3.1153 \(\int \frac{a+b \tan ^{-1}(c x)}{x (d+e x^2)} \, dx\)

Optimal. Leaf size=353 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 d}+\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 d}+\frac{i b \text{PolyLog}(2,-i c x)}{2 d}-\frac{i b \text{PolyLog}(2,i c x)}{2 d}-\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 d}+\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.385525, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {4928, 4848, 2391, 4980, 4856, 2402, 2315, 2447} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 d}+\frac{i b \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 d}+\frac{i b \text{PolyLog}(2,-i c x)}{2 d}-\frac{i b \text{PolyLog}(2,i c x)}{2 d}-\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 d}+\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4928

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTan[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[a,
0])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{a+b \tan ^{-1}(c x)}{x \left (d+e x^2\right )} \, dx &=\int \left (\frac{a+b \tan ^{-1}(c x)}{d x}-\frac{e x \left (a+b \tan ^{-1}(c x)\right )}{d \left (d+e x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{a+b \tan ^{-1}(c x)}{x} \, dx}{d}-\frac{e \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx}{d}\\ &=\frac{a \log (x)}{d}+\frac{(i b) \int \frac{\log (1-i c x)}{x} \, dx}{2 d}-\frac{(i b) \int \frac{\log (1+i c x)}{x} \, dx}{2 d}-\frac{e \int \left (-\frac{a+b \tan ^{-1}(c x)}{2 \sqrt{e} \left (\sqrt{-d}-\sqrt{e} x\right )}+\frac{a+b \tan ^{-1}(c x)}{2 \sqrt{e} \left (\sqrt{-d}+\sqrt{e} x\right )}\right ) \, dx}{d}\\ &=\frac{a \log (x)}{d}+\frac{i b \text{Li}_2(-i c x)}{2 d}-\frac{i b \text{Li}_2(i c x)}{2 d}+\frac{\sqrt{e} \int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}-\sqrt{e} x} \, dx}{2 d}-\frac{\sqrt{e} \int \frac{a+b \tan ^{-1}(c x)}{\sqrt{-d}+\sqrt{e} x} \, dx}{2 d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 d}+\frac{i b \text{Li}_2(-i c x)}{2 d}-\frac{i b \text{Li}_2(i c x)}{2 d}-2 \frac{(b c) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 d}+\frac{(b c) \int \frac{\log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{2 d}+\frac{(b c) \int \frac{\log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{2 d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 d}+\frac{i b \text{Li}_2(-i c x)}{2 d}-\frac{i b \text{Li}_2(i c x)}{2 d}+\frac{i b \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{4 d}+\frac{i b \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{4 d}-2 \frac{(i b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{2 d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 d}+\frac{i b \text{Li}_2(-i c x)}{2 d}-\frac{i b \text{Li}_2(i c x)}{2 d}-\frac{i b \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{2 d}+\frac{i b \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{4 d}+\frac{i b \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.241868, size = 429, normalized size = 1.22 \[ \frac{i b \text{PolyLog}\left (2,\frac{\sqrt{e} (-c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )-i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1-i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )+i b \text{PolyLog}\left (2,\frac{\sqrt{e} (1+i c x)}{\sqrt{e}+i c \sqrt{-d}}\right )-i b \text{PolyLog}\left (2,\frac{\sqrt{e} (c x+i)}{c \sqrt{-d}+i \sqrt{e}}\right )+2 i b \text{PolyLog}(2,-i c x)-2 i b \text{PolyLog}(2,i c x)-2 a \log \left (d+e x^2\right )+4 a \log (x)+i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )-i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}-\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )-i b \log (1-i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}-i \sqrt{e}}\right )+i b \log (1+i c x) \log \left (\frac{c \left (\sqrt{-d}+\sqrt{e} x\right )}{c \sqrt{-d}+i \sqrt{e}}\right )}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.208, size = 736, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d*ln(c^2*e*x^2+c^2*d)+a/d*ln(c*x)-1/2*b*arctan(c*x)/d*ln(c^2*e*x^2+c^2*d)+b*arctan(c*x)/d*ln(c*x)+1/4*I
*b/d*ln(c*x-I)*ln((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2))+1/4
*I*b/d*dilog((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1))+1/2*I*b/
d*ln(c*x)*ln(1+I*c*x)-1/4*I*b/d*ln(c*x+I)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2)-c*x-I)/RootOf(e*_Z^2-2*I
*_Z*e+c^2*d-e,index=2))-1/4*I*b/d*dilog((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e
+c^2*d-e,index=1))-1/4*I*b/d*ln(c*x+I)*ln((RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=1)-c*x-I)/RootOf(e*_Z^2-2*I*_Z
*e+c^2*d-e,index=1))-1/4*I*b/d*ln(c*x-I)*ln(c^2*e*x^2+c^2*d)-1/2*I*b/d*dilog(1-I*c*x)+1/2*I*b/d*dilog(1+I*c*x)
+1/4*I*b/d*ln(c*x+I)*ln(c^2*e*x^2+c^2*d)+1/4*I*b/d*ln(c*x-I)*ln((RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1)-c*x+I
)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=1))-1/2*I*b/d*ln(c*x)*ln(1-I*c*x)-1/4*I*b/d*dilog((RootOf(e*_Z^2-2*I*_Z
*e+c^2*d-e,index=2)-c*x-I)/RootOf(e*_Z^2-2*I*_Z*e+c^2*d-e,index=2))+1/4*I*b/d*dilog((RootOf(e*_Z^2+2*I*_Z*e+c^
2*d-e,index=2)-c*x+I)/RootOf(e*_Z^2+2*I*_Z*e+c^2*d-e,index=2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a{\left (\frac{\log \left (e x^{2} + d\right )}{d} - \frac{2 \, \log \left (x\right )}{d}\right )} + 2 \, b \int \frac{\arctan \left (c x\right )}{2 \,{\left (e x^{3} + d x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(log(e*x^2 + d)/d - 2*log(x)/d) + 2*b*integrate(1/2*arctan(c*x)/(e*x^3 + d*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arctan \left (c x\right ) + a}{e x^{3} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(c*x) + a)/(e*x^3 + d*x), 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))/x/(e*x**2+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\right ) + a}{{\left (e x^{2} + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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