3.18 \(\int \frac {(a+b \tan ^{-1}(c+d x))^3}{(c e+d e x)^2} \, dx\)

Optimal. Leaf size=163 \[ -\frac {3 i b^2 \text {Li}_2\left (\frac {2}{1-i (c+d x)}-1\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^2}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2}+\frac {3 b \log \left (2-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2}+\frac {3 b^3 \text {Li}_3\left (\frac {2}{1-i (c+d x)}-1\right )}{2 d e^2} \]

[Out]

-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)
))/d/e^2

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Rubi [A]  time = 0.30, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5043, 12, 4852, 4924, 4868, 4884, 4992, 6610} \[ -\frac {3 i b^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^2}+\frac {3 b^3 \text {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^2}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}-\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2}+\frac {3 b \log \left (2-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-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)

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 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 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 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 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 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 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 \frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{(c e+d e x)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{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 )^3}{d e^2}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {(3 i b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x (i+x)} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \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 )^3}{d e^2}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\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 e^2}+\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 e^2}\\ &=-\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{d e^2 (c+d x)}+\frac {3 b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^2}-\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 e^2}+\frac {3 b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^2}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 263, normalized size = 1.61 \[ \frac {-\frac {2 a^3}{c+d x}-3 a^2 b \log \left (c^2+2 c d x+d^2 x^2+1\right )+6 a^2 b \log (c+d x)-\frac {6 a^2 b \tan ^{-1}(c+d x)}{c+d x}+6 a b^2 \left (\tan ^{-1}(c+d x) \left (\left (-\frac {1}{c+d x}-i\right ) \tan ^{-1}(c+d x)+2 \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )\right )-i \text {Li}_2\left (e^{2 i \tan ^{-1}(c+d x)}\right )\right )+2 b^3 \left (3 i \tan ^{-1}(c+d x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c+d x)}\right )+\frac {3}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(c+d x)}\right )-\frac {\tan ^{-1}(c+d x)^3}{c+d x}+i \tan ^{-1}(c+d x)^3+3 \tan ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c+d x)}\right )-\frac {i \pi ^3}{8}\right )}{2 d e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-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)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (d x + c\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c\right ) + a^{3}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 0.44, size = 2696, normalized size = 16.54 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*I/d*a*b^2/e^2*ln(I+d*x+c)*ln(1/2*I*(d*x+c-I))+3*I/d*a*b^2/e^2*ln(d*x+c)*ln(1+I*(d*x+c))-3/2*I/d*a*b^2/e^2
*ln(d*x+c-I)*ln(1+(d*x+c)^2)-3*I/d*a*b^2/e^2*ln(d*x+c)*ln(1-I*(d*x+c))-1/d*b^3/e^2/(d*x+c)*arctan(d*x+c)^3-3/d
*b^3/e^2*arctan(d*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)+3/d*b^3/e^2*arctan(d*x+c)^2*ln(1+(1+I*(d*x+c))/(1
+(d*x+c)^2)^(1/2))+3/d*b^3/e^2*arctan(d*x+c)^2*ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3/d*b^3/e^2*arctan(d*x+c)
^2*ln(1-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+3/d*b^3/e^2*ln(d*x+c)*arctan(d*x+c)^2-3/2/d*b^3/e^2*arctan(d*x+c)^2
*ln(1+(d*x+c)^2)+3/d*a^2*b/e^2*ln(d*x+c)-3/2/d*a^2*b/e^2*ln(1+(d*x+c)^2)+3/d*b^3/e^2*arctan(d*x+c)^2*ln(2)-I/d
*b^3/e^2*arctan(d*x+c)^3+3/2*I/d*a*b^2/e^2*ln(I+d*x+c)*ln(1+(d*x+c)^2)-3/2*I/d*b^3/e^2*Pi*csgn(((1+I*(d*x+c))^
2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2+3/2*I/d*b^3/e^2*Pi*csgn(((1+I*(d*x+c))
^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^3*arctan(d*x+c)^2+3/4*I/d*b^3/e^2*arctan(d*x+c)^2*Pi*cs
