3.148 \(\int \frac {(d+e x^2)^{3/2} (a+b \text {sech}^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=409 \[ -\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}+\frac {4 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (c^2 d+2 e\right ) \sqrt {d+e x^2}}{75 x^3}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {d+e x^2}}{75 d x}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (c^2 d+e\right ) \left (8 c^4 d^2+19 c^2 d e+15 e^2\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{75 c d \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{75 d \sqrt {\frac {e x^2}{d}+1}} \]

[Out]

-1/5*(e*x^2+d)^(5/2)*(a+b*arcsech(c*x))/d/x^5+1/25*b*(e*x^2+d)^(3/2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2
+1)^(1/2)/x^5+4/75*b*(c^2*d+2*e)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)*(e*x^2+d)^(1/2)/x^3+1/75*b
*(8*c^4*d^2+23*c^2*d*e+23*e^2)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(-c^2*x^2+1)^(1/2)*(e*x^2+d)^(1/2)/d/x+1/75*b*c
*(8*c^4*d^2+23*c^2*d*e+23*e^2)*EllipticE(c*x,(-e/c^2/d)^(1/2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)*(e*x^2+d)^(1/2)
/d/(1+e*x^2/d)^(1/2)-1/75*b*(c^2*d+e)*(8*c^4*d^2+19*c^2*d*e+15*e^2)*EllipticF(c*x,(-e/c^2/d)^(1/2))*(1/(c*x+1)
)^(1/2)*(c*x+1)^(1/2)*(1+e*x^2/d)^(1/2)/c/d/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 409, 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 = {264, 6301, 12, 474, 580, 583, 524, 426, 424, 421, 419} \[ -\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {d+e x^2}}{75 d x}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (c^2 d+e\right ) \left (8 c^4 d^2+19 c^2 d e+15 e^2\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{75 c d \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{75 d \sqrt {\frac {e x^2}{d}+1}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}+\frac {4 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (c^2 d+2 e\right ) \sqrt {d+e x^2}}{75 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^(3/2)*(a + b*ArcSech[c*x]))/x^6,x]

[Out]

(4*b*(c^2*d + 2*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(75*x^3) + (b*(8*c^4*
d^2 + 23*c^2*d*e + 23*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(75*d*x) + (b
*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*(d + e*x^2)^(3/2))/(25*x^5) - ((d + e*x^2)^(5/2)*(a + b*
ArcSech[c*x]))/(5*d*x^5) + (b*c*(8*c^4*d^2 + 23*c^2*d*e + 23*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[d +
e*x^2]*EllipticE[ArcSin[c*x], -(e/(c^2*d))])/(75*d*Sqrt[1 + (e*x^2)/d]) - (b*(c^2*d + e)*(8*c^4*d^2 + 19*c^2*d
*e + 15*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(75*
c*d*Sqrt[d + e*x^2])

