3.1251 \(\int \frac {(d+e x^2) (a+b \tan ^{-1}(c x))^2}{x} \, dx\)

Optimal. Leaf size=217 \[ \frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-i b d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {a b e x}{c}+\frac {b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (\frac {2}{i c x+1}-1\right )-\frac {b^2 e x \tan ^{-1}(c x)}{c} \]

[Out]

-a*b*e*x/c-b^2*e*x*arctan(c*x)/c+1/2*e*(a+b*arctan(c*x))^2/c^2+1/2*e*x^2*(a+b*arctan(c*x))^2-2*d*(a+b*arctan(c
*x))^2*arctanh(-1+2/(1+I*c*x))+1/2*b^2*e*ln(c^2*x^2+1)/c^2-I*b*d*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))+I*
b*d*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-1/2*b^2*d*polylog(3,1-2/(1+I*c*x))+1/2*b^2*d*polylog(3,-1+2/(1
+I*c*x))

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Rubi [A]  time = 0.44, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {4980, 4850, 4988, 4884, 4994, 6610, 4852, 4916, 4846, 260} \[ -i b d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} b^2 d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+2 d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {a b e x}{c}+\frac {b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {b^2 e x \tan ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*e*x)/c) - (b^2*e*x*ArcTan[c*x])/c + (e*(a + b*ArcTan[c*x])^2)/(2*c^2) + (e*x^2*(a + b*ArcTan[c*x])^2)/2
+ 2*d*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (b^2*e*Log[1 + c^2*x^2])/(2*c^2) - I*b*d*(a + b*ArcT
an[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - (b^2*d*Pol
yLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d*PolyLog[3, -1 + 2/(1 + I*c*x)])/2

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 4850

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4884

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

Rule 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 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 4988

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

Rule 4994

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

Rule 6610

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

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+e x \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+e \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-(4 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(b c e) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+(2 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(2 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {(b e) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}+\frac {(b e) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c}\\ &=-\frac {a b e x}{c}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (i b^2 c d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {\left (b^2 e\right ) \int \tan ^{-1}(c x) \, dx}{c}\\ &=-\frac {a b e x}{c}-\frac {b^2 e x \tan ^{-1}(c x)}{c}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\left (b^2 e\right ) \int \frac {x}{1+c^2 x^2} \, dx\\ &=-\frac {a b e x}{c}-\frac {b^2 e x \tan ^{-1}(c x)}{c}+\frac {e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}+\frac {1}{2} e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\frac {b^2 e \log \left (1+c^2 x^2\right )}{2 c^2}-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 263, normalized size = 1.21 \[ a^2 d \log (x)+\frac {1}{2} a^2 e x^2+\frac {a b e \left (\left (c^2 x^2+1\right ) \tan ^{-1}(c x)-c x\right )}{c^2}+i a b d (\text {Li}_2(-i c x)-\text {Li}_2(i c x))+\frac {b^2 e \left (\log \left (c^2 x^2+1\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x)\right )}{2 c^2}+b^2 d \left (i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+\frac {2}{3} i \tan ^{-1}(c x)^3+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\frac {i \pi ^3}{24}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*e*x^2)/2 + (a*b*e*(-(c*x) + (1 + c^2*x^2)*ArcTan[c*x]))/c^2 + a^2*d*Log[x] + (b^2*e*(-2*c*x*ArcTan[c*x] +
(1 + c^2*x^2)*ArcTan[c*x]^2 + Log[1 + c^2*x^2]))/(2*c^2) + I*a*b*d*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x])
+ b^2*d*((-1/24*I)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - ArcTan[c*
x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog
[2, -E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3, -E^((2*I)*ArcTan[c*x])]/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [C]  time = 4.14, size = 1284, normalized size = 5.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b/c^2*arctan(c*x)*e+1/2*b^2/c^2*arctan(c*x)^2*e+1/2*b^2*arctan(c*x)^2*x^2*e+b^2*arctan(c*x)^2*d*ln(c*x)-b^2/
c^2*e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-b^2*d*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+b^2*d*arctan(c*x)^2*ln(1
+(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2*d*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I*b^2*d*Pi*csgn(((1+I*
c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2+1/2*I*b^2*d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2
+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-1/2*I*b^2*d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)
^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+I*a*b*d*ln(c*x)*ln(1+I*c*x)-1/2*I*b^2*d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+
1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-1/2*I*b^2*d*Pi*csgn(I*((1+
I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I
*b^2*d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1
+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*I*b^2*d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^
2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+a*b*arctan(c*x)*x^2*e+2
*a*b*arctan(c*x)*d*ln(c*x)+1/2*I*b^2*d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+
1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+a^2*d*ln(c*x)+2*b^2*d*polylo
g(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*b^2*d*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*b^2*d*polylog(3,-(1+I*c*x
)^2/(c^2*x^2+1))+I*a*b*d*dilog(1+I*c*x)+1/2*I*b^2*d*Pi*arctan(c*x)^2+I*b^2*d*arctan(c*x)*polylog(2,-(1+I*c*x)^
2/(c^2*x^2+1))+I*b^2/c^2*arctan(c*x)*e-2*I*b^2*d*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*b^2*d*
arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*a*b*d*dilog(1-I*c*x)-I*a*b*d*ln(c*x)*ln(1-I*c*x)-a*b*e*x
/c-b^2*e*x*arctan(c*x)/c+1/2*a^2*x^2*e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, b^{2} e x^{2} \arctan \left (c x\right )^{2} - \frac {1}{32} \, b^{2} e x^{2} \log \left (c^{2} x^{2} + 1\right )^{2} + 12 \, b^{2} c^{2} e \int \frac {x^{4} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} c^{2} e \int \frac {x^{4} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b c^{2} e \int \frac {x^{4} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 2 \, b^{2} c^{2} e \int \frac {x^{4} \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 12 \, b^{2} c^{2} d \int \frac {x^{2} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b c^{2} d \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + \frac {1}{96} \, b^{2} d \log \left (c^{2} x^{2} + 1\right )^{3} + \frac {1}{2} \, a^{2} e x^{2} - 4 \, b^{2} c e \int \frac {x^{3} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 12 \, b^{2} e \int \frac {x^{2} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b e \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 12 \, b^{2} d \int \frac {\arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} d \int \frac {\log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b d \int \frac {\arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + \frac {b^{2} e \log \left (c^{2} x^{2} + 1\right )^{3}}{96 \, c^{2}} + a^{2} d \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*e*x^2*arctan(c*x)^2 - 1/32*b^2*e*x^2*log(c^2*x^2 + 1)^2 + 12*b^2*c^2*e*integrate(1/16*x^4*arctan(c*x)^
2/(c^2*x^3 + x), x) + b^2*c^2*e*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*e*integra
te(1/16*x^4*arctan(c*x)/(c^2*x^3 + x), x) + 2*b^2*c^2*e*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^2*x^3 + x), x)
+ 12*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*d*integrate(1/16*x^2*arctan(c*x
)/(c^2*x^3 + x), x) + 1/96*b^2*d*log(c^2*x^2 + 1)^3 + 1/2*a^2*e*x^2 - 4*b^2*c*e*integrate(1/16*x^3*arctan(c*x)
/(c^2*x^3 + x), x) + 12*b^2*e*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^3 + x), x) + 32*a*b*e*integrate(1/16*x^2
*arctan(c*x)/(c^2*x^3 + x), x) + 12*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*d*integrate(1/1
6*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*d*integrate(1/16*arctan(c*x)/(c^2*x^3 + x), x) + 1/96*b^2*e*lo
g(c^2*x^2 + 1)^3/c^2 + a^2*d*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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