\(\int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx\) [231]

Optimal result

Integrand size = 22, antiderivative size = 118 \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=-\frac {a \sqrt {c+a^2 c x^2}}{6 c x^2}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{3 c x^3}+\frac {2 a^2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 c x}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{6 \sqrt {c}} \] Output:

-1/6*a*(a^2*c*x^2+c)^(1/2)/c/x^2-1/3*(a^2*c*x^2+c)^(1/2)*arctan(a*x)/c/x^3 
+2/3*a^2*(a^2*c*x^2+c)^(1/2)*arctan(a*x)/c/x+5/6*a^3*arctanh((a^2*c*x^2+c) 
^(1/2)/c^(1/2))/c^(1/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.93 \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\frac {-a x \sqrt {c+a^2 c x^2}+2 \left (-1+2 a^2 x^2\right ) \sqrt {c+a^2 c x^2} \arctan (a x)-5 a^3 \sqrt {c} x^3 \log (x)+5 a^3 \sqrt {c} x^3 \log \left (c+\sqrt {c} \sqrt {c+a^2 c x^2}\right )}{6 c x^3} \] Input:

Integrate[ArcTan[a*x]/(x^4*Sqrt[c + a^2*c*x^2]),x]
 

Output:

(-(a*x*Sqrt[c + a^2*c*x^2]) + 2*(-1 + 2*a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTa 
n[a*x] - 5*a^3*Sqrt[c]*x^3*Log[x] + 5*a^3*Sqrt[c]*x^3*Log[c + Sqrt[c]*Sqrt 
[c + a^2*c*x^2]])/(6*c*x^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.27, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {5497, 243, 52, 73, 221, 5479, 243, 73, 221}

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.

\(\displaystyle \int \frac {\arctan (a x)}{x^4 \sqrt {a^2 c x^2+c}} \, dx\)

\(\Big \downarrow \) 5497

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

\(\Big \downarrow \) 243

\(\displaystyle -\frac {2}{3} a^2 \int \frac {\arctan (a x)}{x^2 \sqrt {a^2 c x^2+c}}dx+\frac {1}{6} a \int \frac {1}{x^4 \sqrt {a^2 c x^2+c}}dx^2-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{3 c x^3}\)

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 5479

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[ArcTan[a*x]/(x^4*Sqrt[c + a^2*c*x^2]),x]
 

Output:

-1/3*(Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(c*x^3) - (2*a^2*(-((Sqrt[c + a^2*c 
*x^2]*ArcTan[a*x])/(c*x)) - (a*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/Sqrt[ 
c]))/3 + (a*(-(Sqrt[c + a^2*c*x^2]/(c*x^2)) + (a^2*ArcTanh[Sqrt[c + a^2*c* 
x^2]/Sqrt[c]])/Sqrt[c]))/6
 

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

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 5479

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + 
 b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 1)))   Int[(f*x) 
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, 
c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] 
&& NeQ[m, -1]
 

rule 5497

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) 
+ (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*Ar 
cTan[c*x])^p/(d*f*(m + 1))), x] + (-Simp[b*c*(p/(f*(m + 1)))   Int[(f*x)^(m 
 + 1)*((a + b*ArcTan[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] - Simp[c^2*((m 
+ 2)/(f^2*(m + 1)))   Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/Sqrt[d + e*x 
^2]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[p, 0] 
 && LtQ[m, -1] && NeQ[m, -2]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.41 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {\left (4 x^{2} a^{2} \arctan \left (a x \right )-a x -2 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 c \,x^{3}}-\frac {5 a^{3} \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, c}+\frac {5 a^{3} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, c}\)	\(163\)

Input:

int(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(4*x^2*a^2*arctan(a*x)-a*x-2*arctan(a*x))*(c*(a*x-I)*(a*x+I))^(1/2)/c/ 
x^3-5/6*a^3*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)*(c*(a*x-I)*(a*x+I))^(1/2)/(a 
^2*x^2+1)^(1/2)/c+5/6*a^3*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))*(c*(a*x-I)*(a* 
x+I))^(1/2)/(a^2*x^2+1)^(1/2)/c
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\frac {5 \, a^{3} \sqrt {c} x^{3} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - 2 \, {\left (2 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )\right )}}{12 \, c x^{3}} \] Input:

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{4} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \] Input:

integrate(atan(a*x)/x**4/(a**2*c*x**2+c)**(1/2),x)
 

Output:

Integral(atan(a*x)/(x**4*sqrt(c*(a**2*x**2 + 1))), x)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.69 \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\frac {{\left (5 \, a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\sqrt {a^{2} x^{2} + 1}}{x^{2}}\right )} a + 2 \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{x} - \frac {\sqrt {a^{2} x^{2} + 1}}{x^{3}}\right )} \arctan \left (a x\right )}{6 \, \sqrt {c}} \] Input:

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")
 

Output:

1/6*((5*a^2*arcsinh(1/(a*abs(x))) - sqrt(a^2*x^2 + 1)/x^2)*a + 2*(2*sqrt(a 
^2*x^2 + 1)*a^2/x - sqrt(a^2*x^2 + 1)/x^3)*arctan(a*x))/sqrt(c)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionindex.cc index_m i_lex_is_greater Error: Bad Argument Valu 
e
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{x^4\,\sqrt {c\,a^2\,x^2+c}} \,d x \] Input:

int(atan(a*x)/(x^4*(c + a^2*c*x^2)^(1/2)),x)
 

Output:

int(atan(a*x)/(x^4*(c + a^2*c*x^2)^(1/2)), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.94 \[ \int \frac {\arctan (a x)}{x^4 \sqrt {c+a^2 c x^2}} \, dx=\frac {\sqrt {c}\, \left (4 \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right ) a^{2} x^{2}-2 \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right )-\sqrt {a^{2} x^{2}+1}\, a x -5 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x -1\right ) a^{3} x^{3}+5 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x +1\right ) a^{3} x^{3}\right )}{6 c \,x^{3}} \] Input:

int(atan(a*x)/x^4/(a^2*c*x^2+c)^(1/2),x)
 

Output:

(sqrt(c)*(4*sqrt(a**2*x**2 + 1)*atan(a*x)*a**2*x**2 - 2*sqrt(a**2*x**2 + 1 
)*atan(a*x) - sqrt(a**2*x**2 + 1)*a*x - 5*log(sqrt(a**2*x**2 + 1) + a*x - 
1)*a**3*x**3 + 5*log(sqrt(a**2*x**2 + 1) + a*x + 1)*a**3*x**3))/(6*c*x**3)