3.13 \(\int \frac{(a+b \tan ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=194 \[ \frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{3 d e^4}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac{2 b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2}{3 d e^4 (c+d x)}-\frac{b^2 \tan ^{-1}(c+d x)}{3 d e^4} \]

[Out]

-b^2/(3*d*e^4*(c + d*x)) - (b^2*ArcTan[c + d*x])/(3*d*e^4) - (b*(a + b*ArcTan[c + d*x]))/(3*d*e^4*(c + d*x)^2)
+ ((I/3)*(a + b*ArcTan[c + d*x])^2)/(d*e^4) - (a + b*ArcTan[c + d*x])^2/(3*d*e^4*(c + d*x)^3) - (2*b*(a + b*A
rcTan[c + d*x])*Log[2 - 2/(1 - I*(c + d*x))])/(3*d*e^4) + ((I/3)*b^2*PolyLog[2, -1 + 2/(1 - I*(c + d*x))])/(d*
e^4)

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Rubi [A]  time = 0.257446, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5043, 12, 4852, 4918, 325, 203, 4924, 4868, 2447} \[ \frac{i b^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{3 d e^4}-\frac{b \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^2}{3 d e^4}-\frac{2 b \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{3 d e^4}-\frac{b^2}{3 d e^4 (c+d x)}-\frac{b^2 \tan ^{-1}(c+d x)}{3 d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

-b^2/(3*d*e^4*(c + d*x)) - (b^2*ArcTan[c + d*x])/(3*d*e^4) - (b*(a + b*ArcTan[c + d*x]))/(3*d*e^4*(c + d*x)^2)
+ ((I/3)*(a + b*ArcTan[c + d*x])^2)/(d*e^4) - (a + b*ArcTan[c + d*x])^2/(3*d*e^4*(c + d*x)^3) - (2*b*(a + b*A
rcTan[c + d*x])*Log[2 - 2/(1 - I*(c + d*x))])/(3*d*e^4) + ((I/3)*b^2*PolyLog[2, -1 + 2/(1 - I*(c + d*x))])/(d*
e^4)

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 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

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

Mathematica [A]  time = 0.606005, size = 163, normalized size = 0.84 \[ -\frac{-i b^2 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c+d x)}\right )+\frac{a^2}{(c+d x)^3}+\frac{a b}{(c+d x)^2}+2 a b \log \left (\frac{c+d x}{\sqrt{(c+d x)^2+1}}\right )+b \tan ^{-1}(c+d x) \left (\frac{2 a}{(c+d x)^3}+\frac{b}{(c+d x)^2}+2 b \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )+b\right )+a b+\frac{b^2}{c+d x}+b^2 \left (\frac{1}{(c+d x)^3}-i\right ) \tan ^{-1}(c+d x)^2}{3 d e^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^4,x]

[Out]

-(a*b + a^2/(c + d*x)^3 + (a*b)/(c + d*x)^2 + b^2/(c + d*x) + b^2*(-I + (c + d*x)^(-3))*ArcTan[c + d*x]^2 + b*
ArcTan[c + d*x]*(b + (2*a)/(c + d*x)^3 + b/(c + d*x)^2 + 2*b*Log[1 - E^((2*I)*ArcTan[c + d*x])]) + 2*a*b*Log[(
c + d*x)/Sqrt[1 + (c + d*x)^2]] - I*b^2*PolyLog[2, E^((2*I)*ArcTan[c + d*x])])/(3*d*e^4)

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Maple [B]  time = 0.125, size = 547, normalized size = 2.8 \begin{align*} \text{result too large to display} \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)^4,x)

[Out]

-1/3/d*a^2/e^4/(d*x+c)^3-1/3/d*b^2/e^4/(d*x+c)^3*arctan(d*x+c)^2+1/3/d*b^2/e^4*arctan(d*x+c)*ln(1+(d*x+c)^2)-1
/3/d*b^2/e^4*arctan(d*x+c)/(d*x+c)^2-2/3/d*b^2/e^4*ln(d*x+c)*arctan(d*x+c)-1/6*I/d*b^2/e^4*ln(d*x+c+I)*ln(1+(d
*x+c)^2)-1/6*I/d*b^2/e^4*ln(d*x+c-I)*ln(-1/2*I*(d*x+c+I))-1/3*I/d*b^2/e^4*dilog(1+I*(d*x+c))+1/3*I/d*b^2/e^4*d
ilog(1-I*(d*x+c))+1/3*I/d*b^2/e^4*ln(d*x+c)*ln(1-I*(d*x+c))+1/6*I/d*b^2/e^4*dilog(1/2*I*(d*x+c-I))+1/12*I/d*b^
2/e^4*ln(d*x+c+I)^2-1/12*I/d*b^2/e^4*ln(d*x+c-I)^2-1/3*b^2*arctan(d*x+c)/d/e^4-1/3*b^2/d/e^4/(d*x+c)+1/6*I/d*b
^2/e^4*ln(d*x+c+I)*ln(1/2*I*(d*x+c-I))-1/3*I/d*b^2/e^4*ln(d*x+c)*ln(1+I*(d*x+c))-1/6*I/d*b^2/e^4*dilog(-1/2*I*
(d*x+c+I))+1/6*I/d*b^2/e^4*ln(d*x+c-I)*ln(1+(d*x+c)^2)-2/3/d*a*b/e^4/(d*x+c)^3*arctan(d*x+c)+1/3/d*a*b/e^4*ln(
1+(d*x+c)^2)-1/3/d*a*b/e^4/(d*x+c)^2-2/3/d*a*b/e^4*ln(d*x+c)

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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)^4,x, algorithm="maxima")

[Out]

Timed out

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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^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, 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)^4,x, algorithm="fricas")

[Out]

integral((b^2*arctan(d*x + c)^2 + 2*a*b*arctan(d*x + c) + a^2)/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*
x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{b^{2} \operatorname{atan}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{2 a b \operatorname{atan}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \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)**4,x)

[Out]

(Integral(a**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**2*atan(c +
d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(2*a*b*atan(c + d*x)
/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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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 )}^{4}}\,{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)^4,x, algorithm="giac")

[Out]

integrate((b*arctan(d*x + c) + a)^2/(d*e*x + c*e)^4, x)