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

Optimal. Leaf size=195 \[ \frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (-c x+i)^2}+\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {a \log (x)}{d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}+\frac {i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 d^3}+\frac {5 b}{8 d^3 (-c x+i)}+\frac {i b}{8 d^3 (-c x+i)^2}-\frac {5 b \tan ^{-1}(c x)}{8 d^3} \]

[Out]

1/8*I*b/d^3/(I-c*x)^2+5/8*b/d^3/(I-c*x)-5/8*b*arctan(c*x)/d^3+1/2*(-a-b*arctan(c*x))/d^3/(I-c*x)^2+I*(a+b*arct
an(c*x))/d^3/(I-c*x)+a*ln(x)/d^3+(a+b*arctan(c*x))*ln(2/(1+I*c*x))/d^3+1/2*I*b*polylog(2,-I*c*x)/d^3-1/2*I*b*p
olylog(2,I*c*x)/d^3+1/2*I*b*polylog(2,1-2/(1+I*c*x))/d^3

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Rubi [A]  time = 0.24, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4848, 2391, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac {i b \text {PolyLog}(2,-i c x)}{2 d^3}-\frac {i b \text {PolyLog}(2,i c x)}{2 d^3}+\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (-c x+i)^2}+\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {a \log (x)}{d^3}+\frac {5 b}{8 d^3 (-c x+i)}+\frac {i b}{8 d^3 (-c x+i)^2}-\frac {5 b \tan ^{-1}(c x)}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*b)/(d^3*(I - c*x)^2) + (5*b)/(8*d^3*(I - c*x)) - (5*b*ArcTan[c*x])/(8*d^3) - (a + b*ArcTan[c*x])/(2*d^3
*(I - c*x)^2) + (I*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) + (a*Log[x])/d^3 + ((a + b*ArcTan[c*x])*Log[2/(1 + I*c
*x)])/d^3 + ((I/2)*b*PolyLog[2, (-I)*c*x])/d^3 - ((I/2)*b*PolyLog[2, I*c*x])/d^3 + ((I/2)*b*PolyLog[2, 1 - 2/(
1 + I*c*x)])/d^3

