3.27 \(\int \frac {(d+i c d x)^3 (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=189 \[ \frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-i a c^3 d^3 \log (x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)-\frac {10}{3} b c^3 d^3 \log (x)-\frac {3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac {3 i b c^2 d^3}{2 x}+\frac {5}{3} b c^3 d^3 \log \left (c^2 x^2+1\right )-\frac {b c d^3}{6 x^2} \]

[Out]

-1/6*b*c*d^3/x^2-3/2*I*b*c^2*d^3/x-3/2*I*b*c^3*d^3*arctan(c*x)-1/3*d^3*(a+b*arctan(c*x))/x^3-3/2*I*c*d^3*(a+b*
arctan(c*x))/x^2+3*c^2*d^3*(a+b*arctan(c*x))/x-I*a*c^3*d^3*ln(x)-10/3*b*c^3*d^3*ln(x)+5/3*b*c^3*d^3*ln(c^2*x^2
+1)+1/2*b*c^3*d^3*polylog(2,-I*c*x)-1/2*b*c^3*d^3*polylog(2,I*c*x)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4876, 4852, 266, 44, 325, 203, 36, 29, 31, 4848, 2391} \[ \frac {1}{2} b c^3 d^3 \text {PolyLog}(2,-i c x)-\frac {1}{2} b c^3 d^3 \text {PolyLog}(2,i c x)+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-i a c^3 d^3 \log (x)+\frac {5}{3} b c^3 d^3 \log \left (c^2 x^2+1\right )-\frac {3 i b c^2 d^3}{2 x}-\frac {10}{3} b c^3 d^3 \log (x)-\frac {3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac {b c d^3}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

-(b*c*d^3)/(6*x^2) - (((3*I)/2)*b*c^2*d^3)/x - ((3*I)/2)*b*c^3*d^3*ArcTan[c*x] - (d^3*(a + b*ArcTan[c*x]))/(3*
x^3) - (((3*I)/2)*c*d^3*(a + b*ArcTan[c*x]))/x^2 + (3*c^2*d^3*(a + b*ArcTan[c*x]))/x - I*a*c^3*d^3*Log[x] - (1
0*b*c^3*d^3*Log[x])/3 + (5*b*c^3*d^3*Log[1 + c^2*x^2])/3 + (b*c^3*d^3*PolyLog[2, (-I)*c*x])/2 - (b*c^3*d^3*Pol
yLog[2, I*c*x])/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4876

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=\int \left (\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4}+\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^3 \int \frac {a+b \tan ^{-1}(c x)}{x^4} \, dx+\left (3 i c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx-\left (3 c^2 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (i c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac {1}{3} \left (b c d^3\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (3 i b c^2 d^3\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (b c^3 d^3\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (b c^3 d^3\right ) \int \frac {\log (1+i c x)}{x} \, dx-\left (3 b c^3 d^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {3 i b c^2 d^3}{2 x}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)+\frac {1}{6} \left (b c d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (3 b c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (3 i b c^4 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {3 i b c^2 d^3}{2 x}-\frac {3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)+\frac {1}{6} \left (b c d^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {c^2}{x}+\frac {c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\frac {1}{2} \left (3 b c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (3 b c^5 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^3}{6 x^2}-\frac {3 i b c^2 d^3}{2 x}-\frac {3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)-\frac {10}{3} b c^3 d^3 \log (x)+\frac {5}{3} b c^3 d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.10, size = 170, normalized size = 0.90 \[ \frac {d^3 \left (-6 i a c^3 x^3 \log (x)+18 a c^2 x^2-9 i a c x-2 a+3 b c^3 x^3 \text {Li}_2(-i c x)-3 b c^3 x^3 \text {Li}_2(i c x)-20 b c^3 x^3 \log (x)-9 i b c^2 x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )+18 b c^2 x^2 \tan ^{-1}(c x)+10 b c^3 x^3 \log \left (c^2 x^2+1\right )-b c x-9 i b c x \tan ^{-1}(c x)-2 b \tan ^{-1}(c x)\right )}{6 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

(d^3*(-2*a - (9*I)*a*c*x - b*c*x + 18*a*c^2*x^2 - 2*b*ArcTan[c*x] - (9*I)*b*c*x*ArcTan[c*x] + 18*b*c^2*x^2*Arc
Tan[c*x] - (9*I)*b*c^2*x^2*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] - (6*I)*a*c^3*x^3*Log[x] - 20*b*c^3*x^3
*Log[x] + 10*b*c^3*x^3*Log[1 + c^2*x^2] + 3*b*c^3*x^3*PolyLog[2, (-I)*c*x] - 3*b*c^3*x^3*PolyLog[2, I*c*x]))/(
6*x^3)

