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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.186854, antiderivative size = 119, normalized size of antiderivative = 1., 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 = {5043, 12, 4852, 4924, 4868, 2447} \[ -\frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{d e^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}-\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}+\frac{2 b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^2} \]

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 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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 4924

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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 4868

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 2447

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]]

Rubi steps

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

Mathematica [A]  time = 0.190932, size = 135, normalized size = 1.13 \[ \frac{-i b^2 (c+d x) \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c+d x)}\right )+a \left (2 b (c+d x) \log \left (\frac{c+d x}{\sqrt{(c+d x)^2+1}}\right )-a\right )+2 b \tan ^{-1}(c+d x) \left (-a+b (c+d x) \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )\right )-i b^2 (c+d x-i) \tan ^{-1}(c+d x)^2}{d e^2 (c+d x)} \]

Warning: Unable to verify antiderivative.

[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))

________________________________________________________________________________________

Maple [B]  time = 0.125, size = 471, normalized size = 4. \begin{align*} -{\frac{{a}^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2}\arctan \left ( dx+c \right ) \ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{d{e}^{2}}}+2\,{\frac{{b}^{2}\ln \left ( dx+c \right ) \arctan \left ( dx+c \right ) }{d{e}^{2}}}-{\frac{i{b}^{2}\ln \left ( dx+c \right ) \ln \left ( 1-i \left ( dx+c \right ) \right ) }{d{e}^{2}}}-{\frac{i{b}^{2}{\it dilog} \left ( 1-i \left ( dx+c \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( dx+c-i \right ) \ln \left ( -{\frac{i}{2}} \left ( dx+c+i \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( dx+c+i \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) \ln \left ( dx+c+i \right ) }{d{e}^{2}}}-{\frac{{\frac{i}{2}}{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( dx+c-i \right ) \right ) }{d{e}^{2}}}+{\frac{{\frac{i}{4}}{b}^{2} \left ( \ln \left ( dx+c-i \right ) \right ) ^{2}}{d{e}^{2}}}-{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) \ln \left ( dx+c-i \right ) }{d{e}^{2}}}-{\frac{{\frac{i}{4}}{b}^{2} \left ( \ln \left ( dx+c+i \right ) \right ) ^{2}}{d{e}^{2}}}+{\frac{i{b}^{2}\ln \left ( dx+c \right ) \ln \left ( 1+i \left ( dx+c \right ) \right ) }{d{e}^{2}}}-{\frac{{\frac{i}{2}}{b}^{2}\ln \left ( dx+c+i \right ) \ln \left ({\frac{i}{2}} \left ( dx+c-i \right ) \right ) }{d{e}^{2}}}+{\frac{i{b}^{2}{\it dilog} \left ( 1+i \left ( dx+c \right ) \right ) }{d{e}^{2}}}-2\,{\frac{ab\arctan \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-{\frac{ab\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{d{e}^{2}}}+2\,{\frac{ab\ln \left ( dx+c \right ) }{d{e}^{2}}} \end{align*}

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)

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \arctan \left (d x + c\right )^{2} + 2 \, a b \arctan \left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \end{align*}

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)/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}

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]

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