3.19 \(\int \frac {(d+i c d x)^2 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=171 \[ \frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {1}{60} b c^5 d^2 \log (-c x+i)-\frac {31}{60} b c^5 d^2 \log (c x+i)+\frac {i b c^4 d^2}{2 x}+\frac {4 b c^3 d^2}{15 x^2}-\frac {i b c^2 d^2}{6 x^3}-\frac {b c d^2}{20 x^4} \]

[Out]

-1/20*b*c*d^2/x^4-1/6*I*b*c^2*d^2/x^3+4/15*b*c^3*d^2/x^2+1/2*I*b*c^4*d^2/x-1/5*d^2*(a+b*arctan(c*x))/x^5-1/2*I
*c*d^2*(a+b*arctan(c*x))/x^4+1/3*c^2*d^2*(a+b*arctan(c*x))/x^3+8/15*b*c^5*d^2*ln(x)-1/60*b*c^5*d^2*ln(I-c*x)-3
1/60*b*c^5*d^2*ln(I+c*x)

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Rubi [A]  time = 0.16, antiderivative size = 171, normalized size of antiderivative = 1.00, 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 = {43, 4872, 12, 1802} \[ \frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}+\frac {4 b c^3 d^2}{15 x^2}-\frac {i b c^2 d^2}{6 x^3}+\frac {i b c^4 d^2}{2 x}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {1}{60} b c^5 d^2 \log (-c x+i)-\frac {31}{60} b c^5 d^2 \log (c x+i)-\frac {b c d^2}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-(b*c*d^2)/(20*x^4) - ((I/6)*b*c^2*d^2)/x^3 + (4*b*c^3*d^2)/(15*x^2) + ((I/2)*b*c^4*d^2)/x - (d^2*(a + b*ArcTa
n[c*x]))/(5*x^5) - ((I/2)*c*d^2*(a + b*ArcTan[c*x]))/x^4 + (c^2*d^2*(a + b*ArcTan[c*x]))/(3*x^3) + (8*b*c^5*d^
2*Log[x])/15 - (b*c^5*d^2*Log[I - c*x])/60 - (31*b*c^5*d^2*Log[I + c*x])/60

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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)^2 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-(b c) \int \frac {d^2 \left (-6-15 i c x+10 c^2 x^2\right )}{30 x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {1}{30} \left (b c d^2\right ) \int \frac {-6-15 i c x+10 c^2 x^2}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {1}{30} \left (b c d^2\right ) \int \left (-\frac {6}{x^5}-\frac {15 i c}{x^4}+\frac {16 c^2}{x^3}+\frac {15 i c^3}{x^2}-\frac {16 c^4}{x}+\frac {c^5}{2 (-i+c x)}+\frac {31 c^5}{2 (i+c x)}\right ) \, dx\\ &=-\frac {b c d^2}{20 x^4}-\frac {i b c^2 d^2}{6 x^3}+\frac {4 b c^3 d^2}{15 x^2}+\frac {i b c^4 d^2}{2 x}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^4}+\frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac {8}{15} b c^5 d^2 \log (x)-\frac {1}{60} b c^5 d^2 \log (i-c x)-\frac {31}{60} b c^5 d^2 \log (i+c x)\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 124, normalized size = 0.73 \[ \frac {d^2 \left (20 a c^2 x^2-30 i a c x-12 a+32 b c^5 x^5 \log (x)+16 b c^3 x^3-10 i b c^2 x^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )+2 b \left (10 c^2 x^2-15 i c x-6\right ) \tan ^{-1}(c x)-16 b c^5 x^5 \log \left (c^2 x^2+1\right )-3 b c x\right )}{60 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

(d^2*(-12*a - (30*I)*a*c*x - 3*b*c*x + 20*a*c^2*x^2 + 16*b*c^3*x^3 + 2*b*(-6 - (15*I)*c*x + 10*c^2*x^2)*ArcTan
[c*x] - (10*I)*b*c^2*x^2*Hypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 32*b*c^5*x^5*Log[x] - 16*b*c^5*x^5*Log
[1 + c^2*x^2]))/(60*x^5)

