3.438 \(\int \frac{(3+\sin ^2(x)) \tan ^3(x)}{(-2+\cos ^2(x)) (5-4 \sec ^2(x))^{3/2}} \, dx\)
Optimal. Leaf size=73 \[ -\frac{2}{15 \sqrt{5-4 \sec ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{5-4 \sec ^2(x)}}{\sqrt{3}}\right )}{6 \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{5-4 \sec ^2(x)}}{\sqrt{5}}\right )}{5 \sqrt{5}} \]
[Out]
-ArcTanh[Sqrt[5 - 4*Sec[x]^2]/Sqrt[3]]/(6*Sqrt[3]) - ArcTanh[Sqrt[5 - 4*Sec[x]^2]/Sqrt[5]]/(5*Sqrt[5]) - 2/(15
*Sqrt[5 - 4*Sec[x]^2])
________________________________________________________________________________________
Rubi [A] time = 1.22989, antiderivative size = 73, normalized size of antiderivative
= 1., number of steps used = 16, number of rules used = 12, integrand size = 31,
\(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.387, Rules used = {4373, 6725, 261, 266, 51, 63, 206, 514, 446, 85, 156, 207}
\[ -\frac{2}{15 \sqrt{5-4 \sec ^2(x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{5-4 \sec ^2(x)}}{\sqrt{3}}\right )}{6 \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{5-4 \sec ^2(x)}}{\sqrt{5}}\right )}{5 \sqrt{5}} \]
Antiderivative was successfully verified.
[In]
Int[((3 + Sin[x]^2)*Tan[x]^3)/((-2 + Cos[x]^2)*(5 - 4*Sec[x]^2)^(3/2)),x]
[Out]
-ArcTanh[Sqrt[5 - 4*Sec[x]^2]/Sqrt[3]]/(6*Sqrt[3]) - ArcTanh[Sqrt[5 - 4*Sec[x]^2]/Sqrt[5]]/(5*Sqrt[5]) - 2/(15
*Sqrt[5 - 4*Sec[x]^2])
Rule 4373
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[(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 6725
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]
Rule 261
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 266
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 51
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 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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 514
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 446
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 85
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 156
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 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])
Rubi steps
\begin{align*} \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 &=-\operatorname{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 )\\ &=-\operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{\left (5-\frac{4}{x^2}\right )^{3/2} x} \, dx,x,\cos (x)\right )+2 \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5-4 x} x} \, dx,x,\sec ^2(x)\right )-\frac{1}{4} \operatorname{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} \operatorname{Subst}\left (\int \frac{-6-8 x}{\sqrt{5-4 x} (1-2 x) x} \, dx,x,\sec ^2(x)\right )-\frac{3}{40} \operatorname{Subst}\left (\int \frac{1}{\frac{5}{4}-\frac{x^2}{4}} \, dx,x,\sqrt{5-4 \sec ^2(x)}\right )\\ &=-\frac{3 \tanh ^{-1}\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} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5-4 x} x} \, dx,x,\sec ^2(x)\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5-4 x} (1-2 x)} \, dx,x,\sec ^2(x)\right )\\ &=-\frac{3 \tanh ^{-1}\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} \operatorname{Subst}\left (\int \frac{1}{\frac{5}{4}-\frac{x^2}{4}} \, dx,x,\sqrt{5-4 \sec ^2(x)}\right )+\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{-\frac{3}{2}+\frac{x^2}{2}} \, dx,x,\sqrt{5-4 \sec ^2(x)}\right )\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{5-4 \sec ^2(x)}}{\sqrt{3}}\right )}{6 \sqrt{3}}-\frac{\tanh ^{-1}\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 [B] time = 2.61882, size = 255, normalized size = 3.49 \[ \frac{15 \sin ^2(x) \sqrt{15 \cos (2 x)-9} \tanh ^{-1}\left (\frac{\sqrt{5 \cos (2 x)-3}}{\sqrt{6} \sqrt{\cos ^2(x)}}\right )-2 \left (15 \sqrt{2} \sqrt{\sin ^2(x)} \sqrt{\sin ^2(2 x)}+9 \sqrt{5} \sin ^2(x) \sqrt{5 \cos (2 x)-3} \left (\log \left (10 \sin ^2(x)\right )-\log \left (5 \left (\sqrt{10} \sqrt{\sin ^2(x)} \sqrt{\sin ^2(2 x)}+\sqrt{5 \cos (2 x)-3} \cos (2 x)-\sqrt{5 \cos (2 x)-3}\right )\right )\right )+10 \sqrt{\sin ^2(x)} \sqrt{\sin ^2(2 x)} \sqrt{15 \cos (2 x)-9} \sec (x) \tanh ^{-1}\left (\frac{\sqrt{6} \cos (x)}{\sqrt{5 \cos (2 x)-3}}\right )\right )}{225 \sqrt{\sin ^2(x)} \sqrt{\sin ^2(2 x)} \sqrt{10-8 \sec ^2(x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((3 + Sin[x]^2)*Tan[x]^3)/((-2 + Cos[x]^2)*(5 - 4*Sec[x]^2)^(3/2)),x]
[Out]
(15*ArcTanh[Sqrt[-3 + 5*Cos[2*x]]/(Sqrt[6]*Sqrt[Cos[x]^2])]*Sqrt[-9 + 15*Cos[2*x]]*Sin[x]^2 - 2*(9*Sqrt[5]*Sqr
t[-3 + 5*Cos[2*x]]*(Log[10*Sin[x]^2] - Log[5*(-Sqrt[-3 + 5*Cos[2*x]] + Cos[2*x]*Sqrt[-3 + 5*Cos[2*x]] + Sqrt[1
0]*Sqrt[Sin[x]^2]*Sqrt[Sin[2*x]^2])])*Sin[x]^2 + 15*Sqrt[2]*Sqrt[Sin[x]^2]*Sqrt[Sin[2*x]^2] + 10*ArcTanh[(Sqrt
[6]*Cos[x])/Sqrt[-3 + 5*Cos[2*x]]]*Sqrt[-9 + 15*Cos[2*x]]*Sec[x]*Sqrt[Sin[x]^2]*Sqrt[Sin[2*x]^2]))/(225*Sqrt[1
0 - 8*Sec[x]^2]*Sqrt[Sin[x]^2]*Sqrt[Sin[2*x]^2])
