3.39 \(\int \frac {(d+i c d x)^4 (a+b \tan ^{-1}(c x))}{x^5} \, dx\)

Optimal. Leaf size=227 \[ \frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {3 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+a c^4 d^4 \log (x)+\frac {1}{2} i b c^4 d^4 \text {Li}_2(-i c x)-\frac {1}{2} i b c^4 d^4 \text {Li}_2(i c x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \tan ^{-1}(c x)+\frac {13 b c^3 d^4}{4 x}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {8}{3} i b c^4 d^4 \log \left (c^2 x^2+1\right )-\frac {b c d^4}{12 x^3} \]

[Out]

-1/12*b*c*d^4/x^3-2/3*I*b*c^2*d^4/x^2+13/4*b*c^3*d^4/x+13/4*b*c^4*d^4*arctan(c*x)-1/4*d^4*(a+b*arctan(c*x))/x^
4-4/3*I*c*d^4*(a+b*arctan(c*x))/x^3+3*c^2*d^4*(a+b*arctan(c*x))/x^2+4*I*c^3*d^4*(a+b*arctan(c*x))/x+a*c^4*d^4*
ln(x)-16/3*I*b*c^4*d^4*ln(x)+8/3*I*b*c^4*d^4*ln(c^2*x^2+1)+1/2*I*b*c^4*d^4*polylog(2,-I*c*x)-1/2*I*b*c^4*d^4*p
olylog(2,I*c*x)

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Rubi [A]  time = 0.23, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4876, 4852, 325, 203, 266, 44, 36, 29, 31, 4848, 2391} \[ \frac {1}{2} i b c^4 d^4 \text {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \text {PolyLog}(2,i c x)+\frac {3 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}+a c^4 d^4 \log (x)-\frac {2 i b c^2 d^4}{3 x^2}+\frac {8}{3} i b c^4 d^4 \log \left (c^2 x^2+1\right )+\frac {13 b c^3 d^4}{4 x}-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \tan ^{-1}(c x)-\frac {b c d^4}{12 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

-(b*c*d^4)/(12*x^3) - (((2*I)/3)*b*c^2*d^4)/x^2 + (13*b*c^3*d^4)/(4*x) + (13*b*c^4*d^4*ArcTan[c*x])/4 - (d^4*(
a + b*ArcTan[c*x]))/(4*x^4) - (((4*I)/3)*c*d^4*(a + b*ArcTan[c*x]))/x^3 + (3*c^2*d^4*(a + b*ArcTan[c*x]))/x^2
+ ((4*I)*c^3*d^4*(a + b*ArcTan[c*x]))/x + a*c^4*d^4*Log[x] - ((16*I)/3)*b*c^4*d^4*Log[x] + ((8*I)/3)*b*c^4*d^4
*Log[1 + c^2*x^2] + (I/2)*b*c^4*d^4*PolyLog[2, (-I)*c*x] - (I/2)*b*c^4*d^4*PolyLog[2, I*c*x]

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)^4 \left (a+b \tan ^{-1}(c x)\right )}{x^5} \, dx &=\int \left (\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^5}+\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^4}-\frac {6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {c^4 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^4 \int \frac {a+b \tan ^{-1}(c x)}{x^5} \, dx+\left (4 i c d^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^4} \, dx-\left (6 c^2 d^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx-\left (4 i c^3 d^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx+\left (c^4 d^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac {3 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)+\frac {1}{4} \left (b c d^4\right ) \int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx+\frac {1}{3} \left (4 i b c^2 d^4\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx-\left (3 b c^3 d^4\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (i b c^4 d^4\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b c^4 d^4\right ) \int \frac {\log (1+i c x)}{x} \, dx-\left (4 i b c^4 d^4\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c d^4}{12 x^3}+\frac {3 b c^3 d^4}{x}-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac {3 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)+\frac {1}{2} i b c^4 d^4 \text {Li}_2(-i c x)-\frac {1}{2} i b c^4 d^4 \text {Li}_2(i c x)+\frac {1}{3} \left (2 i b c^2 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{4} \left (b c^3 d^4\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\left (2 i b c^4 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )+\left (3 b c^5 d^4\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^4}{12 x^3}+\frac {13 b c^3 d^4}{4 x}+3 b c^4 d^4 \tan ^{-1}(c x)-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac {3 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)+\frac {1}{2} i b c^4 d^4 \text {Li}_2(-i c x)-\frac {1}{2} i b c^4 d^4 \text {Li}_2(i c x)+\frac {1}{3} \left (2 i b c^2 d^4\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 )-\left (2 i b c^4 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b c^5 d^4\right ) \int \frac {1}{1+c^2 x^2} \, dx+\left (2 i b c^6 d^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^4}{12 x^3}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \tan ^{-1}(c x)-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+\frac {3 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}+a c^4 d^4 \log (x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {8}{3} i b c^4 d^4 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b c^4 d^4 \text {Li}_2(-i c x)-\frac {1}{2} i b c^4 d^4 \text {Li}_2(i c x)\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 227, normalized size = 1.00 \[ \frac {d^4 \left (12 a c^4 x^4 \log (x)+48 i a c^3 x^3+36 a c^2 x^2-16 i a c x-3 a+6 i b c^4 x^4 \text {Li}_2(-i c x)-6 i b c^4 x^4 \text {Li}_2(i c x)-64 i b c^4 x^4 \log (x)+48 i b c^3 x^3 \tan ^{-1}(c x)-b c x \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )-8 i b c^2 x^2+36 b c^2 x^2 \tan ^{-1}(c x)+32 i b c^4 x^4 \log \left (c^2 x^2+1\right )+36 b c^3 x^3 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )-16 i b c x \tan ^{-1}(c x)-3 b \tan ^{-1}(c x)\right )}{12 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

