3.2.57 \(\int \frac {a+b \text {sech}^{-1}(c x)}{x^2 \sqrt {d+e x^2}} \, dx\) [157]

Optimal. Leaf size=221 \[ \frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2} \sqrt {d+e x^2}}{d x}-\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{d x}+\frac {b c \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2} E\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{d \sqrt {1+\frac {e x^2}{d}}}-\frac {b \left (c^2 d+e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{c d \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 221, 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, 486, 21, 434, 437, 435, 432, 430} \begin {gather*} -\frac {\sqrt {d+e x^2} \left (a+b \text {sech}^{-1}(c x)\right )}{d x}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{c d \sqrt {d+e x^2}}+\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {d+e x^2} E\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{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} \sqrt {d+e x^2}}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/(x^2*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])/(d*x) - (Sqrt[d + e*x^2]*(a + b*ArcSe
ch[c*x]))/(d*x) + (b*c*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[d + e*x^2]*EllipticE[ArcSin[c*x], -(e/(c^2*d))]
)/(d*Sqrt[1 + (e*x^2)/d]) - (b*(c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 + (e*x^2)/d]*EllipticF[Ar
cSin[c*x], -(e/(c^2*d))])/(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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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 434

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

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 486

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

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

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Mathematica [C] Result contains complex when optimal does not.
time = 22.42, size = 501, normalized size = 2.27 \begin {gather*} -\frac {a \left (\frac {d}{x}+e x\right )+b c \sqrt {\frac {1-c x}{1+c x}} \left (d+e x^2\right )-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (d+e x^2\right )}{x}+\frac {b \left (d+e x^2\right ) \text {sech}^{-1}(c x)}{x}+\frac {b \sqrt {\frac {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)}} \left (i \sqrt {d}+\sqrt {e} x\right ) \left (\left (c \sqrt {d}-i \sqrt {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 )+2 i \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 )}{\sqrt {-\frac {\left (c \sqrt {d}-i \sqrt {e}\right ) (-1+c x)}{\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)}}}}{d \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(d/x + e*x) + b*c*Sqrt[(1 - c*x)/(1 + c*x)]*(d + e*x^2) - (b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + e*x
^2))/x + (b*(d + e*x^2)*ArcSech[c*x])/x + (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[(c*(Sqrt[d] + I*Sqrt[e]*x))/((c*Sq
rt[d] + I*Sqrt[e])*(1 + c*x))]*(I*Sqrt[d] + Sqrt[e]*x)*((c*Sqrt[d] - I*Sqrt[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)*Sqrt[e]*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)))]*Sqrt[(c*(Sqrt[d] - I*Sqrt[e]*x))/((c*Sqrt[d] - I*Sqrt[e])*(1 + c*x))]))/(d*Sqrt[d +
e*x^2]))

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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \frac {a +b \,\mathrm {arcsech}\left (c x \right )}{x^{2} \sqrt {e \,x^{2}+d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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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))/x^2/(e*x^2+d)^(1/2),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 {a + b \operatorname {asech}{\left (c x \right )}}{x^{2} \sqrt {d + e x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asech(c*x))/(x**2*sqrt(d + e*x**2)), 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))/x^2/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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