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

Optimal. Leaf size=272 \[ -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c e^{5/2}}-\frac {8 b \sqrt {d} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 e^3}-\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 1.28, antiderivative size = 272, 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, 1614, 157, 63, 217, 203, 93, 207} \[ -\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{c e^{5/2}}-\frac {8 b \sqrt {d} \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(3*e^2*(c^2*d + e)*Sqrt[d + e*x^2]) - (d^2*(a + b*
ArcSech[c*x]))/(3*e^3*(d + e*x^2)^(3/2)) + (2*d*(a + b*ArcSech[c*x]))/(e^3*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]
*(a + b*ArcSech[c*x]))/e^3 - (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[
d + e*x^2])])/(c*e^(5/2)) - (8*b*Sqrt[d]*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(Sqrt[d]*S
qrt[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 1614

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[{
Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(b*R*(a + b*x)^(m + 1)
*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e
- a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f
*R*(m + 1) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x], x], x]] /; FreeQ[{a, b,
c, d, e, f, n, p}, x] && PolyQ[Px, x] && LtQ[m, -1] && 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 )^{5/2}} \, dx &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {8 d^2+12 d e x^2+3 e^2 x^4}{3 e^3 x \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx\\ &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {8 d^2+12 d e x^2+3 e^2 x^4}{x \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 e^3}\\ &=-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {8 d^2+12 d e x+3 e^2 x^2}{x \sqrt {1-c^2 x} (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e^3}\\ &=-\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {4 d^2 \left (c^2 d+e\right )+\frac {3}{2} d e \left (c^2 d+e\right ) x}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 d e^3 \left (c^2 d+e\right )}\\ &=-\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (4 b d \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 \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 )}{2 e^2}\\ &=-\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}+\frac {\left (8 b d \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 \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 )}{c^2 e^2}\\ &=-\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {8 b \sqrt {d} \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 \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 )}{c^2 e^2}\\ &=-\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e^2 \left (c^2 d+e\right ) \sqrt {d+e x^2}}-\frac {d^2 \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \text {sech}^{-1}(c x)\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{e^3}-\frac {b \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 )}{c e^{5/2}}-\frac {8 b \sqrt {d} \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 = 1.97, size = 348, normalized size = 1.28 \[ \frac {a \left (c^2 d+e\right ) \left (8 d^2+12 d e x^2+3 e^2 x^4\right )+b \left (c^2 d+e\right ) \text {sech}^{-1}(c x) \left (8 d^2+12 d e x^2+3 e^2 x^4\right )-b d e \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (d+e x^2\right )}{3 e^3 \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac {b \sqrt {\frac {1-c x}{c x+1}} \sqrt {1-c^2 x^2} \left (3 \sqrt {-c^2} \sqrt {e} \sqrt {c^2 (-d)-e} \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sin ^{-1}\left (\frac {c \sqrt {e} \sqrt {1-c^2 x^2}}{\sqrt {-c^2} \sqrt {c^2 (-d)-e}}\right )+8 c^3 \sqrt {d} \sqrt {-d-e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )\right )}{3 c^3 e^3 (c x-1) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*d*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + e*x^2)) + a*(c^2*d + e)*(8*d^2 + 12*d*e*x^2 + 3*e^2*x^4) + b
*(c^2*d + e)*(8*d^2 + 12*d*e*x^2 + 3*e^2*x^4)*ArcSech[c*x])/(3*e^3*(c^2*d + e)*(d + e*x^2)^(3/2)) + (b*Sqrt[(1
- c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*(3*Sqrt[-c^2]*Sqrt[-(c^2*d) - e]*Sqrt[e]*Sqrt[(c^2*(d + e*x^2))/(c^2*d +
e)]*ArcSin[(c*Sqrt[e]*Sqrt[1 - c^2*x^2])/(Sqrt[-c^2]*Sqrt[-(c^2*d) - e])] + 8*c^3*Sqrt[d]*Sqrt[-d - e*x^2]*Arc
Tan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d - e*x^2]]))/(3*c^3*e^3*(-1 + c*x)*Sqrt[d + e*x^2])

