3.40 \(\int \frac{(d+i c d x)^4 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=117 \[ -\frac{d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{11 b c^3 d^4}{10 x^2}-\frac{i b c^2 d^4}{3 x^3}+\frac{3 i b c^4 d^4}{x}+\frac{16}{5} b c^5 d^4 \log (x)-\frac{16}{5} b c^5 d^4 \log (c x+i)-\frac{b c d^4}{20 x^4} \]

[Out]

-(b*c*d^4)/(20*x^4) - ((I/3)*b*c^2*d^4)/x^3 + (11*b*c^3*d^4)/(10*x^2) + ((3*I)*b*c^4*d^4)/x - (d^4*(1 + I*c*x)
^5*(a + b*ArcTan[c*x]))/(5*x^5) + (16*b*c^5*d^4*Log[x])/5 - (16*b*c^5*d^4*Log[I + c*x])/5

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Rubi [A]  time = 0.0972312, antiderivative size = 117, 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} \[ -\frac{d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{11 b c^3 d^4}{10 x^2}-\frac{i b c^2 d^4}{3 x^3}+\frac{3 i b c^4 d^4}{x}+\frac{16}{5} b c^5 d^4 \log (x)-\frac{16}{5} b c^5 d^4 \log (c x+i)-\frac{b c d^4}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(b*c*d^4)/(20*x^4) - ((I/3)*b*c^2*d^4)/x^3 + (11*b*c^3*d^4)/(10*x^2) + ((3*I)*b*c^4*d^4)/x - (d^4*(1 + I*c*x)
^5*(a + b*ArcTan[c*x]))/(5*x^5) + (16*b*c^5*d^4*Log[x])/5 - (16*b*c^5*d^4*Log[I + c*x])/5

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)^4 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-(b c) \int -\frac{i d^4 (i-c x)^4}{5 x^5 (i+c x)} \, dx\\ &=-\frac{d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{1}{5} \left (i b c d^4\right ) \int \frac{(i-c x)^4}{x^5 (i+c x)} \, dx\\ &=-\frac{d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{1}{5} \left (i b c d^4\right ) \int \left (-\frac{i}{x^5}+\frac{5 c}{x^4}+\frac{11 i c^2}{x^3}-\frac{15 c^3}{x^2}-\frac{16 i c^4}{x}+\frac{16 i c^5}{i+c x}\right ) \, dx\\ &=-\frac{b c d^4}{20 x^4}-\frac{i b c^2 d^4}{3 x^3}+\frac{11 b c^3 d^4}{10 x^2}+\frac{3 i b c^4 d^4}{x}-\frac{d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac{16}{5} b c^5 d^4 \log (x)-\frac{16}{5} b c^5 d^4 \log (i+c x)\\ \end{align*}

Mathematica [C]  time = 0.159117, size = 191, normalized size = 1.63 \[ -\frac{d^4 \left (3 \left (-40 i b c^4 x^4 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )+20 a c^4 x^4-40 i a c^3 x^3-40 a c^2 x^2+20 i a c x+4 a-22 b c^3 x^3-64 b c^5 x^5 \log (x)+32 b c^5 x^5 \log \left (c^2 x^2+1\right )+4 b \left (5 c^4 x^4-10 i c^3 x^3-10 c^2 x^2+5 i c x+1\right ) \tan ^{-1}(c x)+b c x\right )+20 i b c^2 x^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-c^2 x^2\right )\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(d^4*((20*I)*b*c^2*x^2*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 3*(4*a + (20*I)*a*c*x + b*c*x - 40*a*c^
2*x^2 - (40*I)*a*c^3*x^3 - 22*b*c^3*x^3 + 20*a*c^4*x^4 + 4*b*(1 + (5*I)*c*x - 10*c^2*x^2 - (10*I)*c^3*x^3 + 5*
c^4*x^4)*ArcTan[c*x] - (40*I)*b*c^4*x^4*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] - 64*b*c^5*x^5*Log[x] + 32
*b*c^5*x^5*Log[1 + c^2*x^2])))/(60*x^5)

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Maple [B]  time = 0.036, size = 230, normalized size = 2. \begin{align*}{\frac{-ic{d}^{4}a}{{x}^{4}}}+{\frac{2\,i{c}^{3}{d}^{4}b\arctan \left ( cx \right ) }{{x}^{2}}}-{\frac{{d}^{4}{c}^{4}a}{x}}-{\frac{{d}^{4}a}{5\,{x}^{5}}}+2\,{\frac{{c}^{2}{d}^{4}a}{{x}^{3}}}+3\,i{c}^{5}{d}^{4}b\arctan \left ( cx \right ) -{\frac{ic{d}^{4}b\arctan \left ( cx \right ) }{{x}^{4}}}-{\frac{b{c}^{4}{d}^{4}\arctan \left ( cx \right ) }{x}}-{\frac{b{d}^{4}\arctan \left ( cx \right ) }{5\,{x}^{5}}}+2\,{\frac{{c}^{2}{d}^{4}b\arctan \left ( cx \right ) }{{x}^{3}}}-{\frac{8\,{c}^{5}{d}^{4}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{5}}+{\frac{2\,i{c}^{3}{d}^{4}a}{{x}^{2}}}-{\frac{{\frac{i}{3}}b{c}^{2}{d}^{4}}{{x}^{3}}}+{\frac{3\,ib{c}^{4}{d}^{4}}{x}}-{\frac{bc{d}^{4}}{20\,{x}^{4}}}+{\frac{11\,b{c}^{3}{d}^{4}}{10\,{x}^{2}}}+{\frac{16\,{c}^{5}{d}^{4}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)^4*(a+b*arctan(c*x))/x^6,x)

