3.159 \(\int \frac {x^5 (a+b \text {sech}^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=278 \[ -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {8 b d^{3/2} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 e^3}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (9 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{6 c^3 e^{5/2}} \]

[Out]

1/3*(e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/e^3+1/6*b*(9*c^2*d-e)*arctan(e^(1/2)*(-c^2*x^2+1)^(1/2)/c/(e*x^2+d)^(1/
2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^3/e^(5/2)+8/3*b*d^(3/2)*arctanh((e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2+1)^(1/
2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/e^3-d^2*(a+b*arcsech(c*x))/e^3/(e*x^2+d)^(1/2)-2*d*(a+b*arcsech(c*x))*(e*x
^2+d)^(1/2)/e^3-1/6*b*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)*(e*x^2+d)^(1/2)/c^2/e^2

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Rubi [A]  time = 1.12, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {266, 43, 6301, 12, 1615, 157, 63, 217, 203, 93, 207} \[ -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {8 b d^{3/2} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 e^3}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (9 c^2 d-e\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{6 c^3 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSech[c*x]))/(d + e*x^2)^(3/2),x]

[Out]

-(b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(6*c^2*e^2) - (d^2*(a + b*ArcSech[c*
x]))/(e^3*Sqrt[d + e*x^2]) - (2*d*Sqrt[d + e*x^2]*(a + b*ArcSech[c*x]))/e^3 + ((d + e*x^2)^(3/2)*(a + b*ArcSec
h[c*x]))/(3*e^3) + (b*(9*c^2*d - e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*S
qrt[d + e*x^2])])/(6*c^3*e^(5/2)) + (8*b*d^(3/2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(S
qrt[d]*Sqrt[1 - c^2*x^2])])/(3*e^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 1615

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[
{q = Expon[Px, x], k = Coeff[Px, x, Expon[Px, x]]}, Simp[(k*(a + b*x)^(m + q - 1)*(c + d*x)^(n + 1)*(e + f*x)^
(p + 1))/(d*f*b^(q - 1)*(m + n + p + q + 1)), x] + Dist[1/(d*f*b^q*(m + n + p + q + 1)), Int[(a + b*x)^m*(c +
d*x)^n*(e + f*x)^p*ExpandToSum[d*f*b^q*(m + n + p + q + 1)*Px - d*f*k*(m + n + p + q + 1)*(a + b*x)^q + k*(a +
b*x)^(q - 2)*(a^2*d*f*(m + n + p + q + 1) - b*(b*c*e*(m + q - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*
(2*(m + q) + n + p) - b*(d*e*(m + q + n) + c*f*(m + q + p)))*x), x], x], x] /; NeQ[m + n + p + q + 1, 0]] /; F
reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && IntegersQ[2*m, 2*n, 2*p]

Rule 6301

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)],
Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] &&
((IGtQ[p, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ
[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-8 d^2-4 d e x^2+e^2 x^4}{3 e^3 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx\\ &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-8 d^2-4 d e x^2+e^2 x^4}{x \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{3 e^3}\\ &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {-8 d^2-4 d e x+e^2 x^2}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{6 e^3}\\ &=-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {8 c^2 d^2 e+\frac {1}{2} \left (9 c^2 d-e\right ) e^2 x}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{6 c^2 e^4}\\ &=-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}-\frac {\left (4 b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 e^3}-\frac {\left (b \left (9 c^2 d-e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{12 c^2 e^2}\\ &=-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}-\frac {\left (8 b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{-d+x^2} \, dx,x,\frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}}\right )}{3 e^3}+\frac {\left (b \left (9 c^2 d-e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+\frac {e}{c^2}-\frac {e x^2}{c^2}}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{6 c^4 e^2}\\ &=-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {8 b d^{3/2} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 e^3}+\frac {\left (b \left (9 c^2 d-e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {e x^2}{c^2}} \, dx,x,\frac {\sqrt {1-c^2 x^2}}{\sqrt {d+e x^2}}\right )}{6 c^4 e^2}\\ &=-\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{6 c^2 e^2}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3}+\frac {b \left (9 c^2 d-e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{6 c^3 e^{5/2}}+\frac {8 b d^{3/2} \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 e^3}\\ \end {align*}

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Mathematica [A]  time = 1.50, size = 436, normalized size = 1.57 \[ \frac {-2 a c^2 \left (8 d^2+4 d e x^2-e^2 x^4\right )-2 b c^2 \text {sech}^{-1}(c x) \left (8 d^2+4 d e x^2-e^2 x^4\right )-b e \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (d+e x^2\right )}{6 c^2 e^3 \sqrt {d+e x^2}}-\frac {b \sqrt {\frac {1-c x}{c x+1}} \sqrt {1-c^2 x^2} \left (\sqrt {-c^2} e^{3/2} \sqrt {c^2 (-d)-e} \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sin ^{-1}\left (\frac {\sqrt {-c^2} \sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {c^2 (-d)-e}}\right )-9 \left (-c^2\right )^{3/2} d \sqrt {e} \sqrt {c^2 (-d)-e} \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sin ^{-1}\left (\frac {c \sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {-c^2} \sqrt {c^2 (-d)-e}}\right )+16 c^5 d^{3/2} \sqrt {-d-e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )\right )}{6 c^5 e^3 (c x-1) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcSech[c*x]))/(d + e*x^2)^(3/2),x]

