3.19 \(\int \frac{(d+i c d x)^2 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=171 \[ \frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{4 b c^3 d^2}{15 x^2}-\frac{i b c^2 d^2}{6 x^3}+\frac{i b c^4 d^2}{2 x}+\frac{8}{15} b c^5 d^2 \log (x)-\frac{1}{60} b c^5 d^2 \log (-c x+i)-\frac{31}{60} b c^5 d^2 \log (c x+i)-\frac{b c d^2}{20 x^4} \]

[Out]

-(b*c*d^2)/(20*x^4) - ((I/6)*b*c^2*d^2)/x^3 + (4*b*c^3*d^2)/(15*x^2) + ((I/2)*b*c^4*d^2)/x - (d^2*(a + b*ArcTa
n[c*x]))/(5*x^5) - ((I/2)*c*d^2*(a + b*ArcTan[c*x]))/x^4 + (c^2*d^2*(a + b*ArcTan[c*x]))/(3*x^3) + (8*b*c^5*d^
2*Log[x])/15 - (b*c^5*d^2*Log[I - c*x])/60 - (31*b*c^5*d^2*Log[I + c*x])/60

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Rubi [A]  time = 0.158001, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {43, 4872, 12, 1802} \[ \frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{4 b c^3 d^2}{15 x^2}-\frac{i b c^2 d^2}{6 x^3}+\frac{i b c^4 d^2}{2 x}+\frac{8}{15} b c^5 d^2 \log (x)-\frac{1}{60} b c^5 d^2 \log (-c x+i)-\frac{31}{60} b c^5 d^2 \log (c x+i)-\frac{b c d^2}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(b*c*d^2)/(20*x^4) - ((I/6)*b*c^2*d^2)/x^3 + (4*b*c^3*d^2)/(15*x^2) + ((I/2)*b*c^4*d^2)/x - (d^2*(a + b*ArcTa
n[c*x]))/(5*x^5) - ((I/2)*c*d^2*(a + b*ArcTan[c*x]))/x^4 + (c^2*d^2*(a + b*ArcTan[c*x]))/(3*x^3) + (8*b*c^5*d^
2*Log[x])/15 - (b*c^5*d^2*Log[I - c*x])/60 - (31*b*c^5*d^2*Log[I + c*x])/60

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 4872

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_.), x_Symbol] :> With[{u = I
ntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^2*x^
2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && IntegerQ[2*m] && ((IGtQ[m, 0] && IGtQ[q, 0
]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac{d^2 \left (-6-15 i c x+10 c^2 x^2\right )}{30 x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{30} \left (b c d^2\right ) \int \frac{-6-15 i c x+10 c^2 x^2}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{30} \left (b c d^2\right ) \int \left (-\frac{6}{x^5}-\frac{15 i c}{x^4}+\frac{16 c^2}{x^3}+\frac{15 i c^3}{x^2}-\frac{16 c^4}{x}+\frac{c^5}{2 (-i+c x)}+\frac{31 c^5}{2 (i+c x)}\right ) \, dx\\ &=-\frac{b c d^2}{20 x^4}-\frac{i b c^2 d^2}{6 x^3}+\frac{4 b c^3 d^2}{15 x^2}+\frac{i b c^4 d^2}{2 x}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{8}{15} b c^5 d^2 \log (x)-\frac{1}{60} b c^5 d^2 \log (i-c x)-\frac{31}{60} b c^5 d^2 \log (i+c x)\\ \end{align*}

Mathematica [C]  time = 0.0862631, size = 124, normalized size = 0.73 \[ \frac{d^2 \left (-10 i b c^2 x^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )+20 a c^2 x^2-30 i a c x-12 a+16 b c^3 x^3+32 b c^5 x^5 \log (x)-16 b c^5 x^5 \log \left (c^2 x^2+1\right )+2 b \left (10 c^2 x^2-15 i c x-6\right ) \tan ^{-1}(c x)-3 b c x\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

(d^2*(-12*a - (30*I)*a*c*x - 3*b*c*x + 20*a*c^2*x^2 + 16*b*c^3*x^3 + 2*b*(-6 - (15*I)*c*x + 10*c^2*x^2)*ArcTan
[c*x] - (10*I)*b*c^2*x^2*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 32*b*c^5*x^5*Log[x] - 16*b*c^5*x^5*Log
[1 + c^2*x^2]))/(60*x^5)

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Maple [A]  time = 0.037, size = 172, normalized size = 1. \begin{align*}{\frac{-{\frac{i}{2}}c{d}^{2}a}{{x}^{4}}}-{\frac{{d}^{2}a}{5\,{x}^{5}}}+{\frac{{c}^{2}{d}^{2}a}{3\,{x}^{3}}}-{\frac{{\frac{i}{2}}c{d}^{2}b\arctan \left ( cx \right ) }{{x}^{4}}}-{\frac{b{d}^{2}\arctan \left ( cx \right ) }{5\,{x}^{5}}}+{\frac{b{c}^{2}{d}^{2}\arctan \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{4\,{c}^{5}{d}^{2}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{15}}+{\frac{i}{2}}{c}^{5}{d}^{2}b\arctan \left ( cx \right ) -{\frac{{\frac{i}{6}}b{c}^{2}{d}^{2}}{{x}^{3}}}+{\frac{{\frac{i}{2}}b{c}^{4}{d}^{2}}{x}}-{\frac{bc{d}^{2}}{20\,{x}^{4}}}+{\frac{4\,b{c}^{3}{d}^{2}}{15\,{x}^{2}}}+{\frac{8\,{c}^{5}{d}^{2}b\ln \left ( cx \right ) }{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x)