[Out]

-I*c*d^4*a/x^4+2*I*c^3*d^4*b*arctan(c*x)/x^2-c^4*d^4*a/x-1/5*d^4*a/x^5+2*c^2*d^4*a/x^3+3*I*c^5*d^4*b*arctan(c*
x)-I*c*d^4*b*arctan(c*x)/x^4-c^4*d^4*b*arctan(c*x)/x-1/5*d^4*b*arctan(c*x)/x^5+2*c^2*d^4*b*arctan(c*x)/x^3-8/5
*c^5*d^4*b*ln(c^2*x^2+1)+2*I*c^3*d^4*a/x^2-1/3*I*b*c^2*d^4/x^3+3*I*b*c^4*d^4/x-1/20*b*c*d^4/x^4+11/10*b*c^3*d^
4/x^2+16/5*c^5*d^4*b*ln(c*x)

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Maxima [B]  time = 1.4758, size = 371, normalized size = 3.17 \begin{align*} -\frac{1}{2} \,{\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^{4} d^{4} + 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^{4} -{\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^{4} - \frac{a c^{4} d^{4}}{x} + \frac{1}{3} 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^{4} - \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^{4} + \frac{2 i \, a c^{3} d^{4}}{x^{2}} + \frac{2 \, a c^{2} d^{4}}{x^{3}} - \frac{i \, a c d^{4}}{x^{4}} - \frac{a d^{4}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.39585, size = 475, normalized size = 4.06 \begin{align*} \frac{192 \, b c^{5} d^{4} x^{5} \log \left (x\right ) - 186 \, b c^{5} d^{4} x^{5} \log \left (\frac{c x + i}{c}\right ) - 6 \, b c^{5} d^{4} x^{5} \log \left (\frac{c x - i}{c}\right ) - 60 \,{\left (a - 3 i \, b\right )} c^{4} d^{4} x^{4} +{\left (120 i \, a + 66 \, b\right )} c^{3} d^{4} x^{3} + 20 \,{\left (6 \, a - i \, b\right )} c^{2} d^{4} x^{2} +{\left (-60 i \, a - 3 \, b\right )} c d^{4} x - 12 \, a d^{4} +{\left (-30 i \, b c^{4} d^{4} x^{4} - 60 \, b c^{3} d^{4} x^{3} + 60 i \, b c^{2} d^{4} x^{2} + 30 \, b c d^{4} x - 6 i \, b d^{4}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

1/60*(192*b*c^5*d^4*x^5*log(x) - 186*b*c^5*d^4*x^5*log((c*x + I)/c) - 6*b*c^5*d^4*x^5*log((c*x - I)/c) - 60*(a
- 3*I*b)*c^4*d^4*x^4 + (120*I*a + 66*b)*c^3*d^4*x^3 + 20*(6*a - I*b)*c^2*d^4*x^2 + (-60*I*a - 3*b)*c*d^4*x -
12*a*d^4 + (-30*I*b*c^4*d^4*x^4 - 60*b*c^3*d^4*x^3 + 60*I*b*c^2*d^4*x^2 + 30*b*c*d^4*x - 6*I*b*d^4)*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)**4*(a+b*atan(c*x))/x**6,x)

[Out]

Timed out

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Giac [B]  time = 1.4451, size = 308, normalized size = 2.63 \begin{align*} -\frac{186 \, b c^{5} d^{4} x^{5} \log \left (c x + i\right ) + 6 \, b c^{5} d^{4} x^{5} \log \left (c x - i\right ) - 192 \, b c^{5} d^{4} x^{5} \log \left (x\right ) - 180 \, b c^{4} d^{4} i x^{4} + 60 \, b c^{4} d^{4} x^{4} \arctan \left (c x\right ) + 60 \, a c^{4} d^{4} x^{4} - 120 \, b c^{3} d^{4} i x^{3} \arctan \left (c x\right ) - 120 \, a c^{3} d^{4} i x^{3} - 66 \, b c^{3} d^{4} x^{3} + 20 \, b c^{2} d^{4} i x^{2} - 120 \, b c^{2} d^{4} x^{2} \arctan \left (c x\right ) - 120 \, a c^{2} d^{4} x^{2} + 60 \, b c d^{4} i x \arctan \left (c x\right ) + 60 \, a c d^{4} i x + 3 \, b c d^{4} x + 12 \, b d^{4} \arctan \left (c x\right ) + 12 \, a d^{4}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

-1/60*(186*b*c^5*d^4*x^5*log(c*x + i) + 6*b*c^5*d^4*x^5*log(c*x - i) - 192*b*c^5*d^4*x^5*log(x) - 180*b*c^4*d^
4*i*x^4 + 60*b*c^4*d^4*x^4*arctan(c*x) + 60*a*c^4*d^4*x^4 - 120*b*c^3*d^4*i*x^3*arctan(c*x) - 120*a*c^3*d^4*i*
x^3 - 66*b*c^3*d^4*x^3 + 20*b*c^2*d^4*i*x^2 - 120*b*c^2*d^4*x^2*arctan(c*x) - 120*a*c^2*d^4*x^2 + 60*b*c*d^4*i
*x*arctan(c*x) + 60*a*c*d^4*i*x + 3*b*c*d^4*x + 12*b*d^4*arctan(c*x) + 12*a*d^4)/x^5