3.29 \(\int \frac{(d+i c d x)^3 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=150 \[ \frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{20 x^4}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{3 b c^3 d^3}{5 x^2}-\frac{i b c^2 d^3}{4 x^3}+\frac{5 i b c^4 d^3}{4 x}+\frac{6}{5} b c^5 d^3 \log (x)-\frac{6}{5} b c^5 d^3 \log (c x+i)-\frac{b c d^3}{20 x^4} \]

[Out]

-(b*c*d^3)/(20*x^4) - ((I/4)*b*c^2*d^3)/x^3 + (3*b*c^3*d^3)/(5*x^2) + (((5*I)/4)*b*c^4*d^3)/x - (d^3*(1 + I*c*
x)^4*(a + b*ArcTan[c*x]))/(5*x^5) + ((I/20)*c*d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x]))/x^4 + (6*b*c^5*d^3*Log[x]
)/5 - (6*b*c^5*d^3*Log[I + c*x])/5

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Rubi [A]  time = 0.106787, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {45, 37, 4872, 12, 148} \[ \frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{20 x^4}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{3 b c^3 d^3}{5 x^2}-\frac{i b c^2 d^3}{4 x^3}+\frac{5 i b c^4 d^3}{4 x}+\frac{6}{5} b c^5 d^3 \log (x)-\frac{6}{5} b c^5 d^3 \log (c x+i)-\frac{b c d^3}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(b*c*d^3)/(20*x^4) - ((I/4)*b*c^2*d^3)/x^3 + (3*b*c^3*d^3)/(5*x^2) + (((5*I)/4)*b*c^4*d^3)/x - (d^3*(1 + I*c*
x)^4*(a + b*ArcTan[c*x]))/(5*x^5) + ((I/20)*c*d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x]))/x^4 + (6*b*c^5*d^3*Log[x]
)/5 - (6*b*c^5*d^3*Log[I + c*x])/5

Rule 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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 148

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{20 x^4}-(b c) \int \frac{d^3 (-4 i-c x) (1+i c x)^3}{20 x^5 (i+c x)} \, dx\\ &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{20 x^4}-\frac{1}{20} \left (b c d^3\right ) \int \frac{(-4 i-c x) (1+i c x)^3}{x^5 (i+c x)} \, dx\\ &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{20 x^4}-\frac{1}{20} \left (b c d^3\right ) \int \left (-\frac{4}{x^5}-\frac{15 i c}{x^4}+\frac{24 c^2}{x^3}+\frac{25 i c^3}{x^2}-\frac{24 c^4}{x}+\frac{24 c^5}{i+c x}\right ) \, dx\\ &=-\frac{b c d^3}{20 x^4}-\frac{i b c^2 d^3}{4 x^3}+\frac{3 b c^3 d^3}{5 x^2}+\frac{5 i b c^4 d^3}{4 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )}{20 x^4}+\frac{6}{5} b c^5 d^3 \log (x)-\frac{6}{5} b c^5 d^3 \log (i+c x)\\ \end{align*}

Mathematica [C]  time = 0.094492, size = 185, normalized size = 1.23 \[ \frac{d^3 \left (10 i b c^4 x^4 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )-5 i b c^2 x^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )+10 i a c^3 x^3+20 a c^2 x^2-15 i a c x-4 a+12 b c^3 x^3+24 b c^5 x^5 \log (x)-12 b c^5 x^5 \log \left (c^2 x^2+1\right )+10 i b c^3 x^3 \tan ^{-1}(c x)+20 b c^2 x^2 \tan ^{-1}(c x)-b c x-15 i b c x \tan ^{-1}(c x)-4 b \tan ^{-1}(c x)\right )}{20 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

(d^3*(-4*a - (15*I)*a*c*x - b*c*x + 20*a*c^2*x^2 + (10*I)*a*c^3*x^3 + 12*b*c^3*x^3 - 4*b*ArcTan[c*x] - (15*I)*
b*c*x*ArcTan[c*x] + 20*b*c^2*x^2*ArcTan[c*x] + (10*I)*b*c^3*x^3*ArcTan[c*x] - (5*I)*b*c^2*x^2*Hypergeometric2F
1[-3/2, 1, -1/2, -(c^2*x^2)] + (10*I)*b*c^4*x^4*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] + 24*b*c^5*x^5*Log
[x] - 12*b*c^5*x^5*Log[1 + c^2*x^2]))/(20*x^5)