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fricas [A]  time = 0.45, size = 168, normalized size = 0.98 \[ \frac {32 \, b c^{5} d^{2} x^{5} \log \relax (x) - 31 \, b c^{5} d^{2} x^{5} \log \left (\frac {c x + i}{c}\right ) - b c^{5} d^{2} x^{5} \log \left (\frac {c x - i}{c}\right ) + 30 i \, b c^{4} d^{2} x^{4} + 16 \, b c^{3} d^{2} x^{3} + 10 \, {\left (2 \, a - i \, b\right )} c^{2} d^{2} x^{2} + {\left (-30 i \, a - 3 \, b\right )} c d^{2} x - 12 \, a d^{2} + {\left (10 i \, b c^{2} d^{2} x^{2} + 15 \, b c d^{2} x - 6 i \, b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{60 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

1/60*(32*b*c^5*d^2*x^5*log(x) - 31*b*c^5*d^2*x^5*log((c*x + I)/c) - b*c^5*d^2*x^5*log((c*x - I)/c) + 30*I*b*c^
4*d^2*x^4 + 16*b*c^3*d^2*x^3 + 10*(2*a - I*b)*c^2*d^2*x^2 + (-30*I*a - 3*b)*c*d^2*x - 12*a*d^2 + (10*I*b*c^2*d
^2*x^2 + 15*b*c*d^2*x - 6*I*b*d^2)*log(-(c*x + I)/(c*x - I)))/x^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.04, size = 172, normalized size = 1.01 \[ \frac {c^{2} d^{2} a}{3 x^{3}}-\frac {i c \,d^{2} a}{2 x^{4}}-\frac {d^{2} a}{5 x^{5}}+\frac {c^{2} d^{2} b \arctan \left (c x \right )}{3 x^{3}}-\frac {i c \,d^{2} b \arctan \left (c x \right )}{2 x^{4}}-\frac {d^{2} b \arctan \left (c x \right )}{5 x^{5}}-\frac {i b \,c^{2} d^{2}}{6 x^{3}}+\frac {i b \,c^{4} d^{2}}{2 x}-\frac {b c \,d^{2}}{20 x^{4}}+\frac {4 b \,c^{3} d^{2}}{15 x^{2}}+\frac {8 c^{5} d^{2} b \ln \left (c x \right )}{15}-\frac {4 c^{5} d^{2} b \ln \left (c^{2} x^{2}+1\right )}{15}+\frac {i c^{5} d^{2} b \arctan \left (c x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x)

[Out]

1/3*c^2*d^2*a/x^3-1/2*I*c*d^2*a/x^4-1/5*d^2*a/x^5+1/3*c^2*d^2*b*arctan(c*x)/x^3-1/2*I*c*d^2*b*arctan(c*x)/x^4-
1/5*d^2*b*arctan(c*x)/x^5-1/6*I*b*c^2*d^2/x^3+1/2*I*b*c^4*d^2/x-1/20*b*c*d^2/x^4+4/15*b*c^3*d^2/x^2+8/15*c^5*d
^2*b*ln(c*x)-4/15*c^5*d^2*b*ln(c^2*x^2+1)+1/2*I*c^5*d^2*b*arctan(c*x)

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maxima [A]  time = 0.42, size = 183, normalized size = 1.07 \[ -\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c^{2} d^{2} + \frac {1}{6} i \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b c d^{2} - \frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{2} + \frac {a c^{2} d^{2}}{3 \, x^{3}} - \frac {i \, a c d^{2}}{2 \, x^{4}} - \frac {a d^{2}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/6*((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^2 + 1/6*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^2 - 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^2 + 1/3*a*c^2*d^2/x^3 - 1/2*I*a*c*d^2/x^4 - 1/5*a*d^2/x^5

