3.1.10 \(\int \frac {(a+b \text {ArcTan}(c+d x))^2}{c e+d e x} \, dx\) [10]

Optimal. Leaf size=183 \[ \frac {2 (a+b \text {ArcTan}(c+d x))^2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {i b (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {i b (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,-1+\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i (c+d x)}\right )}{2 d e} \]

[Out]

-2*(a+b*arctan(d*x+c))^2*arctanh(-1+2/(1+I*(d*x+c)))/d/e-I*b*(a+b*arctan(d*x+c))*polylog(2,1-2/(1+I*(d*x+c)))/
d/e+I*b*(a+b*arctan(d*x+c))*polylog(2,-1+2/(1+I*(d*x+c)))/d/e-1/2*b^2*polylog(3,1-2/(1+I*(d*x+c)))/d/e+1/2*b^2
*polylog(3,-1+2/(1+I*(d*x+c)))/d/e

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Rubi [A]
time = 0.22, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5151, 12, 4942, 5108, 5004, 5114, 6745} \begin {gather*} -\frac {i b \text {Li}_2\left (1-\frac {2}{i (c+d x)+1}\right ) (a+b \text {ArcTan}(c+d x))}{d e}+\frac {i b \text {Li}_2\left (\frac {2}{i (c+d x)+1}-1\right ) (a+b \text {ArcTan}(c+d x))}{d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))^2}{d e}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{i (c+d x)+1}\right )}{2 d e}+\frac {b^2 \text {Li}_3\left (\frac {2}{i (c+d x)+1}-1\right )}{2 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcTan[c + d*x])^2*ArcTanh[1 - 2/(1 + I*(c + d*x))])/(d*e) - (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2,
1 - 2/(1 + I*(c + d*x))])/(d*e) + (I*b*(a + b*ArcTan[c + d*x])*PolyLog[2, -1 + 2/(1 + I*(c + d*x))])/(d*e) -
(b^2*PolyLog[3, 1 - 2/(1 + I*(c + d*x))])/(2*d*e) + (b^2*PolyLog[3, -1 + 2/(1 + I*(c + d*x))])/(2*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 4942

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

Rule 5004

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

Rule 5108

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

Rule 5114

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

Rule 5151

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]

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(4 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {(2 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (2-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}+\frac {(2 b) \text {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {i b \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {i b \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tan ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {i b \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1+i (c+d x)}\right )}{d e}+\frac {i b \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i (c+d x)}\right )}{d e}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+i (c+d x)}\right )}{2 d e}+\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+i (c+d x)}\right )}{2 d e}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 170, normalized size = 0.93 \begin {gather*} \frac {4 (a+b \text {ArcTan}(c+d x))^2 \tanh ^{-1}\left (\frac {i+c+d x}{-i+c+d x}\right )+2 i b (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,-\frac {i+c+d x}{-i+c+d x}\right )-2 i b (a+b \text {ArcTan}(c+d x)) \text {PolyLog}\left (2,\frac {i+c+d x}{-i+c+d x}\right )+b^2 \text {PolyLog}\left (3,-\frac {i+c+d x}{-i+c+d x}\right )-b^2 \text {PolyLog}\left (3,\frac {i+c+d x}{-i+c+d x}\right )}{2 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.71, size = 1362, normalized size = 7.44

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1362\)
default	\(\text {Expression too large to display}\)	\(1362\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^2/(d*e*x+c*e),x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2/e*ln(d*x+c)-I*a*b/e*ln(d*x+c)*ln(1-I*(d*x+c))+b^2/e*arctan(d*x+c)^2*ln(1-(1+I*(d*x+c))/(1+(d*x+c)^2)^
(1/2))+b^2/e*ln(d*x+c)*arctan(d*x+c)^2-b^2/e*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)+b^2/e*arctan(
d*x+c)^2*ln(1+(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+2*a*b/e*ln(d*x+c)*arctan(d*x+c)+I*a*b/e*dilog(1+I*(d*x+c))-I*
a*b/e*dilog(1-I*(d*x+c))+I*b^2/e*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-2*I*b^2/e*arctan(d*x+
c)*polylog(2,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-2*I*b^2/e*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))/(1+(d*x+c)^2)
^(1/2))+1/2*I*b^2/e*Pi*arctan(d*x+c)^2-1/2*I*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c)
)^2/(1+(d*x+c)^2)))*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2*arctan(d*x+c)^
2-1/2*I*b^2/e*Pi*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d
*x+c))^2/(1+(d*x+c)^2)))^2*arctan(d*x+c)^2+1/2*I*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*
x+c))^2/(1+(d*x+c)^2)))*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*arctan(d*x+c
)^2-1/2*I*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*
(d*x+c))^2/(1+(d*x+c)^2)))^2*arctan(d*x+c)^2-1/2*b^2/e*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+2*b^2/e*polyl
og(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+2*b^2/e*polylog(3,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-1/2*I*b^2/e*Pi*c
sgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2*arctan(d*x+c)^2+I*a*b/e*ln(d*x+c)*l
n(1+I*(d*x+c))+1/2*I*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^3*ar
ctan(d*x+c)^2+1/2*I*b^2/e*Pi*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^3*arcta
n(d*x+c)^2+1/2*I*b^2/e*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1))*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*
csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*arctan(d*x+c)^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*e^(-1)*log(d*x*e + c*e)/d + integrate(1/16*(12*b^2*arctan(d*x + c)^2 + b^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1
)^2 + 32*a*b*arctan(d*x + c))/(d*x*e + c*e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctan(d*x + c)^2 + 2*a*b*arctan(d*x + c) + a^2)*e^(-1)/(d*x + c), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c + d x}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c + d x \right )}}{c + d x}\, dx}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c + d*x), x) + Integral(b**2*atan(c + d*x)**2/(c + d*x), x) + Integral(2*a*b*atan(c + d*x)/(c
+ d*x), x))/e

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e),x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^2}{c\,e+d\,e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^2/(c*e + d*e*x),x)

[Out]

int((a + b*atan(c + d*x))^2/(c*e + d*e*x), x)

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