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

Optimal. Leaf size=106 \[ -\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {1}{3} b c^3 d \log (x)-\frac {1}{12} b c^3 d \log (-c x+i)+\frac {5}{12} b c^3 d \log (c x+i)-\frac {i b c^2 d}{2 x}-\frac {b c d}{6 x^2} \]

[Out]

-1/6*b*c*d/x^2-1/2*I*b*c^2*d/x-1/3*d*(a+b*arctan(c*x))/x^3-1/2*I*c*d*(a+b*arctan(c*x))/x^2-1/3*b*c^3*d*ln(x)-1
/12*b*c^3*d*ln(I-c*x)+5/12*b*c^3*d*ln(I+c*x)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {43, 4872, 12, 801} \[ -\frac {i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i b c^2 d}{2 x}-\frac {1}{3} b c^3 d \log (x)-\frac {1}{12} b c^3 d \log (-c x+i)+\frac {5}{12} b c^3 d \log (c x+i)-\frac {b c d}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d)/(6*x^2) - ((I/2)*b*c^2*d)/x - (d*(a + b*ArcTan[c*x]))/(3*x^3) - ((I/2)*c*d*(a + b*ArcTan[c*x]))/x^2 -
(b*c^3*d*Log[x])/3 - (b*c^3*d*Log[I - c*x])/12 + (5*b*c^3*d*Log[I + c*x])/12

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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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) \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-(b c) \int \frac {d (-2-3 i c x)}{6 x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {1}{6} (b c d) \int \frac {-2-3 i c x}{x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {1}{6} (b c d) \int \left (-\frac {2}{x^3}-\frac {3 i c}{x^2}+\frac {2 c^2}{x}+\frac {c^3}{2 (-i+c x)}-\frac {5 c^3}{2 (i+c x)}\right ) \, dx\\ &=-\frac {b c d}{6 x^2}-\frac {i b c^2 d}{2 x}-\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {1}{3} b c^3 d \log (x)-\frac {1}{12} b c^3 d \log (i-c x)+\frac {5}{12} b c^3 d \log (i+c x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 94, normalized size = 0.89 \[ -\frac {d \left (3 i a c x+2 a+2 b c^3 x^3 \log (x)+3 i b c^2 x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )-b c^3 x^3 \log \left (c^2 x^2+1\right )+b c x+b (2+3 i c x) \tan ^{-1}(c x)\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 111, normalized size = 1.05 \[ -\frac {4 \, b c^{3} d x^{3} \log \relax (x) - 5 \, b c^{3} d x^{3} \log \left (\frac {c x + i}{c}\right ) + b c^{3} d x^{3} \log \left (\frac {c x - i}{c}\right ) + 6 i \, b c^{2} d x^{2} - {\left (-6 i \, a - 2 \, b\right )} c d x + 4 \, a d - {\left (3 \, b c d x - 2 i \, b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*b*c^3*d*x^3*log(x) - 5*b*c^3*d*x^3*log((c*x + I)/c) + b*c^3*d*x^3*log((c*x - I)/c) + 6*I*b*c^2*d*x^2
- (-6*I*a - 2*b)*c*d*x + 4*a*d - (3*b*c*d*x - 2*I*b*d)*log(-(c*x + I)/(c*x - I)))/x^3

________________________________________________________________________________________

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

[Out]

sage0*x

________________________________________________________________________________________

maple [A]  time = 0.04, size = 101, normalized size = 0.95 \[ -\frac {d a}{3 x^{3}}-\frac {i c d a}{2 x^{2}}-\frac {d b \arctan \left (c x \right )}{3 x^{3}}-\frac {i c d b \arctan \left (c x \right )}{2 x^{2}}-\frac {i b \,c^{2} d}{2 x}-\frac {b c d}{6 x^{2}}-\frac {c^{3} d b \ln \left (c x \right )}{3}+\frac {c^{3} d b \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {i c^{3} d b \arctan \left (c x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)*(a+b*arctan(c*x))/x^4,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 87, normalized size = 0.82 \[ -\frac {1}{2} i \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c d + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d - \frac {i \, a c d}{2 \, x^{2}} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.88, size = 176, normalized size = 1.66 \[ \frac {b\,c^3\,d\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {\frac {a\,d}{3}-x^5\,\left (\frac {b\,c^5\,d}{6}+\frac {a\,c^5\,d\,1{}\mathrm {i}}{2}\right )+\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {c\,d\,x\,\left (b+a\,3{}\mathrm {i}\right )}{6}+\frac {c^2\,d\,x^2\,\left (2\,a+b\,3{}\mathrm {i}\right )}{6}+\frac {b\,c^4\,d\,x^4\,1{}\mathrm {i}}{2}+\frac {b\,c^2\,d\,x^2\,\mathrm {atan}\left (c\,x\right )}{3}+\frac {b\,c^3\,d\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}+\frac {b\,c\,d\,x\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{2}}{c^2\,x^5+x^3}-\frac {b\,c^3\,d\,\ln \relax (x)}{3}-\frac {b\,d\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {c^2}}\right )\,{\left (c^2\right )}^{3/2}\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i))/x^4,x)

[Out]

(b*c^3*d*log(c^2*x^2 + 1))/6 - (b*d*atan((c^2*x)/(c^2)^(1/2))*(c^2)^(3/2)*1i)/2 - ((a*d)/3 - x^5*((a*c^5*d*1i)
/2 + (b*c^5*d)/6) + (b*d*atan(c*x))/3 + (c*d*x*(a*3i + b))/6 + (c^2*d*x^2*(2*a + b*3i))/6 + (b*c^4*d*x^4*1i)/2
+ (b*c^2*d*x^2*atan(c*x))/3 + (b*c^3*d*x^3*atan(c*x)*1i)/2 + (b*c*d*x*atan(c*x)*1i)/2)/(x^3 + c^2*x^5) - (b*c
^3*d*log(x))/3

________________________________________________________________________________________

sympy [A]  time = 4.63, size = 197, normalized size = 1.86 \[ - \frac {b c^{3} d \log {\left (27 b^{2} c^{7} d^{2} x \right )}}{3} - \frac {b c^{3} d \log {\left (27 b^{2} c^{7} d^{2} x - 27 i b^{2} c^{6} d^{2} \right )}}{12} + \frac {5 b c^{3} d \log {\left (27 b^{2} c^{7} d^{2} x + 27 i b^{2} c^{6} d^{2} \right )}}{12} + \frac {\left (- 3 b c d x + 2 i b d\right ) \log {\left (i c x + 1 \right )}}{12 x^{3}} + \frac {\left (3 b c d x - 2 i b d\right ) \log {\left (- i c x + 1 \right )}}{12 x^{3}} + \frac {- 2 a d - 3 i b c^{2} d x^{2} + x \left (- 3 i a c d - b c d\right )}{6 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*atan(c*x))/x**4,x)

[Out]

-b*c**3*d*log(27*b**2*c**7*d**2*x)/3 - b*c**3*d*log(27*b**2*c**7*d**2*x - 27*I*b**2*c**6*d**2)/12 + 5*b*c**3*d
*log(27*b**2*c**7*d**2*x + 27*I*b**2*c**6*d**2)/12 + (-3*b*c*d*x + 2*I*b*d)*log(I*c*x + 1)/(12*x**3) + (3*b*c*
d*x - 2*I*b*d)*log(-I*c*x + 1)/(12*x**3) + (-2*a*d - 3*I*b*c**2*d*x**2 + x*(-3*I*a*c*d - b*c*d))/(6*x**3)

________________________________________________________________________________________