\(\int \frac {(a+b \arctan (c x^2))^2}{x^5} \, dx\) [80]

Optimal result

Integrand size = 16, antiderivative size = 87 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=-\frac {b c \left (a+b \arctan \left (c x^2\right )\right )}{2 x^2}-\frac {1}{4} c^2 \left (a+b \arctan \left (c x^2\right )\right )^2-\frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{4 x^4}+b^2 c^2 \log (x)-\frac {1}{4} b^2 c^2 \log \left (1+c^2 x^4\right ) \] Output:

-1/2*b*c*(a+b*arctan(c*x^2))/x^2-1/4*c^2*(a+b*arctan(c*x^2))^2-1/4*(a+b*ar 
ctan(c*x^2))^2/x^4+b^2*c^2*ln(x)-1/4*b^2*c^2*ln(c^2*x^4+1)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=-\frac {a^2+2 a b c x^2+2 b \left (a+b c x^2+a c^2 x^4\right ) \arctan \left (c x^2\right )+b^2 \left (1+c^2 x^4\right ) \arctan \left (c x^2\right )^2-4 b^2 c^2 x^4 \log (x)+b^2 c^2 x^4 \log \left (1+c^2 x^4\right )}{4 x^4} \] Input:

Integrate[(a + b*ArcTan[c*x^2])^2/x^5,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5363, 5361, 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.

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

\(\Big \downarrow \) 5363

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5453

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 5419

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

Input:

Int[(a + b*ArcTan[c*x^2])^2/x^5,x]
 

Output:

(-1/2*(a + b*ArcTan[c*x^2])^2/x^4 + b*c*(-((a + b*ArcTan[c*x^2])/x^2) - (c 
*(a + b*ArcTan[c*x^2])^2)/(2*b) + (b*c*(Log[x^4] - Log[1 + c^2*x^4]))/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 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 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 = 0.24 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.36

method	result	size
default	\(-\frac {a^{2}}{4 x^{4}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 x^{4}}-\frac {b^{2} c \arctan \left (c \,x^{2}\right )}{2 x^{2}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2} c^{2}}{4}-\frac {b^{2} c^{2} \ln \left (c^{2} x^{4}+1\right )}{4}+b^{2} c^{2} \ln \left (x \right )-\frac {a b \arctan \left (c \,x^{2}\right )}{2 x^{4}}-\frac {a b c}{2 x^{2}}-\frac {a b \arctan \left (c \,x^{2}\right ) c^{2}}{2}\)	\(118\)
parts	\(-\frac {a^{2}}{4 x^{4}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2}}{4 x^{4}}-\frac {b^{2} c \arctan \left (c \,x^{2}\right )}{2 x^{2}}-\frac {b^{2} \arctan \left (c \,x^{2}\right )^{2} c^{2}}{4}-\frac {b^{2} c^{2} \ln \left (c^{2} x^{4}+1\right )}{4}+b^{2} c^{2} \ln \left (x \right )-\frac {a b \arctan \left (c \,x^{2}\right )}{2 x^{4}}-\frac {a b c}{2 x^{2}}-\frac {a b \arctan \left (c \,x^{2}\right ) c^{2}}{2}\)	\(118\)
parallelrisch	\(\frac {-b^{2} \arctan \left (c \,x^{2}\right )^{2} x^{4} c^{2}+4 b^{2} c^{2} \ln \left (x \right ) x^{4}-b^{2} c^{2} \ln \left (c^{2} x^{4}+1\right ) x^{4}-2 a b \arctan \left (c \,x^{2}\right ) x^{4} c^{2}+a^{2} c^{2} x^{4}-2 b^{2} \arctan \left (c \,x^{2}\right ) x^{2} c -2 a b c \,x^{2}-b^{2} \arctan \left (c \,x^{2}\right )^{2}-2 a b \arctan \left (c \,x^{2}\right )-a^{2}}{4 x^{4}}\)	\(137\)
risch	\(\frac {b^{2} \left (c^{2} x^{4}+1\right ) \ln \left (i c \,x^{2}+1\right )^{2}}{16 x^{4}}+\frac {i b \left (i b \,c^{2} x^{4} \ln \left (-i c \,x^{2}+1\right )+2 b c \,x^{2}+2 a +i b \ln \left (-i c \,x^{2}+1\right )\right ) \ln \left (i c \,x^{2}+1\right )}{8 x^{4}}-\frac {4 i \ln \left (\left (-5 i b c +a c \right ) x^{2}+5 b +i a \right ) a b \,c^{2} x^{4}-4 i \ln \left (\left (5 i b c +a c \right ) x^{2}+5 b -i a \right ) a b \,c^{2} x^{4}-b^{2} c^{2} x^{4} \ln \left (-i c \,x^{2}+1\right )^{2}-16 b^{2} c^{2} \ln \left (x \right ) x^{4}+4 \ln \left (\left (-5 i b c +a c \right ) x^{2}+5 b +i a \right ) b^{2} c^{2} x^{4}+4 \ln \left (\left (5 i b c +a c \right ) x^{2}+5 b -i a \right ) b^{2} c^{2} x^{4}+4 i b^{2} c \,x^{2} \ln \left (-i c \,x^{2}+1\right )+8 a b c \,x^{2}+4 i b \ln \left (-i c \,x^{2}+1\right ) a -b^{2} \ln \left (-i c \,x^{2}+1\right )^{2}+4 a^{2}}{16 x^{4}}\)	\(332\)