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fricas [B]  time = 1.19, size = 2415, normalized size = 8.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(b*c^2*d^3 + (b*c^2*d*e^2 + b*e^3)*x^4 + b*d^2*e + 2*(b*c^2*d^2*e + b*d*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) - 4*(8*b*c^3*d^3 + 8*b*c*d^2*e + 3*(b*c^3*d*e^2 + b*c*e^3)*x
^4 + 12*(b*c^3*d^2*e + b*c*d*e^2)*x^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^3 + b*c*d^2*e + (b*c^3*d*e^2 + b*c*e^3)*x^4 + 2*(b*c^3*d^2*e + b*c*d*e^2)*x^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*(8*a*c^3*d^3 + 8*a*c*d^2*e + 3*(a*c^3*d*e^2 + a*c*e^3)*x^4 + 12*(
a*c^3*d^2*e + a*c*d*e^2)*x^2 - (b*c^2*d*e^2*x^3 + b*c^2*d^2*e*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 +
d))/(c^3*d^3*e^3 + c*d^2*e^4 + (c^3*d*e^5 + c*e^6)*x^4 + 2*(c^3*d^2*e^4 + c*d*e^5)*x^2), -1/6*(3*(b*c^2*d^3 +
(b*c^2*d*e^2 + b*e^3)*x^4 + b*d^2*e + 2*(b*c^2*d^2*e + b*d*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)) - 2*
(8*b*c^3*d^3 + 8*b*c*d^2*e + 3*(b*c^3*d*e^2 + b*c*e^3)*x^4 + 12*(b*c^3*d^2*e + b*c*d*e^2)*x^2)*sqrt(e*x^2 + d)
*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 4*(b*c^3*d^3 + b*c*d^2*e + (b*c^3*d*e^2 + b*c*e^3)*x^4
+ 2*(b*c^3*d^2*e + b*c*d*e^2)*x^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*(8*a*c^3*
d^3 + 8*a*c*d^2*e + 3*(a*c^3*d*e^2 + a*c*e^3)*x^4 + 12*(a*c^3*d^2*e + a*c*d*e^2)*x^2 - (b*c^2*d*e^2*x^3 + b*c^
2*d^2*e*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^3*d^3*e^3 + c*d^2*e^4 + (c^3*d*e^5 + c*e^6)*x^4
+ 2*(c^3*d^2*e^4 + c*d*e^5)*x^2), -1/12*(16*(b*c^3*d^3 + b*c*d^2*e + (b*c^3*d*e^2 + b*c*e^3)*x^4 + 2*(b*c^3*d
^2*e + b*c*d*e^2)*x^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)) + 3*(b*c^2*d^3 + (b*c^2*d*e^2 + b*e^3)*x^4 + b*
d^2*e + 2*(b*c^2*d^2*e + b*d*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) - 4
*(8*b*c^3*d^3 + 8*b*c*d^2*e + 3*(b*c^3*d*e^2 + b*c*e^3)*x^4 + 12*(b*c^3*d^2*e + b*c*d*e^2)*x^2)*sqrt(e*x^2 + d
)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 4*(8*a*c^3*d^3 + 8*a*c*d^2*e + 3*(a*c^3*d*e^2 + a*c*e^
3)*x^4 + 12*(a*c^3*d^2*e + a*c*d*e^2)*x^2 - (b*c^2*d*e^2*x^3 + b*c^2*d^2*e*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*
sqrt(e*x^2 + d))/(c^3*d^3*e^3 + c*d^2*e^4 + (c^3*d*e^5 + c*e^6)*x^4 + 2*(c^3*d^2*e^4 + c*d*e^5)*x^2), -1/6*(8*
(b*c^3*d^3 + b*c*d^2*e + (b*c^3*d*e^2 + b*c*e^3)*x^4 + 2*(b*c^3*d^2*e + b*c*d*e^2)*x^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)) + 3*(b*c^2*d^3 + (b*c^2*d*e^2 + b*e^3)*x^4 + b*d^2*e + 2*(b*c^2*d^2*e + b*d*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)) - 2*(8*b*c^3*d^3 + 8*b*c*d^2*e + 3*(b*c^3*d*e^2 + b*c*e^3)*x^4 + 12*(b*c^3*d^2*
e + b*c*d*e^2)*x^2)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(8*a*c^3*d^3 + 8*a
*c*d^2*e + 3*(a*c^3*d*e^2 + a*c*e^3)*x^4 + 12*(a*c^3*d^2*e + a*c*d*e^2)*x^2 - (b*c^2*d*e^2*x^3 + b*c^2*d^2*e*x
)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^3*d^3*e^3 + c*d^2*e^4 + (c^3*d*e^5 + c*e^6)*x^4 + 2*(c^3
*d^2*e^4 + c*d*e^5)*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 4.58, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (\frac {3 \, x^{4}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} + \frac {12 \, d x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}} + \frac {8 \, d^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{3}}\right )} a + b \int \frac {x^{5} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*x^4/((e*x^2 + d)^(3/2)*e) + 12*d*x^2/((e*x^2 + d)^(3/2)*e^2) + 8*d^2/((e*x^2 + d)^(3/2)*e^3))*a + b*int
egrate(x^5*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(e*x^2 + d)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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