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 \[ -\frac {2}{15 \sqrt {5-4 \sec ^2(x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {3}}\right )}{6 \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {5-4 \sec ^2(x)}}{\sqrt {5}}\right )}{5 \sqrt {5}} \]
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Rubi [A] time = 1.23, antiderivative size = 73, normalized size of antiderivative =
1.00, 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 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 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 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 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 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 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 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 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]
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*}
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Mathematica [B] time = 2.57, 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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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (3+\sin ^2(x)\right ) \tan ^3(x)}{\left (-2+\cos ^2(x)\right ) \left (5-4 \sec ^2(x)\right )^{3/2}} \, dx \]
Verification is Not applicable to the result.
[In]
IntegrateAlgebraic[((3 + Sin[x]^2)*Tan[x]^3)/((-2 + Cos[x]^2)*(5 - 4*Sec[x]^2)^(3/2)),x]
[Out]
Could not integrate
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fricas [B] time = 1.82, size = 257, normalized size = 3.52 \[ -\frac {480 \, \sqrt {\frac {5 \, \cos \relax (x)^{2} - 4}{\cos \relax (x)^{2}}} \cos \relax (x)^{2} - 18 \, {\left (5 \, \sqrt {5} \cos \relax (x)^{2} - 4 \, \sqrt {5}\right )} \log \left (625 \, \cos \relax (x)^{8} - 1000 \, \cos \relax (x)^{6} + 500 \, \cos \relax (x)^{4} - 80 \, \cos \relax (x)^{2} - {\left (125 \, \sqrt {5} \cos \relax (x)^{8} - 150 \, \sqrt {5} \cos \relax (x)^{6} + 50 \, \sqrt {5} \cos \relax (x)^{4} - 4 \, \sqrt {5} \cos \relax (x)^{2}\right )} \sqrt {\frac {5 \, \cos \relax (x)^{2} - 4}{\cos \relax (x)^{2}}} + 2\right ) - 25 \, {\left (5 \, \sqrt {3} \cos \relax (x)^{2} - 4 \, \sqrt {3}\right )} \log \left (\frac {1921 \, \cos \relax (x)^{8} - 3464 \, \cos \relax (x)^{6} + 2040 \, \cos \relax (x)^{4} - 416 \, \cos \relax (x)^{2} - 8 \, {\left (62 \, \sqrt {3} \cos \relax (x)^{8} - 87 \, \sqrt {3} \cos \relax (x)^{6} + 36 \, \sqrt {3} \cos \relax (x)^{4} - 4 \, \sqrt {3} \cos \relax (x)^{2}\right )} \sqrt {\frac {5 \, \cos \relax (x)^{2} - 4}{\cos \relax (x)^{2}}} + 16}{\cos \relax (x)^{8} - 8 \, \cos \relax (x)^{6} + 24 \, \cos \relax (x)^{4} - 32 \, \cos \relax (x)^{2} + 16}\right )}{3600 \, {\left (5 \, \cos \relax (x)^{2} - 4\right )}} \]
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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giac [C] time = 0.93, size = 171, normalized size = 2.34 \[ -\frac {1}{4500} \, \sqrt {15} \sqrt {5} {\left (6 i \, \sqrt {15} \pi + 12 \, \sqrt {15} \log \relax (2) - 25 \, \log \left (-\frac {\sqrt {15} + 5}{\sqrt {15} - 5}\right )\right )} \mathrm {sgn}\left (\cos \relax (x)\right ) - \frac {\sqrt {15} \sqrt {5} \log \left (-\frac {2 \, {\left ({\left (\sqrt {5} \cos \relax (x) - \sqrt {5 \, \cos \relax (x)^{2} - 4}\right )}^{2} - 4 \, \sqrt {15} - 16\right )}}{{\left | 2 \, {\left (\sqrt {5} \cos \relax (x) - \sqrt {5 \, \cos \relax (x)^{2} - 4}\right )}^{2} + 8 \, \sqrt {15} - 32 \right |}}\right )}{180 \, \mathrm {sgn}\left (\cos \relax (x)\right )} + \frac {\sqrt {5} \log \left ({\left (\sqrt {5} \cos \relax (x) - \sqrt {5 \, \cos \relax (x)^{2} - 4}\right )}^{2}\right )}{50 \, \mathrm {sgn}\left (\cos \relax (x)\right )} - \frac {2 \, \cos \relax (x)}{15 \, \sqrt {5 \, \cos \relax (x)^{2} - 4} \mathrm {sgn}\left (\cos \relax (x)\right )} \]
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)))
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maple [B] time = 0.91, size = 1615, normalized size = 22.12