gn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^3-3/4*I/d*b^3/e^2*arctan(d*x+c)^2*Pi*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+
c)^2))^3-3/4*I/d*b^3/e^2*arctan(d*x+c)^2*Pi*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/((1+I*(d*x+c))^2/(1+(d*x+c)^2
)+1)^2)^3+3/2*I/d*b^3/e^2*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)+1))^3*arc
tan(d*x+c)^2+3/2*I/d*a*b^2/e^2*ln(d*x+c-I)*ln(-1/2*I*(I+d*x+c))-1/d*a^3/e^2/(d*x+c)+6/d*b^3/e^2*polylog(3,(1+I
*(d*x+c))/(1+(d*x+c)^2)^(1/2))+6/d*b^3/e^2*polylog(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-3/d*a*b^2/e^2/(d*x+c)
*arctan(d*x+c)^2+6/d*a*b^2/e^2*arctan(d*x+c)*ln(d*x+c)-3/d*a*b^2/e^2*arctan(d*x+c)*ln(1+(d*x+c)^2)-3/d*a^2*b/e
^2/(d*x+c)*arctan(d*x+c)-6*I/d*b^3/e^2*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-3/4*I/d*b^3
/e^2*arctan(d*x+c)^2*Pi*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^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+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)+3/2*I/d*b^3/e^2*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))*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)+1))*arctan(d*x+c)^2-3/4*I/d*b^3/e^2*arctan(d*x+c)^2*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))+3/4*I/d*b^3/e^2*arctan(d*x+c)^2*Pi*csgn(I*((1+I*(d*x+c))^2
/(1+(d*x+c)^2)+1))^2*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)+3/2*I/d*b^3/e^2*arctan(d*x+c)^2*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+3/4*I/d*b^3/e^2*arctan(d*x+c)^2*Pi*csg
n(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+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2
-3/2*I/d*b^3/e^2*arctan(d*x+c)^2*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)^2)^2+3/2*I/d*b^3/e^2*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)+1))*c
sgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))*arctan(d*x+c)^2+3*I/d*a*b^2/e^2*dilog
(1+I*(d*x+c))+3/4*I/d*a*b^2/e^2*ln(d*x+c-I)^2+3/2*I/d*a*b^2/e^2*dilog(-1/2*I*(I+d*x+c))-6*I/d*b^3/e^2*arctan(d
*x+c)*polylog(2,(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-3/4*I/d*a*b^2/e^2*ln(I+d*x+c)^2-3/2*I/d*a*b^2/e^2*dilog(1/2
*I*(d*x+c-I))+3/2*I/d*b^3/e^2*Pi*arctan(d*x+c)^2-3*I/d*a*b^2/e^2*dilog(1-I*(d*x+c))-3/2*I/d*b^3/e^2*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+I*(d*x+c))^2/(1+(d*x+c)^2)+1))
^2*arctan(d*x+c)^2+3/4*I/d*b^3/e^2*arctan(d*x+c)^2*Pi*csgn(I/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)*csgn(I*(1+I*
(d*x+c))^2/(1+(d*x+c)^2)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1)^2)^2-3/2*I/d*b^3/e^2*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+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2-3
/2*I/d*b^3/e^2*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)+1))*csgn(((1+I*(d*x+
c))^2/(1+(d*x+c)^2)-1)/((1+I*(d*x+c))^2/(1+(d*x+c)^2)+1))^2*arctan(d*x+c)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {3}{2} \, {\left (d {\left (\frac {\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}}\right )} + \frac {2 \, \arctan \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} a^{2} b - \frac {a^{3}}{d^{2} e^{2} x + c d e^{2}} - \frac {\frac {15}{2} \, b^{3} \arctan \left (d x + c\right )^{3} - \frac {21}{8} \, b^{3} \arctan \left (d x + c\right ) \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2} - {\left (d^{2} e^{2} x + c d e^{2}\right )} \int \frac {196 \, {\left (b^{3} d^{2} x^{2} + 2 \, b^{3} c d x + b^{3} c^{2} + b^{3}\right )} \arctan \left (d x + c\right )^{3} + 12 \, {\left (64 \, a b^{2} d^{2} x^{2} + 64 \, a b^{2} c^{2} + 15 \, b^{3} c + 64 \, a b^{2} + {\left (128 \, a b^{2} c + 15 \, b^{3}\right )} d x\right )} \arctan \left (d x + c\right )^{2} - 84 \, {\left (b^{3} d^{2} x^{2} + 2 \, b^{3} c d x + b^{3} c^{2}\right )} \arctan \left (d x + c\right ) \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 21 \, {\left (b^{3} d x + b^{3} c - {\left (b^{3} d^{2} x^{2} + 2 \, b^{3} c d x + b^{3} c^{2} + b^{3}\right )} \arctan \left (d x + c\right )\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )^{2}}{8 \, {\left (d^{4} e^{2} x^{4} + 4 \, c d^{3} e^{2} x^{3} + {\left (6 \, c^{2} + 1\right )} d^{2} e^{2} x^{2} + 2 \, {\left (2 \, c^{3} + c\right )} d e^{2} x + {\left (c^{4} + c^{2}\right )} e^{2}\right )}}\,{d x}}{32 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))
/(d^2*e^2*x + c*d*e^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{3}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{2} + 2 c d x + d^{2} x^{2}}\, dx}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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