[Out]

-1/2*I*c*d^2*a/x^4-1/5*d^2*a/x^5+1/3*c^2*d^2*a/x^3-1/2*I*c*d^2*b*arctan(c*x)/x^4-1/5*d^2*b*arctan(c*x)/x^5+1/3
*c^2*d^2*b*arctan(c*x)/x^3-4/15*c^5*d^2*b*ln(c^2*x^2+1)+1/2*I*c^5*d^2*b*arctan(c*x)-1/6*I*b*c^2*d^2/x^3+1/2*I*
b*c^4*d^2/x-1/20*b*c*d^2/x^4+4/15*b*c^3*d^2/x^2+8/15*c^5*d^2*b*ln(c*x)

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Maxima [A]  time = 1.48874, size = 247, normalized size = 1.44 \begin{align*} -\frac{1}{6} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c^{2} d^{2} + \frac{1}{6} i \,{\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac{3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac{3 \, \arctan \left (c x\right )}{x^{4}}\right )} b c d^{2} - \frac{1}{20} \,{\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac{2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac{4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{2} + \frac{a c^{2} d^{2}}{3 \, x^{3}} - \frac{i \, a c d^{2}}{2 \, x^{4}} - \frac{a d^{2}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/6*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*c^2*d^2 + 1/6*I*((3*c^3*arctan(c*
x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*c*d^2 - 1/20*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2
*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*b*d^2 + 1/3*a*c^2*d^2/x^3 - 1/2*I*a*c*d^2/x^4 - 1/5*a*d^2/x^5

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Fricas [A]  time = 2.89968, size = 387, normalized size = 2.26 \begin{align*} \frac{32 \, b c^{5} d^{2} x^{5} \log \left (x\right ) - 31 \, b c^{5} d^{2} x^{5} \log \left (\frac{c x + i}{c}\right ) - b c^{5} d^{2} x^{5} \log \left (\frac{c x - i}{c}\right ) + 30 i \, b c^{4} d^{2} x^{4} + 16 \, b c^{3} d^{2} x^{3} + 10 \,{\left (2 \, a - i \, b\right )} c^{2} d^{2} x^{2} +{\left (-30 i \, a - 3 \, b\right )} c d^{2} x - 12 \, a d^{2} +{\left (10 i \, b c^{2} d^{2} x^{2} + 15 \, b c d^{2} x - 6 i \, b d^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

1/60*(32*b*c^5*d^2*x^5*log(x) - 31*b*c^5*d^2*x^5*log((c*x + I)/c) - b*c^5*d^2*x^5*log((c*x - I)/c) + 30*I*b*c^
4*d^2*x^4 + 16*b*c^3*d^2*x^3 + 10*(2*a - I*b)*c^2*d^2*x^2 + (-30*I*a - 3*b)*c*d^2*x - 12*a*d^2 + (10*I*b*c^2*d
^2*x^2 + 15*b*c*d^2*x - 6*I*b*d^2)*log(-(c*x + I)/(c*x - I)))/x^5

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**2*(a+b*atan(c*x))/x**6,x)

[Out]

Timed out

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Giac [A]  time = 1.38457, size = 228, normalized size = 1.33 \begin{align*} -\frac{31 \, b c^{5} d^{2} x^{5} \log \left (c x + i\right ) + b c^{5} d^{2} x^{5} \log \left (c x - i\right ) - 32 \, b c^{5} d^{2} x^{5} \log \left (x\right ) - 30 \, b c^{4} d^{2} i x^{4} - 16 \, b c^{3} d^{2} x^{3} + 10 \, b c^{2} d^{2} i x^{2} - 20 \, b c^{2} d^{2} x^{2} \arctan \left (c x\right ) - 20 \, a c^{2} d^{2} x^{2} + 30 \, b c d^{2} i x \arctan \left (c x\right ) + 30 \, a c d^{2} i x + 3 \, b c d^{2} x + 12 \, b d^{2} \arctan \left (c x\right ) + 12 \, a d^{2}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

-1/60*(31*b*c^5*d^2*x^5*log(c*x + i) + b*c^5*d^2*x^5*log(c*x - i) - 32*b*c^5*d^2*x^5*log(x) - 30*b*c^4*d^2*i*x
^4 - 16*b*c^3*d^2*x^3 + 10*b*c^2*d^2*i*x^2 - 20*b*c^2*d^2*x^2*arctan(c*x) - 20*a*c^2*d^2*x^2 + 30*b*c*d^2*i*x*
arctan(c*x) + 30*a*c*d^2*i*x + 3*b*c*d^2*x + 12*b*d^2*arctan(c*x) + 12*a*d^2)/x^5