3.16 \(\int \frac {(d+i c d x)^2 (a+b \tan ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)-i b c^2 d^2 \log \left (c^2 x^2+1\right )+2 i b c^2 d^2 \log (x)-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {b c d^2}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391} \[ -\frac {1}{2} i b c^2 d^2 \text {PolyLog}(2,-i c x)+\frac {1}{2} i b c^2 d^2 \text {PolyLog}(2,i c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-i b c^2 d^2 \log \left (c^2 x^2+1\right )+2 i b c^2 d^2 \log (x)-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {b c d^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

-(b*c*d^2)/(2*x) - (b*c^2*d^2*ArcTan[c*x])/2 - (d^2*(a + b*ArcTan[c*x]))/(2*x^2) - ((2*I)*c*d^2*(a + b*ArcTan[
c*x]))/x - a*c^2*d^2*Log[x] + (2*I)*b*c^2*d^2*Log[x] - I*b*c^2*d^2*Log[1 + c^2*x^2] - (I/2)*b*c^2*d^2*PolyLog[
2, (-I)*c*x] + (I/2)*b*c^2*d^2*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 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)^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3} \, dx &=\int \left (\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^2 \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (2 i c d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (c^2 d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)+\frac {1}{2} \left (b c d^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (i b c^2 d^2\right ) \int \frac {\log (1-i c x)}{x} \, dx+\frac {1}{2} \left (i b c^2 d^2\right ) \int \frac {\log (1+i c x)}{x} \, dx+\left (2 i b c^2 d^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c d^2}{2 x}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)+\left (i b c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (b c^3 d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^2}{2 x}-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)+\left (i b c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\left (i b c^4 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^2}{2 x}-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)+2 i b c^2 d^2 \log (x)-i b c^2 d^2 \log \left (1+c^2 x^2\right )-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 139, normalized size = 0.91 \[ -\frac {d^2 \left (2 a c^2 x^2 \log (x)+4 i a c x+a+b c x \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )+i b c^2 x^2 \text {Li}_2(-i c x)-i b c^2 x^2 \text {Li}_2(i c x)-4 i b c^2 x^2 \log (x)+2 i b c^2 x^2 \log \left (c^2 x^2+1\right )+4 i b c x \tan ^{-1}(c x)+b \tan ^{-1}(c x)\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

-1/2*(d^2*(a + (4*I)*a*c*x + b*ArcTan[c*x] + (4*I)*b*c*x*ArcTan[c*x] + b*c*x*Hypergeometric2F1[-1/2, 1, 1/2, -
(c^2*x^2)] + 2*a*c^2*x^2*Log[x] - (4*I)*b*c^2*x^2*Log[x] + (2*I)*b*c^2*x^2*Log[1 + c^2*x^2] + I*b*c^2*x^2*Poly
Log[2, (-I)*c*x] - I*b*c^2*x^2*PolyLog[2, I*c*x]))/x^2

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {2 \, a c^{2} d^{2} x^{2} - 4 i \, a c d^{2} x - 2 \, a d^{2} - {\left (-i \, b c^{2} d^{2} x^{2} - 2 \, b c d^{2} x + i \, b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^3,x, algorithm="fricas")

[Out]

integral(-1/2*(2*a*c^2*d^2*x^2 - 4*I*a*c*d^2*x - 2*a*d^2 - (-I*b*c^2*d^2*x^2 - 2*b*c*d^2*x + I*b*d^2)*log(-(c*
x + I)/(c*x - I)))/x^3, 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((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^3,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.07, size = 217, normalized size = 1.43 \[ -c^{2} d^{2} a \ln \left (c x \right )-\frac {2 i c \,d^{2} a}{x}-\frac {d^{2} a}{2 x^{2}}-c^{2} d^{2} b \ln \left (c x \right ) \arctan \left (c x \right )-\frac {2 i c \,d^{2} b \arctan \left (c x \right )}{x}-\frac {d^{2} b \arctan \left (c x \right )}{2 x^{2}}-\frac {i c^{2} d^{2} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i c^{2} d^{2} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i c^{2} d^{2} b \dilog \left (i c x +1\right )}{2}+\frac {i c^{2} d^{2} b \dilog \left (-i c x +1\right )}{2}-\frac {b c \,d^{2}}{2 x}+2 i c^{2} d^{2} b \ln \left (c x \right )-i b \,c^{2} d^{2} \ln \left (c^{2} x^{2}+1\right )-\frac {b \,c^{2} d^{2} \arctan \left (c x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^3,x)

[Out]

-c^2*d^2*a*ln(c*x)-2*I*c*d^2*a/x-1/2*d^2*a/x^2-c^2*d^2*b*ln(c*x)*arctan(c*x)-2*I*c*d^2*b*arctan(c*x)/x-1/2*d^2
*b*arctan(c*x)/x^2-1/2*I*c^2*d^2*b*ln(c*x)*ln(1+I*c*x)+1/2*I*c^2*d^2*b*ln(c*x)*ln(1-I*c*x)-1/2*I*c^2*d^2*b*dil
og(1+I*c*x)+1/2*I*c^2*d^2*b*dilog(1-I*c*x)-1/2*b*c*d^2/x+2*I*c^2*d^2*b*ln(c*x)-I*b*c^2*d^2*ln(c^2*x^2+1)-1/2*b
*c^2*d^2*arctan(c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^2*(a+b*arctan(c*x))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.74, size = 161, normalized size = 1.06 \[ \left \{\begin {array}{cl} -\frac {a\,d^2}{2\,x^2} & \text {\ if\ \ }c=0\\ b\,d^2\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,2{}\mathrm {i}+\frac {b\,c^2\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {b\,c^2\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {b\,d^2\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-\frac {a\,d^2\,\left (2\,c^2\,x^2\,\ln \relax (x)+1+c\,x\,4{}\mathrm {i}\right )}{2\,x^2}-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}-\frac {b\,c\,d^2\,\mathrm {atan}\left (c\,x\right )\,2{}\mathrm {i}}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^2)/x^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - d^{2} \left (\int \left (- \frac {a}{x^{3}}\right )\, dx + \int \frac {a c^{2}}{x}\, dx + \int \left (- \frac {b \operatorname {atan}{\left (c x \right )}}{x^{3}}\right )\, dx + \int \left (- \frac {2 i a c}{x^{2}}\right )\, dx + \int \frac {b c^{2} \operatorname {atan}{\left (c x \right )}}{x}\, dx + \int \left (- \frac {2 i b c \operatorname {atan}{\left (c x \right )}}{x^{2}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**2*(a+b*atan(c*x))/x**3,x)

[Out]

-d**2*(Integral(-a/x**3, x) + Integral(a*c**2/x, x) + Integral(-b*atan(c*x)/x**3, x) + Integral(-2*I*a*c/x**2,
x) + Integral(b*c**2*atan(c*x)/x, x) + Integral(-2*I*b*c*atan(c*x)/x**2, x))

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