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-2 i \, a c^{3} d^{3} x^{3} - 6 \, a c^{2} d^{3} x^{2} + 6 i \, a c d^{3} x + 2 \, a d^{3} + {\left (b c^{3} d^{3} x^{3} - 3 i \, b c^{2} d^{3} x^{2} - 3 \, b c d^{3} x + i \, b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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^4,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.07, size = 255, normalized size = 1.35 \[ -\frac {d^{3} a}{3 x^{3}}-i c^{3} d^{3} b \arctan \left (c x \right ) \ln \left (c x \right )+\frac {3 c^{2} d^{3} a}{x}-\frac {3 i b \,c^{3} d^{3} \arctan \left (c x \right )}{2}-\frac {d^{3} b \arctan \left (c x \right )}{3 x^{3}}-\frac {3 i c \,d^{3} a}{2 x^{2}}+\frac {3 c^{2} d^{3} b \arctan \left (c x \right )}{x}-\frac {3 i c \,d^{3} b \arctan \left (c x \right )}{2 x^{2}}+\frac {c^{3} d^{3} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {c^{3} d^{3} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {c^{3} d^{3} b \dilog \left (i c x +1\right )}{2}-\frac {c^{3} d^{3} b \dilog \left (-i c x +1\right )}{2}-i c^{3} d^{3} a \ln \left (c x \right )-\frac {b c \,d^{3}}{6 x^{2}}-\frac {10 c^{3} d^{3} b \ln \left (c x \right )}{3}+\frac {5 b \,c^{3} d^{3} \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 i b \,c^{2} d^{3}}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^4,x)

[Out]

-1/3*d^3*a/x^3-I*c^3*d^3*b*arctan(c*x)*ln(c*x)+3*c^2*d^3*a/x-3/2*I*b*c^3*d^3*arctan(c*x)-1/3*d^3*b*arctan(c*x)
/x^3-3/2*I*c*d^3*a/x^2+3*c^2*d^3*b*arctan(c*x)/x-3/2*I*c*d^3*b*arctan(c*x)/x^2+1/2*c^3*d^3*b*ln(c*x)*ln(1+I*c*
x)-1/2*c^3*d^3*b*ln(c*x)*ln(1-I*c*x)+1/2*c^3*d^3*b*dilog(1+I*c*x)-1/2*c^3*d^3*b*dilog(1-I*c*x)-I*c^3*d^3*a*ln(
c*x)-1/6*b*c*d^3/x^2-10/3*c^3*d^3*b*ln(c*x)+5/3*b*c^3*d^3*ln(c^2*x^2+1)-3/2*I*b*c^2*d^3/x

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -i \, b c^{3} d^{3} \int \frac {\arctan \left (c x\right )}{x}\,{d x} - i \, a c^{3} d^{3} \log \relax (x) + \frac {3}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b c^{2} d^{3} - \frac {3}{2} i \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c d^{3} + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{3} + \frac {3 \, a c^{2} d^{3}}{x} - \frac {3 i \, a c d^{3}}{2 \, x^{2}} - \frac {a d^{3}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

-I*b*c^3*d^3*integrate(arctan(c*x)/x, x) - I*a*c^3*d^3*log(x) + 3/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arcta
n(c*x)/x)*b*c^2*d^3 - 3/2*I*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c*d^3 + 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^3 + 3*a*c^2*d^3/x - 3/2*I*a*c*d^3/x^2 - 1/3*a*d^3/x^3

________________________________________________________________________________________

mupad [B]  time = 0.97, size = 221, normalized size = 1.17 \[ \left \{\begin {array}{cl} -\frac {a\,d^3}{3\,x^3} & \text {\ if\ \ }c=0\\ \frac {b\,c^3\,d^3\,\ln \left (-\frac {3\,c^6\,x^2}{2}-\frac {3\,c^4}{2}\right )}{6}-\frac {b\,c^3\,d^3\,\ln \relax (x)}{3}-\frac {b\,c^3\,d^3\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}-3\,b\,c\,d^3\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )-\frac {b\,c\,d^3}{6\,x^2}-\frac {a\,d^3\,\left (2-18\,c^2\,x^2+c\,x\,9{}\mathrm {i}+c^3\,x^3\,\ln \relax (x)\,6{}\mathrm {i}\right )}{6\,x^3}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}+\frac {3\,b\,c^2\,d^3\,\mathrm {atan}\left (c\,x\right )}{x}-\frac {b\,d^3\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )\,3{}\mathrm {i}}{2}-\frac {b\,c\,d^3\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i}}{2\,x^2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^3)/x^4,x)

[Out]

piecewise(c == 0, -(a*d^3)/(3*x^3), c ~= 0, - (b*d^3*(c^3*atan(c*x) + c^2/x)*3i)/2 - (b*c^3*d^3*(dilog(- c*x*1
i + 1) - dilog(c*x*1i + 1)))/2 - (b*c^3*d^3*log(x))/3 + (b*c^3*d^3*log(- (3*c^4)/2 - (3*c^6*x^2)/2))/6 - 3*b*c
*d^3*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2) - (b*c*d^3)/(6*x^2) - (a*d^3*(c*x*9i - 18*c^2*x^2 + c^3*x^3*log(x
)*6i + 2))/(6*x^3) - (b*d^3*atan(c*x))/(3*x^3) - (b*c*d^3*atan(c*x)*3i)/(2*x^2) + (3*b*c^2*d^3*atan(c*x))/x)

________________________________________________________________________________________

sympy [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*atan(c*x))/x**4,x)

[Out]

Timed out

________________________________________________________________________________________