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\(\text {Expression too large to display}\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((3+sin(x)^2)*tan(x)^3/(-2+cos(x)^2)/(5-4*sec(x)^2)^(3/2),x,method=_RETURNVERBOSE)
[Out]
-3/5/(-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))/(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*cos(x)^2-4)*(50*cos(x)*3^(1/2)*2^(1/2)*((5*cos(x)^2-4
)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2
)+4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))+50*cos(x)*3^(1/2)*2^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/
2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((5*c
os(x)^2-4)/(1+cos(x))^2)^(1/2))-25*cos(x)*2^(1/2)*6^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3
^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x
))^2)^(1/2))+25*cos(x)*2^(1/2)*6^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(
1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))+100*co
s(x)*3^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(
x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)-100*cos(x)*3^(1/2)*((5*cos(x)^2-
4)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/
2)-4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))+72*cos(x)*5^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)*arct
anh(1/2*5^(1/2)*cos(x)*4^(1/2)*(-1+cos(x))/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))-50*cos(x)*6^(1/2)*arc
tanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((5*cos(x)^
2-4)/(1+cos(x))^2)^(1/2))*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)-50*cos(x)*6^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^
(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((
5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))+50*3^(1/2)*2^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*c
os(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))*((5*cos(x)^2-4)/(1+cos(x))^
2)^(1/2)+50*3^(1/2)*2^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*
2^(1/2)-4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)-25*2^(1/2)*6^(1/2
)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((5*co
s(x)^2-4)/(1+cos(x))^2)^(1/2))*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)+25*2^(1/2)*6^(1/2)*arctanh(1/2/(2*3^(1/2)-6
^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1
/2))*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)+100*3^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2
)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)
^(1/2))-100*3^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(-1+cos(x))*(5
*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))+72*5^(1/2)*((5*cos(x)^2-4
)/(1+cos(x))^2)^(1/2)*arctanh(1/2*5^(1/2)*cos(x)*4^(1/2)*(-1+cos(x))/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1
/2))-50*6^(1/2)*((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2)+6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos
(x)*2^(1/2)+10*cos(x)+4*2^(1/2)+4)/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))-50*6^(1/2)*((5*cos(x)^2-4)/(1
+cos(x))^2)^(1/2)*arctanh(1/2/(2*3^(1/2)-6^(1/2))*4^(1/2)*(-1+cos(x))*(5*cos(x)*2^(1/2)-10*cos(x)+4*2^(1/2)-4)
/sin(x)^2/((5*cos(x)^2-4)/(1+cos(x))^2)^(1/2))-240*cos(x))/cos(x)^3/((5*cos(x)^2-4)/cos(x)^2)^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (\sin \relax (x)^{2} + 3\right )} \tan \relax (x)^{3}}{{\left (\cos \relax (x)^{2} - 2\right )} {\left (-4 \, \sec \relax (x)^{2} + 5\right )}^{\frac {3}{2}}}\,{d x} \]
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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\relax (x)}^3\,\left ({\sin \relax (x)}^2+3\right )}{\left ({\cos \relax (x)}^2-2\right )\,{\left (5-\frac {4}{{\cos \relax (x)}^2}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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sympy [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((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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