3.41 \(\int \frac {(d+i c d x)^4 (a+b \tan ^{-1}(c x))}{x^7} \, dx\)

Optimal. Leaf size=168 \[ -\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}+\frac {i c d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{30 x^5}+\frac {32}{15} i b c^6 d^4 \log (x)-\frac {32}{15} i b c^6 d^4 \log (c x+i)-\frac {13 b c^5 d^4}{6 x}+\frac {16 i b c^4 d^4}{15 x^2}+\frac {5 b c^3 d^4}{9 x^3}-\frac {i b c^2 d^4}{5 x^4}-\frac {b c d^4}{30 x^5} \]

[Out]

-1/30*b*c*d^4/x^5-1/5*I*b*c^2*d^4/x^4+5/9*b*c^3*d^4/x^3+16/15*I*b*c^4*d^4/x^2-13/6*b*c^5*d^4/x-1/6*d^4*(1+I*c*
x)^5*(a+b*arctan(c*x))/x^6+1/30*I*c*d^4*(1+I*c*x)^5*(a+b*arctan(c*x))/x^5+32/15*I*b*c^6*d^4*ln(x)-32/15*I*b*c^
6*d^4*ln(I+c*x)

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Rubi [A]  time = 0.11, antiderivative size = 168, 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} \[ \frac {i c d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{30 x^5}-\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}+\frac {16 i b c^4 d^4}{15 x^2}+\frac {5 b c^3 d^4}{9 x^3}-\frac {i b c^2 d^4}{5 x^4}-\frac {13 b c^5 d^4}{6 x}+\frac {32}{15} i b c^6 d^4 \log (x)-\frac {32}{15} i b c^6 d^4 \log (c x+i)-\frac {b c d^4}{30 x^5} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-(b*c*d^4)/(30*x^5) - ((I/5)*b*c^2*d^4)/x^4 + (5*b*c^3*d^4)/(9*x^3) + (((16*I)/15)*b*c^4*d^4)/x^2 - (13*b*c^5*
d^4)/(6*x) - (d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/(6*x^6) + ((I/30)*c*d^4*(1 + I*c*x)^5*(a + b*ArcTan[c*x])
)/x^5 + ((32*I)/15)*b*c^6*d^4*Log[x] - ((32*I)/15)*b*c^6*d^4*Log[I + c*x]

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)^4 \left (a+b \tan ^{-1}(c x)\right )}{x^7} \, dx &=-\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}+\frac {i c d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{30 x^5}-(b c) \int \frac {d^4 (i-c x)^4 (-5 i-c x)}{30 x^6 (i+c x)} \, dx\\ &=-\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}+\frac {i c d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{30 x^5}-\frac {1}{30} \left (b c d^4\right ) \int \frac {(i-c x)^4 (-5 i-c x)}{x^6 (i+c x)} \, dx\\ &=-\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}+\frac {i c d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{30 x^5}-\frac {1}{30} \left (b c d^4\right ) \int \left (-\frac {5}{x^6}-\frac {24 i c}{x^5}+\frac {50 c^2}{x^4}+\frac {64 i c^3}{x^3}-\frac {65 c^4}{x^2}-\frac {64 i c^5}{x}+\frac {64 i c^6}{i+c x}\right ) \, dx\\ &=-\frac {b c d^4}{30 x^5}-\frac {i b c^2 d^4}{5 x^4}+\frac {5 b c^3 d^4}{9 x^3}+\frac {16 i b c^4 d^4}{15 x^2}-\frac {13 b c^5 d^4}{6 x}-\frac {d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}+\frac {i c d^4 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )}{30 x^5}+\frac {32}{15} i b c^6 d^4 \log (x)-\frac {32}{15} i b c^6 d^4 \log (i+c x)\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 235, normalized size = 1.40 \[ -\frac {d^4 \left (15 a c^4 x^4-40 i a c^3 x^3-45 a c^2 x^2+24 i a c x+5 a-64 i b c^6 x^6 \log (x)-32 i b c^4 x^4+15 b c^4 x^4 \tan ^{-1}(c x)-40 i b c^3 x^3 \tan ^{-1}(c x)+b c x \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-c^2 x^2\right )+6 i b c^2 x^2-45 b c^2 x^2 \tan ^{-1}(c x)+32 i b c^6 x^6 \log \left (c^2 x^2+1\right )+15 b c^5 x^5 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )-15 b c^3 x^3 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )+24 i b c x \tan ^{-1}(c x)+5 b \tan ^{-1}(c x)\right )}{30 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-1/30*(d^4*(5*a + (24*I)*a*c*x - 45*a*c^2*x^2 + (6*I)*b*c^2*x^2 - (40*I)*a*c^3*x^3 + 15*a*c^4*x^4 - (32*I)*b*c
^4*x^4 + 5*b*ArcTan[c*x] + (24*I)*b*c*x*ArcTan[c*x] - 45*b*c^2*x^2*ArcTan[c*x] - (40*I)*b*c^3*x^3*ArcTan[c*x]
+ 15*b*c^4*x^4*ArcTan[c*x] + b*c*x*Hypergeometric2F1[-5/2, 1, -3/2, -(c^2*x^2)] - 15*b*c^3*x^3*Hypergeometric2
F1[-3/2, 1, -1/2, -(c^2*x^2)] + 15*b*c^5*x^5*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] - (64*I)*b*c^6*x^6*Lo
g[x] + (32*I)*b*c^6*x^6*Log[1 + c^2*x^2]))/x^6

