3.43 \(\int \frac{x^3 (a+b \tan ^{-1}(c x))}{d+i c d x} \, dx\)

Optimal. Leaf size=196 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}+\frac{i a x}{c^3 d}+\frac{i b x^2}{6 c^2 d}-\frac{2 i b \log \left (c^2 x^2+1\right )}{3 c^4 d}-\frac{b x}{2 c^3 d}+\frac{i b x \tan ^{-1}(c x)}{c^3 d}+\frac{b \tan ^{-1}(c x)}{2 c^4 d} \]

[Out]

(I*a*x)/(c^3*d) - (b*x)/(2*c^3*d) + ((I/6)*b*x^2)/(c^2*d) + (b*ArcTan[c*x])/(2*c^4*d) + (I*b*x*ArcTan[c*x])/(c
^3*d) + (x^2*(a + b*ArcTan[c*x]))/(2*c^2*d) - ((I/3)*x^3*(a + b*ArcTan[c*x]))/(c*d) + ((a + b*ArcTan[c*x])*Log
[2/(1 + I*c*x)])/(c^4*d) - (((2*I)/3)*b*Log[1 + c^2*x^2])/(c^4*d) + ((I/2)*b*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c
^4*d)

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Rubi [A]  time = 0.286152, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {4866, 4852, 266, 43, 321, 203, 4846, 260, 4854, 2402, 2315} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}+\frac{i a x}{c^3 d}+\frac{i b x^2}{6 c^2 d}-\frac{2 i b \log \left (c^2 x^2+1\right )}{3 c^4 d}-\frac{b x}{2 c^3 d}+\frac{i b x \tan ^{-1}(c x)}{c^3 d}+\frac{b \tan ^{-1}(c x)}{2 c^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcTan[c*x]))/(d + I*c*d*x),x]

[Out]

(I*a*x)/(c^3*d) - (b*x)/(2*c^3*d) + ((I/6)*b*x^2)/(c^2*d) + (b*ArcTan[c*x])/(2*c^4*d) + (I*b*x*ArcTan[c*x])/(c
^3*d) + (x^2*(a + b*ArcTan[c*x]))/(2*c^2*d) - ((I/3)*x^3*(a + b*ArcTan[c*x]))/(c*d) + ((a + b*ArcTan[c*x])*Log
[2/(1 + I*c*x)])/(c^4*d) - (((2*I)/3)*b*Log[1 + c^2*x^2])/(c^4*d) + ((I/2)*b*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c
^4*d)

Rule 4866

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f)/e, Int[((f*x)^(m - 1)*(a + b*ArcTan[c*x])^p)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && GtQ[m, 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{d+i c d x} \, dx &=\frac{i \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{d+i c d x} \, dx}{c}-\frac{i \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c d}\\ &=-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}-\frac{\int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d+i c d x} \, dx}{c^2}+\frac{(i b) \int \frac{x^3}{1+c^2 x^2} \, dx}{3 d}+\frac{\int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^2 d}\\ &=\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}-\frac{i \int \frac{a+b \tan ^{-1}(c x)}{d+i c d x} \, dx}{c^3}+\frac{(i b) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )}{6 d}+\frac{i \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^3 d}-\frac{b \int \frac{x^2}{1+c^2 x^2} \, dx}{2 c d}\\ &=\frac{i a x}{c^3 d}-\frac{b x}{2 c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d}+\frac{(i b) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 d}+\frac{(i b) \int \tan ^{-1}(c x) \, dx}{c^3 d}+\frac{b \int \frac{1}{1+c^2 x^2} \, dx}{2 c^3 d}-\frac{b \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d}\\ &=\frac{i a x}{c^3 d}-\frac{b x}{2 c^3 d}+\frac{i b x^2}{6 c^2 d}+\frac{b \tan ^{-1}(c x)}{2 c^4 d}+\frac{i b x \tan ^{-1}(c x)}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d}-\frac{i b \log \left (1+c^2 x^2\right )}{6 c^4 d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c^4 d}-\frac{(i b) \int \frac{x}{1+c^2 x^2} \, dx}{c^2 d}\\ &=\frac{i a x}{c^3 d}-\frac{b x}{2 c^3 d}+\frac{i b x^2}{6 c^2 d}+\frac{b \tan ^{-1}(c x)}{2 c^4 d}+\frac{i b x \tan ^{-1}(c x)}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 c d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d}-\frac{2 i b \log \left (1+c^2 x^2\right )}{3 c^4 d}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d}\\ \end{align*}