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Maple [A]  time = 0.038, size = 200, normalized size = 1.3 \begin{align*}{\frac{{\frac{i}{2}}{c}^{3}{d}^{3}a}{{x}^{2}}}-{\frac{{\frac{3\,i}{4}}c{d}^{3}a}{{x}^{4}}}-{\frac{{d}^{3}a}{5\,{x}^{5}}}+{\frac{{c}^{2}{d}^{3}a}{{x}^{3}}}+{\frac{{\frac{i}{2}}{c}^{3}{d}^{3}b\arctan \left ( cx \right ) }{{x}^{2}}}-{\frac{{\frac{3\,i}{4}}c{d}^{3}b\arctan \left ( cx \right ) }{{x}^{4}}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{5\,{x}^{5}}}+{\frac{b{c}^{2}{d}^{3}\arctan \left ( cx \right ) }{{x}^{3}}}-{\frac{3\,{c}^{5}{d}^{3}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{5}}+{\frac{5\,i}{4}}{c}^{5}{d}^{3}b\arctan \left ( cx \right ) -{\frac{{\frac{i}{4}}b{c}^{2}{d}^{3}}{{x}^{3}}}+{\frac{{\frac{5\,i}{4}}b{c}^{4}{d}^{3}}{x}}-{\frac{bc{d}^{3}}{20\,{x}^{4}}}+{\frac{3\,b{c}^{3}{d}^{3}}{5\,{x}^{2}}}+{\frac{6\,{c}^{5}{d}^{3}b\ln \left ( cx \right ) }{5}} \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^6,x)

[Out]

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

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Maxima [A]  time = 1.48809, size = 302, normalized size = 2.01 \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^{3} d^{3} - \frac{1}{2} \,{\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^{3} + \frac{1}{4} 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^{3} - \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^{3} + \frac{i \, a c^{3} d^{3}}{2 \, x^{2}} + \frac{a c^{2} d^{3}}{x^{3}} - \frac{3 i \, a c d^{3}}{4 \, x^{4}} - \frac{a d^{3}}{5 \, x^{5}} \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^6,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.87688, size = 427, normalized size = 2.85 \begin{align*} \frac{48 \, b c^{5} d^{3} x^{5} \log \left (x\right ) - 49 \, b c^{5} d^{3} x^{5} \log \left (\frac{c x + i}{c}\right ) + b c^{5} d^{3} x^{5} \log \left (\frac{c x - i}{c}\right ) + 50 i \, b c^{4} d^{3} x^{4} +{\left (20 i \, a + 24 \, b\right )} c^{3} d^{3} x^{3} + 10 \,{\left (4 \, a - i \, b\right )} c^{2} d^{3} x^{2} +{\left (-30 i \, a - 2 \, b\right )} c d^{3} x - 8 \, a d^{3} -{\left (10 \, b c^{3} d^{3} x^{3} - 20 i \, b c^{2} d^{3} x^{2} - 15 \, b c d^{3} x + 4 i \, b d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{40 \, x^{5}} \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^6,x, algorithm="fricas")

[Out]

1/40*(48*b*c^5*d^3*x^5*log(x) - 49*b*c^5*d^3*x^5*log((c*x + I)/c) + b*c^5*d^3*x^5*log((c*x - I)/c) + 50*I*b*c^
4*d^3*x^4 + (20*I*a + 24*b)*c^3*d^3*x^3 + 10*(4*a - I*b)*c^2*d^3*x^2 + (-30*I*a - 2*b)*c*d^3*x - 8*a*d^3 - (10
*b*c^3*d^3*x^3 - 20*I*b*c^2*d^3*x^2 - 15*b*c*d^3*x + 4*I*b*d^3)*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)**3*(a+b*atan(c*x))/x**6,x)

[Out]

Timed out

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Giac [A]  time = 1.56752, size = 270, normalized size = 1.8 \begin{align*} -\frac{49 \, b c^{5} d^{3} x^{5} \log \left (c x + i\right ) - b c^{5} d^{3} x^{5} \log \left (c x - i\right ) - 48 \, b c^{5} d^{3} x^{5} \log \left (x\right ) - 50 \, b c^{4} d^{3} i x^{4} - 20 \, b c^{3} d^{3} i x^{3} \arctan \left (c x\right ) - 20 \, a c^{3} d^{3} i x^{3} - 24 \, b c^{3} d^{3} x^{3} + 10 \, b c^{2} d^{3} i x^{2} - 40 \, b c^{2} d^{3} x^{2} \arctan \left (c x\right ) - 40 \, a c^{2} d^{3} x^{2} + 30 \, b c d^{3} i x \arctan \left (c x\right ) + 30 \, a c d^{3} i x + 2 \, b c d^{3} x + 8 \, b d^{3} \arctan \left (c x\right ) + 8 \, a d^{3}}{40 \, x^{5}} \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^6,x, algorithm="giac")

[Out]

-1/40*(49*b*c^5*d^3*x^5*log(c*x + i) - b*c^5*d^3*x^5*log(c*x - i) - 48*b*c^5*d^3*x^5*log(x) - 50*b*c^4*d^3*i*x
^4 - 20*b*c^3*d^3*i*x^3*arctan(c*x) - 20*a*c^3*d^3*i*x^3 - 24*b*c^3*d^3*x^3 + 10*b*c^2*d^3*i*x^2 - 40*b*c^2*d^
3*x^2*arctan(c*x) - 40*a*c^2*d^3*x^2 + 30*b*c*d^3*i*x*arctan(c*x) + 30*a*c*d^3*i*x + 2*b*c*d^3*x + 8*b*d^3*arc
tan(c*x) + 8*a*d^3)/x^5