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Maple [B] time = 0.276, size = 1615, normalized size = 22.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((3+sin(x)^2)*tan(x)^3/(cos(x)^2-2)/(5-4*sec(x)^2)^(3/2),x)
[Out]
-3/5/(6+2*5^(1/2)-2^(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))/(2*3^(1/2)-6^(1/2))/(5+2*5^(1/2))/(-5+2*5^(1/2))*(-4+5*cos(x)^2)*(50*3^(1/2)*cos(x)*2^(1/2)*((-4+5*cos(x)
^2)/(cos(x)+1)^2)^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/
2)+4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))+50*3^(1/2)*cos(x)*2^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^
(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((-
4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))-25*cos(x)*2^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*arctanh(1/2/(2*3^(1/
2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2
)^(1/2))*6^(1/2)+25*cos(x)*2^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2
)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))*6^(1/2)+1
00*3^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*cos(x)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*c
os(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))-100*3^(1/2)*((-4+5*cos(x)^
2)/(cos(x)+1)^2)^(1/2)*cos(x)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4
*2^(1/2)-4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))+50*3^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*2^(
1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4
+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))+50*3^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*2^(1/2)*arctanh(1/2/(2*3^(1/
2)-6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2
)^(1/2))+72*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*cos(x)*5^(1/2)*arctanh(1/2*5^(1/2)*cos(x)*4^(1/2)*(cos(x)-1)/
sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))-50*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*cos(x)*arctanh(1/2/(2*3
^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+
1)^2)^(1/2))*6^(1/2)-50*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*cos(x)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(c
os(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))*6^(1/2)-25*((
-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*2^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2
)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))*6^(1/2)+25*((-4+5*cos(x)^2)/(cos(x)+1)
^2)^(1/2)*2^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/
sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))*6^(1/2)+100*3^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*arctan
h(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4+5*cos(x)^2
)/(cos(x)+1)^2)^(1/2))-100*3^(1/2)*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2
)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))+72*((-4+5
*cos(x)^2)/(cos(x)+1)^2)^(1/2)*5^(1/2)*arctanh(1/2*5^(1/2)*cos(x)*4^(1/2)*(cos(x)-1)/sin(x)^2/((-4+5*cos(x)^2)
/(cos(x)+1)^2)^(1/2))-50*((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(cos(x)-
1)*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))*6^(1/2)-50*((-4+5*c
os(x)^2)/(cos(x)+1)^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(cos(x)-1)*(5*cos(x)*2^(1/2)-10*cos(x)+4*
2^(1/2)-4)/sin(x)^2/((-4+5*cos(x)^2)/(cos(x)+1)^2)^(1/2))*6^(1/2)-240*cos(x))/cos(x)^3/((-4+5*cos(x)^2)/cos(x)
^2)^(3/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Fricas [B] time = 4.45508, size = 799, normalized size = 10.95 \begin{align*} -\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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((3+sin(x)**2)*tan(x)**3/(-2+cos(x)**2)/(5-4*sec(x)**2)**(3/2),x)
[Out]
Timed out
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Giac [C] time = 1.35068, size = 231, normalized size = 3.16 \begin{align*} -\frac{1}{4500} \, \sqrt{15} \sqrt{5}{\left (6 i \, \sqrt{15} \pi + 12 \, \sqrt{15} \log \left (2\right ) - 25 \, \log \left (-\frac{\sqrt{15} + 5}{\sqrt{15} - 5}\right )\right )} \mathrm{sgn}\left (\cos \left (x\right )\right ) - \frac{\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 )}{180 \, \mathrm{sgn}\left (\cos \left (x\right )\right )} + \frac{\sqrt{5} \log \left ({\left (\sqrt{5} \cos \left (x\right ) - \sqrt{5 \, \cos \left (x\right )^{2} - 4}\right )}^{2}\right )}{50 \, \mathrm{sgn}\left (\cos \left (x\right )\right )} - \frac{2 \, \cos \left (x\right )}{15 \, \sqrt{5 \, \cos \left (x\right )^{2} - 4} \mathrm{sgn}\left (\cos \left (x\right )\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((3+sin(x)^2)*tan(x)^3/(-2+cos(x)^2)/(5-4*sec(x)^2)^(3/2),x, algorithm="giac")
[Out]
-1/4500*sqrt(15)*sqrt(5)*(6*I*sqrt(15)*pi + 12*sqrt(15)*log(2) - 25*log(-(sqrt(15) + 5)/(sqrt(15) - 5)))*sgn(c
os(x)) - 1/180*sqrt(15)*sqrt(5)*log(-2*((sqrt(5)*cos(x) - sqrt(5*cos(x)^2 - 4))^2 - 4*sqrt(15) - 16)/abs(2*(sq
rt(5)*cos(x) - sqrt(5*cos(x)^2 - 4))^2 + 8*sqrt(15) - 32))/sgn(cos(x)) + 1/50*sqrt(5)*log((sqrt(5)*cos(x) - sq
rt(5*cos(x)^2 - 4))^2)/sgn(cos(x)) - 2/15*cos(x)/(sqrt(5*cos(x)^2 - 4)*sgn(cos(x)))