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

Optimal. Leaf size=178 \[ -\frac {d^4 (1+i c x)^6 \left (a+b \tan ^{-1}(c x)\right )}{6 c^2}+\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}+\frac {b d^4 (-c x+i)^5}{30 c^2}+\frac {i b d^4 (-c x+i)^4}{30 c^2}-\frac {4 b d^4 (-c x+i)^3}{45 c^2}-\frac {4 i b d^4 (-c x+i)^2}{15 c^2}+\frac {32 i b d^4 \log (c x+i)}{15 c^2}-\frac {16 b d^4 x}{15 c} \]

[Out]

-16/15*b*d^4*x/c-4/15*I*b*d^4*(I-c*x)^2/c^2-4/45*b*d^4*(I-c*x)^3/c^2+1/30*I*b*d^4*(I-c*x)^4/c^2+1/30*b*d^4*(I-
c*x)^5/c^2+1/5*d^4*(1+I*c*x)^5*(a+b*arctan(c*x))/c^2-1/6*d^4*(1+I*c*x)^6*(a+b*arctan(c*x))/c^2+32/15*I*b*d^4*l
n(I+c*x)/c^2

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Rubi [A]  time = 0.11, antiderivative size = 178, 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, 77} \[ -\frac {d^4 (1+i c x)^6 \left (a+b \tan ^{-1}(c x)\right )}{6 c^2}+\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}+\frac {b d^4 (-c x+i)^5}{30 c^2}+\frac {i b d^4 (-c x+i)^4}{30 c^2}-\frac {4 b d^4 (-c x+i)^3}{45 c^2}-\frac {4 i b d^4 (-c x+i)^2}{15 c^2}+\frac {32 i b d^4 \log (c x+i)}{15 c^2}-\frac {16 b d^4 x}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b*d^4*x)/(15*c) - (((4*I)/15)*b*d^4*(I - c*x)^2)/c^2 - (4*b*d^4*(I - c*x)^3)/(45*c^2) + ((I/30)*b*d^4*(I
- c*x)^4)/c^2 + (b*d^4*(I - c*x)^5)/(30*c^2) + (d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/(5*c^2) - (d^4*(1 + I*c
*x)^6*(a + b*ArcTan[c*x]))/(6*c^2) + (((32*I)/15)*b*d^4*Log[I + c*x])/c^2