(d^4*(-3*a - (16*I)*a*c*x + 36*a*c^2*x^2 - (8*I)*b*c^2*x^2 + (48*I)*a*c^3*x^3 - 3*b*ArcTan[c*x] - (16*I)*b*c*x
*ArcTan[c*x] + 36*b*c^2*x^2*ArcTan[c*x] + (48*I)*b*c^3*x^3*ArcTan[c*x] - b*c*x*Hypergeometric2F1[-3/2, 1, -1/2
, -(c^2*x^2)] + 36*b*c^3*x^3*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] + 12*a*c^4*x^4*Log[x] - (64*I)*b*c^4*
x^4*Log[x] + (32*I)*b*c^4*x^4*Log[1 + c^2*x^2] + (6*I)*b*c^4*x^4*PolyLog[2, (-I)*c*x] - (6*I)*b*c^4*x^4*PolyLo
g[2, I*c*x]))/(12*x^4)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a c^{4} d^{4} x^{4} - 8 i \, a c^{3} d^{4} x^{3} - 12 \, a c^{2} d^{4} x^{2} + 8 i \, a c d^{4} x + 2 \, a d^{4} + {\left (i \, b c^{4} d^{4} x^{4} + 4 \, b c^{3} d^{4} x^{3} - 6 i \, b c^{2} d^{4} x^{2} - 4 \, b c d^{4} x + i \, b d^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,x, algorithm="fricas")

[Out]

integral(1/2*(2*a*c^4*d^4*x^4 - 8*I*a*c^3*d^4*x^3 - 12*a*c^2*d^4*x^2 + 8*I*a*c*d^4*x + 2*a*d^4 + (I*b*c^4*d^4*
x^4 + 4*b*c^3*d^4*x^3 - 6*I*b*c^2*d^4*x^2 - 4*b*c*d^4*x + I*b*d^4)*log(-(c*x + I)/(c*x - I)))/x^5, x)

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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^5,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.07, size = 298, normalized size = 1.31 \[ \frac {4 i c^{3} d^{4} b \arctan \left (c x \right )}{x}+c^{4} d^{4} a \ln \left (c x \right )-\frac {4 i c \,d^{4} b \arctan \left (c x \right )}{3 x^{3}}-\frac {d^{4} a}{4 x^{4}}+\frac {3 c^{2} d^{4} a}{x^{2}}-\frac {4 i c \,d^{4} a}{3 x^{3}}+c^{4} d^{4} b \ln \left (c x \right ) \arctan \left (c x \right )+\frac {i c^{4} d^{4} b \dilog \left (i c x +1\right )}{2}-\frac {d^{4} b \arctan \left (c x \right )}{4 x^{4}}+\frac {3 c^{2} d^{4} b \arctan \left (c x \right )}{x^{2}}+\frac {4 i c^{3} d^{4} a}{x}-\frac {2 i b \,c^{2} d^{4}}{3 x^{2}}-\frac {b c \,d^{4}}{12 x^{3}}+\frac {13 b \,c^{3} d^{4}}{4 x}-\frac {16 i c^{4} d^{4} b \ln \left (c x \right )}{3}+\frac {13 b \,c^{4} d^{4} \arctan \left (c x \right )}{4}+\frac {8 i b \,c^{4} d^{4} \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {i c^{4} d^{4} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i c^{4} d^{4} b \dilog \left (-i c x +1\right )}{2}-\frac {i c^{4} d^{4} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,x)

[Out]

