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

Optimal. Leaf size=124 \[ -\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i b c^2 d}{6 x^2}+\frac{b c^3 d}{4 x}-\frac{1}{3} i b c^4 d \log (x)+\frac{1}{24} i b c^4 d \log (-c x+i)+\frac{7}{24} i b c^4 d \log (c x+i)-\frac{b c d}{12 x^3} \]

[Out]

-(b*c*d)/(12*x^3) - ((I/6)*b*c^2*d)/x^2 + (b*c^3*d)/(4*x) - (d*(a + b*ArcTan[c*x]))/(4*x^4) - ((I/3)*c*d*(a +
b*ArcTan[c*x]))/x^3 - (I/3)*b*c^4*d*Log[x] + (I/24)*b*c^4*d*Log[I - c*x] + ((7*I)/24)*b*c^4*d*Log[I + c*x]

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Rubi [A]  time = 0.0964044, antiderivative size = 124, 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} \[ -\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i b c^2 d}{6 x^2}+\frac{b c^3 d}{4 x}-\frac{1}{3} i b c^4 d \log (x)+\frac{1}{24} i b c^4 d \log (-c x+i)+\frac{7}{24} i b c^4 d \log (c x+i)-\frac{b c d}{12 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d)/(12*x^3) - ((I/6)*b*c^2*d)/x^2 + (b*c^3*d)/(4*x) - (d*(a + b*ArcTan[c*x]))/(4*x^4) - ((I/3)*c*d*(a +
b*ArcTan[c*x]))/x^3 - (I/3)*b*c^4*d*Log[x] + (I/24)*b*c^4*d*Log[I - c*x] + ((7*I)/24)*b*c^4*d*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 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^5} \, dx &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac{d (-3-4 i c x)}{12 x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{12} (b c d) \int \frac{-3-4 i c x}{x^4 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{12} (b c d) \int \left (-\frac{3}{x^4}-\frac{4 i c}{x^3}+\frac{3 c^2}{x^2}+\frac{4 i c^3}{x}-\frac{i c^4}{2 (-i+c x)}-\frac{7 i c^4}{2 (i+c x)}\right ) \, dx\\ &=-\frac{b c d}{12 x^3}-\frac{i b c^2 d}{6 x^2}+\frac{b c^3 d}{4 x}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{1}{3} i b c^4 d \log (x)+\frac{1}{24} i b c^4 d \log (i-c x)+\frac{7}{24} i b c^4 d \log (i+c x)\\ \end{align*}

Mathematica [C]  time = 0.0513447, size = 99, normalized size = 0.8 \[ -\frac{b c d \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )}{12 x^3}-\frac{i c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{1}{6} i b c^2 d \left (-c^2 \log \left (c^2 x^2+1\right )+2 c^2 \log (x)+\frac{1}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 112, normalized size = 0.9 \begin{align*} -{\frac{da}{4\,{x}^{4}}}-{\frac{{\frac{i}{3}}cda}{{x}^{3}}}-{\frac{db\arctan \left ( cx \right ) }{4\,{x}^{4}}}-{\frac{{\frac{i}{3}}cdb\arctan \left ( cx \right ) }{{x}^{3}}}+{\frac{i}{6}}{c}^{4}db\ln \left ({c}^{2}{x}^{2}+1 \right ) +{\frac{{c}^{4}db\arctan \left ( cx \right ) }{4}}-{\frac{{\frac{i}{6}}b{c}^{2}d}{{x}^{2}}}-{\frac{i}{3}}{c}^{4}db\ln \left ( cx \right ) -{\frac{bcd}{12\,{x}^{3}}}+{\frac{b{c}^{3}d}{4\,x}} \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^5,x)

[Out]

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

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Maxima [A]  time = 1.49316, size = 138, normalized size = 1.11 \begin{align*} \frac{1}{6} 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 + \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 - \frac{i \, a c d}{3 \, x^{3}} - \frac{a d}{4 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/6*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 + 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 - 1/3*I*a*c*d/x^3 - 1/4*a*d/x^4

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Fricas [A]  time = 2.85988, size = 296, normalized size = 2.39 \begin{align*} \frac{-8 i \, b c^{4} d x^{4} \log \left (x\right ) + 7 i \, b c^{4} d x^{4} \log \left (\frac{c x + i}{c}\right ) + i \, b c^{4} d x^{4} \log \left (\frac{c x - i}{c}\right ) + 6 \, b c^{3} d x^{3} - 4 i \, b c^{2} d x^{2} +{\left (-8 i \, a - 2 \, b\right )} c d x - 6 \, a d +{\left (4 \, b c d x - 3 i \, b d\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)*(a+b*arctan(c*x))/x^5,x, algorithm="fricas")

[Out]

1/24*(-8*I*b*c^4*d*x^4*log(x) + 7*I*b*c^4*d*x^4*log((c*x + I)/c) + I*b*c^4*d*x^4*log((c*x - I)/c) + 6*b*c^3*d*
x^3 - 4*I*b*c^2*d*x^2 + (-8*I*a - 2*b)*c*d*x - 6*a*d + (4*b*c*d*x - 3*I*b*d)*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)*(a+b*atan(c*x))/x**5,x)

[Out]

Timed out

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Giac [A]  time = 1.23585, size = 151, normalized size = 1.22 \begin{align*} \frac{7 \, b c^{4} d i x^{4} \log \left (c i x - 1\right ) + b c^{4} d i x^{4} \log \left (-c i x - 1\right ) - 8 \, b c^{4} d i x^{4} \log \left (x\right ) + 6 \, b c^{3} d x^{3} - 4 \, b c^{2} d i x^{2} - 8 \, b c d i x \arctan \left (c x\right ) - 8 \, a c d i x - 2 \, b c d x - 6 \, b d \arctan \left (c x\right ) - 6 \, a d}{24 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

1/24*(7*b*c^4*d*i*x^4*log(c*i*x - 1) + b*c^4*d*i*x^4*log(-c*i*x - 1) - 8*b*c^4*d*i*x^4*log(x) + 6*b*c^3*d*x^3
- 4*b*c^2*d*i*x^2 - 8*b*c*d*i*x*arctan(c*x) - 8*a*c*d*i*x - 2*b*c*d*x - 6*b*d*arctan(c*x) - 6*a*d)/x^4