3.13.0 ∫ (d+ex2)(a+b arctan(cx))2 _____________________________________

Integrand size = 21, antiderivative size = 217 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=-\frac {a b e x}{c}-\frac {b^2 e x \arctan (c x)}{c}+\frac {e (a+b \arctan (c x))^2}{2 c^2}+\frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+\frac {b^2 e \log \left (1+c^2 x^2\right )}{2 c^2}-i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

-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

Rubi [A] (veriﬁed)

Time = 0.32 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {5100, 4942, 5108, 5004, 5114, 6745, 4946, 5036, 4930, 266} \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=2 d \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2+\frac {e (a+b \arctan (c x))^2}{2 c^2}-i b d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))+i b d \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))+\frac {1}{2} e x^2 (a+b \arctan (c x))^2-\frac {a b e x}{c}-\frac {b^2 e x \arctan (c x)}{c}+\frac {b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right ) \]

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

-((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

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

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c

*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0

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

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^

n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),

x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

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]

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]

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])

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L

og[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)))^

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar

cTan[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

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d (a+b \arctan (c x))^2}{x}+e x (a+b \arctan (c x))^2\right ) \, dx \\ & = d \int \frac {(a+b \arctan (c x))^2}{x} \, dx+e \int x (a+b \arctan (c x))^2 \, dx \\ & = \frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-(4 b c d) \int \frac {(a+b \arctan (c x)) \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(b c e) \int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx \\ & = \frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+(2 b c d) \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(2 b c d) \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {(b e) \int (a+b \arctan (c x)) \, dx}{c}+\frac {(b e) \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{c} \\ & = -\frac {a b e x}{c}+\frac {e (a+b \arctan (c x))^2}{2 c^2}+\frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )+\left (i b^2 c d\right ) \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d\right ) \int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {\left (b^2 e\right ) \int \arctan (c x) \, dx}{c} \\ & = -\frac {a b e x}{c}-\frac {b^2 e x \arctan (c x)}{c}+\frac {e (a+b \arctan (c x))^2}{2 c^2}+\frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,-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 \arctan (c x)}{c}+\frac {e (a+b \arctan (c x))^2}{2 c^2}+\frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\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 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.25 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\frac {1}{2} a^2 e x^2+\frac {a b e \left (-c x+\left (1+c^2 x^2\right ) \arctan (c x)\right )}{c^2}+a^2 d \log (x)+\frac {b^2 e \left (-2 c x \arctan (c x)+\left (1+c^2 x^2\right ) \arctan (c x)^2+\log \left (1+c^2 x^2\right )\right )}{2 c^2}+i a b d (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+b^2 d \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \arctan (c x)^3+\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-\arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right ) \]

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

(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)

Maple [C] (warning: unable to verify)

Time = 6.62 (sec) , antiderivative size = 1262, normalized size of antiderivative = 5.82


method	result	size

parts	\(\text {Expression too large to display}\)	\(1262\)
derivativedivides	\(\text {Expression too large to display}\)	\(1263\)
default	\(\text {Expression too large to display}\)	\(1263\)

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

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

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

ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/2*I*c^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*c^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+e*arctan(c*x)*(c*x-I)-1/2*I*c^2*d*Pi*arctan(c*x)^2+ln((1+I*c*x)^2/(c^2*x^2+1

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

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

^(1/2))-2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))*c^2*d+1/2*I*c^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*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+1/2*d*c^

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

tan(c*x))-I*c^2*d*ln(c*x)*ln(1+I*c*x)+I*c^2*d*ln(c*x)*ln(1-I*c*x)-I*c^2*d*dilog(1+I*c*x)+I*c^2*d*dilog(1-I*c*x

Fricas [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

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

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)

Sympy [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )}{x}\, dx \]

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

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

Maxima [F]

\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]

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

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

Giac [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\text {Timed out} \]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (e\,x^2+d\right )}{x} \,d x \]

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

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

\(\int \frac {(d+e x^2) (a+b \arctan (c x))^2}{x} \, dx\) [1251]

Optimal result