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 x (d+i c d x)^4 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {d^4 (1+i c x)^6 \left (a+b \tan ^{-1}(c x)\right )}{6 c^2}-(b c) \int \frac {d^4 (i-c x)^4 (i+5 c x)}{30 c^2 (i+c x)} \, dx\\ &=\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {d^4 (1+i c x)^6 \left (a+b \tan ^{-1}(c x)\right )}{6 c^2}-\frac {\left (b d^4\right ) \int \frac {(i-c x)^4 (i+5 c x)}{i+c x} \, dx}{30 c}\\ &=\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {d^4 (1+i c x)^6 \left (a+b \tan ^{-1}(c x)\right )}{6 c^2}-\frac {\left (b d^4\right ) \int \left (32+5 (i-c x)^4+16 i (-i+c x)-8 (-i+c x)^2-4 i (-i+c x)^3-\frac {64 i}{i+c x}\right ) \, dx}{30 c}\\ &=-\frac {16 b d^4 x}{15 c}-\frac {4 i b d^4 (i-c x)^2}{15 c^2}-\frac {4 b d^4 (i-c x)^3}{45 c^2}+\frac {i b d^4 (i-c x)^4}{30 c^2}+\frac {b d^4 (i-c x)^5}{30 c^2}+\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {d^4 (1+i c x)^6 \left (a+b \tan ^{-1}(c x)\right )}{6 c^2}+\frac {32 i b d^4 \log (i+c x)}{15 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 264, normalized size = 1.48 \[ \frac {1}{6} a c^4 d^4 x^6-\frac {4}{5} i a c^3 d^4 x^5-\frac {3}{2} a c^2 d^4 x^4+\frac {4}{3} i a c d^4 x^3+\frac {1}{2} a d^4 x^2+\frac {1}{6} b c^4 d^4 x^6 \tan ^{-1}(c x)-\frac {1}{30} b c^3 d^4 x^5-\frac {4}{5} i b c^3 d^4 x^5 \tan ^{-1}(c x)+\frac {1}{5} i b c^2 d^4 x^4-\frac {3}{2} b c^2 d^4 x^4 \tan ^{-1}(c x)+\frac {16 i b d^4 \log \left (c^2 x^2+1\right )}{15 c^2}+\frac {13 b d^4 \tan ^{-1}(c x)}{6 c^2}+\frac {5}{9} b c d^4 x^3+\frac {4}{3} i b c d^4 x^3 \tan ^{-1}(c x)+\frac {1}{2} b d^4 x^2 \tan ^{-1}(c x)-\frac {13 b d^4 x}{6 c}-\frac {16}{15} i b d^4 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13*b*d^4*x)/(6*c) + (a*d^4*x^2)/2 - ((16*I)/15)*b*d^4*x^2 + ((4*I)/3)*a*c*d^4*x^3 + (5*b*c*d^4*x^3)/9 - (3*a
*c^2*d^4*x^4)/2 + (I/5)*b*c^2*d^4*x^4 - ((4*I)/5)*a*c^3*d^4*x^5 - (b*c^3*d^4*x^5)/30 + (a*c^4*d^4*x^6)/6 + (13
*b*d^4*ArcTan[c*x])/(6*c^2) + (b*d^4*x^2*ArcTan[c*x])/2 + ((4*I)/3)*b*c*d^4*x^3*ArcTan[c*x] - (3*b*c^2*d^4*x^4
*ArcTan[c*x])/2 - ((4*I)/5)*b*c^3*d^4*x^5*ArcTan[c*x] + (b*c^4*d^4*x^6*ArcTan[c*x])/6 + (((16*I)/15)*b*d^4*Log
[1 + c^2*x^2])/c^2

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fricas [A]  time = 1.36, size = 205, normalized size = 1.15 \[ \frac {30 \, a c^{6} d^{4} x^{6} + {\left (-144 i \, a - 6 \, b\right )} c^{5} d^{4} x^{5} - 18 \, {\left (15 \, a - 2 i \, b\right )} c^{4} d^{4} x^{4} + {\left (240 i \, a + 100 \, b\right )} c^{3} d^{4} x^{3} + 6 \, {\left (15 \, a - 32 i \, b\right )} c^{2} d^{4} x^{2} - 390 \, b c d^{4} x + 387 i \, b d^{4} \log \left (\frac {c x + i}{c}\right ) - 3 i \, b d^{4} \log \left (\frac {c x - i}{c}\right ) + {\left (15 i \, b c^{6} d^{4} x^{6} + 72 \, b c^{5} d^{4} x^{5} - 135 i \, b c^{4} d^{4} x^{4} - 120 \, b c^{3} d^{4} x^{3} + 45 i \, b c^{2} d^{4} x^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{180 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(30*a*c^6*d^4*x^6 + (-144*I*a - 6*b)*c^5*d^4*x^5 - 18*(15*a - 2*I*b)*c^4*d^4*x^4 + (240*I*a + 100*b)*c^3
*d^4*x^3 + 6*(15*a - 32*I*b)*c^2*d^4*x^2 - 390*b*c*d^4*x + 387*I*b*d^4*log((c*x + I)/c) - 3*I*b*d^4*log((c*x -
I)/c) + (15*I*b*c^6*d^4*x^6 + 72*b*c^5*d^4*x^5 - 135*I*b*c^4*d^4*x^4 - 120*b*c^3*d^4*x^3 + 45*I*b*c^2*d^4*x^2
)*log(-(c*x + I)/(c*x - I)))/c^2