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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -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 {a+b \tan ^{-1}(c x)}{x (d+i c d x)^3} \, dx &=\int \left (\frac {a+b \tan ^{-1}(c x)}{d^3 x}+\frac {c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^2}-\frac {c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tan ^{-1}(c x)}{x} \, dx}{d^3}+\frac {(i c) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}+\frac {c \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{d^3}-\frac {c \int \frac {a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {(i b) \int \frac {\log (1-i c x)}{x} \, dx}{2 d^3}-\frac {(i b) \int \frac {\log (1+i c x)}{x} \, dx}{2 d^3}+\frac {(i b c) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}+\frac {(b c) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}-\frac {(b c) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{d^3}+\frac {(i b c) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}+\frac {(b c) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{2 d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}+\frac {i b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {(i b c) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^3}+\frac {(b c) \int \left (-\frac {i}{2 (-i+c x)^3}+\frac {1}{4 (-i+c x)^2}-\frac {1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^3}\\ &=\frac {i b}{8 d^3 (i-c x)^2}+\frac {5 b}{8 d^3 (i-c x)}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}+\frac {i b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 d^3}-\frac {(b c) \int \frac {1}{1+c^2 x^2} \, dx}{8 d^3}-\frac {(b c) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=\frac {i b}{8 d^3 (i-c x)^2}+\frac {5 b}{8 d^3 (i-c x)}-\frac {5 b \tan ^{-1}(c x)}{8 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \text {Li}_2(-i c x)}{2 d^3}-\frac {i b \text {Li}_2(i c x)}{2 d^3}+\frac {i b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 162, normalized size = 0.83 \[ \frac {-\frac {8 i \left (a+b \tan ^{-1}(c x)\right )}{c x-i}-\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{(c x-i)^2}+8 \log \left (\frac {2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )+8 a \log (x)+4 i b \text {Li}_2(-i c x)-4 i b \text {Li}_2(i c x)+4 i b \text {Li}_2\left (\frac {c x+i}{c x-i}\right )+\frac {5 b}{-c x+i}+\frac {i b}{(c x-i)^2}-5 b \tan ^{-1}(c x)}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b)/(I - c*x) + (I*b)/(-I + c*x)^2 - 5*b*ArcTan[c*x] - (4*(a + b*ArcTan[c*x]))/(-I + c*x)^2 - ((8*I)*(a + b
*ArcTan[c*x]))/(-I + c*x) + 8*a*Log[x] + 8*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + (4*I)*b*PolyLog[2, (-I)*
c*x] - (4*I)*b*PolyLog[2, I*c*x] + (4*I)*b*PolyLog[2, (I + c*x)/(-I + c*x)])/(8*d^3)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c^{3} d^{3} x^{4} - 6 i \, c^{2} d^{3} x^{3} - 6 \, c d^{3} x^{2} + 2 i \, d^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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maple [A]  time = 0.07, size = 327, normalized size = 1.68 \[ \frac {a \ln \left (c x \right )}{d^{3}}-\frac {a}{2 d^{3} \left (c x -i\right )^{2}}-\frac {i a}{d^{3} \left (c x -i\right )}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {i b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}+\frac {b \ln \left (c x \right ) \arctan \left (c x \right )}{d^{3}}-\frac {b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}+\frac {i b}{8 d^{3} \left (c x -i\right )^{2}}-\frac {b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}-\frac {5 b \arctan \left (c x \right )}{8 d^{3}}-\frac {i b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {5 b}{8 d^{3} \left (c x -i\right )}-\frac {i b \dilog \left (-i c x \right )}{2 d^{3}}+\frac {i b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}-\frac {i b \ln \left (c x -i\right )^{2}}{4 d^{3}}-\frac {i b \dilog \left (-i \left (c x +i\right )\right )}{2 d^{3}}-\frac {i b \ln \left (c x \right ) \ln \left (-i \left (c x +i\right )\right )}{2 d^{3}}+\frac {i b \ln \left (c x \right ) \ln \left (-i \left (-c x +i\right )\right )}{2 d^{3}}-\frac {i a \arctan \left (c x \right )}{d^{3}}-\frac {i b \ln \left (-i c x \right ) \ln \left (-i \left (-c x +i\right )\right )}{2 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/d^3*ln(c*x)-1/2*a/d^3/(c*x-I)^2-I*a/d^3/(c*x-I)-1/2*a/d^3*ln(c^2*x^2+1)+1/2*I*b/d^3*ln(c*x-I)*ln(-1/2*I*(I+c
*x))+b/d^3*ln(c*x)*arctan(c*x)-1/2*b/d^3*arctan(c*x)/(c*x-I)^2+1/8*I*b/d^3/(c*x-I)^2-b/d^3*arctan(c*x)*ln(c*x-
I)-5/8*b*arctan(c*x)/d^3-I*b/d^3*arctan(c*x)/(c*x-I)-5/8*b/d^3/(c*x-I)-1/4*I*b/d^3*ln(c*x-I)^2-1/2*I*b/d^3*ln(
c*x)*ln(-I*(I+c*x))-1/2*I*b/d^3*dilog(-I*c*x)+1/2*I*b/d^3*dilog(-1/2*I*(I+c*x))+1/2*I*b/d^3*ln(c*x)*ln(-I*(-c*
x+I))-I*a/d^3*arctan(c*x)-1/2*I*b/d^3*dilog(-I*(I+c*x))-1/2*I*b/d^3*ln(-I*c*x)*ln(-I*(-c*x+I))

