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

Optimal. Leaf size=237 \[ -\frac {d^2 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^3}+\frac {d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}+\frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e}-\frac {a d x}{e^2}+\frac {b d \log \left (c^2 x^2+1\right )}{2 c e^2}+\frac {b \tan ^{-1}(c x)}{2 c^2 e}+\frac {i b d^2 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e^3}-\frac {i b d^2 \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^3}-\frac {b d x \tan ^{-1}(c x)}{e^2}-\frac {b x}{2 c e} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {4876, 4846, 260, 4852, 321, 203, 4856, 2402, 2315, 2447} \[ \frac {i b d^2 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e^3}-\frac {i b d^2 \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^3}-\frac {d^2 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^3}+\frac {d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}+\frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e}-\frac {a d x}{e^2}+\frac {b d \log \left (c^2 x^2+1\right )}{2 c e^2}+\frac {b \tan ^{-1}(c x)}{2 c^2 e}-\frac {b d x \tan ^{-1}(c x)}{e^2}-\frac {b x}{2 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 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 4876

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

Rubi steps

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

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Mathematica [A]  time = 1.68, size = 404, normalized size = 1.70 \[ \frac {2 a d^2 \log (d+e x)-2 a d e x+a e^2 x^2-\frac {b d e \sqrt {\frac {c^2 d^2}{e^2}+1} \tan ^{-1}(c x)^2 e^{i \tan ^{-1}\left (\frac {c d}{e}\right )}}{c}+\frac {1}{2} \pi b d^2 \log \left (c^2 x^2+1\right )+\frac {b d e \log \left (c^2 x^2+1\right )}{c}+\frac {b e^2 \tan ^{-1}(c x)}{c^2}-i b d^2 \text {Li}_2\left (e^{2 i \left (\tan ^{-1}\left (\frac {c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-2 i b d^2 \tan ^{-1}(c x) \tan ^{-1}\left (\frac {c d}{e}\right )+2 b d^2 \tan ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac {c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )+2 b d^2 \tan ^{-1}(c x) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac {c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-2 b d^2 \tan ^{-1}\left (\frac {c d}{e}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac {c d}{e}\right )+\tan ^{-1}(c x)\right )\right )+i b d^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+i b d^2 \tan ^{-1}(c x)^2+i \pi b d^2 \tan ^{-1}(c x)+\pi b d^2 \log \left (1+e^{-2 i \tan ^{-1}(c x)}\right )-2 b d^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+\frac {b d e \tan ^{-1}(c x)^2}{c}-2 b d e x \tan ^{-1}(c x)+b e^2 x^2 \tan ^{-1}(c x)-\frac {b e^2 x}{c}}{2 e^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \arctan \left (c x\right ) + a x^{2}}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.08, size = 305, normalized size = 1.29 \[ \frac {a \,x^{2}}{2 e}-\frac {a d x}{e^{2}}+\frac {a \,d^{2} \ln \left (c e x +d c \right )}{e^{3}}+\frac {b \arctan \left (c x \right ) x^{2}}{2 e}-\frac {b d x \arctan \left (c x \right )}{e^{2}}+\frac {b \arctan \left (c x \right ) d^{2} \ln \left (c e x +d c \right )}{e^{3}}+\frac {i b \,d^{2} \ln \left (c e x +d c \right ) \ln \left (\frac {-c e x +i e}{d c +i e}\right )}{2 e^{3}}-\frac {i b \,d^{2} \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +i e}{-d c +i e}\right )}{2 e^{3}}+\frac {i b \,d^{2} \dilog \left (\frac {-c e x +i e}{d c +i e}\right )}{2 e^{3}}-\frac {i b \,d^{2} \dilog \left (\frac {c e x +i e}{-d c +i e}\right )}{2 e^{3}}+\frac {b d \ln \left (c^{2} d^{2}-2 \left (c e x +d c \right ) c d +\left (c e x +d c \right )^{2}+e^{2}\right )}{2 c \,e^{2}}+\frac {b \arctan \left (c x \right )}{2 c^{2} e}-\frac {b x}{2 c e}-\frac {b d}{2 c \,e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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