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mupad [B]  time = 0.92, size = 244, normalized size = 1.43 \[ \frac {8\,b\,c^5\,d^2\,\ln \relax (x)}{15}-\frac {4\,b\,c^5\,d^2\,\ln \left (c^2\,x^2+1\right )}{15}-\frac {\frac {a\,d^2}{5}+\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {4\,b\,c^5\,d^2\,x^5}{15}-\frac {b\,c^6\,d^2\,x^6\,1{}\mathrm {i}}{2}-\frac {c^4\,d^2\,x^4\,\left (a+b\,1{}\mathrm {i}\right )}{3}+\frac {c\,d^2\,x\,\left (b+a\,10{}\mathrm {i}\right )}{20}-\frac {c^2\,d^2\,x^2\,\left (4\,a-b\,5{}\mathrm {i}\right )}{30}+\frac {c^3\,d^2\,x^3\,\left (-13\,b+a\,30{}\mathrm {i}\right )}{60}-\frac {2\,b\,c^2\,d^2\,x^2\,\mathrm {atan}\left (c\,x\right )}{15}+\frac {b\,c^3\,d^2\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}-\frac {b\,c^4\,d^2\,x^4\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {b\,c\,d^2\,x\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}}{c^2\,x^7+x^5}+\frac {b\,c^8\,d^2\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {c^2}}\right )\,1{}\mathrm {i}}{2\,{\left (c^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^2)/x^6,x)

[Out]

(8*b*c^5*d^2*log(x))/15 - (4*b*c^5*d^2*log(c^2*x^2 + 1))/15 - ((a*d^2)/5 + (b*d^2*atan(c*x))/5 - (4*b*c^5*d^2*
x^5)/15 - (b*c^6*d^2*x^6*1i)/2 - (c^4*d^2*x^4*(a + b*1i))/3 + (c*d^2*x*(a*10i + b))/20 - (c^2*d^2*x^2*(4*a - b
*5i))/30 + (c^3*d^2*x^3*(a*30i - 13*b))/60 - (2*b*c^2*d^2*x^2*atan(c*x))/15 + (b*c^3*d^2*x^3*atan(c*x)*1i)/2 -
(b*c^4*d^2*x^4*atan(c*x))/3 + (b*c*d^2*x*atan(c*x)*1i)/2)/(x^5 + c^2*x^7) + (b*c^8*d^2*atan((c^2*x)/(c^2)^(1/
2))*1i)/(2*(c^2)^(3/2))

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sympy [A]  time = 26.07, size = 287, normalized size = 1.68 \[ \frac {8 b c^{5} d^{2} \log {\left (10395 b^{2} c^{11} d^{4} x \right )}}{15} - \frac {b c^{5} d^{2} \log {\left (10395 b^{2} c^{11} d^{4} x - 10395 i b^{2} c^{10} d^{4} \right )}}{60} - \frac {31 b c^{5} d^{2} \log {\left (10395 b^{2} c^{11} d^{4} x + 10395 i b^{2} c^{10} d^{4} \right )}}{60} + \frac {\left (- 10 i b c^{2} d^{2} x^{2} - 15 b c d^{2} x + 6 i b d^{2}\right ) \log {\left (i c x + 1 \right )}}{60 x^{5}} + \frac {\left (10 i b c^{2} d^{2} x^{2} + 15 b c d^{2} x - 6 i b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{60 x^{5}} - \frac {12 a d^{2} - 30 i b c^{4} d^{2} x^{4} - 16 b c^{3} d^{2} x^{3} + x^{2} \left (- 20 a c^{2} d^{2} + 10 i b c^{2} d^{2}\right ) + x \left (30 i a c d^{2} + 3 b c d^{2}\right )}{60 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**2*(a+b*atan(c*x))/x**6,x)

[Out]

8*b*c**5*d**2*log(10395*b**2*c**11*d**4*x)/15 - b*c**5*d**2*log(10395*b**2*c**11*d**4*x - 10395*I*b**2*c**10*d
**4)/60 - 31*b*c**5*d**2*log(10395*b**2*c**11*d**4*x + 10395*I*b**2*c**10*d**4)/60 + (-10*I*b*c**2*d**2*x**2 -
15*b*c*d**2*x + 6*I*b*d**2)*log(I*c*x + 1)/(60*x**5) + (10*I*b*c**2*d**2*x**2 + 15*b*c*d**2*x - 6*I*b*d**2)*l
og(-I*c*x + 1)/(60*x**5) - (12*a*d**2 - 30*I*b*c**4*d**2*x**4 - 16*b*c**3*d**2*x**3 + x**2*(-20*a*c**2*d**2 +
10*I*b*c**2*d**2) + x*(30*I*a*c*d**2 + 3*b*c*d**2))/(60*x**5)

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