\(\int \frac {(a+b \arctan (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^3}{1-c^2 x^2} \, dx\) [32]

Optimal result

Integrand size = 40, antiderivative size = 431 \[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=-\frac {2 \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \text {arctanh}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 i b \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 i b \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^2 \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \operatorname {PolyLog}\left (3,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b^2 \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \operatorname {PolyLog}\left (3,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c}+\frac {3 i b^3 \operatorname {PolyLog}\left (4,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c} \] Output:

2*(a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3*arctanh(-1+2/(1+I*(-c*x+1)^ 
(1/2)/(c*x+1)^(1/2)))/c+3/2*I*b*(a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))) 
^2*polylog(2,1-2/(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-3/2*I*b*(a+b*arctan 
((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2*polylog(2,-1+2/(1+I*(-c*x+1)^(1/2)/(c*x+ 
1)^(1/2)))/c+3/2*b^2*(a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))*polylog(3, 
1-2/(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-3/2*b^2*(a+b*arctan((-c*x+1)^(1/ 
2)/(c*x+1)^(1/2)))*polylog(3,-1+2/(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-3/ 
4*I*b^3*polylog(4,1-2/(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c+3/4*I*b^3*poly 
log(4,-1+2/(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx \] Input:

Integrate[(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]
 

Output:

Integrate[(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2), x]
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {7232, 5357, 5523, 5529, 5533, 7164}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{1-c^2 x^2} \, dx\)

\(\Big \downarrow \) 7232

\(\displaystyle -\frac {\int \frac {\sqrt {c x+1} \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{\sqrt {1-c x}}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}{c}\)

\(\Big \downarrow \) 5357

\(\displaystyle -\frac {2 \text {arctanh}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-6 b \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2 \text {arctanh}\left (1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}{c}\)

\(\Big \downarrow \) 5523

\(\displaystyle -\frac {2 \text {arctanh}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-6 b \left (\frac {1}{2} \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2 \log \left (2-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}-\frac {1}{2} \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2 \log \left (\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}\)

\(\Big \downarrow \) 5529

\(\displaystyle -\frac {2 \text {arctanh}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-6 b \left (\frac {1}{2} \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )+\frac {1}{2} \left (i b \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right ) \operatorname {PolyLog}\left (2,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2\right )\right )}{c}\)

\(\Big \downarrow \) 5533

\(\displaystyle -\frac {2 \text {arctanh}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-6 b \left (\frac {1}{2} \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (3,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )+\frac {1}{2} \left (i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (3,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (3,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right )}{\frac {1-c x}{c x+1}+1}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2\right )\right )}{c}\)

\(\Big \downarrow \) 7164

\(\displaystyle -\frac {2 \text {arctanh}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-6 b \left (\frac {1}{2} \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (3,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )+\frac {1}{4} b \operatorname {PolyLog}\left (4,1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )\right )\right )+\frac {1}{2} \left (i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (3,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )+\frac {1}{4} b \operatorname {PolyLog}\left (4,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right )\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2\right )\right )}{c}\)

Input:

Int[(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]
 

Output:

-((2*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3*ArcTanh[1 - 2/(1 + (I*S 
qrt[1 - c*x])/Sqrt[1 + c*x])] - 6*b*(((I/2)*(a + b*ArcTan[Sqrt[1 - c*x]/Sq 
rt[1 + c*x]])^2*PolyLog[2, 1 - 2/(1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])] - 
I*b*((I/2)*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[3, 1 - 2/(1 
 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])] + (b*PolyLog[4, 1 - 2/(1 + (I*Sqrt[1 
- c*x])/Sqrt[1 + c*x])])/4))/2 + ((-1/2*I)*(a + b*ArcTan[Sqrt[1 - c*x]/Sqr 
t[1 + c*x]])^2*PolyLog[2, -1 + 2/(1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])] + 
I*b*((I/2)*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[3, -1 + 2/( 
1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])] + (b*PolyLog[4, -1 + 2/(1 + (I*Sqrt[ 
1 - c*x])/Sqrt[1 + c*x])])/4))/2))/c)
 

Defintions of rubi rules used

rule 5357

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

rule 5523

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

rule 5529

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 
), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) 
, x] + Simp[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 5533

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

rule 7164

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]
 

rule 7232

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.) 
*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^2), x_Symbol] :> Simp[2*e*(g/(C*(e*f - d 
*g)))   Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x]], 
x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && 
EqQ[e*f + d*g, 0] && IGtQ[n, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1663 vs. \(2 (360 ) = 720\).

