\(\int \frac {a+b \arctan (c x^2)}{x^6} \, dx\) [73]

Optimal result

Integrand size = 14, antiderivative size = 125 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=-\frac {2 b c}{15 x^3}-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}+\frac {b c^{5/2} \arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{5 \sqrt {2}}-\frac {b c^{5/2} \arctan \left (1+\sqrt {2} \sqrt {c} x\right )}{5 \sqrt {2}}-\frac {b c^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} x}{1+c x^2}\right )}{5 \sqrt {2}} \] Output:

-2/15*b*c/x^3-1/5*(a+b*arctan(c*x^2))/x^5-1/10*b*c^(5/2)*arctan(-1+2^(1/2) 
*c^(1/2)*x)*2^(1/2)-1/10*b*c^(5/2)*arctan(1+2^(1/2)*c^(1/2)*x)*2^(1/2)-1/1 
0*b*c^(5/2)*arctanh(2^(1/2)*c^(1/2)*x/(c*x^2+1))*2^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=-\frac {a}{5 x^5}-\frac {2 b c}{15 x^3}-\frac {b \arctan \left (c x^2\right )}{5 x^5}-\frac {b c^{5/2} \arctan \left (\frac {-\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{5 \sqrt {2}}-\frac {b c^{5/2} \arctan \left (\frac {\sqrt {2}+2 \sqrt {c} x}{\sqrt {2}}\right )}{5 \sqrt {2}}+\frac {b c^{5/2} \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{10 \sqrt {2}}-\frac {b c^{5/2} \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{10 \sqrt {2}} \] Input:

Integrate[(a + b*ArcTan[c*x^2])/x^6,x]
 

Output:

-1/5*a/x^5 - (2*b*c)/(15*x^3) - (b*ArcTan[c*x^2])/(5*x^5) - (b*c^(5/2)*Arc 
Tan[(-Sqrt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(5*Sqrt[2]) - (b*c^(5/2)*ArcTan[(Sq 
rt[2] + 2*Sqrt[c]*x)/Sqrt[2]])/(5*Sqrt[2]) + (b*c^(5/2)*Log[1 - Sqrt[2]*Sq 
rt[c]*x + c*x^2])/(10*Sqrt[2]) - (b*c^(5/2)*Log[1 + Sqrt[2]*Sqrt[c]*x + c* 
x^2])/(10*Sqrt[2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.37, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5361, 847, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {2}{5} b c \int \frac {1}{x^4 \left (c^2 x^4+1\right )}dx-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {2}{5} b c \left (c^2 \left (-\int \frac {1}{c^2 x^4+1}dx\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \int \frac {c x^2+1}{c^2 x^4+1}dx\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}+\frac {\int \frac {1}{x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (1-\sqrt {2} \sqrt {c} x\right )^2-1}d\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{-\left (\sqrt {2} \sqrt {c} x+1\right )^2-1}d\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \int \frac {1-c x^2}{c^2 x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {c} x}{\sqrt {c} \left (x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {c} \left (x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {c} x}{\sqrt {c} \left (x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {c} \left (x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}\right )}dx}{2 \sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {c} x}{x^2-\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 \sqrt {2} c}+\frac {\int \frac {\sqrt {2} \sqrt {c} x+1}{x^2+\frac {\sqrt {2} x}{\sqrt {c}}+\frac {1}{c}}dx}{2 c}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2}{5} b c \left (-\left (c^2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}\right )+\frac {1}{2} \left (\frac {\log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}-\frac {\log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}\right )\right )\right )-\frac {1}{3 x^3}\right )-\frac {a+b \arctan \left (c x^2\right )}{5 x^5}\)

Input:

Int[(a + b*ArcTan[c*x^2])/x^6,x]
 

Output:

-1/5*(a + b*ArcTan[c*x^2])/x^5 + (2*b*c*(-1/3*1/x^3 - c^2*((-(ArcTan[1 - S 
qrt[2]*Sqrt[c]*x]/(Sqrt[2]*Sqrt[c])) + ArcTan[1 + Sqrt[2]*Sqrt[c]*x]/(Sqrt 
[2]*Sqrt[c]))/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[c]*x + c*x^2]/(Sqrt[2]*Sqrt[c 
]) + Log[1 + Sqrt[2]*Sqrt[c]*x + c*x^2]/(2*Sqrt[2]*Sqrt[c]))/2)))/5
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

