsumQ[u] && (EqQ[F, Tan] || EqQ[F, tan])

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-x^2\right ) \left (4-x^2\right )}{\left (5-\frac {4}{x^2}\right )^{3/2} x^3 \left (-2+x^2\right )} \, dx,x,\cos (x)\right ) \\ & = -\text {Subst}\left (\int \left (-\frac {2}{\left (5-\frac {4}{x^2}\right )^{3/2} x^3}+\frac {3}{2 \left (5-\frac {4}{x^2}\right )^{3/2} x}-\frac {x}{2 \left (5-\frac {4}{x^2}\right )^{3/2} \left (-2+x^2\right )}\right ) \, dx,x,\cos (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x}{\left (5-\frac {4}{x^2}\right )^{3/2} \left (-2+x^2\right )} \, dx,x,\cos (x)\right )-\frac {3}{2} \text {Subst}\left (\int \frac {1}{\left (5-\frac {4}{x^2}\right )^{3/2} x} \, dx,x,\cos (x)\right )+2 \text {Subst}\left (\int \frac {1}{\left (5-\frac {4}{x^2}\right )^{3/2} x^3} \, dx,x,\cos (x)\right ) \\ & = -\frac {1}{2 \sqrt {5-4 \sec ^2(x)}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\left (5-\frac {4}{x^2}\right )^{3/2} \left (1-\frac {2}{x^2}\right ) x} \, dx,x,\cos (x)\right )+\frac {3}{4} \text {Subst}\left (\int \frac {1}{(5-4 x)^{3/2} x} \, dx,x,\sec ^2(x)\right ) \\ & = -\frac {1}{5 \sqrt {5-4 \sec ^2(x)}}+\frac {3}{20} \text {Subst}\left (\int \frac {1}{\sqrt {5-4 x} x} \, dx,x,\sec ^2(x)\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{(5-4 x)^{3/2} (1-2 x) x} \, dx,x,\sec ^2(x)\right ) \\ & = -\frac {2}{15 \sqrt {5-4 \sec ^2(x)}}+\frac {1}{120} \text {Subst}\left (\int \frac {-6-8 x}{\sqrt {5-4 x} (1-2 x) x} \, dx,x,\sec ^2(x)\right )-\frac {3}{40} \text {Subst}\left (\int \frac {1}{\frac {5}{4}-\frac {x^2}{4}} \, dx,x,\sqrt {5-4 \sec ^2(x)}\right ) \\ & = -\frac {3 \text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {5}}\right )}{10 \sqrt {5}}-\frac {2}{15 \sqrt {5-4 \sec ^2(x)}}-\frac {1}{20} \text {Subst}\left (\int \frac {1}{\sqrt {5-4 x} x} \, dx,x,\sec ^2(x)\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{\sqrt {5-4 x} (1-2 x)} \, dx,x,\sec ^2(x)\right ) \\ & = -\frac {3 \text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {5}}\right )}{10 \sqrt {5}}-\frac {2}{15 \sqrt {5-4 \sec ^2(x)}}+\frac {1}{40} \text {Subst}\left (\int \frac {1}{\frac {5}{4}-\frac {x^2}{4}} \, dx,x,\sqrt {5-4 \sec ^2(x)}\right )+\frac {1}{12} \text {Subst}\left (\int \frac {1}{-\frac {3}{2}+\frac {x^2}{2}} \, dx,x,\sqrt {5-4 \sec ^2(x)}\right ) \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\text {arctanh}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {5}}\right )}{5 \sqrt {5}}-\frac {2}{15 \sqrt {5-4 \sec ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.77 \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {\sqrt {-\cos ^2(x)} \left (60 \sqrt {-\cos ^2(x)}+9 \text {arcsinh}\left (\frac {1}{2} \sqrt {5} \sqrt {-\cos ^2(x)}\right ) \sqrt {30-50 \cos (2 x)}+25 \text {arctanh}\left (\frac {\sqrt {3} \sqrt {-\cos ^2(x)}}{\sqrt {-1+5 \sin ^2(x)}}\right ) \sqrt {-3+15 \sin ^2(x)}\right ) \sqrt {\sec ^2(x)-5 \tan ^2(x)}}{450 \left (-1+5 \sin ^2(x)\right )} \]

[In]

Integrate[((3 + Sin[x]^2)*Tan[x]^3)/((-2 + Cos[x]^2)*(5 - 4*Sec[x]^2)^(3/2)),x]

[Out]

-1/450*(Sqrt[-Cos[x]^2]*(60*Sqrt[-Cos[x]^2] + 9*ArcSinh[(Sqrt[5]*Sqrt[-Cos[x]^2])/2]*Sqrt[30 - 50*Cos[2*x]] +
25*ArcTanh[(Sqrt[3]*Sqrt[-Cos[x]^2])/Sqrt[-1 + 5*Sin[x]^2]]*Sqrt[-3 + 15*Sin[x]^2])*Sqrt[Sec[x]^2 - 5*Tan[x]^2
])/(-1 + 5*Sin[x]^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1497\) vs. \(2(55)=110\).

