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

Optimal. Leaf size=501 \[ \frac{b c \text{PolyLog}\left (2,\frac{\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt{-c^2} e}-\frac{b c \text{PolyLog}\left (2,\frac{\sqrt{-\sqrt{-c^2}} (d+e x)}{\sqrt{-\sqrt{-c^2}} d-e}\right )}{2 \sqrt{-c^2} e}+\frac{b c \text{PolyLog}\left (2,\frac{\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt{-c^2} e}-\frac{b c \text{PolyLog}\left (2,\frac{\sqrt{-\sqrt{-c^2}} (d+e x)}{\sqrt{-\sqrt{-c^2}} d+e}\right )}{2 \sqrt{-c^2} e}+\frac{\log (d+e x) \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{e}+\frac{b c \log (d+e x) \log \left (\frac{e \left (1-\sqrt [4]{-c^2} x\right )}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt{-c^2} e}+\frac{b c \log (d+e x) \log \left (-\frac{e \left (\sqrt [4]{-c^2} x+1\right )}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt{-c^2} e}-\frac{b c \log (d+e x) \log \left (\frac{e \left (1-\sqrt{-\sqrt{-c^2}} x\right )}{\sqrt{-\sqrt{-c^2}} d+e}\right )}{2 \sqrt{-c^2} e}-\frac{b c \log (d+e x) \log \left (-\frac{e \left (\sqrt{-\sqrt{-c^2}} x+1\right )}{\sqrt{-\sqrt{-c^2}} d-e}\right )}{2 \sqrt{-c^2} e} \]

[Out]

((a + b*ArcTan[c*x^2])*Log[d + e*x])/e + (b*c*Log[(e*(1 - (-c^2)^(1/4)*x))/((-c^2)^(1/4)*d + e)]*Log[d + e*x])
/(2*Sqrt[-c^2]*e) + (b*c*Log[-((e*(1 + (-c^2)^(1/4)*x))/((-c^2)^(1/4)*d - e))]*Log[d + e*x])/(2*Sqrt[-c^2]*e)
- (b*c*Log[(e*(1 - Sqrt[-Sqrt[-c^2]]*x))/(Sqrt[-Sqrt[-c^2]]*d + e)]*Log[d + e*x])/(2*Sqrt[-c^2]*e) - (b*c*Log[
-((e*(1 + Sqrt[-Sqrt[-c^2]]*x))/(Sqrt[-Sqrt[-c^2]]*d - e))]*Log[d + e*x])/(2*Sqrt[-c^2]*e) + (b*c*PolyLog[2, (
(-c^2)^(1/4)*(d + e*x))/((-c^2)^(1/4)*d - e)])/(2*Sqrt[-c^2]*e) - (b*c*PolyLog[2, (Sqrt[-Sqrt[-c^2]]*(d + e*x)
)/(Sqrt[-Sqrt[-c^2]]*d - e)])/(2*Sqrt[-c^2]*e) + (b*c*PolyLog[2, ((-c^2)^(1/4)*(d + e*x))/((-c^2)^(1/4)*d + e)
])/(2*Sqrt[-c^2]*e) - (b*c*PolyLog[2, (Sqrt[-Sqrt[-c^2]]*(d + e*x))/(Sqrt[-Sqrt[-c^2]]*d + e)])/(2*Sqrt[-c^2]*
e)

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Rubi [F]  time = 0.0637799, 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 = {} \[ \int \frac{a+b \tan ^{-1}\left (c x^2\right )}{d+e x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 32.123, size = 326, normalized size = 0.65 \[ \frac{a \log (d+e x)}{e}+\frac{b \left (2 \tan ^{-1}\left (c x^2\right ) \log (d+e x)+i \left (\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt [4]{-1} e}\right )+\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt [4]{-1} e}\right )-\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-(-1)^{3/4} e}\right )-\text{PolyLog}\left (2,\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+(-1)^{3/4} e}\right )+\log (d+e x) \log \left (1-\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt [4]{-1} e}\right )+\log (d+e x) \log \left (1-\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt [4]{-1} e}\right )-\log (d+e x) \log \left (1-\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-(-1)^{3/4} e}\right )-\log (d+e x) \log \left (1-\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+(-1)^{3/4} e}\right )\right )\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[d + e*x])/e + (b*(2*ArcTan[c*x^2]*Log[d + e*x] + I*(Log[d + e*x]*Log[1 - (Sqrt[c]*(d + e*x))/(Sqrt[c]*d
- (-1)^(1/4)*e)] + Log[d + e*x]*Log[1 - (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + (-1)^(1/4)*e)] - Log[d + e*x]*Log[1
- (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - (-1)^(3/4)*e)] - Log[d + e*x]*Log[1 - (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + (-1)
^(3/4)*e)] + PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - (-1)^(1/4)*e)] + PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt
[c]*d + (-1)^(1/4)*e)] - PolyLog[2, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - (-1)^(3/4)*e)] - PolyLog[2, (Sqrt[c]*(d +
e*x))/(Sqrt[c]*d + (-1)^(3/4)*e)])))/(2*e)

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Maple [C]  time = 0.125, size = 138, normalized size = 0.3 \begin{align*}{\frac{a\ln \left ( ex+d \right ) }{e}}+{\frac{b\ln \left ( ex+d \right ) \arctan \left ( c{x}^{2} \right ) }{e}}-{\frac{be}{2\,c}\sum _{{\it \_R1}={\it RootOf} \left ({c}^{2}{{\it \_Z}}^{4}-4\,{c}^{2}d{{\it \_Z}}^{3}+6\,{c}^{2}{d}^{2}{{\it \_Z}}^{2}-4\,{c}^{2}{d}^{3}{\it \_Z}+{c}^{2}{d}^{4}+{e}^{4} \right ) }{\frac{1}{{{\it \_R1}}^{2}-2\,{\it \_R1}\,d+{d}^{2}} \left ( \ln \left ( ex+d \right ) \ln \left ({\frac{-ex+{\it \_R1}-d}{{\it \_R1}}} \right ) +{\it dilog} \left ({\frac{-ex+{\it \_R1}-d}{{\it \_R1}}} \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln(e*x+d)/e+b*ln(e*x+d)/e*arctan(c*x^2)-1/2*b*e/c*sum(1/(_R1^2-2*_R1*d+d^2)*(ln(e*x+d)*ln((-e*x+_R1-d)/_R1)+
dilog((-e*x+_R1-d)/_R1)),_R1=RootOf(_Z^4*c^2-4*_Z^3*c^2*d+6*_Z^2*c^2*d^2-4*_Z*c^2*d^3+c^2*d^4+e^4))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, b \int \frac{\arctan \left (c x^{2}\right )}{2 \,{\left (e x + d\right )}}\,{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*arctan(c*x^2))/(e*x+d),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arctan \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*arctan(c*x^2))/(e*x+d),x, algorithm="fricas")

[Out]

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

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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*atan(c*x**2))/(e*x+d),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arctan \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*arctan(c*x^2))/(e*x+d),x, algorithm="giac")

[Out]

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