3.26 \(\int \frac{a+b \tanh ^{-1}(c x^2)}{d+e x} \, dx\)

Optimal. Leaf size=325 \[ -\frac{b \text{PolyLog}\left (2,\frac{\sqrt{-c} (d+e x)}{\sqrt{-c} d-e}\right )}{2 e}+\frac{b \text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-e}\right )}{2 e}-\frac{b \text{PolyLog}\left (2,\frac{\sqrt{-c} (d+e x)}{\sqrt{-c} d+e}\right )}{2 e}+\frac{b \text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+e}\right )}{2 e}+\frac{\log (d+e x) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{e}-\frac{b \log (d+e x) \log \left (\frac{e \left (1-\sqrt{-c} x\right )}{\sqrt{-c} d+e}\right )}{2 e}-\frac{b \log (d+e x) \log \left (-\frac{e \left (\sqrt{-c} x+1\right )}{\sqrt{-c} d-e}\right )}{2 e}+\frac{b \log (d+e x) \log \left (\frac{e \left (1-\sqrt{c} x\right )}{\sqrt{c} d+e}\right )}{2 e}+\frac{b \log (d+e x) \log \left (-\frac{e \left (\sqrt{c} x+1\right )}{\sqrt{c} d-e}\right )}{2 e} \]

[Out]

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

________________________________________________________________________________________

Rubi [F]  time = 0.0626453, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{a+b \tanh ^{-1}\left (c x^2\right )}{d+e x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 17.4703, size = 285, normalized size = 0.88 \[ \frac{a \log (d+e x)}{e}+\frac{b \left (\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-e}\right )-\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-i e}\right )-\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+i e}\right )+\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+e}\right )+2 \tanh ^{-1}\left (c x^2\right ) \log (d+e x)-\log (d+e x) \log \left (\frac{e \left (-\sqrt{c} x+i\right )}{\sqrt{c} d+i e}\right )-\log (d+e x) \log \left (-\frac{e \left (\sqrt{c} x+i\right )}{\sqrt{c} d-i e}\right )+\log (d+e x) \log \left (-\frac{e \left (\sqrt{c} x+1\right )}{\sqrt{c} d-e}\right )+\log (d+e x) \log \left (\frac{e-\sqrt{c} e x}{\sqrt{c} d+e}\right )\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[d + e*x])/e + (b*(2*ArcTanh[c*x^2]*Log[d + e*x] - Log[(e*(I - Sqrt[c]*x))/(Sqrt[c]*d + I*e)]*Log[d + e*
x] - Log[-((e*(I + Sqrt[c]*x))/(Sqrt[c]*d - I*e))]*Log[d + e*x] + Log[-((e*(1 + Sqrt[c]*x))/(Sqrt[c]*d - e))]*
Log[d + e*x] + Log[d + e*x]*Log[(e - Sqrt[c]*e*x)/(Sqrt[c]*d + e)] + PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d
- e)] - PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - I*e)] - PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + I*e)]
+ PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + e)]))/(2*e)

________________________________________________________________________________________

Maple [A]  time = 0.07, size = 362, normalized size = 1.1 \begin{align*}{\frac{a\ln \left ( ex+d \right ) }{e}}+{\frac{b\ln \left ( ex+d \right ){\it Artanh} \left ( c{x}^{2} \right ) }{e}}-{\frac{b\ln \left ( ex+d \right ) }{2\,e}\ln \left ({ \left ( e\sqrt{-c}- \left ( ex+d \right ) c+cd \right ) \left ( e\sqrt{-c}+cd \right ) ^{-1}} \right ) }-{\frac{b\ln \left ( ex+d \right ) }{2\,e}\ln \left ({ \left ( e\sqrt{-c}+ \left ( ex+d \right ) c-cd \right ) \left ( e\sqrt{-c}-cd \right ) ^{-1}} \right ) }-{\frac{b}{2\,e}{\it dilog} \left ({ \left ( e\sqrt{-c}- \left ( ex+d \right ) c+cd \right ) \left ( e\sqrt{-c}+cd \right ) ^{-1}} \right ) }-{\frac{b}{2\,e}{\it dilog} \left ({ \left ( e\sqrt{-c}+ \left ( ex+d \right ) c-cd \right ) \left ( e\sqrt{-c}-cd \right ) ^{-1}} \right ) }+{\frac{b\ln \left ( ex+d \right ) }{2\,e}\ln \left ({ \left ( e\sqrt{c}- \left ( ex+d \right ) c+cd \right ) \left ( e\sqrt{c}+cd \right ) ^{-1}} \right ) }+{\frac{b\ln \left ( ex+d \right ) }{2\,e}\ln \left ({ \left ( e\sqrt{c}+ \left ( ex+d \right ) c-cd \right ) \left ( e\sqrt{c}-cd \right ) ^{-1}} \right ) }+{\frac{b}{2\,e}{\it dilog} \left ({ \left ( e\sqrt{c}- \left ( ex+d \right ) c+cd \right ) \left ( e\sqrt{c}+cd \right ) ^{-1}} \right ) }+{\frac{b}{2\,e}{\it dilog} \left ({ \left ( e\sqrt{c}+ \left ( ex+d \right ) c-cd \right ) \left ( e\sqrt{c}-cd \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln(e*x+d)/e+b*ln(e*x+d)/e*arctanh(c*x^2)-1/2*b/e*ln(e*x+d)*ln((e*(-c)^(1/2)-(e*x+d)*c+c*d)/(e*(-c)^(1/2)+c*d
))-1/2*b/e*ln(e*x+d)*ln((e*(-c)^(1/2)+(e*x+d)*c-c*d)/(e*(-c)^(1/2)-c*d))-1/2*b/e*dilog((e*(-c)^(1/2)-(e*x+d)*c
+c*d)/(e*(-c)^(1/2)+c*d))-1/2*b/e*dilog((e*(-c)^(1/2)+(e*x+d)*c-c*d)/(e*(-c)^(1/2)-c*d))+1/2*b/e*ln(e*x+d)*ln(
(e*c^(1/2)-(e*x+d)*c+c*d)/(e*c^(1/2)+c*d))+1/2*b/e*ln(e*x+d)*ln((e*c^(1/2)+(e*x+d)*c-c*d)/(e*c^(1/2)-c*d))+1/2
*b/e*dilog((e*c^(1/2)-(e*x+d)*c+c*d)/(e*c^(1/2)+c*d))+1/2*b/e*dilog((e*c^(1/2)+(e*x+d)*c-c*d)/(e*c^(1/2)-c*d))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, b \int \frac{\log \left (c x^{2} + 1\right ) - \log \left (-c x^{2} + 1\right )}{e x + d}\,{d x} + \frac{a \log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{artanh}\left (c x^{2}\right ) + a}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctanh(c*x^2) + a)/(e*x + d), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{artanh}\left (c x^{2}\right ) + a}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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