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

Optimal. Leaf size=229 \[ \frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )}{e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )}{e}-\frac {\log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{e}-\frac {b \text {Li}_2\left (-\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}-\frac {b \text {Li}_2\left (-\frac {\left (e+\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}+\frac {b \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right )}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.93, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6287, 2518} \[ -\frac {b \text {PolyLog}\left (2,-\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}-\frac {b \text {PolyLog}\left (2,-\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}+\frac {b \text {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(c x)}\right )}{2 e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )}{e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )}{e}-\frac {\log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/(d + e*x),x]

[Out]

-(((a + b*ArcSech[c*x])*Log[1 + E^(-2*ArcSech[c*x])])/e) + ((a + b*ArcSech[c*x])*Log[1 + (e - Sqrt[-(c^2*d^2)
+ e^2])/(c*d*E^ArcSech[c*x])])/e + ((a + b*ArcSech[c*x])*Log[1 + (e + Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c
*x])])/e + (b*PolyLog[2, -E^(-2*ArcSech[c*x])])/(2*e) - (b*PolyLog[2, -((e - Sqrt[-(c^2*d^2) + e^2])/(c*d*E^Ar
cSech[c*x]))])/e - (b*PolyLog[2, -((e + Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x]))])/e

Rule 2518

Int[Log[v_]*(u_), x_Symbol] :> With[{w = DerivativeDivides[v, u*(1 - v), x]}, Simp[w*PolyLog[2, 1 - v], x] /;
!FalseQ[w]]

Rule 6287

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*ArcSech[c*x])*Log[1 +
(e - Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])])/e, x] + (Dist[b/e, Int[(Sqrt[(1 - c*x)/(1 + c*x)]*Log[1 +
(e - Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])])/(x*(1 - c*x)), x], x] + Dist[b/e, Int[(Sqrt[(1 - c*x)/(1 +
c*x)]*Log[1 + (e + Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])])/(x*(1 - c*x)), x], x] - Dist[b/e, Int[(Sqrt
[(1 - c*x)/(1 + c*x)]*Log[1 + 1/E^(2*ArcSech[c*x])])/(x*(1 - c*x)), x], x] + Simp[((a + b*ArcSech[c*x])*Log[1
+ (e + Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])])/e, x] - Simp[((a + b*ArcSech[c*x])*Log[1 + 1/E^(2*ArcSec
h[c*x])])/e, x]) /; FreeQ[{a, b, c, d, e}, x]

Rubi steps

\begin {align*} \int \frac {a+b \text {sech}^{-1}(c x)}{d+e x} \, dx &=-\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+e^{-2 \text {sech}^{-1}(c x)}\right )}{e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}-\frac {b \int \frac {\sqrt {\frac {1-c x}{1+c x}} \log \left (1+e^{-2 \text {sech}^{-1}(c x)}\right )}{x (1-c x)} \, dx}{e}+\frac {b \int \frac {\sqrt {\frac {1-c x}{1+c x}} \log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{x (1-c x)} \, dx}{e}+\frac {b \int \frac {\sqrt {\frac {1-c x}{1+c x}} \log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{x (1-c x)} \, dx}{e}\\ &=-\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+e^{-2 \text {sech}^{-1}(c x)}\right )}{e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}+\frac {\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}+\frac {b \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right )}{2 e}-\frac {b \text {Li}_2\left (-\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}-\frac {b \text {Li}_2\left (-\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )}{e}\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 393, normalized size = 1.72 \[ \frac {a \log (d+e x)}{e}+\frac {b \left (\text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right )-2 \left (\text {Li}_2\left (\frac {\left (\sqrt {e^2-c^2 d^2}-e\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )+\text {Li}_2\left (-\frac {\left (e+\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}\right )-\text {sech}^{-1}(c x) \log \left (\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )-\text {sech}^{-1}(c x) \log \left (\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )+2 i \sin ^{-1}\left (\frac {\sqrt {\frac {e}{c d}+1}}{\sqrt {2}}\right ) \log \left (\frac {\left (e-\sqrt {e^2-c^2 d^2}\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )-2 i \sin ^{-1}\left (\frac {\sqrt {\frac {e}{c d}+1}}{\sqrt {2}}\right ) \log \left (\frac {\left (\sqrt {e^2-c^2 d^2}+e\right ) e^{-\text {sech}^{-1}(c x)}}{c d}+1\right )-4 i \sin ^{-1}\left (\frac {\sqrt {\frac {e}{c d}+1}}{\sqrt {2}}\right ) \tanh ^{-1}\left (\frac {(e-c d) \tanh \left (\frac {1}{2} \text {sech}^{-1}(c x)\right )}{\sqrt {e^2-c^2 d^2}}\right )+\text {sech}^{-1}(c x) \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )\right )\right )}{2 e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Log[d + e*x])/e + (b*(PolyLog[2, -E^(-2*ArcSech[c*x])] - 2*((-4*I)*ArcSin[Sqrt[1 + e/(c*d)]/Sqrt[2]]*ArcTan
h[((-(c*d) + e)*Tanh[ArcSech[c*x]/2])/Sqrt[-(c^2*d^2) + e^2]] + ArcSech[c*x]*Log[1 + E^(-2*ArcSech[c*x])] - Ar
cSech[c*x]*Log[1 + (e - Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])] + (2*I)*ArcSin[Sqrt[1 + e/(c*d)]/Sqrt[2]
]*Log[1 + (e - Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])] - ArcSech[c*x]*Log[1 + (e + Sqrt[-(c^2*d^2) + e^2
])/(c*d*E^ArcSech[c*x])] - (2*I)*ArcSin[Sqrt[1 + e/(c*d)]/Sqrt[2]]*Log[1 + (e + Sqrt[-(c^2*d^2) + e^2])/(c*d*E
^ArcSech[c*x])] + PolyLog[2, (-e + Sqrt[-(c^2*d^2) + e^2])/(c*d*E^ArcSech[c*x])] + PolyLog[2, -((e + Sqrt[-(c^
2*d^2) + e^2])/(c*d*E^ArcSech[c*x]))])))/(2*e)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsech}\left (c x\right ) + a}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsech(c*x) + a)/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.68, size = 514, normalized size = 2.24 \[ \frac {a \ln \left (c x e +c d \right )}{e}+\frac {b \,\mathrm {arcsech}\left (c x \right ) \ln \left (\frac {-c d \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \,\mathrm {arcsech}\left (c x \right ) \ln \left (\frac {c d \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \dilog \left (\frac {c d \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b \dilog \left (\frac {-c d \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{e}-\frac {b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{e}-\frac {b \dilog \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{e}-\frac {b \dilog \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln(c*e*x+c*d)/e+b/e*arcsech(c*x)*ln((-c*d*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))+(-c^2*d^2+e^2)^(1/2)-e)/(
-e+(-c^2*d^2+e^2)^(1/2)))+b/e*arcsech(c*x)*ln((c*d*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))+(-c^2*d^2+e^2)^(1/
2)+e)/(e+(-c^2*d^2+e^2)^(1/2)))+b/e*dilog((c*d*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))+(-c^2*d^2+e^2)^(1/2)+e
)/(e+(-c^2*d^2+e^2)^(1/2)))+b/e*dilog((-c*d*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))+(-c^2*d^2+e^2)^(1/2)-e)/(
-e+(-c^2*d^2+e^2)^(1/2)))-b/e*arcsech(c*x)*ln(1+I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))-b/e*arcsech(c*x)*l
n(1-I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))-b/e*dilog(1+I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))-b/e*di
log(1-I*(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(e*x + d), x) + a*log(e*x + d)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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