3.2.38 \(\int \frac {\sqrt {d+e x^2} (a+b \text {sech}^{-1}(c x))}{x^4} \, dx\) [138]

Optimal. Leaf size=312 \[ \frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 x^3}+\frac {2 b \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 d x}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 b c \left (c^2 d+2 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2} E\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{9 d \sqrt {1+\frac {e x^2}{d}}}-\frac {b \left (c^2 d+e\right ) \left (2 c^2 d+3 e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{9 c d \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {270, 6436, 12, 485, 597, 538, 437, 435, 432, 430} \begin {gather*} -\frac {\left (d+e x^2\right )^{3/2} \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} \left (c^2 d+e\right ) \left (2 c^2 d+3 e\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{9 c d \sqrt {d+e x^2}}+\frac {2 b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (c^2 d+2 e\right ) \sqrt {d+e x^2} E\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{9 d \sqrt {\frac {e x^2}{d}+1}}+\frac {2 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}}{9 d x}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(9*x^3) + (2*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])/(9*d*x) - ((d + e*x^2)^(3/2)*(a + b*ArcSech[c*x]
))/(3*d*x^3) + (2*b*c*(c^2*d + 2*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*Sqrt[1 + (e*x^2)/d]) - (b*(c^2*d + e)*(2*c^2*d + 3*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*S
qrt[1 + (e*x^2)/d]*EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(9*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 270

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 430

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

Rule 432

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

Rule 435

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

Rule 437

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

Rule 485

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 538

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 597

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 6436

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 {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{x^4} \, dx &=-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int -\frac {\left (d+e x^2\right )^{3/2}}{3 d x^4 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\left (d+e x^2\right )^{3/2}}{x^4 \sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {2 d \left (c^2 d+2 e\right )+e \left (c^2 d+3 e\right ) x^2}{x^2 \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{9 d}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 x^3}+\frac {2 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}}{9 d x}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d e \left (c^2 d+3 e\right )+2 c^2 d e \left (c^2 d+2 e\right ) 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 x^3}+\frac {2 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}}{9 d x}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {\left (2 b c^2 \left (c^2 d+2 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}-\frac {\left (b \left (c^2 d+e\right ) \left (2 c^2 d+3 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}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{9 x^3}+\frac {2 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}}{9 d x}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {\left (2 b c^2 \left (c^2 d+2 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 \sqrt {1+\frac {e x^2}{d}}}-\frac {\left (b \left (c^2 d+e\right ) \left (2 c^2 d+3 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 \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 x^3}+\frac {2 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}}{9 d x}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \text {sech}^{-1}(c x)\right )}{3 d x^3}+\frac {2 b c \left (c^2 d+2 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 \sqrt {1+\frac {e x^2}{d}}}-\frac {b \left (c^2 d+e\right ) \left (2 c^2 d+3 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 \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 22.11, size = 576, normalized size = 1.85 \begin {gather*} \frac {\frac {b \sqrt {\frac {1-c x}{1+c x}} \left (d+e x^2\right )}{x^3}+\frac {b c \sqrt {\frac {1-c x}{1+c x}} \left (d+e x^2\right )}{x^2}+\frac {2 b \left (c^2 d+2 e\right ) \sqrt {\frac {1-c x}{1+c x}} \left (d+e x^2\right )}{d x}-\frac {3 a \left (d+e x^2\right )^2}{d x^3}-\frac {3 b \left (d+e x^2\right )^2 \text {sech}^{-1}(c x)}{d x^3}-\frac {2 i b \left (c \sqrt {d}-i \sqrt {e}\right )^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \sqrt {\frac {c \left (\sqrt {d}-i \sqrt {e} x\right )}{\left (c \sqrt {d}-i \sqrt {e}\right ) (1+c x)}} \sqrt {\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{\left (c \sqrt {d}+i \sqrt {e}\right ) (1+c x)}} \left (\left (c^2 d+2 e\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (c^2 d+e\right ) (1-c x)}{\left (c \sqrt {d}+i \sqrt {e}\right )^2 (1+c x)}}\right )|\frac {\left (c \sqrt {d}+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )+\left (2 i c \sqrt {d}-3 \sqrt {e}\right ) \sqrt {e} F\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (c^2 d+e\right ) (1-c x)}{\left (c \sqrt {d}+i \sqrt {e}\right )^2 (1+c x)}}\right )|\frac {\left (c \sqrt {d}+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )\right )}{c d \sqrt {-\frac {\left (c \sqrt {d}-i \sqrt {e}\right ) (-1+c x)}{\left (c \sqrt {d}+i \sqrt {e}\right ) (1+c x)}}}}{9 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2))/x^3 + (b*c*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2))/x^2 + (2*b*(c^2*d
+ 2*e)*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2))/(d*x) - (3*a*(d + e*x^2)^2)/(d*x^3) - (3*b*(d + e*x^2)^2*ArcSec
h[c*x])/(d*x^3) - ((2*I)*b*(c*Sqrt[d] - I*Sqrt[e])^2*Sqrt[(1 - c*x)/(1 + c*x)]*(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))]*((c^2*d + 2*e)*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*I)*c*Sqrt[d] - 3*Sqrt[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]))/(c*d*Sqrt[-(((c*Sqrt[d] - I*Sqrt[e])*(-1 + c*x))/((c*Sqrt[d] + I*Sqrt[e])*(1 + c*x)))]))
/(9*Sqrt[d + e*x^2])

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Maple [F]
time = 0.38, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,\mathrm {arcsech}\left (c x \right )\right ) \sqrt {e \,x^{2}+d}}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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