3.79 \(\int \frac{(a+b \tan ^{-1}(c x^2))^2}{x^3} \, dx\)

Optimal. Leaf size=97 \[ -\frac{1}{2} i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x^2}\right )-\frac{1}{2} i c \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2-\frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{2 x^2}+b c \log \left (2-\frac{2}{1-i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [B]  time = 0.65346, antiderivative size = 290, normalized size of antiderivative = 2.99, number of steps used = 24, number of rules used = 13, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.812, Rules used = {5035, 2454, 2397, 2392, 2391, 2395, 36, 29, 31, 2439, 2416, 2394, 2393} \[ \frac{1}{2} i b^2 c \text{PolyLog}\left (2,-i c x^2\right )-\frac{1}{2} i b^2 c \text{PolyLog}\left (2,i c x^2\right )-\frac{1}{4} i b^2 c \text{PolyLog}\left (2,\frac{1}{2} \left (1-i c x^2\right )\right )+\frac{1}{4} i b^2 c \text{PolyLog}\left (2,\frac{1}{2} \left (1+i c x^2\right )\right )+\frac{1}{4} i b c \log \left (\frac{1}{2} \left (1+i c x^2\right )\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac{b \log \left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )}{4 x^2}-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+2 a b c \log (x)+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{4} i b^2 c \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTan[c*x^2])^2/x^3,x]

[Out]

2*a*b*c*Log[x] - ((1 - I*c*x^2)*(2*a + I*b*Log[1 - I*c*x^2])^2)/(8*x^2) + (I/4)*b*c*((2*I)*a - b*Log[1 - I*c*x
^2])*Log[(1 + I*c*x^2)/2] + (I/4)*b^2*c*Log[(1 - I*c*x^2)/2]*Log[1 + I*c*x^2] + (b*((2*I)*a - b*Log[1 - I*c*x^
2])*Log[1 + I*c*x^2])/(4*x^2) + (b^2*(1 + I*c*x^2)*Log[1 + I*c*x^2]^2)/(8*x^2) + (I/2)*b^2*c*PolyLog[2, (-I)*c
*x^2] - (I/2)*b^2*c*PolyLog[2, I*c*x^2] - (I/4)*b^2*c*PolyLog[2, (1 - I*c*x^2)/2] + (I/4)*b^2*c*PolyLog[2, (1
+ I*c*x^2)/2]

Rule 5035

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

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2397

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_))^2, x_Symbol] :> Simp[((d +
e*x)*(a + b*Log[c*(d + e*x)^n])^p)/((e*f - d*g)*(f + g*x)), x] - Dist[(b*e*n*p)/(e*f - d*g), Int[(a + b*Log[c*
(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0
]

Rule 2392

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

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 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 36

