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

Optimal. Leaf size=176 \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (-c x+i)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (-c x+i)^2}-\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}+\frac{7 i b}{8 c^3 d^3 (-c x+i)}+\frac{b}{8 c^3 d^3 (-c x+i)^2}-\frac{7 i b \tan ^{-1}(c x)}{8 c^3 d^3} \]

[Out]

b/(8*c^3*d^3*(I - c*x)^2) + (((7*I)/8)*b)/(c^3*d^3*(I - c*x)) - (((7*I)/8)*b*ArcTan[c*x])/(c^3*d^3) + ((I/2)*(
a + b*ArcTan[c*x]))/(c^3*d^3*(I - c*x)^2) - (2*(a + b*ArcTan[c*x]))/(c^3*d^3*(I - c*x)) - (I*(a + b*ArcTan[c*x
])*Log[2/(1 + I*c*x)])/(c^3*d^3) + (b*PolyLog[2, 1 - 2/(1 + I*c*x)])/(2*c^3*d^3)

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Rubi [A]  time = 0.21721, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4876, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (-c x+i)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (-c x+i)^2}-\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}+\frac{7 i b}{8 c^3 d^3 (-c x+i)}+\frac{b}{8 c^3 d^3 (-c x+i)^2}-\frac{7 i b \tan ^{-1}(c x)}{8 c^3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b/(8*c^3*d^3*(I - c*x)^2) + (((7*I)/8)*b)/(c^3*d^3*(I - c*x)) - (((7*I)/8)*b*ArcTan[c*x])/(c^3*d^3) + ((I/2)*(
a + b*ArcTan[c*x]))/(c^3*d^3*(I - c*x)^2) - (2*(a + b*ArcTan[c*x]))/(c^3*d^3*(I - c*x)) - (I*(a + b*ArcTan[c*x
])*Log[2/(1 + I*c*x)])/(c^3*d^3) + (b*PolyLog[2, 1 - 2/(1 + I*c*x)])/(2*c^3*d^3)

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])

Rule 4862

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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^2 \left (a+b \tan ^{-1}(c x)\right )}{(d+i c d x)^3} \, dx &=\int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^3 (-i+c x)^3}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^3 (-i+c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^3 (-i+c x)}\right ) \, dx\\ &=-\frac{i \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{c^2 d^3}+\frac{i \int \frac{a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{c^2 d^3}-\frac{2 \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^2 d^3}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{(i b) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 c^2 d^3}+\frac{(i b) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d^3}-\frac{(2 b) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^2 d^3}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c^3 d^3}-\frac{(i b) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{2 c^2 d^3}-\frac{(2 b) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{c^2 d^3}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}-\frac{(i b) \int \left (-\frac{i}{2 (-i+c x)^3}+\frac{1}{4 (-i+c x)^2}-\frac{1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 c^2 d^3}-\frac{(2 b) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=\frac{b}{8 c^3 d^3 (i-c x)^2}+\frac{7 i b}{8 c^3 d^3 (i-c x)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}+\frac{(i b) \int \frac{1}{1+c^2 x^2} \, dx}{8 c^2 d^3}-\frac{(i b) \int \frac{1}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac{b}{8 c^3 d^3 (i-c x)^2}+\frac{7 i b}{8 c^3 d^3 (i-c x)}-\frac{7 i b \tan ^{-1}(c x)}{8 c^3 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}\\ \end{align*}

