3.18 \(\int \frac{(d+i c d x)^2 (a+b \tan ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=161 \[ \frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i b c^2 d^2}{3 x^2}+\frac{3 b c^3 d^2}{4 x}-\frac{2}{3} i b c^4 d^2 \log (x)-\frac{1}{24} i b c^4 d^2 \log (-c x+i)+\frac{17}{24} i b c^4 d^2 \log (c x+i)-\frac{b c d^2}{12 x^3} \]

[Out]

-(b*c*d^2)/(12*x^3) - ((I/3)*b*c^2*d^2)/x^2 + (3*b*c^3*d^2)/(4*x) - (d^2*(a + b*ArcTan[c*x]))/(4*x^4) - (((2*I
)/3)*c*d^2*(a + b*ArcTan[c*x]))/x^3 + (c^2*d^2*(a + b*ArcTan[c*x]))/(2*x^2) - ((2*I)/3)*b*c^4*d^2*Log[x] - (I/
24)*b*c^4*d^2*Log[I - c*x] + ((17*I)/24)*b*c^4*d^2*Log[I + c*x]

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Rubi [A]  time = 0.149881, antiderivative size = 161, 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} \[ \frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i b c^2 d^2}{3 x^2}+\frac{3 b c^3 d^2}{4 x}-\frac{2}{3} i b c^4 d^2 \log (x)-\frac{1}{24} i b c^4 d^2 \log (-c x+i)+\frac{17}{24} i b c^4 d^2 \log (c x+i)-\frac{b c d^2}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

-(b*c*d^2)/(12*x^3) - ((I/3)*b*c^2*d^2)/x^2 + (3*b*c^3*d^2)/(4*x) - (d^2*(a + b*ArcTan[c*x]))/(4*x^4) - (((2*I
)/3)*c*d^2*(a + b*ArcTan[c*x]))/x^3 + (c^2*d^2*(a + b*ArcTan[c*x]))/(2*x^2) - ((2*I)/3)*b*c^4*d^2*Log[x] - (I/
24)*b*c^4*d^2*Log[I - c*x] + ((17*I)/24)*b*c^4*d^2*Log[I + c*x]

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^5} \, dx &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-(b c) \int \frac{d^2 \left (-3-8 i c x+6 c^2 x^2\right )}{12 x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{1}{12} \left (b c d^2\right ) \int \frac{-3-8 i c x+6 c^2 x^2}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{1}{12} \left (b c d^2\right ) \int \left (-\frac{3}{x^4}-\frac{8 i c}{x^3}+\frac{9 c^2}{x^2}+\frac{8 i c^3}{x}+\frac{i c^4}{2 (-i+c x)}-\frac{17 i c^4}{2 (i+c x)}\right ) \, dx\\ &=-\frac{b c d^2}{12 x^3}-\frac{i b c^2 d^2}{3 x^2}+\frac{3 b c^3 d^2}{4 x}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2}{3} i b c^4 d^2 \log (x)-\frac{1}{24} i b c^4 d^2 \log (i-c x)+\frac{17}{24} i b c^4 d^2 \log (i+c x)\\ \end{align*}

Mathematica [C]  time = 0.0679123, size = 152, normalized size = 0.94 \[ \frac{d^2 \left (6 b c^3 x^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )-b c x \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )+6 a c^2 x^2-8 i a c x-3 a-4 i b c^2 x^2-8 i b c^4 x^4 \log (x)+4 i b c^4 x^4 \log \left (c^2 x^2+1\right )+6 b c^2 x^2 \tan ^{-1}(c x)-8 i b c x \tan ^{-1}(c x)-3 b \tan ^{-1}(c x)\right )}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

(d^2*(-3*a - (8*I)*a*c*x + 6*a*c^2*x^2 - (4*I)*b*c^2*x^2 - 3*b*ArcTan[c*x] - (8*I)*b*c*x*ArcTan[c*x] + 6*b*c^2
*x^2*ArcTan[c*x] - b*c*x*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 6*b*c^3*x^3*Hypergeometric2F1[-1/2, 1,
1/2, -(c^2*x^2)] - (8*I)*b*c^4*x^4*Log[x] + (4*I)*b*c^4*x^4*Log[1 + c^2*x^2]))/(12*x^4)

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Maple [A]  time = 0.038, size = 160, normalized size = 1. \begin{align*}{\frac{{c}^{2}{d}^{2}a}{2\,{x}^{2}}}-{\frac{{d}^{2}a}{4\,{x}^{4}}}-{\frac{{\frac{2\,i}{3}}c{d}^{2}a}{{x}^{3}}}+{\frac{b{c}^{2}{d}^{2}\arctan \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{b{d}^{2}\arctan \left ( cx \right ) }{4\,{x}^{4}}}-{\frac{{\frac{2\,i}{3}}c{d}^{2}b\arctan \left ( cx \right ) }{{x}^{3}}}+{\frac{i}{3}}{c}^{4}{d}^{2}b\ln \left ({c}^{2}{x}^{2}+1 \right ) +{\frac{3\,b{c}^{4}{d}^{2}\arctan \left ( cx \right ) }{4}}-{\frac{{\frac{i}{3}}b{c}^{2}{d}^{2}}{{x}^{2}}}-{\frac{2\,i}{3}}{c}^{4}{d}^{2}b\ln \left ( cx \right ) -{\frac{bc{d}^{2}}{12\,{x}^{3}}}+{\frac{3\,b{c}^{3}{d}^{2}}{4\,x}} \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^5,x)