Mathematica [A]  time = 0.442145, size = 166, normalized size = 0.85 \[ -\frac{i \left (3 b \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x) \left (6 a+b \left (2 c^3 x^3+3 i c^2 x^2-6 c x+3 i\right )+6 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+2 a c^3 x^3+3 i a c^2 x^2-3 i a \log \left (c^2 x^2+1\right )-6 a c x-b c^2 x^2+4 b \log \left (c^2 x^2+1\right )-3 i b c x+6 b \tan ^{-1}(c x)^2-b\right )}{6 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I/6)*(-b - 6*a*c*x - (3*I)*b*c*x + (3*I)*a*c^2*x^2 - b*c^2*x^2 + 2*a*c^3*x^3 + 6*b*ArcTan[c*x]^2 + ArcTan[c
*x]*(6*a + b*(3*I - 6*c*x + (3*I)*c^2*x^2 + 2*c^3*x^3) + (6*I)*b*Log[1 + E^((2*I)*ArcTan[c*x])]) - (3*I)*a*Log
[1 + c^2*x^2] + 4*b*Log[1 + c^2*x^2] + 3*b*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(c^4*d)

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Maple [B]  time = 0.053, size = 353, normalized size = 1.8 \begin{align*}{\frac{-{\frac{i}{3}}a{x}^{3}}{dc}}+{\frac{{\frac{i}{6}}b{x}^{2}}{{c}^{2}d}}+{\frac{a{x}^{2}}{2\,{c}^{2}d}}-{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,d{c}^{4}}}-{\frac{ia\arctan \left ( cx \right ) }{d{c}^{4}}}+{\frac{iax}{d{c}^{3}}}+{\frac{{\frac{i}{2}}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{d{c}^{4}}}+{\frac{b{x}^{2}\arctan \left ( cx \right ) }{2\,{c}^{2}d}}-{\frac{b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{d{c}^{4}}}-{\frac{{\frac{i}{3}}b\arctan \left ( cx \right ){x}^{3}}{dc}}+{\frac{ibx\arctan \left ( cx \right ) }{d{c}^{3}}}+{\frac{{\frac{i}{2}}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{d{c}^{4}}}-{\frac{bx}{2\,d{c}^{3}}}-{\frac{{\frac{i}{4}}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{d{c}^{4}}}-{\frac{{\frac{11\,i}{24}}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{d{c}^{4}}}+{\frac{{\frac{2\,i}{3}}b}{d{c}^{4}}}+{\frac{5\,b}{24\,d{c}^{4}}\arctan \left ({\frac{cx}{2}} \right ) }-{\frac{5\,b}{24\,d{c}^{4}}\arctan \left ({\frac{{c}^{3}{x}^{3}}{6}}+{\frac{7\,cx}{6}} \right ) }-{\frac{5\,b}{12\,d{c}^{4}}\arctan \left ({\frac{cx}{2}}-{\frac{i}{2}} \right ) }-{\frac{{\frac{5\,i}{48}}b\ln \left ({c}^{4}{x}^{4}+10\,{c}^{2}{x}^{2}+9 \right ) }{d{c}^{4}}}+{\frac{11\,b\arctan \left ( cx \right ) }{12\,d{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I/c*a/d*x^3+1/6*I*b*x^2/c^2/d+1/2/c^2*a/d*x^2-1/2/c^4*a/d*ln(c^2*x^2+1)-I/c^4*a/d*arctan(c*x)+I*a*x/d/c^3
+1/2*I/c^4*b/d*dilog(-1/2*I*(c*x+I))+1/2/c^2*b/d*arctan(c*x)*x^2-1/c^4*b/d*arctan(c*x)*ln(c*x-I)-1/3*I/c*b/d*a
rctan(c*x)*x^3+I*b*x*arctan(c*x)/d/c^3+1/2*I/c^4*b/d*ln(-1/2*I*(c*x+I))*ln(c*x-I)-1/2*b*x/d/c^3-1/4*I/c^4*b/d*
ln(c*x-I)^2-11/24*I/c^4*b/d*ln(c^2*x^2+1)+2/3*I/c^4*b/d+5/24/c^4*b/d*arctan(1/2*c*x)-5/24/c^4*b/d*arctan(1/6*c
^3*x^3+7/6*c*x)-5/12/c^4*b/d*arctan(1/2*c*x-1/2*I)-5/48*I/c^4*b/d*ln(c^4*x^4+10*c^2*x^2+9)+11/12*b*arctan(c*x)