Mathematica [A]  time = 0.155007, size = 187, normalized size = 1.06 \[ -\frac{i \left (4 i b (c x-i)^2 \text{PolyLog}\left (2,\frac{c x+i}{c x-i}\right )+8 a c^2 x^2 \log \left (\frac{2 i}{-c x+i}\right )+16 i a c x-16 i a c x \log \left (\frac{2 i}{-c x+i}\right )-8 a \log \left (\frac{2 i}{-c x+i}\right )+12 a+b \left (7 c^2 x^2+2 i c x+8 (c x-i)^2 \log \left (\frac{2 i}{-c x+i}\right )+5\right ) \tan ^{-1}(c x)+7 b c x-6 i b\right )}{8 c^3 d^3 (c x-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*(12*a - (6*I)*b + (16*I)*a*c*x + 7*b*c*x - 8*a*Log[(2*I)/(I - c*x)] - (16*I)*a*c*x*Log[(2*I)/(I - c*x)
] + 8*a*c^2*x^2*Log[(2*I)/(I - c*x)] + b*ArcTan[c*x]*(5 + (2*I)*c*x + 7*c^2*x^2 + 8*(-I + c*x)^2*Log[(2*I)/(I
- c*x)]) + (4*I)*b*(-I + c*x)^2*PolyLog[2, (I + c*x)/(-I + c*x)]))/(c^3*d^3*(-I + c*x)^2)

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Maple [B]  time = 0.059, size = 349, normalized size = 2. \begin{align*}{\frac{{\frac{7\,i}{32}}b}{{c}^{3}{d}^{3}}\arctan \left ({\frac{cx}{2}} \right ) }-{\frac{{\frac{7\,i}{32}}b}{{c}^{3}{d}^{3}}\arctan \left ({\frac{{c}^{3}{x}^{3}}{6}}+{\frac{7\,cx}{6}} \right ) }-{\frac{a\arctan \left ( cx \right ) }{{c}^{3}{d}^{3}}}+2\,{\frac{a}{{c}^{3}{d}^{3} \left ( cx-i \right ) }}-{\frac{{\frac{7\,i}{16}}b\arctan \left ( cx \right ) }{{c}^{3}{d}^{3}}}+{\frac{{\frac{i}{2}}a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{3}{d}^{3}}}+2\,{\frac{b\arctan \left ( cx \right ) }{{c}^{3}{d}^{3} \left ( cx-i \right ) }}+{\frac{7\,b\ln \left ({c}^{4}{x}^{4}+10\,{c}^{2}{x}^{2}+9 \right ) }{64\,{c}^{3}{d}^{3}}}+{\frac{{\frac{i}{2}}a}{{c}^{3}{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{ib\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{c}^{3}{d}^{3}}}+{\frac{{\frac{i}{2}}b\arctan \left ( cx \right ) }{{c}^{3}{d}^{3} \left ( cx-i \right ) ^{2}}}-{\frac{{\frac{7\,i}{8}}b}{{c}^{3}{d}^{3} \left ( cx-i \right ) }}+{\frac{b}{8\,{c}^{3}{d}^{3} \left ( cx-i \right ) ^{2}}}-{\frac{7\,b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{32\,{c}^{3}{d}^{3}}}-{\frac{{\frac{7\,i}{16}}b}{{c}^{3}{d}^{3}}\arctan \left ({\frac{cx}{2}}-{\frac{i}{2}} \right ) }+{\frac{b\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{2\,{c}^{3}{d}^{3}}}+{\frac{b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{2\,{c}^{3}{d}^{3}}}-{\frac{b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{4\,{c}^{3}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/32*I/c^3*b/d^3*arctan(1/2*c*x)-7/32*I/c^3*b/d^3*arctan(1/6*c^3*x^3+7/6*c*x)-1/c^3*a/d^3*arctan(c*x)+2/c^3*a/
d^3/(c*x-I)-7/16*I/c^3*b/d^3*arctan(c*x)+1/2*I/c^3*a/d^3*ln(c^2*x^2+1)+2/c^3*b/d^3*arctan(c*x)/(c*x-I)+7/64/c^
3*b/d^3*ln(c^4*x^4+10*c^2*x^2+9)+1/2*I/c^3*a/d^3/(c*x-I)^2+I/c^3*b/d^3*arctan(c*x)*ln(c*x-I)+1/2*I/c^3*b/d^3*a
rctan(c*x)/(c*x-I)^2-7/8*I/c^3*b/d^3/(c*x-I)+1/8/c^3*b/d^3/(c*x-I)^2-7/32/c^3*b/d^3*ln(c^2*x^2+1)-7/16*I/c^3*b
/d^3*arctan(1/2*c*x-1/2*I)+1/2/c^3*b/d^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))+1/2/c^3*b/d^3*dilog(-1/2*I*(c*x+I))-1/4/
c^3*b/d^3*ln(c*x-I)^2