[Out]

1/2*c^2*d^2*a/x^2-1/4*d^2*a/x^4-2/3*I*c*d^2*a/x^3+1/2*c^2*d^2*b*arctan(c*x)/x^2-1/4*d^2*b*arctan(c*x)/x^4-2/3*
I*c*d^2*b*arctan(c*x)/x^3+1/3*I*c^4*d^2*b*ln(c^2*x^2+1)+3/4*b*c^4*d^2*arctan(c*x)-1/3*I*b*c^2*d^2/x^2-2/3*I*c^
4*d^2*b*ln(c*x)-1/12*b*c*d^2/x^3+3/4*b*c^3*d^2/x

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Maxima [A]  time = 1.47861, size = 205, normalized size = 1.27 \begin{align*} \frac{1}{2} \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b c^{2} d^{2} + \frac{1}{3} i \,{\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 d^{2} + \frac{1}{12} \,{\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 d^{2} + \frac{a c^{2} d^{2}}{2 \, x^{2}} - \frac{2 i \, a c d^{2}}{3 \, x^{3}} - \frac{a d^{2}}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c^2*d^2 + 1/3*I*((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*d^2 + 1/12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d
^2 + 1/2*a*c^2*d^2/x^2 - 2/3*I*a*c*d^2/x^3 - 1/4*a*d^2/x^4

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Fricas [A]  time = 2.81822, size = 366, normalized size = 2.27 \begin{align*} \frac{-16 i \, b c^{4} d^{2} x^{4} \log \left (x\right ) + 17 i \, b c^{4} d^{2} x^{4} \log \left (\frac{c x + i}{c}\right ) - i \, b c^{4} d^{2} x^{4} \log \left (\frac{c x - i}{c}\right ) + 18 \, b c^{3} d^{2} x^{3} + 4 \,{\left (3 \, a - 2 i \, b\right )} c^{2} d^{2} x^{2} +{\left (-16 i \, a - 2 \, b\right )} c d^{2} x - 6 \, a d^{2} +{\left (6 i \, b c^{2} d^{2} x^{2} + 8 \, b c d^{2} x - 3 i \, b d^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{24 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

1/24*(-16*I*b*c^4*d^2*x^4*log(x) + 17*I*b*c^4*d^2*x^4*log((c*x + I)/c) - I*b*c^4*d^2*x^4*log((c*x - I)/c) + 18
*b*c^3*d^2*x^3 + 4*(3*a - 2*I*b)*c^2*d^2*x^2 + (-16*I*a - 2*b)*c*d^2*x - 6*a*d^2 + (6*I*b*c^2*d^2*x^2 + 8*b*c*
d^2*x - 3*I*b*d^2)*log(-(c*x + I)/(c*x - I)))/x^4

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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**5,x)

[Out]

Timed out

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Giac [A]  time = 1.20183, size = 217, normalized size = 1.35 \begin{align*} \frac{17 \, b c^{4} d^{2} i x^{4} \log \left (c i x - 1\right ) - b c^{4} d^{2} i x^{4} \log \left (-c i x - 1\right ) - 16 \, b c^{4} d^{2} i x^{4} \log \left (x\right ) + 18 \, b c^{3} d^{2} x^{3} - 8 \, b c^{2} d^{2} i x^{2} + 12 \, b c^{2} d^{2} x^{2} \arctan \left (c x\right ) + 12 \, a c^{2} d^{2} x^{2} - 16 \, b c d^{2} i x \arctan \left (c x\right ) - 16 \, a c d^{2} i x - 2 \, b c d^{2} x - 6 \, b d^{2} \arctan \left (c x\right ) - 6 \, a d^{2}}{24 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

1/24*(17*b*c^4*d^2*i*x^4*log(c*i*x - 1) - b*c^4*d^2*i*x^4*log(-c*i*x - 1) - 16*b*c^4*d^2*i*x^4*log(x) + 18*b*c
^3*d^2*x^3 - 8*b*c^2*d^2*i*x^2 + 12*b*c^2*d^2*x^2*arctan(c*x) + 12*a*c^2*d^2*x^2 - 16*b*c*d^2*i*x*arctan(c*x)
- 16*a*c*d^2*i*x - 2*b*c*d^2*x - 6*b*d^2*arctan(c*x) - 6*a*d^2)/x^4