\(\int x^3 (a+b \arctan (\frac {c}{x}))^2 \, dx\) [140]

Optimal result

Integrand size = 16, antiderivative size = 122 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{12} b^2 c^2 x^2-\frac {1}{2} b c^3 x \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )+\frac {1}{6} b c x^3 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )-\frac {1}{4} c^4 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2+\frac {1}{4} x^4 \left (a+b \cot ^{-1}\left (\frac {x}{c}\right )\right )^2-\frac {1}{3} b^2 c^4 \log \left (1+\frac {c^2}{x^2}\right )-\frac {2}{3} b^2 c^4 \log (x) \] Output:

1/12*b^2*c^2*x^2-1/2*b*c^3*x*(a+b*arccot(x/c))+1/6*b*c*x^3*(a+b*arccot(x/c 
))-1/4*c^4*(a+b*arccot(x/c))^2+1/4*x^4*(a+b*arccot(x/c))^2-1/3*b^2*c^4*ln( 
1+c^2/x^2)-2/3*b^2*c^4*ln(x)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.91 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{12} \left (x \left (b^2 c^2 x+3 a^2 x^3+2 a b c \left (-3 c^2+x^2\right )\right )+2 b \left (b c x \left (-3 c^2+x^2\right )+3 a \left (-c^4+x^4\right )\right ) \arctan \left (\frac {c}{x}\right )+3 b^2 \left (-c^4+x^4\right ) \arctan \left (\frac {c}{x}\right )^2-4 b^2 c^4 \log \left (c^2+x^2\right )\right ) \] Input:

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

Output:

(x*(b^2*c^2*x + 3*a^2*x^3 + 2*a*b*c*(-3*c^2 + x^2)) + 2*b*(b*c*x*(-3*c^2 + 
 x^2) + 3*a*(-c^4 + x^4))*ArcTan[c/x] + 3*b^2*(-c^4 + x^4)*ArcTan[c/x]^2 - 
 4*b^2*c^4*Log[c^2 + x^2])/12
 

Rubi [A] (warning: unable to verify)

Time = 0.93 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5363, 5361, 5453, 5361, 243, 54, 2009, 5453, 5361, 243, 47, 14, 16, 5419}

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.

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

\(\Big \downarrow \) 5363

\(\displaystyle -\int x^5 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2d\frac {1}{x}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \int \frac {x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (\int x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )d\frac {1}{x}-c^2 \int \frac {x^2 \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \int \frac {x^2 \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}+\frac {1}{3} b c \int \frac {x^3}{\frac {c^2}{x^2}+1}d\frac {1}{x}-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \int \frac {x^2 \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}+\frac {1}{6} b c \int \frac {x^2}{\frac {c^2}{x^2}+1}d\frac {1}{x^2}-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )\)

\(\Big \downarrow \) 54

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \int \frac {x^2 \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}+\frac {1}{6} b c \int \left (\frac {c^4}{\frac {c^2}{x^2}+1}-x c^2+x^2\right )d\frac {1}{x^2}-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \int \frac {x^2 \left (a+b \arctan \left (\frac {c}{x}\right )\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (\int x^2 \left (a+b \arctan \left (\frac {c}{x}\right )\right )d\frac {1}{x}-c^2 \int \frac {a+b \arctan \left (\frac {c}{x}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (c^2 \left (-\int \frac {a+b \arctan \left (\frac {c}{x}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )+b c \int \frac {x}{\frac {c^2}{x^2}+1}d\frac {1}{x}-x \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (c^2 \left (-\int \frac {a+b \arctan \left (\frac {c}{x}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )+\frac {1}{2} b c \int \frac {x}{\frac {c^2}{x^2}+1}d\frac {1}{x^2}-x \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 47

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (c^2 \left (-\int \frac {a+b \arctan \left (\frac {c}{x}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )+\frac {1}{2} b c \left (\int xd\frac {1}{x^2}-c^2 \int \frac {1}{\frac {c^2}{x^2}+1}d\frac {1}{x^2}\right )-x \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (c^2 \left (-\int \frac {a+b \arctan \left (\frac {c}{x}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )+\frac {1}{2} b c \left (\log \left (\frac {1}{x^2}\right )-c^2 \int \frac {1}{\frac {c^2}{x^2}+1}d\frac {1}{x^2}\right )-x \left (a+b \arctan \left (\frac {c}{x}\right )\right )\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (c^2 \left (-\int \frac {a+b \arctan \left (\frac {c}{x}\right )}{\frac {c^2}{x^2}+1}d\frac {1}{x}\right )-x \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{2} b c \left (\log \left (\frac {1}{x^2}\right )-\log \left (\frac {c^2}{x^2}+1\right )\right )\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