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]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {a}{5 x^{5}}+b \left (-\frac {\arctan \left (c \,x^{2}\right )}{5 x^{5}}+\frac {2 c \left (-\frac {1}{3 x^{3}}-\frac {c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8}\right )}{5}\right )\)	\(118\)
parts	\(-\frac {a}{5 x^{5}}+b \left (-\frac {\arctan \left (c \,x^{2}\right )}{5 x^{5}}+\frac {2 c \left (-\frac {1}{3 x^{3}}-\frac {c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{8}\right )}{5}\right )\)	\(118\)

Input:

int((a+b*arctan(c*x^2))/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/5*a/x^5+b*(-1/5/x^5*arctan(c*x^2)+2/5*c*(-1/3/x^3-1/8*c^2*(1/c^2)^(1/4) 
*2^(1/2)*(ln((x^2+(1/c^2)^(1/4)*x*2^(1/2)+(1/c^2)^(1/2))/(x^2-(1/c^2)^(1/4 
)*x*2^(1/2)+(1/c^2)^(1/2)))+2*arctan(2^(1/2)/(1/c^2)^(1/4)*x+1)+2*arctan(2 
^(1/2)/(1/c^2)^(1/4)*x-1))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=-\frac {6 \, \sqrt {2} b c^{\frac {5}{2}} x^{5} \arctan \left (\sqrt {2} \sqrt {c} x + 1\right ) + 6 \, \sqrt {2} b c^{\frac {5}{2}} x^{5} \arctan \left (\sqrt {2} \sqrt {c} x - 1\right ) + 3 \, \sqrt {2} b c^{\frac {5}{2}} x^{5} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right ) - 3 \, \sqrt {2} b c^{\frac {5}{2}} x^{5} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right ) + 8 \, b c x^{2} + 12 \, b \arctan \left (c x^{2}\right ) + 12 \, a}{60 \, x^{5}} \] Input:

integrate((a+b*arctan(c*x^2))/x^6,x, algorithm="fricas")
 

Output:

-1/60*(6*sqrt(2)*b*c^(5/2)*x^5*arctan(sqrt(2)*sqrt(c)*x + 1) + 6*sqrt(2)*b 
*c^(5/2)*x^5*arctan(sqrt(2)*sqrt(c)*x - 1) + 3*sqrt(2)*b*c^(5/2)*x^5*log(c 
*x^2 + sqrt(2)*sqrt(c)*x + 1) - 3*sqrt(2)*b*c^(5/2)*x^5*log(c*x^2 - sqrt(2 
)*sqrt(c)*x + 1) + 8*b*c*x^2 + 12*b*arctan(c*x^2) + 12*a)/x^5
 

Sympy [A] (verification not implemented)

Time = 28.66 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.24 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=\begin {cases} - \frac {a}{5 x^{5}} - \frac {b c^{4} \left (- \frac {1}{c^{2}}\right )^{\frac {3}{4}} \operatorname {atan}{\left (c x^{2} \right )}}{5} + \frac {b c^{3} \sqrt [4]{- \frac {1}{c^{2}}} \log {\left (x - \sqrt [4]{- \frac {1}{c^{2}}} \right )}}{5} - \frac {b c^{3} \sqrt [4]{- \frac {1}{c^{2}}} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{10} - \frac {b c^{3} \sqrt [4]{- \frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {x}{\sqrt [4]{- \frac {1}{c^{2}}}} \right )}}{5} - \frac {2 b c}{15 x^{3}} - \frac {b \operatorname {atan}{\left (c x^{2} \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a}{5 x^{5}} & \text {otherwise} \end {cases} \] Input:

integrate((a+b*atan(c*x**2))/x**6,x)
 

Output:

Piecewise((-a/(5*x**5) - b*c**4*(-1/c**2)**(3/4)*atan(c*x**2)/5 + b*c**3*( 
-1/c**2)**(1/4)*log(x - (-1/c**2)**(1/4))/5 - b*c**3*(-1/c**2)**(1/4)*log( 
x**2 + sqrt(-1/c**2))/10 - b*c**3*(-1/c**2)**(1/4)*atan(x/(-1/c**2)**(1/4) 
)/5 - 2*b*c/(15*x**3) - b*atan(c*x**2)/(5*x**5), Ne(c, 0)), (-a/(5*x**5), 
True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=-\frac {1}{60} \, {\left ({\left (6 \, \sqrt {2} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right ) + 6 \, \sqrt {2} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right ) + 3 \, \sqrt {2} c^{\frac {3}{2}} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right ) - 3 \, \sqrt {2} c^{\frac {3}{2}} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right ) + \frac {8}{x^{3}}\right )} c + \frac {12 \, \arctan \left (c x^{2}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \] Input:

integrate((a+b*arctan(c*x^2))/x^6,x, algorithm="maxima")
 

