3.1.0 ∫ (a+b arctan(c+dx))3 ______________________________

Integrand size = 23, antiderivative size = 163 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=-\frac {i (a+b \arctan (c+d x))^3}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {3 i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2}+\frac {3 b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^2} \]

-I*(a+b*arctan(d*x+c))^3/d/e^2-(a+b*arctan(d*x+c))^3/d/e^2/(d*x+c)+3*b*(a+b*arctan(d*x+c))^2*ln(2-2/(1-I*(d*x+

c)))/d/e^2-3*I*b^2*(a+b*arctan(d*x+c))*polylog(2,-1+2/(1-I*(d*x+c)))/d/e^2+3/2*b^3*polylog(3,-1+2/(1-I*(d*x+c)

Rubi [A] (veriﬁed)

Time = 0.21 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5151, 12, 4946, 5044, 4988, 5004, 5112, 6745} \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=-\frac {3 i b^2 \operatorname {PolyLog}\left (2,\frac {2}{1-i (c+d x)}-1\right ) (a+b \arctan (c+d x))}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}-\frac {i (a+b \arctan (c+d x))^3}{d e^2}+\frac {3 b \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2}{d e^2}+\frac {3 b^3 \operatorname {PolyLog}\left (3,\frac {2}{1-i (c+d x)}-1\right )}{2 d e^2} \]

Int[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^2,x]

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

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

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

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

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]

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]

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]

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]

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

n[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

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

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+b \arctan (x))^3}{e^2 x^2} \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b \arctan (x))^3}{x^2} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {(3 b) \text {Subst}\left (\int \frac {(a+b \arctan (x))^2}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^3}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {(3 i b) \text {Subst}\left (\int \frac {(a+b \arctan (x))^2}{x (i+x)} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^3}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {\left (6 b^2\right ) \text {Subst}\left (\int \frac {(a+b \arctan (x)) \log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^3}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {3 i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2}+\frac {\left (3 i b^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {i (a+b \arctan (c+d x))^3}{d e^2}-\frac {(a+b \arctan (c+d x))^3}{d e^2 (c+d x)}+\frac {3 b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {3 i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^2}+\frac {3 b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^2} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.59 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\frac {-\frac {2 a^3}{c+d x}-\frac {6 a^2 b \arctan (c+d x)}{c+d x}+6 a^2 b \log (c+d x)-3 a^2 b \log \left (1+c^2+2 c d x+d^2 x^2\right )+6 a b^2 \left (\arctan (c+d x) \left (\left (-i-\frac {1}{c+d x}\right ) \arctan (c+d x)+2 \log \left (1-e^{2 i \arctan (c+d x)}\right )\right )-i \operatorname {PolyLog}\left (2,e^{2 i \arctan (c+d x)}\right )\right )+2 b^3 \left (-\frac {i \pi ^3}{8}+i \arctan (c+d x)^3-\frac {\arctan (c+d x)^3}{c+d x}+3 \arctan (c+d x)^2 \log \left (1-e^{-2 i \arctan (c+d x)}\right )+3 i \arctan (c+d x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c+d x)}\right )+\frac {3}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c+d x)}\right )\right )}{2 d e^2} \]

Integrate[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^2,x]

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

*x + d^2*x^2] + 6*a*b^2*(ArcTan[c + d*x]*((-I - (c + d*x)^(-1))*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*((-1/8*I)*Pi^3 + I*ArcTan[c + d*x]^3 - ArcTan[c +

d*x]^3/(c + d*x) + 3*ArcTan[c + d*x]^2*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*e^2)

Maple [C] (warning: unable to verify)

Time = 1.33 (sec) , antiderivative size = 2104, normalized size of antiderivative = 12.91


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(2104\)
default	\(\text {Expression too large to display}\)	\(2104\)
parts	\(\text {Expression too large to display}\)	\(2112\)

int((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^2,x,method=_RETURNVERBOSE)

1/d*(-a^3/e^2/(d*x+c)+b^3/e^2*(-1/(d*x+c)*arctan(d*x+c)^3+3*ln(d*x+c)*arctan(d*x+c)^2-3/2*arctan(d*x+c)^2*ln(1

+(d*x+c)^2)+3*arctan(d*x+c)^2*ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-3*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d

*x+c)^2)-1)-I*arctan(d*x+c)^3+3/4*(2*I*Pi*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))*csgn(I*(1+I*(d*x+c))^2/(1+

