3.4 \(\int \frac {a+b \tan ^{-1}(c+d x)}{c e+d e x} \, dx\)

Optimal. Leaf size=63 \[ \frac {a \log (c+d x)}{d e}+\frac {i b \text {Li}_2(-i (c+d x))}{2 d e}-\frac {i b \text {Li}_2(i (c+d x))}{2 d e} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5043, 12, 4848, 2391} \[ \frac {i b \text {PolyLog}(2,-i (c+d x))}{2 d e}-\frac {i b \text {PolyLog}(2,i (c+d x))}{2 d e}+\frac {a \log (c+d x)}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])/(c*e + d*e*x),x]

[Out]

(a*Log[c + d*x])/(d*e) + ((I/2)*b*PolyLog[2, (-I)*(c + d*x)])/(d*e) - ((I/2)*b*PolyLog[2, I*(c + d*x)])/(d*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 5043

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0
] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.83 \[ \frac {a \log (c+d x)+\frac {1}{2} i b \text {Li}_2(-i (c+d x))-\frac {1}{2} i b \text {Li}_2(i (c+d x))}{d e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c + d*x])/(c*e + d*e*x),x]

[Out]

(a*Log[c + d*x] + (I/2)*b*PolyLog[2, (-I)*(c + d*x)] - (I/2)*b*PolyLog[2, I*(c + d*x)])/(d*e)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (d x + c\right ) + a}{d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(d*x + c) + a)/(d*e*x + c*e), 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(d*x+c))/(d*e*x+c*e),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.06, size = 132, normalized size = 2.10 \[ \frac {a \ln \left (d x +c \right )}{d e}+\frac {b \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{d e}+\frac {i b \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{2 d e}-\frac {i b \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{2 d e}+\frac {i b \dilog \left (1+i \left (d x +c \right )\right )}{2 d e}-\frac {i b \dilog \left (1-i \left (d x +c \right )\right )}{2 d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))/(d*e*x+c*e),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {atan}\left (c+d\,x\right )}{c\,e+d\,e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))/(c*e + d*e*x),x)

[Out]

int((a + b*atan(c + d*x))/(c*e + d*e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))/(d*e*x+c*e),x)

[Out]

(Integral(a/(c + d*x), x) + Integral(b*atan(c + d*x)/(c + d*x), x))/e

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