Time = 2.00 (sec) , antiderivative size = 1498, normalized size of antiderivative = 20.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1498\)

[In]

int((3+sin(x)^2)*tan(x)^3/(cos(x)^2-2)/(5-4*sec(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

3/5*sec(x)^3*(5*cos(x)^2-4)*(50*3^(1/2)*2^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh((5*cos(x)*2^(1/2)+
4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)+6^(1/2)))*cos(x)+50*3^(1/2)*2
^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos
(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*cos(x)-25*6^(1/2)*2^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2
)*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)+6
^(1/2)))*cos(x)+25*6^(1/2)*2^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)-10*
cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*cos(x)+100*3^(1/2)*((5*cos(x)^2-
4)/(cos(x)+1)^2)^(1/2)*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^
2)^(1/2)/(2*3^(1/2)+6^(1/2)))*cos(x)-100*3^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh((5*cos(x)*2^(1/2)
+4*2^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*cos(x)-50*6^(1/2)*
((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5*cos(x)^2-4
)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)+6^(1/2)))*cos(x)-50*6^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh((5*co
s(x)*2^(1/2)+4*2^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*cos(x)
+72*5^(1/2)*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*arctanh(cos(x)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*
5^(1/2))*cos(x)+50*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(
1/2)/(2*3^(1/2)+6^(1/2)))*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*2^(1/2)*3^(1/2)+50*arctanh((5*cos(x)*2^(1/2)+4*2
^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*((5*cos(x)^2-4)/(cos(x
)+1)^2)^(1/2)*2^(1/2)*3^(1/2)-25*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5*cos(x)^2-4)/(
cos(x)+1)^2)^(1/2)/(2*3^(1/2)+6^(1/2)))*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*2^(1/2)*6^(1/2)+25*arctanh((5*cos(
x)*2^(1/2)+4*2^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*((5*cos(
x)^2-4)/(cos(x)+1)^2)^(1/2)*2^(1/2)*6^(1/2)+100*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((
5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)+6^(1/2)))*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*3^(1/2)-100*arctanh
((5*cos(x)*2^(1/2)+4*2^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*
((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*3^(1/2)-50*arctanh((5*cos(x)*2^(1/2)+4*2^(1/2)+10*cos(x)+4)/(cos(x)+1)/((5
*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)+6^(1/2)))*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*6^(1/2)-50*arctanh((
5*cos(x)*2^(1/2)+4*2^(1/2)-10*cos(x)-4)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)/(2*3^(1/2)-6^(1/2)))*((
5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*6^(1/2)+72*arctanh(cos(x)/(cos(x)+1)/((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*5^(
1/2))*((5*cos(x)^2-4)/(cos(x)+1)^2)^(1/2)*5^(1/2)+240*cos(x))/(5-4*sec(x)^2)^(3/2)/(-5+2*5^(1/2))/(5+2*5^(1/2)
)/(6+2*5^(1/2)+2^(1/2))/(6-2*5^(1/2)+2^(1/2))/(2*3^(1/2)+6^(1/2))/(-6-2*5^(1/2)+2^(1/2))/(-6+2*5^(1/2)+2^(1/2)
)/(2*3^(1/2)-6^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (55) = 110\).

Time = 0.35 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.52 \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {480 \, \sqrt {\frac {5 \, \cos \left (x\right )^{2} - 4}{\cos \left (x\right )^{2}}} \cos \left (x\right )^{2} - 18 \, {\left (5 \, \sqrt {5} \cos \left (x\right )^{2} - 4 \, \sqrt {5}\right )} \log \left (625 \, \cos \left (x\right )^{8} - 1000 \, \cos \left (x\right )^{6} + 500 \, \cos \left (x\right )^{4} - 80 \, \cos \left (x\right )^{2} - {\left (125 \, \sqrt {5} \cos \left (x\right )^{8} - 150 \, \sqrt {5} \cos \left (x\right )^{6} + 50 \, \sqrt {5} \cos \left (x\right )^{4} - 4 \, \sqrt {5} \cos \left (x\right )^{2}\right )} \sqrt {\frac {5 \, \cos \left (x\right )^{2} - 4}{\cos \left (x\right )^{2}}} + 2\right ) - 25 \, {\left (5 \, \sqrt {3} \cos \left (x\right )^{2} - 4 \, \sqrt {3}\right )} \log \left (\frac {1921 \, \cos \left (x\right )^{8} - 3464 \, \cos \left (x\right )^{6} + 2040 \, \cos \left (x\right )^{4} - 416 \, \cos \left (x\right )^{2} - 8 \, {\left (62 \, \sqrt {3} \cos \left (x\right )^{8} - 87 \, \sqrt {3} \cos \left (x\right )^{6} + 36 \, \sqrt {3} \cos \left (x\right )^{4} - 4 \, \sqrt {3} \cos \left (x\right )^{2}\right )} \sqrt {\frac {5 \, \cos \left (x\right )^{2} - 4}{\cos \left (x\right )^{2}}} + 16}{\cos \left (x\right )^{8} - 8 \, \cos \left (x\right )^{6} + 24 \, \cos \left (x\right )^{4} - 32 \, \cos \left (x\right )^{2} + 16}\right )}{3600 \, {\left (5 \, \cos \left (x\right )^{2} - 4\right )}} \]