(d*x+c)^2))^2-2*I*Pi*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+2*I*Pi+I*Pi*c

sgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)

^2+2*I*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^3-I*Pi*csgn(I*(1+I*(d*x+

c))/(1+(d*x+c)^2)^(1/2))^2*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))-I*Pi*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))

^2)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2

))^2)-I*Pi*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))^3-I*Pi*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^

2/(1+(d*x+c)^2))^2)^3+2*I*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1))*csgn(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)

))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))-2*I*Pi*csgn(I/(1+(1+I*(d*x+c))^

2/(1+(d*x+c)^2)))*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2-2*I*Pi*csgn(I*

((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1

+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+I*Pi*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2*csgn(I*(1+(1+I*(d*x+c))^2/

(1+(d*x+c)^2))^2)+2*I*Pi*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^3+I*Pi*csgn

(I/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))

^2)^2+2*I*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(((1+I*(d*x+c))^2

/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))-2*I*Pi*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*csgn(I*(

1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^2-2*I*Pi*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1))*csgn(I*((1+I*(d*x+c))^2

/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2+I*Pi*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2))^2)^3+4*ln

(2))*arctan(d*x+c)^2+3*arctan(d*x+c)^2*ln(1+(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-6*I*arctan(d*x+c)*polylog(2,-(1

+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6*polylog(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3*arctan(d*x+c)^2*ln(1-(1+I*(

d*x+c))/(1+(d*x+c)^2)^(1/2))-6*I*arctan(d*x+c)*polylog(2,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6*polylog(3,(1+I*(

d*x+c))/(1+(d*x+c)^2)^(1/2)))+3*a*b^2/e^2*(-1/(d*x+c)*arctan(d*x+c)^2+2*ln(d*x+c)*arctan(d*x+c)-arctan(d*x+c)*

ln(1+(d*x+c)^2)-1/2*I*(ln(d*x+c-I)*ln(1+(d*x+c)^2)-1/2*ln(d*x+c-I)^2-dilog(-1/2*I*(d*x+c+I))-ln(d*x+c-I)*ln(-1

/2*I*(d*x+c+I)))+1/2*I*(ln(d*x+c+I)*ln(1+(d*x+c)^2)-1/2*ln(d*x+c+I)^2-dilog(1/2*I*(d*x+c-I))-ln(d*x+c+I)*ln(1/

2*I*(d*x+c-I)))+I*ln(d*x+c)*ln(1+I*(d*x+c))-I*ln(d*x+c)*ln(1-I*(d*x+c))+I*dilog(1+I*(d*x+c))-I*dilog(1-I*(d*x+

c)))+3*a^2*b/e^2*(-1/(d*x+c)*arctan(d*x+c)+ln(d*x+c)-1/2*ln(1+(d*x+c)^2)))

Fricas [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="fricas")

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^2*e^2*x^2 + 2*

Sympy [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]

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

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

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

Maxima [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{2}} \,d x } \]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="maxima")

-3/2*(d*(log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2)) + 2*arctan(d*x + c)/(d^2*e^2*x

+ c*d*e^2))*a^2*b - a^3/(d^2*e^2*x + c*d*e^2) - 1/32*(4*b^3*arctan(d*x + c)^3 - 3*b^3*arctan(d*x + c)*log(d^2

*x^2 + 2*c*d*x + c^2 + 1)^2 - 32*(d^2*e^2*x + c*d*e^2)*integrate(1/32*(28*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2

+ b^3)*arctan(d*x + c)^3 + 12*(8*a*b^2*d^2*x^2 + 8*a*b^2*c^2 + b^3*c + 8*a*b^2 + (16*a*b^2*c + b^3)*d*x)*arct

an(d*x + c)^2 - 12*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2)*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 3*

(b^3*d*x + b^3*c - (b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2 + b^3)*arctan(d*x + c))*log(d^2*x^2 + 2*c*d*x + c^2 +

1)^2)/(d^4*e^2*x^4 + 4*c*d^3*e^2*x^3 + (6*c^2 + 1)*d^2*e^2*x^2 + 2*(2*c^3 + c)*d*e^2*x + (c^4 + c^2)*e^2), x))

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\text {Timed out} \]

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^2,x, algorithm="giac")

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]

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

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

\(\int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^2} \, dx\) [18]

Optimal result