\(\Big \downarrow \) 5419

\(\displaystyle \frac {1}{4} x^4 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2-\frac {1}{2} b c \left (-c^2 \left (-\frac {c \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2}{2 b}-x \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{2} b c \left (\log \left (\frac {1}{x^2}\right )-\log \left (\frac {c^2}{x^2}+1\right )\right )\right )-\frac {1}{3} x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )+\frac {1}{6} b c \left (c^2 \log \left (\frac {c^2}{x^2}+1\right )-c^2 \log \left (\frac {1}{x^2}\right )-x\right )\right )\)

Input:

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

Output:

(x^4*(a + b*ArcTan[c/x])^2)/4 - (b*c*(-1/3*(x^3*(a + b*ArcTan[c/x])) + (b* 
c*(-x + c^2*Log[1 + c^2/x^2] - c^2*Log[x^(-2)]))/6 - c^2*(-(x*(a + b*ArcTa 
n[c/x])) - (c*(a + b*ArcTan[c/x])^2)/(2*b) + (b*c*(-Log[1 + c^2/x^2] + Log 
[x^(-2)]))/2)))/2
 

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 47

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c 
 - a*d)   Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*x), x 
], x] /; FreeQ[{a, b, c, d}, x]
 

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5361

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] - Simp[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]
 

rule 5363

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTan[c*x])^p, x], 
 x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simplif 
y[(m + 1)/n]]
 

rule 5419

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> 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 5453

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], 
 x] - Simp[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]
 

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.16

method	result	size
parts	\(\frac {a^{2} x^{4}}{4}-b^{2} c^{4} \left (-\frac {x^{4} \arctan \left (\frac {c}{x}\right )^{2}}{4 c^{4}}+\frac {\arctan \left (\frac {c}{x}\right )^{2}}{4}-\frac {x^{3} \arctan \left (\frac {c}{x}\right )}{6 c^{3}}+\frac {x \arctan \left (\frac {c}{x}\right )}{2 c}+\frac {\ln \left (1+\frac {c^{2}}{x^{2}}\right )}{3}-\frac {x^{2}}{12 c^{2}}-\frac {2 \ln \left (\frac {c}{x}\right )}{3}\right )+\frac {x^{4} \arctan \left (\frac {c}{x}\right ) a b}{2}+\frac {a b \,c^{4} \arctan \left (\frac {x}{c}\right )}{2}+\frac {a b c \,x^{3}}{6}-\frac {a b \,c^{3} x}{2}\)	\(141\)
derivativedivides	\(-c^{4} \left (-\frac {a^{2} x^{4}}{4 c^{4}}+b^{2} \left (-\frac {x^{4} \arctan \left (\frac {c}{x}\right )^{2}}{4 c^{4}}+\frac {\arctan \left (\frac {c}{x}\right )^{2}}{4}-\frac {x^{3} \arctan \left (\frac {c}{x}\right )}{6 c^{3}}+\frac {x \arctan \left (\frac {c}{x}\right )}{2 c}+\frac {\ln \left (1+\frac {c^{2}}{x^{2}}\right )}{3}-\frac {x^{2}}{12 c^{2}}-\frac {2 \ln \left (\frac {c}{x}\right )}{3}\right )+2 a b \left (-\frac {x^{4} \arctan \left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\arctan \left (\frac {c}{x}\right )}{4}-\frac {x^{3}}{12 c^{3}}+\frac {x}{4 c}\right )\right )\)	\(144\)
default	\(-c^{4} \left (-\frac {a^{2} x^{4}}{4 c^{4}}+b^{2} \left (-\frac {x^{4} \arctan \left (\frac {c}{x}\right )^{2}}{4 c^{4}}+\frac {\arctan \left (\frac {c}{x}\right )^{2}}{4}-\frac {x^{3} \arctan \left (\frac {c}{x}\right )}{6 c^{3}}+\frac {x \arctan \left (\frac {c}{x}\right )}{2 c}+\frac {\ln \left (1+\frac {c^{2}}{x^{2}}\right )}{3}-\frac {x^{2}}{12 c^{2}}-\frac {2 \ln \left (\frac {c}{x}\right )}{3}\right )+2 a b \left (-\frac {x^{4} \arctan \left (\frac {c}{x}\right )}{4 c^{4}}+\frac {\arctan \left (\frac {c}{x}\right )}{4}-\frac {x^{3}}{12 c^{3}}+\frac {x}{4 c}\right )\right )\)	\(144\)
parallelrisch	\(\frac {x^{4} \arctan \left (\frac {c}{x}\right )^{2} b^{2}}{4}-\frac {\arctan \left (\frac {c}{x}\right )^{2} b^{2} c^{4}}{4}-\frac {b^{2} c^{4} \ln \left (c^{2}+x^{2}\right )}{3}+\frac {x^{4} \arctan \left (\frac {c}{x}\right ) a b}{2}+\frac {x^{3} \arctan \left (\frac {c}{x}\right ) b^{2} c}{6}-\frac {x \arctan \left (\frac {c}{x}\right ) b^{2} c^{3}}{2}-\frac {\arctan \left (\frac {c}{x}\right ) a b \,c^{4}}{2}+\frac {a^{2} x^{4}}{4}+\frac {a b c \,x^{3}}{6}+\frac {b^{2} c^{2} x^{2}}{12}-\frac {a b \,c^{3} x}{2}-\frac {b^{2} c^{4}}{12}\)	\(149\)
risch	\(\text {Expression too large to display}\)	\(14738\)