/d/c^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{6} \, a{\left (\frac{i \,{\left (2 \, c^{2} x^{3} + 3 i \, c x^{2} - 6 \, x\right )}}{c^{3} d} + \frac{6 \, \log \left (i \, c x + 1\right )}{c^{4} d}\right )} - \frac{-\frac{1}{2} \,{\left (-12 i \,{\left (2 \,{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{7} d} + \frac{3 \, \arctan \left (c x\right )}{c^{8} d}\right )} \arctan \left (c x\right ) - \frac{c^{2} x^{2} + 3 \, \arctan \left (c x\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right )}{c^{8} d}\right )} c^{8} d - 36 \, c^{8} d \int \frac{x^{4} \log \left (c^{2} x^{2} + 1\right )}{c^{5} d x^{2} + c^{3} d}\,{d x} - 9 i \,{\left (2 \,{\left (\frac{x^{2}}{c^{5} d} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{7} d}\right )} \log \left (c^{2} x^{2} + 1\right ) - \frac{2 \, c^{2} x^{2} - \log \left (c^{2} x^{2} + 1\right )^{2} - 2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{7} d}\right )} c^{7} d + 72 \, c^{7} d \int \frac{x^{3} \arctan \left (c x\right )}{c^{5} d x^{2} + c^{3} d}\,{d x} - 72 \, c^{5} d \int \frac{x \arctan \left (c x\right )}{c^{5} d x^{2} + c^{3} d}\,{d x} - 8 \, c^{3} x^{3} + 36 \, c^{4} d \int \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{5} d x^{2} + c^{3} d}\,{d x} - 6 i \, c^{2} x^{2} + 60 \, c x - 2 \,{\left (12 i \, c^{3} x^{3} - 18 \, c^{2} x^{2} - 36 i \, c x + 30\right )} \arctan \left (c x\right ) - 36 i \, \arctan \left (c x\right )^{2} + 2 \,{\left (6 \, c^{3} x^{3} + 9 i \, c^{2} x^{2} - 18 \, c x + 3 i\right )} \log \left (c^{2} x^{2} + 1\right ) - 9 i \, \log \left (c^{2} x^{2} + 1\right )^{2} - 36 i \, \log \left (12 \, c^{5} d x^{2} + 12 \, c^{3} d\right )\right )} b}{72 \, c^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*(I*(2*c^2*x^3 + 3*I*c*x^2 - 6*x)/(c^3*d) + 6*log(I*c*x + 1)/(c^4*d)) - 1/72*(432*I*c^8*d*integrate(1/12
*x^4*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) + 216*c^8*d*integrate(1/12*x^4*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d),
x) - 432*c^7*d*integrate(1/12*x^3*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) + 216*I*c^7*d*integrate(1/12*x^3*log(c^2
*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 432*c^5*d*integrate(1/12*x*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) - 216*I*c^5
*d*integrate(1/12*x*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 4*c^3*x^3 - 216*c^4*d*integrate(1/12*log(c^2*x^
2 + 1)/(c^5*d*x^2 + c^3*d), x) + 3*I*c^2*x^2 - 30*c*x + (12*I*c^3*x^3 - 18*c^2*x^2 - 36*I*c*x + 30)*arctan(c*x
) + 18*I*arctan(c*x)^2 - (6*c^3*x^3 + 9*I*c^2*x^2 - 18*c*x + 3*I)*log(c^2*x^2 + 1) + 9*I*log(c^2*x^2 + 1)^2 +
18*I*log(12*c^5*d*x^2 + 12*c^3*d))*b/(c^4*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3*log(-(c*x + I)/(c*x - I)) - 2*I*a*x^3)/(2*c*d*x - 2*I*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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