[Out]

(-(b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + e*x^2)) - 2*a*c^2*(8*d^2 + 4*d*e*x^2 - e^2*x^4) - 2*b*c^2*(8*d
^2 + 4*d*e*x^2 - e^2*x^4)*ArcSech[c*x])/(6*c^2*e^3*Sqrt[d + e*x^2]) - (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^
2*x^2]*(-9*(-c^2)^(3/2)*d*Sqrt[-(c^2*d) - e]*Sqrt[e]*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSin[(c*Sqrt[e]*Sqr
t[1 - c^2*x^2])/(Sqrt[-c^2]*Sqrt[-(c^2*d) - e])] + Sqrt[-c^2]*Sqrt[-(c^2*d) - e]*e^(3/2)*Sqrt[(c^2*(d + e*x^2)
)/(c^2*d + e)]*ArcSin[(Sqrt[-c^2]*Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[-(c^2*d) - e])] + 16*c^5*d^(3/2)*Sqrt[-d
- e*x^2]*ArcTan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d - e*x^2]]))/(6*c^5*e^3*(-1 + c*x)*Sqrt[d + e*x^2])

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fricas [B]  time = 2.59, size = 1771, normalized size = 6.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[1/24*((9*b*c^2*d^2 - b*d*e + (9*b*c^2*d*e - b*e^2)*x^2)*sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*
(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(-(c^2*x^2 - 1)/(c^
2*x^2)) + e^2) + 8*(b*c^3*e^2*x^4 - 4*b*c^3*d*e*x^2 - 8*b*c^3*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1
)/(c^2*x^2)) + 1)/(c*x)) + 16*(b*c^3*d*e*x^2 + b*c^3*d^2)*sqrt(d)*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^
2*d^2 - d*e)*x^2 - 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*
d^2)/x^4) + 4*(2*a*c^3*e^2*x^4 - 8*a*c^3*d*e*x^2 - 16*a*c^3*d^2 - (b*c^2*e^2*x^3 + b*c^2*d*e*x)*sqrt(-(c^2*x^2
- 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^3*e^4*x^2 + c^3*d*e^3), 1/12*((9*b*c^2*d^2 - b*d*e + (9*b*c^2*d*e - b*e^
2)*x^2)*sqrt(e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)
)/(c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)) + 4*(b*c^3*e^2*x^4 - 4*b*c^3*d*e*x^2 - 8*b*c^3*d^2)*sqrt(e*x^2 +
d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 8*(b*c^3*d*e*x^2 + b*c^3*d^2)*sqrt(d)*log(((c^4*d^2 -
6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 - 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt(-
(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2)/x^4) + 2*(2*a*c^3*e^2*x^4 - 8*a*c^3*d*e*x^2 - 16*a*c^3*d^2 - (b*c^2*e^2*x^3
+ b*c^2*d*e*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^3*e^4*x^2 + c^3*d*e^3), 1/24*(32*(b*c^3*d*e
*x^2 + b*c^3*d^2)*sqrt(-d)*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt(-(c^2*x^2 -
1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (9*b*c^2*d^2 - b*d*e + (9*b*c^2*d*e - b*e^2)*x^2)*
sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e
)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + e^2) + 8*(b*c^3*e^2*x^4 - 4*b*c^3*d*e*x^2 - 8*b
*c^3*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 4*(2*a*c^3*e^2*x^4 - 8*a*c^3*d
*e*x^2 - 16*a*c^3*d^2 - (b*c^2*e^2*x^3 + b*c^2*d*e*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^3*e^
4*x^2 + c^3*d*e^3), 1/12*(16*(b*c^3*d*e*x^2 + b*c^3*d^2)*sqrt(-d)*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sq
rt(e*x^2 + d)*sqrt(-d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (9*b*c^2*d^
2 - b*d*e + (9*b*c^2*d*e - b*e^2)*x^2)*sqrt(e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2 + d)*sqrt(e
)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)) + 4*(b*c^3*e^2*x^4 - 4*b*c^3*d*e*x
^2 - 8*b*c^3*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(2*a*c^3*e^2*x^4 - 8
*a*c^3*d*e*x^2 - 16*a*c^3*d^2 - (b*c^2*e^2*x^3 + b*c^2*d*e*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))
/(c^3*e^4*x^2 + c^3*d*e^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)*x^5/(e*x^2 + d)^(3/2), x)

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maple [F]  time = 4.65, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (\frac {x^{4}}{\sqrt {e x^{2} + d} e} - \frac {4 \, d x^{2}}{\sqrt {e x^{2} + d} e^{2}} - \frac {8 \, d^{2}}{\sqrt {e x^{2} + d} e^{3}}\right )} a + b \int \frac {x^{5} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

1/3*(x^4/(sqrt(e*x^2 + d)*e) - 4*d*x^2/(sqrt(e*x^2 + d)*e^2) - 8*d^2/(sqrt(e*x^2 + d)*e^3))*a + b*integrate(x^
5*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(e*x^2 + d)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(3/2),x)

[Out]

int((x^5*(a + b*acosh(1/(c*x))))/(d + e*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(x**5*(a+b*asech(c*x))/(e*x**2+d)**(3/2),x)

[Out]

Timed out

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