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fricas [A]  time = 0.46, size = 215, normalized size = 1.28 \[ \frac {384 i \, b c^{6} d^{4} x^{6} \log \relax (x) - 387 i \, b c^{6} d^{4} x^{6} \log \left (\frac {c x + i}{c}\right ) + 3 i \, b c^{6} d^{4} x^{6} \log \left (\frac {c x - i}{c}\right ) - 390 \, b c^{5} d^{4} x^{5} - 6 \, {\left (15 \, a - 32 i \, b\right )} c^{4} d^{4} x^{4} + {\left (240 i \, a + 100 \, b\right )} c^{3} d^{4} x^{3} + 18 \, {\left (15 \, a - 2 i \, b\right )} c^{2} d^{4} x^{2} + {\left (-144 i \, a - 6 \, b\right )} c d^{4} x - 30 \, a d^{4} + {\left (-45 i \, b c^{4} d^{4} x^{4} - 120 \, b c^{3} d^{4} x^{3} + 135 i \, b c^{2} d^{4} x^{2} + 72 \, b c d^{4} x - 15 i \, b d^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{180 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^7,x, algorithm="fricas")

[Out]

1/180*(384*I*b*c^6*d^4*x^6*log(x) - 387*I*b*c^6*d^4*x^6*log((c*x + I)/c) + 3*I*b*c^6*d^4*x^6*log((c*x - I)/c)
- 390*b*c^5*d^4*x^5 - 6*(15*a - 32*I*b)*c^4*d^4*x^4 + (240*I*a + 100*b)*c^3*d^4*x^3 + 18*(15*a - 2*I*b)*c^2*d^
4*x^2 + (-144*I*a - 6*b)*c*d^4*x - 30*a*d^4 + (-45*I*b*c^4*d^4*x^4 - 120*b*c^3*d^4*x^3 + 135*I*b*c^2*d^4*x^2 +
72*b*c*d^4*x - 15*I*b*d^4)*log(-(c*x + I)/(c*x - I)))/x^6

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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)^4*(a+b*arctan(c*x))/x^7,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.07, size = 243, normalized size = 1.45 \[ \frac {4 i c^{3} d^{4} a}{3 x^{3}}+\frac {3 c^{2} d^{4} a}{2 x^{4}}-\frac {4 i c \,d^{4} b \arctan \left (c x \right )}{5 x^{5}}-\frac {c^{4} d^{4} a}{2 x^{2}}-\frac {d^{4} a}{6 x^{6}}-\frac {4 i c \,d^{4} a}{5 x^{5}}+\frac {3 c^{2} d^{4} b \arctan \left (c x \right )}{2 x^{4}}+\frac {4 i c^{3} d^{4} b \arctan \left (c x \right )}{3 x^{3}}-\frac {c^{4} d^{4} b \arctan \left (c x \right )}{2 x^{2}}-\frac {d^{4} b \arctan \left (c x \right )}{6 x^{6}}+\frac {16 i b \,c^{4} d^{4}}{15 x^{2}}-\frac {16 i c^{6} d^{4} b \ln \left (c^{2} x^{2}+1\right )}{15}-\frac {i b \,c^{2} d^{4}}{5 x^{4}}-\frac {b c \,d^{4}}{30 x^{5}}+\frac {5 b \,c^{3} d^{4}}{9 x^{3}}-\frac {13 b \,c^{5} d^{4}}{6 x}+\frac {32 i c^{6} d^{4} b \ln \left (c x \right )}{15}-\frac {13 c^{6} d^{4} b \arctan \left (c x \right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^7,x)