Input:

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

Output:

1/4*a^2*x^4-b^2*c^4*(-1/4/c^4*x^4*arctan(c/x)^2+1/4*arctan(c/x)^2-1/6/c^3* 
x^3*arctan(c/x)+1/2/c*x*arctan(c/x)+1/3*ln(1+c^2/x^2)-1/12/c^2*x^2-2/3*ln( 
c/x))+1/2*x^4*arctan(c/x)*a*b+1/2*a*b*c^4*arctan(x/c)+1/6*a*b*c*x^3-1/2*a* 
b*c^3*x
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{2} \, a b c^{4} \arctan \left (\frac {x}{c}\right ) - \frac {1}{3} \, b^{2} c^{4} \log \left (c^{2} + x^{2}\right ) - \frac {1}{2} \, a b c^{3} x + \frac {1}{12} \, b^{2} c^{2} x^{2} + \frac {1}{6} \, a b c x^{3} + \frac {1}{4} \, a^{2} x^{4} - \frac {1}{4} \, {\left (b^{2} c^{4} - b^{2} x^{4}\right )} \arctan \left (\frac {c}{x}\right )^{2} - \frac {1}{6} \, {\left (3 \, b^{2} c^{3} x - b^{2} c x^{3} - 3 \, a b x^{4}\right )} \arctan \left (\frac {c}{x}\right ) \] Input:

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

Output:

1/2*a*b*c^4*arctan(x/c) - 1/3*b^2*c^4*log(c^2 + x^2) - 1/2*a*b*c^3*x + 1/1 
2*b^2*c^2*x^2 + 1/6*a*b*c*x^3 + 1/4*a^2*x^4 - 1/4*(b^2*c^4 - b^2*x^4)*arct 
an(c/x)^2 - 1/6*(3*b^2*c^3*x - b^2*c*x^3 - 3*a*b*x^4)*arctan(c/x)
 

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.18 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^{2} x^{4}}{4} - \frac {a b c^{4} \operatorname {atan}{\left (\frac {c}{x} \right )}}{2} - \frac {a b c^{3} x}{2} + \frac {a b c x^{3}}{6} + \frac {a b x^{4} \operatorname {atan}{\left (\frac {c}{x} \right )}}{2} - \frac {b^{2} c^{4} \log {\left (c^{2} + x^{2} \right )}}{3} - \frac {b^{2} c^{4} \operatorname {atan}^{2}{\left (\frac {c}{x} \right )}}{4} - \frac {b^{2} c^{3} x \operatorname {atan}{\left (\frac {c}{x} \right )}}{2} + \frac {b^{2} c^{2} x^{2}}{12} + \frac {b^{2} c x^{3} \operatorname {atan}{\left (\frac {c}{x} \right )}}{6} + \frac {b^{2} x^{4} \operatorname {atan}^{2}{\left (\frac {c}{x} \right )}}{4} \] Input:

