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

Optimal. Leaf size=501 \[ \frac {\log (d+e x) \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{e}+\frac {b c \text {Li}_2\left (\frac {\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \text {Li}_2\left (\frac {\sqrt {-\sqrt {-c^2}} (d+e x)}{\sqrt {-\sqrt {-c^2}} d-e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \text {Li}_2\left (\frac {\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \text {Li}_2\left (\frac {\sqrt {-\sqrt {-c^2}} (d+e x)}{\sqrt {-\sqrt {-c^2}} d+e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \log (d+e x) \log \left (\frac {e \left (1-\sqrt [4]{-c^2} x\right )}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \log (d+e x) \log \left (-\frac {e \left (\sqrt [4]{-c^2} x+1\right )}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \log (d+e x) \log \left (\frac {e \left (1-\sqrt {-\sqrt {-c^2}} x\right )}{\sqrt {-\sqrt {-c^2}} d+e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \log (d+e x) \log \left (-\frac {e \left (\sqrt {-\sqrt {-c^2}} x+1\right )}{\sqrt {-\sqrt {-c^2}} d-e}\right )}{2 \sqrt {-c^2} e} \]

[Out]

(a+b*arctan(c*x^2))*ln(e*x+d)/e+1/2*b*c*ln(e*(1-(-c^2)^(1/4)*x)/((-c^2)^(1/4)*d+e))*ln(e*x+d)/e/(-c^2)^(1/2)+1
/2*b*c*ln(-e*(1+(-c^2)^(1/4)*x)/((-c^2)^(1/4)*d-e))*ln(e*x+d)/e/(-c^2)^(1/2)-1/2*b*c*ln(e*x+d)*ln(e*(1-x*(-(-c
^2)^(1/2))^(1/2))/(e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)-1/2*b*c*ln(e*x+d)*ln(-e*(1+x*(-(-c^2)^(1/2))^(1/
2))/(-e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)+1/2*b*c*polylog(2,(-c^2)^(1/4)*(e*x+d)/((-c^2)^(1/4)*d-e))/e/
(-c^2)^(1/2)+1/2*b*c*polylog(2,(-c^2)^(1/4)*(e*x+d)/((-c^2)^(1/4)*d+e))/e/(-c^2)^(1/2)-1/2*b*c*polylog(2,(e*x+
d)*(-(-c^2)^(1/2))^(1/2)/(-e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)-1/2*b*c*polylog(2,(e*x+d)*(-(-c^2)^(1/2)
)^(1/2)/(e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)

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Rubi [F]  time = 0.06, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [C]  time = 34.98, size = 326, normalized size = 0.65 \[ \frac {a \log (d+e x)}{e}+\frac {b \left (2 \tan ^{-1}\left (c x^2\right ) \log (d+e x)+i \left (\text {Li}_2\left (\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt [4]{-1} e}\right )+\text {Li}_2\left (\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt [4]{-1} e}\right )-\text {Li}_2\left (\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-(-1)^{3/4} e}\right )-\text {Li}_2\left (\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+(-1)^{3/4} e}\right )+\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt [4]{-1} e}\right )+\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt [4]{-1} e}\right )-\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-(-1)^{3/4} e}\right )-\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+(-1)^{3/4} e}\right )\right )\right )}{2 e} \]

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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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x^{2}\right ) + a}{e x + d}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arctan \left (c x^{2}\right ) + a}{e x + d}\,{d x} \]

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)

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maple [C]  time = 0.14, size = 138, normalized size = 0.28 \[ \frac {a \ln \left (e x +d \right )}{e}+\frac {b \ln \left (e x +d \right ) \arctan \left (c \,x^{2}\right )}{e}-\frac {b e \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} \textit {\_Z}^{4}-4 c^{2} d \,\textit {\_Z}^{3}+6 c^{2} d^{2} \textit {\_Z}^{2}-4 c^{2} d^{3} \textit {\_Z} +c^{2} d^{4}+e^{4}\right )}{\sum }\frac {\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\dilog \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} d +d^{2}}\right )}{2 c} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ 2 \, b \int \frac {\arctan \left (c x^{2}\right )}{2 \, {\left (e x + d\right )}}\,{d x} + \frac {a \log \left (e x + d\right )}{e} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x^2\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))/(d + e*x),x)

[Out]

int((a + b*atan(c*x^2))/(d + e*x), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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