3.28 \(\int \frac{(d+i c d x)^3 (a+b \tan ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=103 \[ -\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i b c^2 d^3}{2 x^2}+\frac{7 b c^3 d^3}{4 x}-2 i b c^4 d^3 \log (x)+2 i b c^4 d^3 \log (c x+i)-\frac{b c d^3}{12 x^3} \]

[Out]

-(b*c*d^3)/(12*x^3) - ((I/2)*b*c^2*d^3)/x^2 + (7*b*c^3*d^3)/(4*x) - (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x]))/(4
*x^4) - (2*I)*b*c^4*d^3*Log[x] + (2*I)*b*c^4*d^3*Log[I + c*x]

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Rubi [A]  time = 0.0906254, antiderivative size = 103, 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 = {37, 4872, 12, 88} \[ -\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i b c^2 d^3}{2 x^2}+\frac{7 b c^3 d^3}{4 x}-2 i b c^4 d^3 \log (x)+2 i b c^4 d^3 \log (c x+i)-\frac{b c d^3}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

-(b*c*d^3)/(12*x^3) - ((I/2)*b*c^2*d^3)/x^2 + (7*b*c^3*d^3)/(4*x) - (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x]))/(4
*x^4) - (2*I)*b*c^4*d^3*Log[x] + (2*I)*b*c^4*d^3*Log[I + c*x]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )}{x^5} \, dx &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-(b c) \int \frac{d^3 (i-c x)^3}{4 x^4 (i+c x)} \, dx\\ &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{1}{4} \left (b c d^3\right ) \int \frac{(i-c x)^3}{x^4 (i+c x)} \, dx\\ &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{1}{4} \left (b c d^3\right ) \int \left (-\frac{1}{x^4}-\frac{4 i c}{x^3}+\frac{7 c^2}{x^2}+\frac{8 i c^3}{x}-\frac{8 i c^4}{i+c x}\right ) \, dx\\ &=-\frac{b c d^3}{12 x^3}-\frac{i b c^2 d^3}{2 x^2}+\frac{7 b c^3 d^3}{4 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-2 i b c^4 d^3 \log (x)+2 i b c^4 d^3 \log (i+c x)\\ \end{align*}

Mathematica [C]  time = 0.122469, size = 165, normalized size = 1.6 \[ \frac{d^3 \left (-b c x \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )-3 i \left (6 i b c^3 x^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )-4 a c^3 x^3+6 i a c^2 x^2+4 a c x-i a+2 b c^2 x^2+8 b c^4 x^4 \log (x)-4 b c^4 x^4 \log \left (c^2 x^2+1\right )+b \left (-4 c^3 x^3+6 i c^2 x^2+4 c x-i\right ) \tan ^{-1}(c x)\right )\right )}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

(d^3*(-(b*c*x*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)]) - (3*I)*((-I)*a + 4*a*c*x + (6*I)*a*c^2*x^2 + 2*b*
c^2*x^2 - 4*a*c^3*x^3 + b*(-I + 4*c*x + (6*I)*c^2*x^2 - 4*c^3*x^3)*ArcTan[c*x] + (6*I)*b*c^3*x^3*Hypergeometri
c2F1[-1/2, 1, 1/2, -(c^2*x^2)] + 8*b*c^4*x^4*Log[x] - 4*b*c^4*x^4*Log[1 + c^2*x^2])))/(12*x^4)

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Maple [B]  time = 0.036, size = 190, normalized size = 1.8 \begin{align*}{\frac{3\,{c}^{2}{d}^{3}a}{2\,{x}^{2}}}-{\frac{{d}^{3}a}{4\,{x}^{4}}}+{\frac{i{c}^{3}{d}^{3}a}{x}}-{\frac{ic{d}^{3}a}{{x}^{3}}}+{\frac{3\,b{c}^{2}{d}^{3}\arctan \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{4\,{x}^{4}}}+{\frac{i{c}^{3}{d}^{3}b\arctan \left ( cx \right ) }{x}}-{\frac{ic{d}^{3}b\arctan \left ( cx \right ) }{{x}^{3}}}+i{c}^{4}{d}^{3}b\ln \left ({c}^{2}{x}^{2}+1 \right ) +{\frac{7\,b{c}^{4}{d}^{3}\arctan \left ( cx \right ) }{4}}-{\frac{{\frac{i}{2}}{c}^{2}{d}^{3}b}{{x}^{2}}}-2\,i{c}^{4}{d}^{3}b\ln \left ( cx \right ) -{\frac{bc{d}^{3}}{12\,{x}^{3}}}+{\frac{7\,b{c}^{3}{d}^{3}}{4\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^5,x)

[Out]