Rule 12

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 474

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)
*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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 {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{x^6} \, dx &=-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int -\frac {\left (d+e x^2\right )^{5/2}}{5 d x^6 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\left (d+e x^2\right )^{5/2}}{x^6 \sqrt {1-c^2 x^2}} \, dx}{5 d}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2} \left (4 d \left (c^2 d+2 e\right )+e \left (c^2 d+5 e\right ) x^2\right )}{x^4 \sqrt {1-c^2 x^2}} \, dx}{25 d}\\ &=\frac {4 b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 x^3}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {d \left (8 c^4 d^2+23 c^2 d e+23 e^2\right )+e \left (4 c^4 d^2+11 c^2 d e+15 e^2\right ) x^2}{x^2 \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{75 d}\\ &=\frac {4 b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 x^3}+\frac {b \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 d x}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d e \left (4 c^4 d^2+11 c^2 d e+15 e^2\right )+c^2 d e \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) x^2}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{75 d^2}\\ &=\frac {4 b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 x^3}+\frac {b \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 d x}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}-\frac {\left (b \left (c^2 d+e\right ) \left (8 c^4 d^2+19 c^2 d e+15 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{75 d}+\frac {\left (b c^2 \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}} \, dx}{75 d}\\ &=\frac {4 b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 x^3}+\frac {b \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 d x}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}+\frac {\left (b c^2 \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2}\right ) \int \frac {\sqrt {1+\frac {e x^2}{d}}}{\sqrt {1-c^2 x^2}} \, dx}{75 d \sqrt {1+\frac {e x^2}{d}}}-\frac {\left (b \left (c^2 d+e\right ) \left (8 c^4 d^2+19 c^2 d e+15 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{75 d \sqrt {d+e x^2}}\\ &=\frac {4 b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 x^3}+\frac {b \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{75 d x}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{25 x^5}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \text {sech}^{-1}(c x)\right )}{5 d x^5}+\frac {b c \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{75 d \sqrt {1+\frac {e x^2}{d}}}-\frac {b \left (c^2 d+e\right ) \left (8 c^4 d^2+19 c^2 d e+15 e^2\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{75 c d \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 6.37, size = 620, normalized size = 1.52 \[ \frac {-\frac {15 a \left (d+e x^2\right )^3}{x^5}+\frac {b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (d+e x^2\right ) \left (d e x^2 \left (23 c^2 x^2+11\right )+d^2 \left (8 c^4 x^4+4 c^2 x^2+3\right )+23 e^2 x^4\right )}{x^5}+\frac {b \sqrt {\frac {1-c x}{c x+1}} \left (-\left (c^2 \left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) \left (d+e x^2\right )\right )-\frac {i (c x+1) \left (c \sqrt {d}-i \sqrt {e}\right )^2 \sqrt {\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{(c x+1) \left (c \sqrt {d}-i \sqrt {e}\right )}} \sqrt {\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}} \left (\left (8 c^4 d^2+23 c^2 d e+23 e^2\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (d c^2+e\right ) (1-c x)}{\left (\sqrt {d} c+i \sqrt {e}\right )^2 (c x+1)}}\right )|\frac {\left (\sqrt {d} c+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )+2 \sqrt {e} \left (8 i c^3 d^{3/2}-12 c^2 d \sqrt {e}+7 i c \sqrt {d} e-15 e^{3/2}\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (d c^2+e\right ) (1-c x)}{\left (\sqrt {d} c+i \sqrt {e}\right )^2 (c x+1)}}\right )|\frac {\left (\sqrt {d} c+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )\right )}{\sqrt {-\frac {(c x-1) \left (c \sqrt {d}-i \sqrt {e}\right )}{(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}}}\right )}{c}-\frac {15 b \text {sech}^{-1}(c x) \left (d+e x^2\right )^3}{x^5}}{75 d \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^(3/2)*(a + b*ArcSech[c*x]))/x^6,x]

[Out]

((-15*a*(d + e*x^2)^3)/x^5 + (b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + e*x^2)*(23*e^2*x^4 + d*e*x^2*(11 + 23
*c^2*x^2) + d^2*(3 + 4*c^2*x^2 + 8*c^4*x^4)))/x^5 - (15*b*(d + e*x^2)^3*ArcSech[c*x])/x^5 + (b*Sqrt[(1 - c*x)/
(1 + c*x)]*(-(c^2*(8*c^4*d^2 + 23*c^2*d*e + 23*e^2)*(d + e*x^2)) - (I*(c*Sqrt[d] - I*Sqrt[e])^2*(1 + c*x)*Sqrt
[(c*(Sqrt[d] - I*Sqrt[e]*x))/((c*Sqrt[d] - I*Sqrt[e])*(1 + c*x))]*Sqrt[(c*(Sqrt[d] + I*Sqrt[e]*x))/((c*Sqrt[d]
+ I*Sqrt[e])*(1 + c*x))]*((8*c^4*d^2 + 23*c^2*d*e + 23*e^2)*EllipticE[I*ArcSinh[Sqrt[((c^2*d + e)*(1 - c*x))/
((c*Sqrt[d] + I*Sqrt[e])^2*(1 + c*x))]], (c*Sqrt[d] + I*Sqrt[e])^2/(c*Sqrt[d] - I*Sqrt[e])^2] + 2*Sqrt[e]*((8*
I)*c^3*d^(3/2) - 12*c^2*d*Sqrt[e] + (7*I)*c*Sqrt[d]*e - 15*e^(3/2))*EllipticF[I*ArcSinh[Sqrt[((c^2*d + e)*(1 -
c*x))/((c*Sqrt[d] + I*Sqrt[e])^2*(1 + c*x))]], (c*Sqrt[d] + I*Sqrt[e])^2/(c*Sqrt[d] - I*Sqrt[e])^2]))/Sqrt[-(
((c*Sqrt[d] - I*Sqrt[e])*(-1 + c*x))/((c*Sqrt[d] + I*Sqrt[e])*(1 + c*x)))]))/c)/(75*d*Sqrt[d + e*x^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \operatorname {arsech}\left (c x\right )\right )} \sqrt {e x^{2} + d}}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/x^6,x, algorithm="fricas")