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*(x_)^(r_.), x_Symbol] :> Simp[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]))/(r +
1), x] + (-Dist[(g*j*m)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[(b*e*n*
p)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f + g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /
; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (EqQ[p, 1] || GtQ[r, 0]) && N
eQ[r, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{x^3} \, dx &=\int \left (\frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{4 x^3}+\frac{b \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{2 x^3}-\frac{b^2 \log ^2\left (1+i c x^2\right )}{4 x^3}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{x^3} \, dx+\frac{1}{2} b \int \frac{\left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{x^3} \, dx-\frac{1}{4} b^2 \int \frac{\log ^2\left (1+i c x^2\right )}{x^3} \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{(2 a+i b \log (1-i c x))^2}{x^2} \, dx,x,x^2\right )+\frac{1}{4} b \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log (1+i c x)}{x^2} \, dx,x,x^2\right )-\frac{1}{8} b^2 \operatorname{Subst}\left (\int \frac{\log ^2(1+i c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 x^2}+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{4} (i b c) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{x (1+i c x)} \, dx,x,x^2\right )+\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{2 a+i b \log (1-i c x)}{x} \, dx,x,x^2\right )-\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x} \, dx,x,x^2\right )-\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x (1-i c x)} \, dx,x,x^2\right )\\ &=a b c \log (x)-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 x^2}+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{4} i b^2 c \text{Li}_2\left (-i c x^2\right )+\frac{1}{4} (i b c) \operatorname{Subst}\left (\int \left (\frac{-2 i a+b \log (1-i c x)}{x}-\frac{c (-2 i a+b \log (1-i c x))}{-i+c x}\right ) \, dx,x,x^2\right )+\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-i c x)}{x} \, dx,x,x^2\right )-\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{\log (1+i c x)}{x}-\frac{c \log (1+i c x)}{i+c x}\right ) \, dx,x,x^2\right )\\ &=a b c \log (x)-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 x^2}+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{4} i b^2 c \text{Li}_2\left (-i c x^2\right )-\frac{1}{4} i b^2 c \text{Li}_2\left (i c x^2\right )+\frac{1}{4} (i b c) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{x} \, dx,x,x^2\right )-\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{x} \, dx,x,x^2\right )-\frac{1}{4} \left (i b c^2\right ) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^2\right )+\frac{1}{4} \left (i b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{i+c x} \, dx,x,x^2\right )\\ &=2 a b c \log (x)-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+\frac{1}{4} i b c \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )+\frac{1}{4} i b^2 c \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 x^2}+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{2} i b^2 c \text{Li}_2\left (-i c x^2\right )-\frac{1}{4} i b^2 c \text{Li}_2\left (i c x^2\right )+\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-i c x)}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^2\right )\\ &=2 a b c \log (x)-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+\frac{1}{4} i b c \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )+\frac{1}{4} i b^2 c \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 x^2}+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{2} i b^2 c \text{Li}_2\left (-i c x^2\right )-\frac{1}{2} i b^2 c \text{Li}_2\left (i c x^2\right )+\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-i c x^2\right )-\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+i c x^2\right )\\ &=2 a b c \log (x)-\frac{\left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 x^2}+\frac{1}{4} i b c \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )+\frac{1}{4} i b^2 c \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 x^2}+\frac{b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 x^2}+\frac{1}{2} i b^2 c \text{Li}_2\left (-i c x^2\right )-\frac{1}{2} i b^2 c \text{Li}_2\left (i c x^2\right )-\frac{1}{4} i b^2 c \text{Li}_2\left (\frac{1}{2} \left (1-i c x^2\right )\right )+\frac{1}{4} i b^2 c \text{Li}_2\left (\frac{1}{2} \left (1+i c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.148633, size = 127, normalized size = 1.31 \[ \frac{1}{2} b^2 c \left (-i \left (\tan ^{-1}\left (c x^2\right )^2+\text{PolyLog}\left (2,e^{2 i \tan ^{-1}\left (c x^2\right )}\right )\right )-\frac{\tan ^{-1}\left (c x^2\right )^2}{c x^2}+2 \tan ^{-1}\left (c x^2\right ) \log \left (1-e^{2 i \tan ^{-1}\left (c x^2\right )}\right )\right )-\frac{a^2}{2 x^2}+a b c \left (-\frac{1}{2} \log \left (c^2 x^4+1\right )+\log \left (c x^2\right )-\frac{\tan ^{-1}\left (c x^2\right )}{c x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c*x^2])^2/x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.337, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arctan \left ( c{x}^{2} \right ) \right ) ^{2}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^2))^2/x^3,x)

[Out]

int((a+b*arctan(c*x^2))^2/x^3,x)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^2/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (c x^{2}\right )^{2} + 2 \, a b \arctan \left (c x^{2}\right ) + a^{2}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^2/x^3,x, algorithm="fricas")

[Out]

integral((b^2*arctan(c*x^2)^2 + 2*a*b*arctan(c*x^2) + a^2)/x^3, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x^{2} \right )}\right )^{2}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))**2/x**3,x)

[Out]

Integral((a + b*atan(c*x**2))**2/x**3, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))^2/x^3,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^2) + a)^2/x^3, x)