[In]

integrate((3+sin(x)^2)*tan(x)^3/(-2+cos(x)^2)/(5-4*sec(x)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/3600*(480*sqrt((5*cos(x)^2 - 4)/cos(x)^2)*cos(x)^2 - 18*(5*sqrt(5)*cos(x)^2 - 4*sqrt(5))*log(625*cos(x)^8 -
1000*cos(x)^6 + 500*cos(x)^4 - 80*cos(x)^2 - (125*sqrt(5)*cos(x)^8 - 150*sqrt(5)*cos(x)^6 + 50*sqrt(5)*cos(x)
^4 - 4*sqrt(5)*cos(x)^2)*sqrt((5*cos(x)^2 - 4)/cos(x)^2) + 2) - 25*(5*sqrt(3)*cos(x)^2 - 4*sqrt(3))*log((1921*
cos(x)^8 - 3464*cos(x)^6 + 2040*cos(x)^4 - 416*cos(x)^2 - 8*(62*sqrt(3)*cos(x)^8 - 87*sqrt(3)*cos(x)^6 + 36*sq
rt(3)*cos(x)^4 - 4*sqrt(3)*cos(x)^2)*sqrt((5*cos(x)^2 - 4)/cos(x)^2) + 16)/(cos(x)^8 - 8*cos(x)^6 + 24*cos(x)^
4 - 32*cos(x)^2 + 16)))/(5*cos(x)^2 - 4)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((3+sin(x)**2)*tan(x)**3/(-2+cos(x)**2)/(5-4*sec(x)**2)**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=\int { \frac {{\left (\sin \left (x\right )^{2} + 3\right )} \tan \left (x\right )^{3}}{{\left (\cos \left (x\right )^{2} - 2\right )} {\left (-4 \, \sec \left (x\right )^{2} + 5\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((3+sin(x)^2)*tan(x)^3/(-2+cos(x)^2)/(5-4*sec(x)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((sin(x)^2 + 3)*tan(x)^3/((cos(x)^2 - 2)*(-4*sec(x)^2 + 5)^(3/2)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (55) = 110\).

Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.66 \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {15} \sqrt {5} \log \left (-\frac {2 \, {\left ({\left (\sqrt {5} \cos \left (x\right ) - \sqrt {5 \, \cos \left (x\right )^{2} - 4}\right )}^{2} - 4 \, \sqrt {15} - 16\right )}}{{\left | 2 \, {\left (\sqrt {5} \cos \left (x\right ) - \sqrt {5 \, \cos \left (x\right )^{2} - 4}\right )}^{2} + 8 \, \sqrt {15} - 32 \right |}}\right ) - 18 \, \sqrt {5} \log \left ({\left (\sqrt {5} \cos \left (x\right ) - \sqrt {5 \, \cos \left (x\right )^{2} - 4}\right )}^{2}\right ) + \frac {120 \, \cos \left (x\right )}{\sqrt {5 \, \cos \left (x\right )^{2} - 4}}}{900 \, \mathrm {sgn}\left (\cos \left (x\right )\right )} \]

[In]

integrate((3+sin(x)^2)*tan(x)^3/(-2+cos(x)^2)/(5-4*sec(x)^2)^(3/2),x, algorithm="giac")

[Out]

-1/900*(5*sqrt(15)*sqrt(5)*log(-2*((sqrt(5)*cos(x) - sqrt(5*cos(x)^2 - 4))^2 - 4*sqrt(15) - 16)/abs(2*(sqrt(5)
*cos(x) - sqrt(5*cos(x)^2 - 4))^2 + 8*sqrt(15) - 32)) - 18*sqrt(5)*log((sqrt(5)*cos(x) - sqrt(5*cos(x)^2 - 4))
^2) + 120*cos(x)/sqrt(5*cos(x)^2 - 4))/sgn(cos(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (x\right )}^3\,\left ({\sin \left (x\right )}^2+3\right )}{\left ({\cos \left (x\right )}^2-2\right )\,{\left (5-\frac {4}{{\cos \left (x\right )}^2}\right )}^{3/2}} \,d x \]

[In]

int((tan(x)^3*(sin(x)^2 + 3))/((cos(x)^2 - 2)*(5 - 4/cos(x)^2)^(3/2)),x)

[Out]

int((tan(x)^3*(sin(x)^2 + 3))/((cos(x)^2 - 2)*(5 - 4/cos(x)^2)^(3/2)), x)