3.8 \(\int \frac{(d+i c d x) (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=106 \[ -\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{i b c^2 d}{2 x}-\frac{1}{3} b c^3 d \log (x)-\frac{1}{12} b c^3 d \log (-c x+i)+\frac{5}{12} b c^3 d \log (c x+i)-\frac{b c d}{6 x^2} \]

[Out]

-(b*c*d)/(6*x^2) - ((I/2)*b*c^2*d)/x - (d*(a + b*ArcTan[c*x]))/(3*x^3) - ((I/2)*c*d*(a + b*ArcTan[c*x]))/x^2 -
(b*c^3*d*Log[x])/3 - (b*c^3*d*Log[I - c*x])/12 + (5*b*c^3*d*Log[I + c*x])/12

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Rubi [A]  time = 0.0908906, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {43, 4872, 12, 801} \[ -\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{i b c^2 d}{2 x}-\frac{1}{3} b c^3 d \log (x)-\frac{1}{12} b c^3 d \log (-c x+i)+\frac{5}{12} b c^3 d \log (c x+i)-\frac{b c d}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

-(b*c*d)/(6*x^2) - ((I/2)*b*c^2*d)/x - (d*(a + b*ArcTan[c*x]))/(3*x^3) - ((I/2)*c*d*(a + b*ArcTan[c*x]))/x^2 -
(b*c^3*d*Log[x])/3 - (b*c^3*d*Log[I - c*x])/12 + (5*b*c^3*d*Log[I + c*x])/12

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [C]  time = 0.0520788, size = 94, normalized size = 0.89 \[ -\frac{d \left (3 i b c^2 x^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )+3 i a c x+2 a+2 b c^3 x^3 \log (x)-b c^3 x^3 \log \left (c^2 x^2+1\right )+b c x+b (2+3 i c x) \tan ^{-1}(c x)\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

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

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Maple [A]  time = 0.035, size = 101, normalized size = 1. \begin{align*}{\frac{-{\frac{i}{2}}cda}{{x}^{2}}}-{\frac{da}{3\,{x}^{3}}}-{\frac{{\frac{i}{2}}cdb\arctan \left ( cx \right ) }{{x}^{2}}}-{\frac{db\arctan \left ( cx \right ) }{3\,{x}^{3}}}+{\frac{{c}^{3}db\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6}}-{\frac{i}{2}}{c}^{3}db\arctan \left ( cx \right ) -{\frac{{\frac{i}{2}}{c}^{2}bd}{x}}-{\frac{bcd}{6\,{x}^{2}}}-{\frac{{c}^{3}db\ln \left ( cx \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)*(a+b*arctan(c*x))/x^4,x)

[Out]

-1/2*I*c*d*a/x^2-1/3*d*a/x^3-1/2*I*c*d*b*arctan(c*x)/x^2-1/3*d*b*arctan(c*x)/x^3+1/6*c^3*d*b*ln(c^2*x^2+1)-1/2
*I*c^3*d*b*arctan(c*x)-1/2*I*b*c^2*d/x-1/6*b*c*d/x^2-1/3*c^3*d*b*ln(c*x)

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Maxima [A]  time = 1.48848, size = 117, normalized size = 1.1 \begin{align*} -\frac{1}{2} i \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b c d + \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 d - \frac{i \, a c d}{2 \, x^{2}} - \frac{a d}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.95781, size = 266, normalized size = 2.51 \begin{align*} -\frac{4 \, b c^{3} d x^{3} \log \left (x\right ) - 5 \, b c^{3} d x^{3} \log \left (\frac{c x + i}{c}\right ) + b c^{3} d x^{3} \log \left (\frac{c x - i}{c}\right ) + 6 i \, b c^{2} d x^{2} -{\left (-6 i \, a - 2 \, b\right )} c d x + 4 \, a d -{\left (3 \, b c d x - 2 i \, b d\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/12*(4*b*c^3*d*x^3*log(x) - 5*b*c^3*d*x^3*log((c*x + I)/c) + b*c^3*d*x^3*log((c*x - I)/c) + 6*I*b*c^2*d*x^2
- (-6*I*a - 2*b)*c*d*x + 4*a*d - (3*b*c*d*x - 2*I*b*d)*log(-(c*x + I)/(c*x - I)))/x^3

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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)*(a+b*atan(c*x))/x**4,x)

[Out]

Timed out

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Giac [A]  time = 1.19741, size = 134, normalized size = 1.26 \begin{align*} \frac{5 \, b c^{3} d x^{3} \log \left (c x + i\right ) - b c^{3} d x^{3} \log \left (c x - i\right ) - 4 \, b c^{3} d x^{3} \log \left (x\right ) - 6 \, b c^{2} d i x^{2} - 6 \, b c d i x \arctan \left (c x\right ) - 6 \, a c d i x - 2 \, b c d x - 4 \, b d \arctan \left (c x\right ) - 4 \, a d}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))/x^4,x, algorithm="giac")

[Out]

1/12*(5*b*c^3*d*x^3*log(c*x + i) - b*c^3*d*x^3*log(c*x - i) - 4*b*c^3*d*x^3*log(x) - 6*b*c^2*d*i*x^2 - 6*b*c*d
*i*x*arctan(c*x) - 6*a*c*d*i*x - 2*b*c*d*x - 4*b*d*arctan(c*x) - 4*a*d)/x^3