3.19 \(\int \frac{(a+b \tan ^{-1}(c+d x))^3}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=180 \[ -\frac{3 i b^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^3}+\frac{3 b^2 \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3}-\frac{3 b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac{3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d e^3} \]

[Out]

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

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Rubi [A]  time = 0.31859, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {5043, 12, 4852, 4918, 4924, 4868, 2447, 4884} \[ -\frac{3 i b^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i (c+d x)}\right )}{2 d e^3}+\frac{3 b^2 \log \left (2-\frac{2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^3}-\frac{3 b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)}-\frac{3 i b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^3}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d e^3 (c+d x)^2}-\frac{\left (a+b \tan ^{-1}(c+d x)\right )^3}{2 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^3,x]

[Out]

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

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 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]]

Rule 4884

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]

Rubi steps

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

Mathematica [A]  time = 0.225925, size = 225, normalized size = 1.25 \[ -\frac{3 b^3 (c+d x) \left (i (c+d x) \left (\tan ^{-1}(c+d x)^2+\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c+d x)}\right )\right )+\tan ^{-1}(c+d x)^2-2 (c+d x) \tan ^{-1}(c+d x) \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )\right )+3 a^2 b \left (\left ((c+d x)^2+1\right ) \tan ^{-1}(c+d x)+c+d x\right )+a^3+3 a b^2 \left (-2 (c+d x)^2 \log \left (\frac{c+d x}{\sqrt{(c+d x)^2+1}}\right )+2 (c+d x) \tan ^{-1}(c+d x)+\left ((c+d x)^2+1\right ) \tan ^{-1}(c+d x)^2\right )+b^3 \left (c^2+2 c d x+d^2 x^2+1\right ) \tan ^{-1}(c+d x)^3}{2 d e^3 (c+d x)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^3,x]

[Out]

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

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Maple [B]  time = 0.138, size = 631, normalized size = 3.5 \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))^3/(d*e*x+c*e)^3,x)

[Out]

-3/2/d*a*b^2/e^3*arctan(d*x+c)^2-3/2/d*a*b^2/e^3*ln(1+(d*x+c)^2)+3/d*a*b^2/e^3*ln(d*x+c)-3/2/d*a^2*b/e^3*arcta
n(d*x+c)-3/4*I/d*b^3/e^3*dilog(1/2*I*(d*x+c-I))-3/8*I/d*b^3/e^3*ln(d*x+c+I)^2+3/2*I/d*b^3/e^3*dilog(1+I*(d*x+c
))+3/8*I/d*b^3/e^3*ln(d*x+c-I)^2+3/4*I/d*b^3/e^3*dilog(-1/2*I*(d*x+c+I))-3/2*I/d*b^3/e^3*dilog(1-I*(d*x+c))-3/
2/d*a^2*b/e^3/(d*x+c)-1/2/d*b^3/e^3/(d*x+c)^2*arctan(d*x+c)^3-3/2/d*b^3/e^3*arctan(d*x+c)^2/(d*x+c)-3/2/d*b^3/
e^3*arctan(d*x+c)*ln(1+(d*x+c)^2)+3/d*b^3/e^3*ln(d*x+c)*arctan(d*x+c)-3/4*I/d*b^3/e^3*ln(d*x+c+I)*ln(1/2*I*(d*
x+c-I))-3/2/d*a^2*b/e^3/(d*x+c)^2*arctan(d*x+c)+3/4*I/d*b^3/e^3*ln(d*x+c-I)*ln(-1/2*I*(d*x+c+I))+3/4*I/d*b^3/e
^3*ln(1+(d*x+c)^2)*ln(d*x+c+I)+3/2*I/d*b^3/e^3*ln(d*x+c)*ln(1+I*(d*x+c))-3/4*I/d*b^3/e^3*ln(1+(d*x+c)^2)*ln(d*
x+c-I)-3/2*I/d*b^3/e^3*ln(d*x+c)*ln(1-I*(d*x+c))-3/2/d*a*b^2/e^3/(d*x+c)^2*arctan(d*x+c)^2-3/d*a*b^2/e^3*arcta
n(d*x+c)/(d*x+c)-1/2/d*b^3/e^3*arctan(d*x+c)^3-1/2/d*a^3/e^3/(d*x+c)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="maxima")

[Out]

-3/2*(d*(1/(d^3*e^3*x + c*d^2*e^3) + arctan((d^2*x + c*d)/d)/(d^2*e^3)) + arctan(d*x + c)/(d^3*e^3*x^2 + 2*c*d
^2*e^3*x + c^2*d*e^3))*a^2*b - 3/2*(2*d*(1/(d^3*e^3*x + c*d^2*e^3) + arctan((d^2*x + c*d)/d)/(d^2*e^3))*arctan
(d*x + c) - (arctan(d*x + c)^2 - log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 2*log(d*x + c))/(d*e^3))*a*b^2 - 3/2*a*b^2
*arctan(d*x + c)^2/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3) - 1/32*(8*(d^2*x^2 + 2*c*d*x + c^2 + 1)*arctan(d*
x + c)^3 + 12*(d*x + c)*arctan(d*x + c)^2 - 3*(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - 32*(d^3*e^3*x^2 +
2*c*d^2*e^3*x + c^2*d*e^3)*integrate(1/32*(16*(d^2*x^2 + 2*c*d*x + c^2 + 1)*arctan(d*x + c)^3 + 12*(d^3*x^3 +
3*c*d^2*x^2 + c^3 + (3*c^2 + 1)*d*x + c)*arctan(d*x + c)^2 + 3*(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2 + 1)*d*x
+ c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 + 24*(d^2*x^2 + 2*c*d*x + c^2)*arctan(d*x + c) - 12*(d^3*x^3 + 3*c*d^
2*x^2 + 3*c^2*d*x + c^3)*log(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^5*e^3*x^5 + 5*c*d^4*e^3*x^4 + (10*c^2 + 1)*d^3*e
^3*x^3 + (10*c^3 + 3*c)*d^2*e^3*x^2 + (5*c^4 + 3*c^2)*d*e^3*x + (c^5 + c^3)*e^3), x))*b^3/(d^3*e^3*x^2 + 2*c*d
^2*e^3*x + c^2*d*e^3) - 1/2*a^3/(d^3*e^3*x^2 + 2*c*d^2*e^3*x + c^2*d*e^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arctan \left (d x + c\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c\right ) + a^{3}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))**3/(d*e*x+c*e)**3,x)

[Out]

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

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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 )}^{3}}{{\left (d e x + c e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arctan(d*x + c) + a)^3/(d*e*x + c*e)^3, x)