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

Output:

a**2*x**4/4 - a*b*c**4*atan(c/x)/2 - a*b*c**3*x/2 + a*b*c*x**3/6 + a*b*x** 
4*atan(c/x)/2 - b**2*c**4*log(c**2 + x**2)/3 - b**2*c**4*atan(c/x)**2/4 - 
b**2*c**3*x*atan(c/x)/2 + b**2*c**2*x**2/12 + b**2*c*x**3*atan(c/x)/6 + b* 
*2*x**4*atan(c/x)**2/4
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.10 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \arctan \left (\frac {c}{x}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{6} \, {\left (3 \, x^{4} \arctan \left (\frac {c}{x}\right ) + {\left (3 \, c^{3} \arctan \left (\frac {x}{c}\right ) - 3 \, c^{2} x + x^{3}\right )} c\right )} a b + \frac {1}{12} \, {\left ({\left (3 \, c^{2} \arctan \left (\frac {x}{c}\right )^{2} - 4 \, c^{2} \log \left (c^{2} + x^{2}\right ) + x^{2}\right )} c^{2} + 2 \, {\left (3 \, c^{3} \arctan \left (\frac {x}{c}\right ) - 3 \, c^{2} x + x^{3}\right )} c \arctan \left (\frac {c}{x}\right )\right )} b^{2} \] Input:

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

Output:

1/4*b^2*x^4*arctan(c/x)^2 + 1/4*a^2*x^4 + 1/6*(3*x^4*arctan(c/x) + (3*c^3* 
arctan(x/c) - 3*c^2*x + x^3)*c)*a*b + 1/12*((3*c^2*arctan(x/c)^2 - 4*c^2*l 
og(c^2 + x^2) + x^2)*c^2 + 2*(3*c^3*arctan(x/c) - 3*c^2*x + x^3)*c*arctan( 
c/x))*b^2
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.15 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}-\frac {b^2\,c^4\,{\mathrm {atan}\left (\frac {c}{x}\right )}^2}{4}-\frac {b^2\,c^4\,\ln \left (c^2+x^2\right )}{3}+\frac {b^2\,x^4\,{\mathrm {atan}\left (\frac {c}{x}\right )}^2}{4}+\frac {b^2\,c^2\,x^2}{12}+\frac {b^2\,c\,x^3\,\mathrm {atan}\left (\frac {c}{x}\right )}{6}-\frac {b^2\,c^3\,x\,\mathrm {atan}\left (\frac {c}{x}\right )}{2}+\frac {a\,b\,c\,x^3}{6}-\frac {a\,b\,c^3\,x}{2}-\frac {a\,b\,c^4\,\mathrm {atan}\left (\frac {c}{x}\right )}{2}+\frac {a\,b\,x^4\,\mathrm {atan}\left (\frac {c}{x}\right )}{2} \] Input:

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

Output:

(a^2*x^4)/4 - (b^2*c^4*atan(c/x)^2)/4 - (b^2*c^4*log(c^2 + x^2))/3 + (b^2* 
x^4*atan(c/x)^2)/4 + (b^2*c^2*x^2)/12 + (b^2*c*x^3*atan(c/x))/6 - (b^2*c^3 
*x*atan(c/x))/2 + (a*b*c*x^3)/6 - (a*b*c^3*x)/2 - (a*b*c^4*atan(c/x))/2 + 
(a*b*x^4*atan(c/x))/2
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.15 \[ \int x^3 \left (a+b \arctan \left (\frac {c}{x}\right )\right )^2 \, dx=-\frac {\mathit {atan} \left (\frac {c}{x}\right )^{2} b^{2} c^{4}}{4}+\frac {\mathit {atan} \left (\frac {c}{x}\right )^{2} b^{2} x^{4}}{4}-\frac {\mathit {atan} \left (\frac {c}{x}\right ) a b \,c^{4}}{2}+\frac {\mathit {atan} \left (\frac {c}{x}\right ) a b \,x^{4}}{2}-\frac {\mathit {atan} \left (\frac {c}{x}\right ) b^{2} c^{3} x}{2}+\frac {\mathit {atan} \left (\frac {c}{x}\right ) b^{2} c \,x^{3}}{6}-\frac {\mathrm {log}\left (c^{2}+x^{2}\right ) b^{2} c^{4}}{3}+\frac {a^{2} x^{4}}{4}-\frac {a b \,c^{3} x}{2}+\frac {a b c \,x^{3}}{6}+\frac {b^{2} c^{2} x^{2}}{12} \] Input:

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

Output:

( - 3*atan(c/x)**2*b**2*c**4 + 3*atan(c/x)**2*b**2*x**4 - 6*atan(c/x)*a*b* 
c**4 + 6*atan(c/x)*a*b*x**4 - 6*atan(c/x)*b**2*c**3*x + 2*atan(c/x)*b**2*c 
*x**3 - 4*log(c**2 + x**2)*b**2*c**4 + 3*a**2*x**4 - 6*a*b*c**3*x + 2*a*b* 
c*x**3 + b**2*c**2*x**2)/12