[Out]

4/3*I*c^3*d^4*a/x^3+3/2*c^2*d^4*a/x^4-4/5*I*c*d^4*b*arctan(c*x)/x^5-1/2*c^4*d^4*a/x^2-1/6*d^4*a/x^6-4/5*I*c*d^
4*a/x^5+3/2*c^2*d^4*b*arctan(c*x)/x^4+4/3*I*c^3*d^4*b*arctan(c*x)/x^3-1/2*c^4*d^4*b*arctan(c*x)/x^2-1/6*d^4*b*
arctan(c*x)/x^6+16/15*I*b*c^4*d^4/x^2-16/15*I*c^6*d^4*b*ln(c^2*x^2+1)-1/5*I*b*c^2*d^4/x^4-1/30*b*c*d^4/x^5+5/9
*b*c^3*d^4/x^3-13/6*b*c^5*d^4/x+32/15*I*c^6*d^4*b*ln(c*x)-13/6*c^6*d^4*b*arctan(c*x)

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maxima [B]  time = 0.42, size = 290, normalized size = 1.73 \[ -\frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c^{4} d^{4} - \frac {2}{3} i \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c^{3} d^{4} - \frac {1}{2} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b c^{2} d^{4} - \frac {1}{5} i \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b c d^{4} - \frac {a c^{4} d^{4}}{2 \, x^{2}} - \frac {1}{90} \, {\left ({\left (15 \, c^{5} \arctan \left (c x\right ) + \frac {15 \, c^{4} x^{4} - 5 \, c^{2} x^{2} + 3}{x^{5}}\right )} c + \frac {15 \, \arctan \left (c x\right )}{x^{6}}\right )} b d^{4} + \frac {4 i \, a c^{3} d^{4}}{3 \, x^{3}} + \frac {3 \, a c^{2} d^{4}}{2 \, x^{4}} - \frac {4 i \, a c d^{4}}{5 \, x^{5}} - \frac {a d^{4}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^7,x, algorithm="maxima")

[Out]

-1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c^4*d^4 - 2/3*I*((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^3*d^4 - 1/2*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b
*c^2*d^4 - 1/5*I*((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*c*d
^4 - 1/2*a*c^4*d^4/x^2 - 1/90*((15*c^5*arctan(c*x) + (15*c^4*x^4 - 5*c^2*x^2 + 3)/x^5)*c + 15*arctan(c*x)/x^6)
*b*d^4 + 4/3*I*a*c^3*d^4/x^3 + 3/2*a*c^2*d^4/x^4 - 4/5*I*a*c*d^4/x^5 - 1/6*a*d^4/x^6

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mupad [B]  time = 0.90, size = 208, normalized size = 1.24 \[ -\frac {d^4\,\left (195\,b\,c^5\,\mathrm {atan}\left (x\,\sqrt {c^2}\right )\,\sqrt {c^2}+b\,c^6\,\ln \left (c^2\,x^2+1\right )\,96{}\mathrm {i}-b\,c^6\,\ln \relax (x)\,192{}\mathrm {i}\right )}{90}-\frac {\frac {d^4\,\left (15\,a+15\,b\,\mathrm {atan}\left (c\,x\right )\right )}{90}+\frac {d^4\,x\,\left (a\,c\,72{}\mathrm {i}+3\,b\,c+b\,c\,\mathrm {atan}\left (c\,x\right )\,72{}\mathrm {i}\right )}{90}+\frac {d^4\,x^4\,\left (45\,a\,c^4+45\,b\,c^4\,\mathrm {atan}\left (c\,x\right )-b\,c^4\,96{}\mathrm {i}\right )}{90}-\frac {d^4\,x^2\,\left (135\,a\,c^2+135\,b\,c^2\,\mathrm {atan}\left (c\,x\right )-b\,c^2\,18{}\mathrm {i}\right )}{90}-\frac {d^4\,x^3\,\left (a\,c^3\,120{}\mathrm {i}+50\,b\,c^3+b\,c^3\,\mathrm {atan}\left (c\,x\right )\,120{}\mathrm {i}\right )}{90}+\frac {13\,b\,c^5\,d^4\,x^5}{6}}{x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^4)/x^7,x)