Output:

-1/60*((6*sqrt(2)*c^(3/2)*arctan(1/2*sqrt(2)*(2*c*x + sqrt(2)*sqrt(c))/sqr 
t(c)) + 6*sqrt(2)*c^(3/2)*arctan(1/2*sqrt(2)*(2*c*x - sqrt(2)*sqrt(c))/sqr 
t(c)) + 3*sqrt(2)*c^(3/2)*log(c*x^2 + sqrt(2)*sqrt(c)*x + 1) - 3*sqrt(2)*c 
^(3/2)*log(c*x^2 - sqrt(2)*sqrt(c)*x + 1) + 8/x^3)*c + 12*arctan(c*x^2)/x^ 
5)*b - 1/5*a/x^5
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=-\frac {1}{20} \, b c^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} + \frac {\sqrt {2} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}}\right )} - \frac {2 \, b c x^{2} + 3 \, b \arctan \left (c x^{2}\right ) + 3 \, a}{15 \, x^{5}} \] Input:

integrate((a+b*arctan(c*x^2))/x^6,x, algorithm="giac")
 

Output:

-1/20*b*c^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)/sqrt(abs(c)))*sqr 
t(abs(c)))/sqrt(abs(c)) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)/sqrt 
(abs(c)))*sqrt(abs(c)))/sqrt(abs(c)) + sqrt(2)*log(x^2 + sqrt(2)*x/sqrt(ab 
s(c)) + 1/abs(c))/sqrt(abs(c)) - sqrt(2)*log(x^2 - sqrt(2)*x/sqrt(abs(c)) 
+ 1/abs(c))/sqrt(abs(c))) - 1/15*(2*b*c*x^2 + 3*b*arctan(c*x^2) + 3*a)/x^5
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.50 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=-\frac {\frac {2\,b\,c\,x^2}{3}+a}{5\,x^5}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{5\,x^5}+\frac {{\left (-1\right )}^{1/4}\,b\,c^{5/2}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )\,1{}\mathrm {i}}{5}+\frac {{\left (-1\right )}^{1/4}\,b\,c^{5/2}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right )}{5} \] Input:

int((a + b*atan(c*x^2))/x^6,x)
 

Output:

((-1)^(1/4)*b*c^(5/2)*atan((-1)^(1/4)*c^(1/2)*x)*1i)/5 - (b*atan(c*x^2))/( 
5*x^5) - (a + (2*b*c*x^2)/3)/(5*x^5) + ((-1)^(1/4)*b*c^(5/2)*atan((-1)^(1/ 
4)*c^(1/2)*x*1i))/5
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^6} \, dx=\frac {12 \sqrt {c}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {c}\, \sqrt {2}-2 c x}{\sqrt {c}\, \sqrt {2}}\right ) b \,c^{2} x^{5}+6 \sqrt {c}\, \sqrt {2}\, \mathit {atan} \left (c \,x^{2}\right ) b \,c^{2} x^{5}-12 \mathit {atan} \left (c \,x^{2}\right ) b +3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (-\sqrt {c}\, \sqrt {2}\, x +c \,x^{2}+1\right ) b \,c^{2} x^{5}-3 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {c}\, \sqrt {2}\, x +c \,x^{2}+1\right ) b \,c^{2} x^{5}-12 a -8 b c \,x^{2}}{60 x^{5}} \] Input:

int((a+b*atan(c*x^2))/x^6,x)
 

Output:

(12*sqrt(c)*sqrt(2)*atan((sqrt(c)*sqrt(2) - 2*c*x)/(sqrt(c)*sqrt(2)))*b*c* 
*2*x**5 + 6*sqrt(c)*sqrt(2)*atan(c*x**2)*b*c**2*x**5 - 12*atan(c*x**2)*b + 
 3*sqrt(c)*sqrt(2)*log( - sqrt(c)*sqrt(2)*x + c*x**2 + 1)*b*c**2*x**5 - 3* 
sqrt(c)*sqrt(2)*log(sqrt(c)*sqrt(2)*x + c*x**2 + 1)*b*c**2*x**5 - 12*a - 8 
*b*c*x**2)/(60*x**5)