3.1252 \(\int \frac {(d+e x^2) (a+b \tan ^{-1}(c x))^2}{x^2} \, dx\)

Optimal. Leaf size=172 \[ -i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {i e \left (a+b \tan ^{-1}(c x)\right )^2}{c}+e x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 b e \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-i b^2 c d \text {Li}_2\left (\frac {2}{1-i c x}-1\right )+\frac {i b^2 e \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {4980, 4846, 4920, 4854, 2402, 2315, 4852, 4924, 4868, 2447} \[ -i b^2 c d \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+\frac {i b^2 e \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c}-i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {i e \left (a+b \tan ^{-1}(c x)\right )^2}{c}+e x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 b e \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]]

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 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 4854

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

Rule 4868

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

Rule 4920

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

Rule 4924

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 204, normalized size = 1.19 \[ \frac {-a^2 c d+a^2 c e x^2+a b c d \left (c x \left (2 \log (c x)-\log \left (c^2 x^2+1\right )\right )-2 \tan ^{-1}(c x)\right )+a b e x \left (2 c x \tan ^{-1}(c x)-\log \left (c^2 x^2+1\right )\right )-b^2 c d \left (i c x \left (\tan ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )\right )+\tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )+b^2 e x \left (\tan ^{-1}(c x) \left ((c x-i) \tan ^{-1}(c x)+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-i \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )\right )}{c x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2*c*d) + a^2*c*e*x^2 + a*b*c*d*(-2*ArcTan[c*x] + c*x*(2*Log[c*x] - Log[1 + c^2*x^2])) + a*b*e*x*(2*c*x*Ar
cTan[c*x] - Log[1 + c^2*x^2]) + b^2*e*x*(ArcTan[c*x]*((-I + c*x)*ArcTan[c*x] + 2*Log[1 + E^((2*I)*ArcTan[c*x])
]) - I*PolyLog[2, -E^((2*I)*ArcTan[c*x])]) - b^2*c*d*(ArcTan[c*x]^2 - 2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTa
n[c*x])] + I*c*x*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*x])])))/(c*x)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} e x^{2} + a^{2} d + {\left (b^{2} e x^{2} + b^{2} d\right )} \arctan \left (c x\right )^{2} + 2 \, {\left (a b e x^{2} + a b d\right )} \arctan \left (c x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.14, size = 597, normalized size = 3.47 \[ a^{2} e x -\frac {a^{2} d}{x}-i c \,b^{2} d \ln \left (c x \right ) \ln \left (-i c x +1\right )+\frac {i c \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right ) d}{2}+\frac {i c \,b^{2} \ln \left (c^{2} x^{2}+1\right ) \ln \left (c x +i\right ) d}{2}-\frac {i c \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right ) d}{2}+\frac {i c \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right ) d}{2}-i c \,b^{2} d \dilog \left (-i c x +1\right )-\frac {i c \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right ) d}{2}+\frac {i b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right ) e}{2 c}+i c \,b^{2} d \dilog \left (i c x +1\right )-\frac {i b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right ) e}{2 c}+\frac {i c \,b^{2} \ln \left (c x -i\right )^{2} d}{4}-\frac {i c \,b^{2} \ln \left (c x +i\right )^{2} d}{4}+\frac {i b^{2} \ln \left (c x -i\right )^{2} e}{4 c}-\frac {i b^{2} \ln \left (c x +i\right )^{2} e}{4 c}+\frac {i b^{2} \ln \left (c^{2} x^{2}+1\right ) \ln \left (c x +i\right ) e}{2 c}-\frac {i b^{2} \ln \left (c^{2} x^{2}+1\right ) \ln \left (c x -i\right ) e}{2 c}+\frac {i b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right ) e}{2 c}+i c \,b^{2} d \ln \left (c x \right ) \ln \left (i c x +1\right )-\frac {i c \,b^{2} \ln \left (c^{2} x^{2}+1\right ) \ln \left (c x -i\right ) d}{2}-\frac {i b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right ) e}{2 c}-\frac {b^{2} \arctan \left (c x \right )^{2} d}{x}+b^{2} \arctan \left (c x \right )^{2} e x +2 c a b d \ln \left (c x \right )-c a b \ln \left (c^{2} x^{2}+1\right ) d +2 c \,b^{2} \arctan \left (c x \right ) d \ln \left (c x \right )-c \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right ) d -\frac {b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right ) e}{c}-\frac {a b \ln \left (c^{2} x^{2}+1\right ) e}{c}+2 a b \arctan \left (c x \right ) e x -\frac {2 a b \arctan \left (c x \right ) d}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*e*x-a^2*d/x-b^2*arctan(c*x)^2*d/x+b^2*arctan(c*x)^2*e*x+1/2*I*b^2/c*ln(c^2*x^2+1)*ln(I+c*x)*e-1/2*I*b^2/c*
ln(c^2*x^2+1)*ln(c*x-I)*e+1/2*I*b^2/c*ln(c*x-I)*ln(-1/2*I*(I+c*x))*e+I*c*b^2*d*ln(c*x)*ln(1+I*c*x)-1/2*I*c*b^2
*ln(c^2*x^2+1)*ln(c*x-I)*d-1/2*I*b^2/c*ln(I+c*x)*ln(1/2*I*(c*x-I))*e-I*c*b^2*d*ln(c*x)*ln(1-I*c*x)+1/2*I*c*b^2
*ln(c*x-I)*ln(-1/2*I*(I+c*x))*d+1/2*I*c*b^2*ln(c^2*x^2+1)*ln(I+c*x)*d-1/2*I*c*b^2*ln(I+c*x)*ln(1/2*I*(c*x-I))*
d+2*c*a*b*d*ln(c*x)-c*a*b*ln(c^2*x^2+1)*d+2*c*b^2*arctan(c*x)*d*ln(c*x)-c*b^2*arctan(c*x)*ln(c^2*x^2+1)*d-1/2*
I*b^2/c*dilog(1/2*I*(c*x-I))*e+1/2*I*c*b^2*dilog(-1/2*I*(I+c*x))*d+1/4*I*c*b^2*ln(c*x-I)^2*d-1/4*I*c*b^2*ln(I+
c*x)^2*d-I*c*b^2*d*dilog(1-I*c*x)-1/2*I*c*b^2*dilog(1/2*I*(c*x-I))*d+1/4*I*b^2/c*ln(c*x-I)^2*e+1/2*I*b^2/c*dil
og(-1/2*I*(I+c*x))*e-1/4*I*b^2/c*ln(I+c*x)^2*e+I*c*b^2*d*dilog(1+I*c*x)-b^2/c*arctan(c*x)*ln(c^2*x^2+1)*e-a*b/
c*ln(c^2*x^2+1)*e+2*a*b*arctan(c*x)*e*x-2*a*b*arctan(c*x)*d/x

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maxima [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^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (e\,x^2+d\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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