3.1.11 \(\int \frac {(a+b \text {ArcTan}(c+d x))^2}{(c e+d e x)^2} \, dx\) [11]

Optimal. Leaf size=119 \[ -\frac {i (a+b \text {ArcTan}(c+d x))^2}{d e^2}-\frac {(a+b \text {ArcTan}(c+d x))^2}{d e^2 (c+d x)}+\frac {2 b (a+b \text {ArcTan}(c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {i b^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5151, 12, 4946, 5044, 4988, 2497} \begin {gather*} -\frac {(a+b \text {ArcTan}(c+d x))^2}{d e^2 (c+d x)}-\frac {i (a+b \text {ArcTan}(c+d x))^2}{d e^2}+\frac {2 b \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))}{d e^2}-\frac {i b^2 \text {Li}_2\left (\frac {2}{1-i (c+d x)}-1\right )}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4988

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])
^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 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 5044

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

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 135, normalized size = 1.13 \begin {gather*} \frac {-i b^2 (-i+c+d x) \text {ArcTan}(c+d x)^2+2 b \text {ArcTan}(c+d x) \left (-a+b (c+d x) \log \left (1-e^{2 i \text {ArcTan}(c+d x)}\right )\right )+a \left (-a+2 b (c+d x) \log \left (\frac {c+d x}{\sqrt {1+(c+d x)^2}}\right )\right )-i b^2 (c+d x) \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c+d x)}\right )}{d e^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*b^2*(-I + c + d*x)*ArcTan[c + d*x]^2 + 2*b*ArcTan[c + d*x]*(-a + b*(c + d*x)*Log[1 - E^((2*I)*ArcTan[c +
d*x])]) + a*(-a + 2*b*(c + d*x)*Log[(c + d*x)/Sqrt[1 + (c + d*x)^2]]) - I*b^2*(c + d*x)*PolyLog[2, E^((2*I)*A
rcTan[c + d*x])])/(d*e^2*(c + d*x))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (115 ) = 230\).
time = 0.60, size = 418, normalized size = 3.51

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctan \left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{e^{2}}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{e^{2}}-\frac {i b^{2} \ln \left (d x +c +i\right )^{2}}{4 e^{2}}-\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{2 e^{2}}+\frac {i b^{2} \ln \left (d x +c -i\right )^{2}}{4 e^{2}}-\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{e^{2}}-\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{2 e^{2}}+\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{e^{2}}+\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{2 e^{2}}+\frac {i b^{2} \dilog \left (1+i \left (d x +c \right )\right )}{e^{2}}+\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{2 e^{2}}-\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{2}}-\frac {i b^{2} \dilog \left (1-i \left (d x +c \right )\right )}{e^{2}}+\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{2}}-\frac {2 a b \arctan \left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right )}{e^{2}}+\frac {2 a b \ln \left (d x +c \right )}{e^{2}}}{d}\)	\(418\)
default	\(\frac {-\frac {a^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctan \left (d x +c \right )^{2}}{e^{2} \left (d x +c \right )}-\frac {b^{2} \arctan \left (d x +c \right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{e^{2}}+\frac {2 b^{2} \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{e^{2}}-\frac {i b^{2} \ln \left (d x +c +i\right )^{2}}{4 e^{2}}-\frac {i b^{2} \dilog \left (\frac {i \left (d x +c -i\right )}{2}\right )}{2 e^{2}}+\frac {i b^{2} \ln \left (d x +c -i\right )^{2}}{4 e^{2}}-\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{e^{2}}-\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (\frac {i \left (d x +c -i\right )}{2}\right )}{2 e^{2}}+\frac {i b^{2} \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{e^{2}}+\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{2 e^{2}}+\frac {i b^{2} \dilog \left (1+i \left (d x +c \right )\right )}{e^{2}}+\frac {i b^{2} \dilog \left (-\frac {i \left (d x +c +i\right )}{2}\right )}{2 e^{2}}-\frac {i b^{2} \ln \left (d x +c -i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{2}}-\frac {i b^{2} \dilog \left (1-i \left (d x +c \right )\right )}{e^{2}}+\frac {i b^{2} \ln \left (d x +c +i\right ) \ln \left (1+\left (d x +c \right )^{2}\right )}{2 e^{2}}-\frac {2 a b \arctan \left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {a b \ln \left (1+\left (d x +c \right )^{2}\right )}{e^{2}}+\frac {2 a b \ln \left (d x +c \right )}{e^{2}}}{d}\)	\(418\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="maxima")

[Out]

-(d*(e^(-2)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2 - 2*e^(-2)*log(d*x + c)/d^2) + 2*arctan(d*x + c)/(d^2*x*e^2 +
c*d*e^2))*a*b - 1/16*(4*arctan(d*x + c)^2 - 16*(d^2*x*e^2 + c*d*e^2)*integrate(1/16*(12*(d^2*x^2 + 2*c*d*x +
c^2 + 1)*arctan(d*x + c)^2 + (d^2*x^2 + 2*c*d*x + c^2 + 1)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 8*(d*x + c)*ar
ctan(d*x + c) - 4*(d^2*x^2 + 2*c*d*x + c^2)*log(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^4*x^4*e^2 + 4*c*d^3*x^3*e^2 +
(6*c^2*e^2 + e^2)*d^2*x^2 + c^4*e^2 + 2*(2*c^3*e^2 + c*e^2)*d*x + c^2*e^2), x) - log(d^2*x^2 + 2*c*d*x + c^2
+ 1)^2)*b^2/(d^2*x*e^2 + c*d*e^2) - a^2/(d^2*x*e^2 + c*d*e^2)

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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)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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)^2,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}{{\left (c\,e+d\,e\,x\right )}^2} \,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)^2,x)

[Out]

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

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