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Maxima [A]  time = 1.23644, size = 393, normalized size = 2.23 \begin{align*} -\frac{-7 i \, b c^{2} x^{2} \arctan \left (1, c x\right ) -{\left (b{\left (14 \, \arctan \left (1, c x\right ) - 14 i\right )} + 32 \, a\right )} c x + 4 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \arctan \left (c x\right )^{2} +{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} +{\left (-4 i \, b c^{2} x^{2} - 8 \, b c x + 4 i \, b\right )} \arctan \left (c x\right ) \log \left (\frac{1}{4} \, c^{2} x^{2} + \frac{1}{4}\right ) + b{\left (7 i \, \arctan \left (1, c x\right ) + 12\right )} +{\left ({\left (16 \, a + 7 i \, b\right )} c^{2} x^{2} +{\left (-32 i \, a - 18 \, b\right )} c x - 16 \, a + 17 i \, b\right )} \arctan \left (c x\right ) - 8 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )}{\rm Li}_2\left (\frac{1}{2} i \, c x + \frac{1}{2}\right ) +{\left (-8 i \, a c^{2} x^{2} - 16 \, a c x - 2 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \log \left (\frac{1}{4} \, c^{2} x^{2} + \frac{1}{4}\right ) + 8 i \, a\right )} \log \left (c^{2} x^{2} + 1\right ) + 24 i \, a}{16 \, c^{5} d^{3} x^{2} - 32 i \, c^{4} d^{3} x - 16 \, c^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-7*I*b*c^2*x^2*arctan2(1, c*x) - (b*(14*arctan2(1, c*x) - 14*I) + 32*a)*c*x + 4*(b*c^2*x^2 - 2*I*b*c*x - b)*
arctan(c*x)^2 + (b*c^2*x^2 - 2*I*b*c*x - b)*log(c^2*x^2 + 1)^2 + (-4*I*b*c^2*x^2 - 8*b*c*x + 4*I*b)*arctan(c*x
)*log(1/4*c^2*x^2 + 1/4) + b*(7*I*arctan2(1, c*x) + 12) + ((16*a + 7*I*b)*c^2*x^2 + (-32*I*a - 18*b)*c*x - 16*
a + 17*I*b)*arctan(c*x) - 8*(b*c^2*x^2 - 2*I*b*c*x - b)*dilog(1/2*I*c*x + 1/2) + (-8*I*a*c^2*x^2 - 16*a*c*x -
2*(b*c^2*x^2 - 2*I*b*c*x - b)*log(1/4*c^2*x^2 + 1/4) + 8*I*a)*log(c^2*x^2 + 1) + 24*I*a)/(16*c^5*d^3*x^2 - 32*
I*c^4*d^3*x - 16*c^3*d^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b x^{2} \log \left (-\frac{c x + i}{c x - i}\right ) - 2 i \, a x^{2}}{2 \, c^{3} d^{3} x^{3} - 6 i \, c^{2} d^{3} x^{2} - 6 \, c d^{3} x + 2 i \, d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*x^2*log(-(c*x + I)/(c*x - I)) - 2*I*a*x^2)/(2*c^3*d^3*x^3 - 6*I*c^2*d^3*x^2 - 6*c*d^3*x + 2*I*d^3
), 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**2*(a+b*atan(c*x))/(d+I*c*d*x)**3,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^{2}}{{\left (i \, c d x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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