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

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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]))

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]))

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*}

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Mathematica [C]  time = 0.11, size = 185, normalized size = 1.23 \[ \frac {d^3 \left (10 i a c^3 x^3+20 a c^2 x^2-15 i a c x-4 a+24 b c^5 x^5 \log (x)+12 b c^3 x^3+10 i b c^3 x^3 \tan ^{-1}(c x)-5 i b c^2 x^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )+20 b c^2 x^2 \tan ^{-1}(c x)-12 b c^5 x^5 \log \left (c^2 x^2+1\right )+10 i b c^4 x^4 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )-b c x-15 i b c x \tan ^{-1}(c x)-4 b \tan ^{-1}(c x)\right )}{20 x^5} \]

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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fricas [A]  time = 0.53, size = 185, normalized size = 1.23 \[ \frac {48 \, b c^{5} d^{3} x^{5} \log \relax (x) - 49 \, b c^{5} d^{3} x^{5} \log \left (\frac {c x + i}{c}\right ) + b c^{5} d^{3} x^{5} \log \left (\frac {c x - i}{c}\right ) + 50 i \, b c^{4} d^{3} x^{4} + {\left (20 i \, a + 24 \, b\right )} c^{3} d^{3} x^{3} + 10 \, {\left (4 \, a - i \, b\right )} c^{2} d^{3} x^{2} + {\left (-30 i \, a - 2 \, b\right )} c d^{3} x - 8 \, a d^{3} - {\left (10 \, b c^{3} d^{3} x^{3} - 20 i \, b c^{2} d^{3} x^{2} - 15 \, b c d^{3} x + 4 i \, b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{40 \, x^{5}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.06, size = 200, normalized size = 1.33 \[ \frac {c^{2} d^{3} a}{x^{3}}-\frac {3 i c \,d^{3} a}{4 x^{4}}-\frac {d^{3} a}{5 x^{5}}+\frac {i c^{3} d^{3} a}{2 x^{2}}+\frac {c^{2} d^{3} b \arctan \left (c x \right )}{x^{3}}-\frac {3 i c \,d^{3} b \arctan \left (c x \right )}{4 x^{4}}-\frac {d^{3} b \arctan \left (c x \right )}{5 x^{5}}+\frac {i c^{3} d^{3} b \arctan \left (c x \right )}{2 x^{2}}-\frac {i b \,c^{2} d^{3}}{4 x^{3}}+\frac {5 i b \,c^{4} d^{3}}{4 x}-\frac {b c \,d^{3}}{20 x^{4}}+\frac {3 b \,c^{3} d^{3}}{5 x^{2}}+\frac {6 c^{5} d^{3} b \ln \left (c x \right )}{5}-\frac {3 c^{5} d^{3} b \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 i c^{5} d^{3} b \arctan \left (c x \right )}{4} \]

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]

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

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maxima [A]  time = 0.42, size = 224, normalized size = 1.49 \[ \frac {1}{2} i \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c^{3} d^{3} - \frac {1}{2} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c^{2} d^{3} + \frac {1}{4} i \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b c d^{3} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{3} + \frac {i \, a c^{3} d^{3}}{2 \, x^{2}} + \frac {a c^{2} d^{3}}{x^{3}} - \frac {3 i \, a c d^{3}}{4 \, x^{4}} - \frac {a d^{3}}{5 \, x^{5}} \]

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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mupad [B]  time = 0.95, size = 174, normalized size = 1.16 \[ \frac {d^3\,\left (24\,b\,c^5\,\ln \relax (x)-12\,b\,c^5\,\ln \left (c^2\,x^2+1\right )+b\,c^4\,\mathrm {atan}\left (x\,\sqrt {c^2}\right )\,\sqrt {c^2}\,25{}\mathrm {i}\right )}{20}+\frac {-\frac {d^3\,\left (4\,a+4\,b\,\mathrm {atan}\left (c\,x\right )\right )}{20}-\frac {d^3\,x\,\left (a\,c\,15{}\mathrm {i}+b\,c+b\,c\,\mathrm {atan}\left (c\,x\right )\,15{}\mathrm {i}\right )}{20}+\frac {d^3\,x^3\,\left (a\,c^3\,10{}\mathrm {i}+12\,b\,c^3+b\,c^3\,\mathrm {atan}\left (c\,x\right )\,10{}\mathrm {i}\right )}{20}+\frac {d^3\,x^2\,\left (20\,a\,c^2+20\,b\,c^2\,\mathrm {atan}\left (c\,x\right )-b\,c^2\,5{}\mathrm {i}\right )}{20}+\frac {b\,c^4\,d^3\,x^4\,5{}\mathrm {i}}{4}}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^3)/x^6,x)

[Out]

(d^3*(24*b*c^5*log(x) - 12*b*c^5*log(c^2*x^2 + 1) + b*c^4*atan(x*(c^2)^(1/2))*(c^2)^(1/2)*25i))/20 + ((d^3*x^3
*(a*c^3*10i + 12*b*c^3 + b*c^3*atan(c*x)*10i))/20 - (d^3*x*(a*c*15i + b*c + b*c*atan(c*x)*15i))/20 - (d^3*(4*a
+ 4*b*atan(c*x)))/20 + (d^3*x^2*(20*a*c^2 - b*c^2*5i + 20*b*c^2*atan(c*x)))/20 + (b*c^4*d^3*x^4*5i)/4)/x^5

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sympy [B]  time = 48.02, size = 326, normalized size = 2.17 \[ \frac {6 b c^{5} d^{3} \log {\left (113975 b^{2} c^{11} d^{6} x \right )}}{5} + \frac {b c^{5} d^{3} \log {\left (113975 b^{2} c^{11} d^{6} x - 113975 i b^{2} c^{10} d^{6} \right )}}{40} - \frac {49 b c^{5} d^{3} \log {\left (113975 b^{2} c^{11} d^{6} x + 113975 i b^{2} c^{10} d^{6} \right )}}{40} + \frac {\left (- 10 b c^{3} d^{3} x^{3} + 20 i b c^{2} d^{3} x^{2} + 15 b c d^{3} x - 4 i b d^{3}\right ) \log {\left (- i c x + 1 \right )}}{40 x^{5}} + \frac {\left (10 b c^{3} d^{3} x^{3} - 20 i b c^{2} d^{3} x^{2} - 15 b c d^{3} x + 4 i b d^{3}\right ) \log {\left (i c x + 1 \right )}}{40 x^{5}} - \frac {4 a d^{3} - 25 i b c^{4} d^{3} x^{4} + x^{3} \left (- 10 i a c^{3} d^{3} - 12 b c^{3} d^{3}\right ) + x^{2} \left (- 20 a c^{2} d^{3} + 5 i b c^{2} d^{3}\right ) + x \left (15 i a c d^{3} + b c d^{3}\right )}{20 x^{5}} \]

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]

6*b*c**5*d**3*log(113975*b**2*c**11*d**6*x)/5 + b*c**5*d**3*log(113975*b**2*c**11*d**6*x - 113975*I*b**2*c**10
*d**6)/40 - 49*b*c**5*d**3*log(113975*b**2*c**11*d**6*x + 113975*I*b**2*c**10*d**6)/40 + (-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(-I*c*x + 1)/(40*x**5) + (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(I*c*x + 1)/(40*x**5) - (4*a*d**3 - 25*I*b*c**4*d**3*x**4 +
x**3*(-10*I*a*c**3*d**3 - 12*b*c**3*d**3) + x**2*(-20*a*c**2*d**3 + 5*I*b*c**2*d**3) + x*(15*I*a*c*d**3 + b*c
*d**3))/(20*x**5)

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