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

[Out]

sage0*x

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maple [A]  time = 0.03, size = 224, normalized size = 1.26 \[ \frac {c^{4} d^{4} a \,x^{6}}{6}-\frac {4 i c^{3} d^{4} b \arctan \left (c x \right ) x^{5}}{5}-\frac {3 c^{2} d^{4} a \,x^{4}}{2}-\frac {4 i c^{3} d^{4} a \,x^{5}}{5}+\frac {d^{4} a \,x^{2}}{2}+\frac {c^{4} d^{4} b \arctan \left (c x \right ) x^{6}}{6}+\frac {4 i c \,d^{4} b \arctan \left (c x \right ) x^{3}}{3}-\frac {3 c^{2} d^{4} b \arctan \left (c x \right ) x^{4}}{2}+\frac {i c^{2} d^{4} b \,x^{4}}{5}+\frac {d^{4} b \arctan \left (c x \right ) x^{2}}{2}-\frac {13 b \,d^{4} x}{6 c}-\frac {c^{3} d^{4} b \,x^{5}}{30}+\frac {4 i c \,d^{4} a \,x^{3}}{3}+\frac {5 c \,d^{4} b \,x^{3}}{9}-\frac {16 i d^{4} b \,x^{2}}{15}+\frac {16 i d^{4} b \ln \left (c^{2} x^{2}+1\right )}{15 c^{2}}+\frac {13 d^{4} b \arctan \left (c x \right )}{6 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c^4*d^4*a*x^6-4/5*I*c^3*d^4*b*arctan(c*x)*x^5-3/2*c^2*d^4*a*x^4-4/5*I*c^3*d^4*a*x^5+1/2*d^4*a*x^2+1/6*c^4*
d^4*b*arctan(c*x)*x^6+4/3*I*c*d^4*b*arctan(c*x)*x^3-3/2*c^2*d^4*b*arctan(c*x)*x^4+1/5*I*c^2*d^4*b*x^4+1/2*d^4*
b*arctan(c*x)*x^2-13/6*b*d^4*x/c-1/30*c^3*d^4*b*x^5+4/3*I*c*d^4*a*x^3+5/9*c*d^4*b*x^3-16/15*I*d^4*b*x^2+16/15*
I/c^2*d^4*b*ln(c^2*x^2+1)+13/6/c^2*d^4*b*arctan(c*x)

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maxima [B]  time = 0.41, size = 290, normalized size = 1.63 \[ \frac {1}{6} \, a c^{4} d^{4} x^{6} - \frac {4}{5} i \, a c^{3} d^{4} x^{5} - \frac {3}{2} \, a c^{2} d^{4} x^{4} + \frac {1}{90} \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{4} d^{4} - \frac {1}{5} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{4} + \frac {4}{3} i \, a c d^{4} x^{3} - \frac {1}{2} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c^{2} d^{4} + \frac {2}{3} i \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c d^{4} + \frac {1}{2} \, a d^{4} x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^4*d^4*x^6 - 4/5*I*a*c^3*d^4*x^5 - 3/2*a*c^2*d^4*x^4 + 1/90*(15*x^6*arctan(c*x) - c*((3*c^4*x^5 - 5*c^2
*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))*b*c^4*d^4 - 1/5*I*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*lo
g(c^2*x^2 + 1)/c^6))*b*c^3*d^4 + 4/3*I*a*c*d^4*x^3 - 1/2*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arcta
n(c*x)/c^5))*b*c^2*d^4 + 2/3*I*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*c*d^4 + 1/2*a*d^4*x^
2 + 1/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b*d^4