[Out]

- (d^4*(b*c^6*log(c^2*x^2 + 1)*96i - b*c^6*log(x)*192i + 195*b*c^5*atan(x*(c^2)^(1/2))*(c^2)^(1/2)))/90 - ((d^
4*(15*a + 15*b*atan(c*x)))/90 + (d^4*x*(a*c*72i + 3*b*c + b*c*atan(c*x)*72i))/90 + (d^4*x^4*(45*a*c^4 - b*c^4*
96i + 45*b*c^4*atan(c*x)))/90 - (d^4*x^2*(135*a*c^2 - b*c^2*18i + 135*b*c^2*atan(c*x)))/90 - (d^4*x^3*(a*c^3*1
20i + 50*b*c^3 + b*c^3*atan(c*x)*120i))/90 + (13*b*c^5*d^4*x^5)/6)/x^6

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sympy [B]  time = 138.24, size = 388, normalized size = 2.31 \[ \frac {32 i b c^{6} d^{4} \log {\left (2121535 b^{2} c^{13} d^{8} x \right )}}{15} + \frac {i b c^{6} d^{4} \log {\left (2121535 b^{2} c^{13} d^{8} x - 2121535 i b^{2} c^{12} d^{8} \right )}}{60} - \frac {43 i b c^{6} d^{4} \log {\left (2121535 b^{2} c^{13} d^{8} x + 2121535 i b^{2} c^{12} d^{8} \right )}}{20} + \frac {\left (- 15 i b c^{4} d^{4} x^{4} - 40 b c^{3} d^{4} x^{3} + 45 i b c^{2} d^{4} x^{2} + 24 b c d^{4} x - 5 i b d^{4}\right ) \log {\left (- i c x + 1 \right )}}{60 x^{6}} + \frac {\left (15 i b c^{4} d^{4} x^{4} + 40 b c^{3} d^{4} x^{3} - 45 i b c^{2} d^{4} x^{2} - 24 b c d^{4} x + 5 i b d^{4}\right ) \log {\left (i c x + 1 \right )}}{60 x^{6}} + \frac {- 15 a d^{4} - 195 b c^{5} d^{4} x^{5} + x^{4} \left (- 45 a c^{4} d^{4} + 96 i b c^{4} d^{4}\right ) + x^{3} \left (120 i a c^{3} d^{4} + 50 b c^{3} d^{4}\right ) + x^{2} \left (135 a c^{2} d^{4} - 18 i b c^{2} d^{4}\right ) + x \left (- 72 i a c d^{4} - 3 b c d^{4}\right )}{90 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**4*(a+b*atan(c*x))/x**7,x)

[Out]

32*I*b*c**6*d**4*log(2121535*b**2*c**13*d**8*x)/15 + I*b*c**6*d**4*log(2121535*b**2*c**13*d**8*x - 2121535*I*b
**2*c**12*d**8)/60 - 43*I*b*c**6*d**4*log(2121535*b**2*c**13*d**8*x + 2121535*I*b**2*c**12*d**8)/20 + (-15*I*b
*c**4*d**4*x**4 - 40*b*c**3*d**4*x**3 + 45*I*b*c**2*d**4*x**2 + 24*b*c*d**4*x - 5*I*b*d**4)*log(-I*c*x + 1)/(6
0*x**6) + (15*I*b*c**4*d**4*x**4 + 40*b*c**3*d**4*x**3 - 45*I*b*c**2*d**4*x**2 - 24*b*c*d**4*x + 5*I*b*d**4)*l
og(I*c*x + 1)/(60*x**6) + (-15*a*d**4 - 195*b*c**5*d**4*x**5 + x**4*(-45*a*c**4*d**4 + 96*I*b*c**4*d**4) + x**
3*(120*I*a*c**3*d**4 + 50*b*c**3*d**4) + x**2*(135*a*c**2*d**4 - 18*I*b*c**2*d**4) + x*(-72*I*a*c*d**4 - 3*b*c
*d**4))/(90*x**6)

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