4*I*c^3*d^4*b*arctan(c*x)/x+c^4*d^4*a*ln(c*x)-4/3*I*c*d^4*b*arctan(c*x)/x^3-1/4*d^4*a/x^4+3*c^2*d^4*a/x^2-4/3*
I*c*d^4*a/x^3+c^4*d^4*b*ln(c*x)*arctan(c*x)-1/2*I*c^4*d^4*b*dilog(1-I*c*x)-1/4*d^4*b*arctan(c*x)/x^4+3*c^2*d^4
*b*arctan(c*x)/x^2+1/2*I*c^4*d^4*b*dilog(1+I*c*x)+4*I*c^3*d^4*a/x-1/12*b*c*d^4/x^3+13/4*b*c^3*d^4/x-2/3*I*b*c^
2*d^4/x^2+13/4*b*c^4*d^4*arctan(c*x)-16/3*I*c^4*d^4*b*ln(c*x)+8/3*I*b*c^4*d^4*ln(c^2*x^2+1)+1/2*I*c^4*d^4*b*ln
(c*x)*ln(1+I*c*x)-1/2*I*c^4*d^4*b*ln(c*x)*ln(1-I*c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b c^{4} d^{4} \int \frac {\arctan \left (c x\right )}{x}\,{d x} + a c^{4} d^{4} \log \relax (x) + 2 i \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b c^{3} d^{4} + 3 \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b c^{2} d^{4} + \frac {2}{3} i \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b c d^{4} + \frac {4 i \, a c^{3} d^{4}}{x} + \frac {1}{12} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d^{4} + \frac {3 \, a c^{2} d^{4}}{x^{2}} - \frac {4 i \, a c d^{4}}{3 \, x^{3}} - \frac {a d^{4}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,x, algorithm="maxima")

[Out]

b*c^4*d^4*integrate(arctan(c*x)/x, x) + a*c^4*d^4*log(x) + 2*I*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x
)/x)*b*c^3*d^4 + 3*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c^2*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*d^4 + 4*I*a*c^3*d^4/x + 1/12*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1
)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^4 + 3*a*c^2*d^4/x^2 - 4/3*I*a*c*d^4/x^3 - 1/4*a*d^4/x^4

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mupad [B]  time = 0.94, size = 298, normalized size = 1.31 \[ \left \{\begin {array}{cl} -\frac {a\,d^4}{4\,x^4} & \text {\ if\ \ }c=0\\ 3\,b\,c\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )-\frac {b\,d^4\,\left (\frac {\frac {c^2}{3}-c^4\,x^2}{x^3}-c^5\,\mathrm {atan}\left (c\,x\right )\right )}{4\,c}-\frac {b\,c^4\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,c^4\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-b\,c^2\,d^4\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,4{}\mathrm {i}+\frac {a\,d^4\,\left (36\,c^2\,x^2+12\,c^4\,x^4\,\ln \relax (x)-3-c\,x\,16{}\mathrm {i}+c^3\,x^3\,48{}\mathrm {i}\right )}{12\,x^4}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{4\,x^4}-\frac {b\,d^4\,\left (c^4\,\ln \relax (x)-\frac {c^4\,\ln \left (-\frac {c^4\,\left (3\,c^2\,x^2+1\right )}{2}-c^4\right )}{2}+\frac {c^2}{2\,x^2}\right )\,4{}\mathrm {i}}{3}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{3\,x^3}+\frac {3\,b\,c^2\,d^4\,\mathrm {atan}\left (c\,x\right )}{x^2}+\frac {b\,c^3\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^4)/x^5,x)

[Out]

piecewise(c == 0, -(a*d^4)/(4*x^4), c ~= 0, - (b*d^4*(c^4*log(x) - (c^4*log(- (c^4*(3*c^2*x^2 + 1))/2 - c^4))/
2 + c^2/(2*x^2))*4i)/3 - (b*d^4*((c^2/3 - c^4*x^2)/x^3 - c^5*atan(c*x)))/(4*c) - (b*c^4*d^4*dilog(- c*x*1i + 1
)*1i)/2 + (b*c^4*d^4*dilog(c*x*1i + 1)*1i)/2 - b*c^2*d^4*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2)*4i + (a*d^4*(
- c*x*16i + 36*c^2*x^2 + c^3*x^3*48i + 12*c^4*x^4*log(x) - 3))/(12*x^4) - (b*d^4*atan(c*x))/(4*x^4) + 3*b*c*d^
4*(c^3*atan(c*x) + c^2/x) - (b*c*d^4*atan(c*x)*4i)/(3*x^3) + (3*b*c^2*d^4*atan(c*x))/x^2 + (b*c^3*d^4*atan(c*x
)*4i)/x)

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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)**4*(a+b*atan(c*x))/x**5,x)

[Out]

Timed out

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