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mupad [B]  time = 0.79, size = 191, normalized size = 1.07 \[ \frac {\frac {d^4\,\left (195\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,96{}\mathrm {i}\right )}{90}-\frac {13\,b\,c\,d^4\,x}{6}}{c^2}+\frac {d^4\,\left (45\,a\,x^2+45\,b\,x^2\,\mathrm {atan}\left (c\,x\right )-b\,x^2\,96{}\mathrm {i}\right )}{90}+\frac {c^4\,d^4\,\left (15\,a\,x^6+15\,b\,x^6\,\mathrm {atan}\left (c\,x\right )\right )}{90}+\frac {c\,d^4\,\left (a\,x^3\,120{}\mathrm {i}+50\,b\,x^3+b\,x^3\,\mathrm {atan}\left (c\,x\right )\,120{}\mathrm {i}\right )}{90}-\frac {c^3\,d^4\,\left (a\,x^5\,72{}\mathrm {i}+3\,b\,x^5+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,72{}\mathrm {i}\right )}{90}-\frac {c^2\,d^4\,\left (135\,a\,x^4+135\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,18{}\mathrm {i}\right )}{90} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^4*(195*b*atan(c*x) + b*log(c^2*x^2 + 1)*96i))/90 - (13*b*c*d^4*x)/6)/c^2 + (d^4*(45*a*x^2 - b*x^2*96i + 45
*b*x^2*atan(c*x)))/90 + (c^4*d^4*(15*a*x^6 + 15*b*x^6*atan(c*x)))/90 + (c*d^4*(a*x^3*120i + 50*b*x^3 + b*x^3*a
tan(c*x)*120i))/90 - (c^3*d^4*(a*x^5*72i + 3*b*x^5 + b*x^5*atan(c*x)*72i))/90 - (c^2*d^4*(135*a*x^4 - b*x^4*18
i + 135*b*x^4*atan(c*x)))/90

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sympy [B]  time = 5.47, size = 360, normalized size = 2.02 \[ \frac {a c^{4} d^{4} x^{6}}{6} - \frac {13 b d^{4} x}{6 c} + \frac {b d^{4} \left (- \frac {i \log {\left (709 b c d^{4} x - 709 i b d^{4} \right )}}{60} + \frac {117 i \log {\left (709 b c d^{4} x + 709 i b d^{4} \right )}}{70}\right )}{c^{2}} + x^{5} \left (- \frac {4 i a c^{3} d^{4}}{5} - \frac {b c^{3} d^{4}}{30}\right ) + x^{4} \left (- \frac {3 a c^{2} d^{4}}{2} + \frac {i b c^{2} d^{4}}{5}\right ) + x^{3} \left (\frac {4 i a c d^{4}}{3} + \frac {5 b c d^{4}}{9}\right ) + x^{2} \left (\frac {a d^{4}}{2} - \frac {16 i b d^{4}}{15}\right ) + \left (- \frac {i b c^{4} d^{4} x^{6}}{12} - \frac {2 b c^{3} d^{4} x^{5}}{5} + \frac {3 i b c^{2} d^{4} x^{4}}{4} + \frac {2 b c d^{4} x^{3}}{3} - \frac {i b d^{4} x^{2}}{4}\right ) \log {\left (i c x + 1 \right )} - \frac {\left (- 35 i b c^{6} d^{4} x^{6} - 168 b c^{5} d^{4} x^{5} + 315 i b c^{4} d^{4} x^{4} + 280 b c^{3} d^{4} x^{3} - 105 i b c^{2} d^{4} x^{2} - 201 i b d^{4}\right ) \log {\left (- i c x + 1 \right )}}{420 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**4*d**4*x**6/6 - 13*b*d**4*x/(6*c) + b*d**4*(-I*log(709*b*c*d**4*x - 709*I*b*d**4)/60 + 117*I*log(709*b*c*
d**4*x + 709*I*b*d**4)/70)/c**2 + x**5*(-4*I*a*c**3*d**4/5 - b*c**3*d**4/30) + x**4*(-3*a*c**2*d**4/2 + I*b*c*
*2*d**4/5) + x**3*(4*I*a*c*d**4/3 + 5*b*c*d**4/9) + x**2*(a*d**4/2 - 16*I*b*d**4/15) + (-I*b*c**4*d**4*x**6/12
- 2*b*c**3*d**4*x**5/5 + 3*I*b*c**2*d**4*x**4/4 + 2*b*c*d**4*x**3/3 - I*b*d**4*x**2/4)*log(I*c*x + 1) - (-35*
I*b*c**6*d**4*x**6 - 168*b*c**5*d**4*x**5 + 315*I*b*c**4*d**4*x**4 + 280*b*c**3*d**4*x**3 - 105*I*b*c**2*d**4*
x**2 - 201*I*b*d**4)*log(-I*c*x + 1)/(420*c**2)

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