3.158 \(\int \frac {a+b \text {sech}^{-1}(c x)}{x^4 \sqrt {d+e x^2}} \, dx\)

Optimal. Leaf size=346 \[ \frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (2 c^2 d-5 e\right ) \sqrt {d+e x^2}}{9 d^2 x}-\frac {2 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (c^2 d-3 e\right ) \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{9 c d^2 \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (2 c^2 d-5 e\right ) \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{9 d^2 \sqrt {\frac {e x^2}{d}+1}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {271, 264, 6301, 12, 580, 583, 524, 426, 424, 421, 419} \[ \frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (2 c^2 d-5 e\right ) \sqrt {d+e x^2}}{9 d^2 x}-\frac {2 b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (c^2 d-3 e\right ) \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{9 c d^2 \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (2 c^2 d-5 e\right ) \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{9 d^2 \sqrt {\frac {e x^2}{d}+1}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/(x^4*Sqrt[d + e*x^2]),x]

[Out]

(b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(9*d*x^3) + (b*(2*c^2*d - 5*e)*Sqrt[(
1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(9*d^2*x) - (Sqrt[d + e*x^2]*(a + b*ArcSech[c*
x]))/(3*d*x^3) + (2*e*Sqrt[d + e*x^2]*(a + b*ArcSech[c*x]))/(3*d^2*x) + (b*c*(2*c^2*d - 5*e)*Sqrt[(1 + c*x)^(-
1)]*Sqrt[1 + c*x]*Sqrt[d + e*x^2]*EllipticE[ArcSin[c*x], -(e/(c^2*d))])/(9*d^2*Sqrt[1 + (e*x^2)/d]) - (2*b*(c^
2*d - 3*e)*(c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcSin[c*x], -(e/(c^2*
d))])/(9*c*d^2*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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(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 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 {a+b \text {sech}^{-1}(c x)}{x^4 \sqrt {d+e x^2}} \, dx &=-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{3 d^2 x^4 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{x^4 \sqrt {1-c^2 x^2}} \, dx}{3 d^2}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d \left (2 c^2 d-5 e\right )-\left (c^2 d-6 e\right ) e x^2}{x^2 \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{9 d^2}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3}+\frac {b \left (2 c^2 d-5 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {d \left (c^2 d-6 e\right ) e-c^2 d \left (2 c^2 d-5 e\right ) e x^2}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{9 d^3}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3}+\frac {b \left (2 c^2 d-5 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b c^2 \left (2 c^2 d-5 e\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}{9 d^2}-\frac {\left (2 b \left (c^2 d-3 e\right ) \left (c^2 d+e\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}{9 d^2}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3}+\frac {b \left (2 c^2 d-5 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b c^2 \left (2 c^2 d-5 e\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}{9 d^2 \sqrt {1+\frac {e x^2}{d}}}-\frac {\left (2 b \left (c^2 d-3 e\right ) \left (c^2 d+e\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}{9 d^2 \sqrt {d+e x^2}}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x^3}+\frac {b \left (2 c^2 d-5 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d^2 x}+\frac {b c \left (2 c^2 d-5 e\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 )}{9 d^2 \sqrt {1+\frac {e x^2}{d}}}-\frac {2 b \left (c^2 d-3 e\right ) \left (c^2 d+e\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 )}{9 c d^2 \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 5.03, size = 612, normalized size = 1.77 \[ \frac {-\frac {3 a \left (d-2 e x^2\right ) \left (d+e x^2\right )}{x^3}+\frac {b \sqrt {\frac {1-c x}{c x+1}} \left (2 c^2 d-5 e\right ) \left (d+e x^2\right )}{x}-\frac {b \sqrt {\frac {1-c x}{c x+1}} \left (\sqrt {e} x+i \sqrt {d}\right ) \sqrt {\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}} \left (2 \sqrt {e} \left (2 i c^2 d-c \sqrt {d} \sqrt {e}-6 i e\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (d c^2+e\right ) (1-c x)}{\left (\sqrt {d} c+i \sqrt {e}\right )^2 (c x+1)}}\right )|\frac {\left (\sqrt {d} c+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )+\left (2 c^3 d^{3/2}-2 i c^2 d \sqrt {e}-5 c \sqrt {d} e+5 i e^{3/2}\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (d c^2+e\right ) (1-c x)}{\left (\sqrt {d} c+i \sqrt {e}\right )^2 (c x+1)}}\right )|\frac {\left (\sqrt {d} c+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )\right )}{\sqrt {-\frac {(c x-1) \left (c \sqrt {d}-i \sqrt {e}\right )}{(c x+1) \left (c \sqrt {d}+i \sqrt {e}\right )}} \sqrt {\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{(c x+1) \left (c \sqrt {d}-i \sqrt {e}\right )}}}+\frac {b c d \sqrt {\frac {1-c x}{c x+1}} \left (d+e x^2\right )}{x^2}+\frac {b d \sqrt {\frac {1-c x}{c x+1}} \left (d+e x^2\right )}{x^3}-\frac {3 b \text {sech}^{-1}(c x) \left (d-2 e x^2\right ) \left (d+e x^2\right )}{x^3}}{9 d^2 \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])/(x^4*Sqrt[d + e*x^2]),x]

[Out]

((b*d*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2))/x^3 + (b*c*d*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2))/x^2 + (b*(2*c
^2*d - 5*e)*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2))/x - (3*a*(d - 2*e*x^2)*(d + e*x^2))/x^3 - (3*b*(d - 2*e*x^2
)*(d + e*x^2)*ArcSech[c*x])/x^3 - (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[(c*(Sqrt[d] + I*Sqrt[e]*x))/((c*Sqrt[d] +
I*Sqrt[e])*(1 + c*x))]*(I*Sqrt[d] + Sqrt[e]*x)*((2*c^3*d^(3/2) - (2*I)*c^2*d*Sqrt[e] - 5*c*Sqrt[d]*e + (5*I)*e
^(3/2))*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*((2*I)*c^2*d - c*Sqrt[d]*Sqrt[e] - (6*I)*e)*Sqrt[e]*EllipticF[I*A
rcSinh[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)))]*Sqrt[
(c*(Sqrt[d] - I*Sqrt[e]*x))/((c*Sqrt[d] - I*Sqrt[e])*(1 + c*x))]))/(9*d^2*Sqrt[d + e*x^2])

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} {\left (b \operatorname {arsech}\left (c x\right ) + a\right )}}{e x^{6} + d x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x^4/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x^2 + d)*(b*arcsech(c*x) + a)/(e*x^6 + d*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsech}\left (c x\right ) + a}{\sqrt {e x^{2} + d} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x^4/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)/(sqrt(e*x^2 + d)*x^4), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arcsech}\left (c x \right )}{x^{4} \sqrt {e \,x^{2}+d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))/x^4/(e*x^2+d)^(1/2),x)

[Out]

int((a+b*arcsech(c*x))/x^4/(e*x^2+d)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x^4/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{x^4\,\sqrt {e\,x^2+d}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))/(x^4*(d + e*x^2)^(1/2)),x)

[Out]

int((a + b*acosh(1/(c*x)))/(x^4*(d + e*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asech}{\left (c x \right )}}{x^{4} \sqrt {d + e x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))/x**4/(e*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asech(c*x))/(x**4*sqrt(d + e*x**2)), x)

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