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsech(c*x))*sqrt(e*x^2 + d)/x^6, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/x^6,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*arcsech(c*x) + a)/x^6, x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/x^6,x)

[Out]

int((e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/x^6,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{15} \, b {\left (\frac {{\left (2 \, e^{2} x^{5} - d e x^{3} - 3 \, d^{2} x - 5 \, {\left (e^{2} x^{3} + d e x\right )} x^{2}\right )} \sqrt {e x^{2} + d} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right )}{d x^{6}} - 15 \, \int \frac {{\left (15 \, c^{2} d^{2} x^{2} \log \relax (c) + 15 \, {\left (c^{2} d e x^{2} \log \relax (c) - d e \log \relax (c)\right )} x^{2} - 15 \, d^{2} \log \relax (c) + {\left (2 \, c^{2} e^{2} x^{6} - c^{2} d e x^{4} + 3 \, {\left (5 \, d^{2} \log \relax (c) - d^{2}\right )} c^{2} x^{2} - 5 \, {\left (c^{2} e^{2} x^{4} - {\left (3 \, d e \log \relax (c) - d e\right )} c^{2} x^{2} + 3 \, d e \log \relax (c)\right )} x^{2} - 15 \, d^{2} \log \relax (c) + 30 \, {\left (c^{2} d^{2} x^{2} + {\left (c^{2} d e x^{2} - d e\right )} x^{2} - d^{2}\right )} \log \left (\sqrt {x}\right )\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} + 30 \, {\left (c^{2} d^{2} x^{2} + {\left (c^{2} d e x^{2} - d e\right )} x^{2} - d^{2}\right )} \log \left (\sqrt {x}\right )\right )} \sqrt {e x^{2} + d}}{15 \, {\left ({\left (c^{2} d x^{2} - d\right )} x^{6} + {\left (c^{2} d x^{2} - d\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right ) + 6 \, \log \relax (x)\right )}\right )}}\,{d x}\right )} - \frac {{\left (e x^{2} + d\right )}^{\frac {5}{2}} a}{5 \, d x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*arcsech(c*x))/x^6,x, algorithm="maxima")

[Out]

1/15*b*((2*e^2*x^5 - d*e*x^3 - 3*d^2*x - 5*(e^2*x^3 + d*e*x)*x^2)*sqrt(e*x^2 + d)*log(sqrt(c*x + 1)*sqrt(-c*x
+ 1) + 1)/(d*x^6) - 15*integrate(1/15*(15*c^2*d^2*x^2*log(c) + 15*(c^2*d*e*x^2*log(c) - d*e*log(c))*x^2 - 15*d
^2*log(c) + (2*c^2*e^2*x^6 - c^2*d*e*x^4 + 3*(5*d^2*log(c) - d^2)*c^2*x^2 - 5*(c^2*e^2*x^4 - (3*d*e*log(c) - d
*e)*c^2*x^2 + 3*d*e*log(c))*x^2 - 15*d^2*log(c) + 30*(c^2*d^2*x^2 + (c^2*d*e*x^2 - d*e)*x^2 - d^2)*log(sqrt(x)
))*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)) + 30*(c^2*d^2*x^2 + (c^2*d*e*x^2 - d*e)*x^2 - d^2)*log(sqrt(x)))*s
qrt(e*x^2 + d)/((c^2*d*x^2 - d)*x^6 + (c^2*d*x^2 - d)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1) + 6*log(x))), x)
) - 1/5*(e*x^2 + d)^(5/2)*a/(d*x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)^(3/2)*(a + b*acosh(1/(c*x))))/x^6,x)

[Out]

int(((d + e*x^2)^(3/2)*(a + b*acosh(1/(c*x))))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(3/2)*(a+b*asech(c*x))/x**6,x)

[Out]

Integral((a + b*asech(c*x))*(d + e*x**2)**(3/2)/x**6, x)

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