3/2*c^2*d^3*a/x^2-1/4*d^3*a/x^4+I*c^3*d^3*a/x-I*c*d^3*a/x^3+3/2*c^2*d^3*b*arctan(c*x)/x^2-1/4*d^3*b*arctan(c*x
)/x^4+I*c^3*d^3*b*arctan(c*x)/x-I*c*d^3*b*arctan(c*x)/x^3+I*c^4*d^3*b*ln(c^2*x^2+1)+7/4*b*c^4*d^3*arctan(c*x)-
1/2*I*b*c^2*d^3/x^2-2*I*c^4*d^3*b*ln(c*x)-1/12*b*c*d^3/x^3+7/4*b*c^3*d^3/x

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Maxima [B]  time = 1.49499, size = 273, normalized size = 2.65 \begin{align*} \frac{1}{2} i \,{\left (c{\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac{2 \, \arctan \left (c x\right )}{x}\right )} b c^{3} d^{3} + \frac{3}{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^{3} + \frac{1}{2} 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^{3} + \frac{i \, a c^{3} d^{3}}{x} + \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^{3} + \frac{3 \, a c^{2} d^{3}}{2 \, x^{2}} - \frac{i \, a c d^{3}}{x^{3}} - \frac{a d^{3}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.9531, size = 410, normalized size = 3.98 \begin{align*} \frac{-48 i \, b c^{4} d^{3} x^{4} \log \left (x\right ) + 45 i \, b c^{4} d^{3} x^{4} \log \left (\frac{c x + i}{c}\right ) + 3 i \, b c^{4} d^{3} x^{4} \log \left (\frac{c x - i}{c}\right ) +{\left (24 i \, a + 42 \, b\right )} c^{3} d^{3} x^{3} + 12 \,{\left (3 \, a - i \, b\right )} c^{2} d^{3} x^{2} +{\left (-24 i \, a - 2 \, b\right )} c d^{3} x - 6 \, a d^{3} -{\left (12 \, b c^{3} d^{3} x^{3} - 18 i \, b c^{2} d^{3} x^{2} - 12 \, b c d^{3} x + 3 i \, b d^{3}\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)^3*(a+b*arctan(c*x))/x^5,x, algorithm="fricas")

[Out]

1/24*(-48*I*b*c^4*d^3*x^4*log(x) + 45*I*b*c^4*d^3*x^4*log((c*x + I)/c) + 3*I*b*c^4*d^3*x^4*log((c*x - I)/c) +
(24*I*a + 42*b)*c^3*d^3*x^3 + 12*(3*a - I*b)*c^2*d^3*x^2 + (-24*I*a - 2*b)*c*d^3*x - 6*a*d^3 - (12*b*c^3*d^3*x
^3 - 18*I*b*c^2*d^3*x^2 - 12*b*c*d^3*x + 3*I*b*d^3)*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)**3*(a+b*atan(c*x))/x**5,x)

[Out]

Timed out

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Giac [B]  time = 1.2576, size = 258, normalized size = 2.5 \begin{align*} \frac{45 \, b c^{4} d^{3} i x^{4} \log \left (c i x - 1\right ) + 3 \, b c^{4} d^{3} i x^{4} \log \left (-c i x - 1\right ) - 48 \, b c^{4} d^{3} i x^{4} \log \left (x\right ) + 24 \, b c^{3} d^{3} i x^{3} \arctan \left (c x\right ) + 24 \, a c^{3} d^{3} i x^{3} + 42 \, b c^{3} d^{3} x^{3} - 12 \, b c^{2} d^{3} i x^{2} + 36 \, b c^{2} d^{3} x^{2} \arctan \left (c x\right ) + 36 \, a c^{2} d^{3} x^{2} - 24 \, b c d^{3} i x \arctan \left (c x\right ) - 24 \, a c d^{3} i x - 2 \, b c d^{3} x - 6 \, b d^{3} \arctan \left (c x\right ) - 6 \, a d^{3}}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^5,x, algorithm="giac")

[Out]

1/24*(45*b*c^4*d^3*i*x^4*log(c*i*x - 1) + 3*b*c^4*d^3*i*x^4*log(-c*i*x - 1) - 48*b*c^4*d^3*i*x^4*log(x) + 24*b
*c^3*d^3*i*x^3*arctan(c*x) + 24*a*c^3*d^3*i*x^3 + 42*b*c^3*d^3*x^3 - 12*b*c^2*d^3*i*x^2 + 36*b*c^2*d^3*x^2*arc
tan(c*x) + 36*a*c^2*d^3*x^2 - 24*b*c*d^3*i*x*arctan(c*x) - 24*a*c*d^3*i*x - 2*b*c*d^3*x - 6*b*d^3*arctan(c*x)
- 6*a*d^3)/x^4