3.1.91 \(\int \frac {(d+e x^2) (a+b \text {sech}^{-1}(c x))}{x^2} \, dx\) [91]

Optimal. Leaf size=96 \[ \frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{x}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \text {ArcSin}(c x)}{c} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {14, 6436, 462, 222} \begin {gather*} -\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \text {ArcSin}(c x)}{c}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 222

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

Rule 462

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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 {\left (d+e x^2\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{x^2} \, dx &=-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text {sech}^{-1}(c x)\right )+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-d+e x^2}{x^2 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{x}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text {sech}^{-1}(c x)\right )+\left (b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b d \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{x}-\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text {sech}^{-1}(c x)\right )+\frac {b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sin ^{-1}(c x)}{c}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 124, normalized size = 1.29 \begin {gather*} -\frac {a d}{x}+a e x+b d \left (c+\frac {1}{x}\right ) \sqrt {\frac {1-c x}{1+c x}}-\frac {b d \text {sech}^{-1}(c x)}{x}+b e x \text {sech}^{-1}(c x)-\frac {2 b e \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \text {ArcTan}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c-c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.19, size = 114, normalized size = 1.19

method	result	size
derivativedivides	\(c \left (\frac {a \left (e c x -\frac {d c}{x}\right )}{c^{2}}+\frac {b \left (\mathrm {arcsech}\left (c x \right ) e c x -\frac {\mathrm {arcsech}\left (c x \right ) d c}{x}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (\sqrt {-c^{2} x^{2}+1}\, c^{2} d +\arcsin \left (c x \right ) e c x \right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}\right )\)	\(114\)
default	\(c \left (\frac {a \left (e c x -\frac {d c}{x}\right )}{c^{2}}+\frac {b \left (\mathrm {arcsech}\left (c x \right ) e c x -\frac {\mathrm {arcsech}\left (c x \right ) d c}{x}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (\sqrt {-c^{2} x^{2}+1}\, c^{2} d +\arcsin \left (c x \right ) e c x \right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}\right )\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 68, normalized size = 0.71 \begin {gather*} {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} b d + a x e + \frac {{\left (c x \operatorname {arsech}\left (c x\right ) - \arctan \left (\sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )\right )} b e}{c} - \frac {a d}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (56) = 112\).
time = 0.47, size = 224, normalized size = 2.33 \begin {gather*} \frac {b c^{2} d x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + a c x^{2} \cosh \left (1\right ) + a c x^{2} \sinh \left (1\right ) - a c d - 2 \, {\left (b x \cosh \left (1\right ) + b x \sinh \left (1\right )\right )} \arctan \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) + {\left (b c d x - b c x \cosh \left (1\right ) - b c x \sinh \left (1\right )\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + {\left (b c d x - b c d + {\left (b c x^{2} - b c x\right )} \cosh \left (1\right ) + {\left (b c x^{2} - b c x\right )} \sinh \left (1\right )\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c^2*d*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + a*c*x^2*cosh(1) + a*c*x^2*sinh(1) - a*c*d - 2*(b*x*cosh(1) + b*x*s
inh(1))*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/(c*x)) + (b*c*d*x - b*c*x*cosh(1) - b*c*x*sinh(1))*log
((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) + (b*c*d*x - b*c*d + (b*c*x^2 - b*c*x)*cosh(1) + (b*c*x^2 - b*c*x
)*sinh(1))*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)))/(c*x)

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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 ) \left (d + e x^{2}\right )}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [B]
time = 1.81, size = 98, normalized size = 1.02 \begin {gather*} a\,e\,x-\frac {a\,d}{x}+b\,c\,d\,\left (\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}-\frac {\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{c\,x}\right )+\frac {b\,e\,\mathrm {atan}\left (\frac {1}{\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}}\right )}{c}+b\,e\,x\,\mathrm {acosh}\left (\frac {1}{c\,x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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