Time = 1.54 (sec) , antiderivative size = 1664, normalized size of antiderivative = 3.86

method	result	size
default	\(\text {Expression too large to display}\)	\(1664\)
parts	\(\text {Expression too large to display}\)	\(1664\)

Input:

int((a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x,method=_RE 
TURNVERBOSE)
 

Output:

-1/2*a^3/c*ln(c*x-1)+1/2*a^3/c*ln(c*x+1)-b^3*(1/c*arctan((-c*x+1)^(1/2)/(c 
*x+1)^(1/2))^3*ln(1+(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1 
)^(1/2))-3*I/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-(1+I*(-c* 
x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))+6/c*arctan((-c*x+1)^ 
(1/2)/(c*x+1)^(1/2))*polylog(3,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+ 
1)/(c*x+1)+1)^(1/2))+6*I/c*polylog(4,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/( 
(-c*x+1)/(c*x+1)+1)^(1/2))-1/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(( 
1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1)+1)+3/2*I/c*arctan 
((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1 
/2))^2/((-c*x+1)/(c*x+1)+1))-3/2/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*po 
lylog(3,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))-3/4*I/ 
c*polylog(4,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))+1/ 
c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(1-(1+I*(-c*x+1)^(1/2)/(c*x+1)^ 
(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))-3*I/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/ 
2))^2*polylog(2,(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1 
/2))+6/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,(1+I*(-c*x+1)^(1/2 
)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))+6*I/c*polylog(4,(1+I*(-c*x+1) 
^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2)))-3*a*b^2*(1/c*arctan((-c 
*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c* 
x+1)/(c*x+1)+1)^(1/2))-2*I/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polyl...
 

Fricas [F]

\[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int { -\frac {{\left (b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1} \,d x } \] Input:

integrate((a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, alg 
orithm="fricas")
 

Output:

integral(-(b^3*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))^3 + 3*a*b^2*arctan(sqr 
t(-c*x + 1)/sqrt(c*x + 1))^2 + 3*a^2*b*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1) 
) + a^3)/(c^2*x^2 - 1), x)
 

Sympy [F]

\[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=- \int \frac {a^{3}}{c^{2} x^{2} - 1}\, dx - \int \frac {b^{3} \operatorname {atan}^{3}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {3 a^{2} b \operatorname {atan}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx \] Input:

integrate((a+b*atan((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**3/(-c**2*x**2+1),x)
 

Output:

-Integral(a**3/(c**2*x**2 - 1), x) - Integral(b**3*atan(sqrt(-c*x + 1)/sqr 
t(c*x + 1))**3/(c**2*x**2 - 1), x) - Integral(3*a*b**2*atan(sqrt(-c*x + 1) 
/sqrt(c*x + 1))**2/(c**2*x**2 - 1), x) - Integral(3*a**2*b*atan(sqrt(-c*x 
+ 1)/sqrt(c*x + 1))/(c**2*x**2 - 1), x)
 

Maxima [F]

\[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int { -\frac {{\left (b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1} \,d x } \] Input:

integrate((a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, alg 
orithm="maxima")
 

Output:

1/2*a^3*(log(c*x + 1)/c - log(c*x - 1)/c) + 1/64*(4*(b^3*log(c*x + 1) - b^ 
3*log(-c*x + 1))*arctan2(sqrt(-c*x + 1), sqrt(c*x + 1))^3 - 3*(b^3*log(2)^ 
2*log(c*x + 1) - b^3*log(2)^2*log(-c*x + 1))*arctan2(sqrt(-c*x + 1), sqrt( 
c*x + 1)) - 64*c*integrate(1/128*(112*b^3*arctan2(sqrt(-c*x + 1), sqrt(c*x 
 + 1))^3 + 384*a*b^2*arctan2(sqrt(-c*x + 1), sqrt(c*x + 1))^2 - 3*(b^3*log 
(2)^2*log(c*x + 1) - b^3*log(2)^2*log(-c*x + 1) - 4*(b^3*log(c*x + 1) - b^ 
3*log(-c*x + 1))*arctan2(sqrt(-c*x + 1), sqrt(c*x + 1))^2)*sqrt(c*x + 1)*s 
qrt(-c*x + 1) + 12*(b^3*log(2)^2 + 32*a^2*b)*arctan2(sqrt(-c*x + 1), sqrt( 
c*x + 1)))/(c^2*x^2 - 1), x))/c
 

Giac [F]

\[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int { -\frac {{\left (b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1} \,d x } \] Input:

integrate((a+b*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, alg 
orithm="giac")
 

Output:

integrate(-(b*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^3/(c^2*x^2 - 1), x 
)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int -\frac {{\left (a+b\,\mathrm {atan}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3}{c^2\,x^2-1} \,d x \] Input:

int(-(a + b*atan((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1),x)
 

Output:

int(-(a + b*atan((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {\left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\frac {-6 \left (\int \frac {\mathit {atan} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )}{c^{2} x^{2}-1}d x \right ) a^{2} b c -2 \left (\int \frac {\mathit {atan} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3}}{c^{2} x^{2}-1}d x \right ) b^{3} c -6 \left (\int \frac {\mathit {atan} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{c^{2} x^{2}-1}d x \right ) a \,b^{2} c -\mathrm {log}\left (c^{2} x -c \right ) a^{3}+\mathrm {log}\left (c^{2} x +c \right ) a^{3}}{2 c} \] Input:

int((a+b*atan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x)
 

Output:

( - 6*int(atan(sqrt( - c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1),x)*a**2*b*c 
 - 2*int(atan(sqrt( - c*x + 1)/sqrt(c*x + 1))**3/(c**2*x**2 - 1),x)*b**3*c 
 - 6*int(atan(sqrt( - c*x + 1)/sqrt(c*x + 1))**2/(c**2*x**2 - 1),x)*a*b**2 
*c - log(c**2*x - c)*a**3 + log(c**2*x + c)*a**3)/(2*c)