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

Optimal. Leaf size=143 \[ \frac{3 i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1+i (c+d x)}\right )}{2 d}+\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{3 b \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.214405, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {5039, 4846, 4920, 4854, 4884, 4994, 6610} \[ \frac{3 i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1+i (c+d x)}\right )}{2 d}+\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{3 b \log \left (\frac{2}{1+i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^3,x]

[Out]

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

Rule 5039

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcTan[x])^p, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rule 4846

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

Rule 4920

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

Rule 4854

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[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 6610

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 \left (a+b \tan ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \tan ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{x \left (a+b \tan ^{-1}(x)\right )^2}{1+x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right )^2}{i-x} \, dx,x,c+d x\right )}{d}\\ &=\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d}-\frac{\left (6 b^2\right ) \operatorname{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}\\ &=\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d}-\frac{\left (3 i b^3\right ) \operatorname{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}\\ &=\frac{i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tan ^{-1}(c+d x)\right )^3}{d}+\frac{3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1+i (c+d x)}\right )}{d}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1+i (c+d x)}\right )}{d}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1+i (c+d x)}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.114604, size = 212, normalized size = 1.48 \[ \frac{6 a b^2 \left (\tan ^{-1}(c+d x) \left ((c+d x-i) \tan ^{-1}(c+d x)+2 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )\right )+2 b^3 \left (-3 i \tan ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c+d x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c+d x)}\right )+\tan ^{-1}(c+d x)^2 \left ((c+d x-i) \tan ^{-1}(c+d x)+3 \log \left (1+e^{2 i \tan ^{-1}(c+d x)}\right )\right )\right )-3 a^2 b \log \left ((c+d x)^2+1\right )+6 a^2 b (c+d x) \tan ^{-1}(c+d x)+2 a^3 (c+d x)}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^3*(c + d*x) + 6*a^2*b*(c + d*x)*ArcTan[c + d*x] - 3*a^2*b*Log[1 + (c + d*x)^2] + 6*a*b^2*(ArcTan[c + d*x]
*((-I + c + d*x)*ArcTan[c + d*x] + 2*Log[1 + E^((2*I)*ArcTan[c + d*x])]) - I*PolyLog[2, -E^((2*I)*ArcTan[c + d
*x])]) + 2*b^3*(ArcTan[c + d*x]^2*((-I + c + d*x)*ArcTan[c + d*x] + 3*Log[1 + E^((2*I)*ArcTan[c + d*x])]) - (3
*I)*ArcTan[c + d*x]*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])] + (3*PolyLog[3, -E^((2*I)*ArcTan[c + d*x])])/2))/(2
*d)

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Maple [B]  time = 0.155, size = 359, normalized size = 2.5 \begin{align*} x{a}^{3}+{\frac{{a}^{3}c}{d}}-{\frac{3\,i{b}^{3}\arctan \left ( dx+c \right ) }{d}{\it polylog} \left ( 2,-{\frac{ \left ( 1+i \left ( dx+c \right ) \right ) ^{2}}{1+ \left ( dx+c \right ) ^{2}}} \right ) }+ \left ( \arctan \left ( dx+c \right ) \right ) ^{3}x{b}^{3}+{\frac{ \left ( \arctan \left ( dx+c \right ) \right ) ^{3}{b}^{3}c}{d}}+3\,{\frac{{b}^{3} \left ( \arctan \left ( dx+c \right ) \right ) ^{2}}{d}\ln \left ({\frac{ \left ( 1+i \left ( dx+c \right ) \right ) ^{2}}{1+ \left ( dx+c \right ) ^{2}}}+1 \right ) }-{\frac{3\,i \left ( \arctan \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{d}}+{\frac{3\,{b}^{3}}{2\,d}{\it polylog} \left ( 3,-{\frac{ \left ( 1+i \left ( dx+c \right ) \right ) ^{2}}{1+ \left ( dx+c \right ) ^{2}}} \right ) }-{\frac{3\,ia{b}^{2}}{d}{\it polylog} \left ( 2,-{\frac{ \left ( 1+i \left ( dx+c \right ) \right ) ^{2}}{1+ \left ( dx+c \right ) ^{2}}} \right ) }+3\, \left ( \arctan \left ( dx+c \right ) \right ) ^{2}xa{b}^{2}+3\,{\frac{ \left ( \arctan \left ( dx+c \right ) \right ) ^{2}a{b}^{2}c}{d}}+6\,{\frac{\arctan \left ( dx+c \right ) a{b}^{2}}{d}\ln \left ({\frac{ \left ( 1+i \left ( dx+c \right ) \right ) ^{2}}{1+ \left ( dx+c \right ) ^{2}}}+1 \right ) }-{\frac{i{b}^{3} \left ( \arctan \left ( dx+c \right ) \right ) ^{3}}{d}}+3\,\arctan \left ( dx+c \right ) x{a}^{2}b+3\,{\frac{\arctan \left ( dx+c \right ){a}^{2}bc}{d}}-{\frac{3\,{a}^{2}b\ln \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^3,x)

[Out]

x*a^3+1/d*a^3*c-3*I/d*b^3*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))+arctan(d*x+c)^3*x*b^3+1/d*ar
ctan(d*x+c)^3*b^3*c+3/d*b^3*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)-3*I/d*arctan(d*x+c)^2*a*b^2+3/
2/d*b^3*polylog(3,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))-3*I/d*polylog(2,-(1+I*(d*x+c))^2/(1+(d*x+c)^2))*a*b^2+3*arct
an(d*x+c)^2*x*a*b^2+3/d*arctan(d*x+c)^2*a*b^2*c+6/d*arctan(d*x+c)*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)*a*b^2-I/
d*b^3*arctan(d*x+c)^3+3*arctan(d*x+c)*x*a^2*b+3/d*arctan(d*x+c)*a^2*b*c-3/2/d*a^2*b*ln(1+(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,x, algorithm="maxima")

[Out]

7/8*b^3*c^2*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 1/8*b^3*x*arctan(d*x + c)^3 + 3*a*b^2*c^2*arctan(d*x
+ c)^2*arctan((d^2*x + c*d)/d)/d - 3/32*b^3*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - (3*arctan(
d*x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^2 - 7/32*(6*arctan(d*x + c)^2*arct
an((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3*c^2
+ 7/8*b^3*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 28*b^3*d^2*integrate(1/32*x^2*arctan(d*x + c)^3/(d^2*
x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*d^2*integrate(1/32*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2
/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 96*a*b^2*d^2*integrate(1/32*x^2*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^
2 + 1), x) + 56*b^3*c*d*integrate(1/32*x*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*d^2*inte
grate(1/32*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 6*b^3*c*d*
integrate(1/32*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 192*a*
b^2*c*d*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*c*d*integrate(1/32*x*arc
tan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*c^2*integrate(1/32*arc
tan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*a*b^2*arctan(d*x + c)^2*
arctan((d^2*x + c*d)/d)/d - 12*b^3*d*integrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*
b^3*d*integrate(1/32*x*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - (3*arctan(d*x +
c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2 - 7/32*(6*arctan(d*x + c)^2*arctan((d^2*x
+ c*d)/d)^2/d - 4*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3 + a^3*x + 3*b
^3*integrate(1/32*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3/2*(
2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*a^2*b/d

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (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}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^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, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atan(c + d*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (d x + c\right ) + a\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,x, algorithm="giac")

[Out]

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