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maxima [B]  time = 0.42, size = 399, normalized size = 2.05 \[ -\frac {{\left (16 i \, a + 10 \, b\right )} c x + {\left (4 i \, b c^{2} x^{2} + 8 \, b c x - 4 i \, b\right )} \arctan \left (c x\right )^{2} + {\left (i \, b c^{2} x^{2} + 2 \, b c x - i \, b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + {\left (4 \, b c^{2} x^{2} - 8 i \, b c x - 4 \, b\right )} \arctan \left (c x\right ) \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) - {\left (16 \, b c^{2} x^{2} - 32 i \, b c x - 16 \, b\right )} \arctan \left (c x\right ) \log \left (c x\right ) + {\left ({\left (16 i \, a + 5 \, b\right )} c^{2} x^{2} + 2 \, {\left (16 \, a + 3 i \, b\right )} c x - 16 i \, a + 19 \, b\right )} \arctan \left (c x\right ) - {\left (5 \, b c^{2} x^{2} - 10 i \, b c x - 5 \, b\right )} \arctan \left (c x, -1\right ) + {\left (8 i \, b c^{2} x^{2} + 16 \, b c x - 8 i \, b\right )} {\rm Li}_2\left (i \, c x + 1\right ) + {\left (-8 i \, b c^{2} x^{2} - 16 \, b c x + 8 i \, b\right )} {\rm Li}_2\left (\frac {1}{2} i \, c x + \frac {1}{2}\right ) + {\left (-8 i \, b c^{2} x^{2} - 16 \, b c x + 8 i \, b\right )} {\rm Li}_2\left (-i \, c x + 1\right ) + {\left (4 \, {\left (\pi b + 2 \, a\right )} c^{2} x^{2} + {\left (-8 i \, \pi b - 16 i \, a\right )} c x - 4 \, \pi b + {\left (-2 i \, b c^{2} x^{2} - 4 \, b c x + 2 i \, b\right )} \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) - 8 \, a\right )} \log \left (c^{2} x^{2} + 1\right ) - 16 \, {\left (a c^{2} x^{2} - 2 i \, a c x - a\right )} \log \relax (x) + 24 \, a - 12 i \, b}{16 \, c^{2} d^{3} x^{2} - 32 i \, c d^{3} x - 16 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((16*I*a + 10*b)*c*x + (4*I*b*c^2*x^2 + 8*b*c*x - 4*I*b)*arctan(c*x)^2 + (I*b*c^2*x^2 + 2*b*c*x - I*b)*log(c^
2*x^2 + 1)^2 + (4*b*c^2*x^2 - 8*I*b*c*x - 4*b)*arctan(c*x)*log(1/4*c^2*x^2 + 1/4) - (16*b*c^2*x^2 - 32*I*b*c*x
- 16*b)*arctan(c*x)*log(c*x) + ((16*I*a + 5*b)*c^2*x^2 + 2*(16*a + 3*I*b)*c*x - 16*I*a + 19*b)*arctan(c*x) -
(5*b*c^2*x^2 - 10*I*b*c*x - 5*b)*arctan2(c*x, -1) + (8*I*b*c^2*x^2 + 16*b*c*x - 8*I*b)*dilog(I*c*x + 1) + (-8*
I*b*c^2*x^2 - 16*b*c*x + 8*I*b)*dilog(1/2*I*c*x + 1/2) + (-8*I*b*c^2*x^2 - 16*b*c*x + 8*I*b)*dilog(-I*c*x + 1)
+ (4*(pi*b + 2*a)*c^2*x^2 + (-8*I*pi*b - 16*I*a)*c*x - 4*pi*b + (-2*I*b*c^2*x^2 - 4*b*c*x + 2*I*b)*log(1/4*c^
2*x^2 + 1/4) - 8*a)*log(c^2*x^2 + 1) - 16*(a*c^2*x^2 - 2*I*a*c*x - a)*log(x) + 24*a - 12*I*b)/(16*c^2*d^3*x^2
- 32*I*c*d^3*x - 16*d^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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