Input:

int((a+b*arctan(c*x^2))^2/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/4*a^2/x^4-1/4*b^2*arctan(c*x^2)^2/x^4-1/2*b^2*c*arctan(c*x^2)/x^2-1/4*b 
^2*arctan(c*x^2)^2*c^2-1/4*b^2*c^2*ln(c^2*x^4+1)+b^2*c^2*ln(x)-1/2*a*b*arc 
tan(c*x^2)/x^4-1/2*a*b*c/x^2-1/2*a*b*arctan(c*x^2)*c^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=-\frac {b^{2} c^{2} x^{4} \log \left (c^{2} x^{4} + 1\right ) - 4 \, b^{2} c^{2} x^{4} \log \left (x\right ) + 2 \, a b c x^{2} + {\left (b^{2} c^{2} x^{4} + b^{2}\right )} \arctan \left (c x^{2}\right )^{2} + a^{2} + 2 \, {\left (a b c^{2} x^{4} + b^{2} c x^{2} + a b\right )} \arctan \left (c x^{2}\right )}{4 \, x^{4}} \] Input:

integrate((a+b*arctan(c*x^2))^2/x^5,x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (80) = 160\).

Time = 21.86 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.93 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=\begin {cases} - \frac {a^{2}}{4 x^{4}} - \frac {a b c^{2} \operatorname {atan}{\left (c x^{2} \right )}}{2} - \frac {a b c}{2 x^{2}} - \frac {a b \operatorname {atan}{\left (c x^{2} \right )}}{2 x^{4}} - \frac {b^{2} c^{3} \sqrt {- \frac {1}{c^{2}}} \operatorname {atan}{\left (c x^{2} \right )}}{2} + b^{2} c^{2} \log {\left (x \right )} - \frac {b^{2} c^{2} \log {\left (x^{2} + \sqrt {- \frac {1}{c^{2}}} \right )}}{2} - \frac {b^{2} c^{2} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4} - \frac {b^{2} c \operatorname {atan}{\left (c x^{2} \right )}}{2 x^{2}} - \frac {b^{2} \operatorname {atan}^{2}{\left (c x^{2} \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a^{2}}{4 x^{4}} & \text {otherwise} \end {cases} \] Input:

integrate((a+b*atan(c*x**2))**2/x**5,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=-\frac {1}{2} \, {\left ({\left (c \arctan \left (c x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {\arctan \left (c x^{2}\right )}{x^{4}}\right )} a b + \frac {1}{4} \, {\left ({\left (\arctan \left (c x^{2}\right )^{2} - \log \left (c^{2} x^{4} + 1\right ) + 4 \, \log \left (x\right )\right )} c^{2} - 2 \, {\left (c \arctan \left (c x^{2}\right ) + \frac {1}{x^{2}}\right )} c \arctan \left (c x^{2}\right )\right )} b^{2} - \frac {b^{2} \arctan \left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \] Input:

integrate((a+b*arctan(c*x^2))^2/x^5,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=\int { \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctan(c*x^2))^2/x^5,x, algorithm="giac")
 

Output:

integrate((b*arctan(c*x^2) + a)^2/x^5, x)
 

Mupad [B] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.75 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=b^2\,c^2\,\ln \left (x\right )-\frac {b^2\,c^2\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4}-\frac {b^2\,{\mathrm {atan}\left (c\,x^2\right )}^2}{4\,x^4}-\frac {b^2\,c^2\,\ln \left (c^2\,x^4+1\right )}{4}-\frac {a^2}{4\,x^4}-\frac {b^2\,c\,\mathrm {atan}\left (c\,x^2\right )}{2\,x^2}-\frac {a\,b\,c}{2\,x^2}-\frac {a\,b\,c^2\,\mathrm {atan}\left (\frac {a^2\,c\,x^2}{a^2+25\,b^2}+\frac {25\,b^2\,c\,x^2}{a^2+25\,b^2}\right )}{2}-\frac {a\,b\,\mathrm {atan}\left (c\,x^2\right )}{2\,x^4} \] Input:

int((a + b*atan(c*x^2))^2/x^5,x)
 

Output:

b^2*c^2*log(x) - (b^2*c^2*atan(c*x^2)^2)/4 - (b^2*atan(c*x^2)^2)/(4*x^4) - 
 (b^2*c^2*log(c^2*x^4 + 1))/4 - a^2/(4*x^4) - (b^2*c*atan(c*x^2))/(2*x^2) 
- (a*b*c)/(2*x^2) - (a*b*c^2*atan((a^2*c*x^2)/(a^2 + 25*b^2) + (25*b^2*c*x 
^2)/(a^2 + 25*b^2)))/2 - (a*b*atan(c*x^2))/(2*x^4)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a+b \arctan \left (c x^2\right )\right )^2}{x^5} \, dx=\frac {-\mathit {atan} \left (c \,x^{2}\right )^{2} b^{2} c^{2} x^{4}-\mathit {atan} \left (c \,x^{2}\right )^{2} b^{2}-2 \mathit {atan} \left (c \,x^{2}\right ) a b \,c^{2} x^{4}-2 \mathit {atan} \left (c \,x^{2}\right ) a b -2 \mathit {atan} \left (c \,x^{2}\right ) b^{2} c \,x^{2}-\mathrm {log}\left (-\sqrt {c}\, \sqrt {2}\, x +c \,x^{2}+1\right ) b^{2} c^{2} x^{4}-\mathrm {log}\left (\sqrt {c}\, \sqrt {2}\, x +c \,x^{2}+1\right ) b^{2} c^{2} x^{4}+4 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{4}-a^{2}-2 a b c \,x^{2}}{4 x^{4}} \] Input:

int((a+b*atan(c*x^2))^2/x^5,x)
 

Output:

( - atan(c*x**2)**2*b**2*c**2*x**4 - atan(c*x**2)**2*b**2 - 2*atan(c*x**2) 
*a*b*c**2*x**4 - 2*atan(c*x**2)*a*b - 2*atan(c*x**2)*b**2*c*x**2 - log( - 
sqrt(c)*sqrt(2)*x + c*x**2 + 1)*b**2*c**2*x**4 - log(sqrt(c)*sqrt(2)*x + c 
*x**2 + 1)*b**2*c**2*x**4 + 4*log(x)*b**2*c**2*x**4 - a**2 - 2*a*b*c*x**2) 
/(4*x**4)