\(\int \frac {a+b \arctan (c x^2)}{x^5} \, dx\) [66]

Optimal result

Integrand size = 14, antiderivative size = 41 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=-\frac {b c}{4 x^2}-\frac {1}{4} b c^2 \arctan \left (c x^2\right )-\frac {a+b \arctan \left (c x^2\right )}{4 x^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4946, 281, 331, 209} \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=-\frac {a+b \arctan \left (c x^2\right )}{4 x^4}-\frac {1}{4} b c^2 \arctan \left (c x^2\right )-\frac {b c}{4 x^2} \]

[In]

Int[(a + b*ArcTan[c*x^2])/x^5,x]

[Out]

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

Rule 209

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

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 4946

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \arctan \left (c x^2\right )}{4 x^4}+\frac {1}{2} (b c) \int \frac {1}{x^3 \left (1+c^2 x^4\right )} \, dx \\ & = -\frac {a+b \arctan \left (c x^2\right )}{4 x^4}+\frac {1}{4} (b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {b c}{4 x^2}-\frac {a+b \arctan \left (c x^2\right )}{4 x^4}-\frac {1}{4} \left (b c^3\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,x^2\right ) \\ & = -\frac {b c}{4 x^2}-\frac {1}{4} b c^2 \arctan \left (c x^2\right )-\frac {a+b \arctan \left (c x^2\right )}{4 x^4} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.17 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=-\frac {a}{4 x^4}-\frac {b \arctan \left (c x^2\right )}{4 x^4}-\frac {b c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^4\right )}{4 x^2} \]

[In]

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

[Out]

-1/4*a/x^4 - (b*ArcTan[c*x^2])/(4*x^4) - (b*c*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^4)])/(4*x^2)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {a}{4 x^{4}}-\frac {b \arctan \left (c \,x^{2}\right )}{4 x^{4}}-\frac {b c}{4 x^{2}}-\frac {b \,c^{2} \arctan \left (c \,x^{2}\right )}{4}\)	\(39\)
parts	\(-\frac {a}{4 x^{4}}-\frac {b \arctan \left (c \,x^{2}\right )}{4 x^{4}}-\frac {b c}{4 x^{2}}-\frac {b \,c^{2} \arctan \left (c \,x^{2}\right )}{4}\)	\(39\)
parallelrisch	\(-\frac {\arctan \left (c \,x^{2}\right ) b \,c^{2} x^{4}-a \,c^{2} x^{4}+b c \,x^{2}+b \arctan \left (c \,x^{2}\right )+a}{4 x^{4}}\)	\(45\)
risch	\(\frac {i b \ln \left (i c \,x^{2}+1\right )}{8 x^{4}}-\frac {-i b \,c^{2} \ln \left (c \,x^{2}-i\right ) x^{4}+i b \,c^{2} \ln \left (c \,x^{2}+i\right ) x^{4}+2 b c \,x^{2}+i b \ln \left (-i c \,x^{2}+1\right )+2 a}{8 x^{4}}\)	\(87\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=-\frac {b c x^{2} + {\left (b c^{2} x^{4} + b\right )} \arctan \left (c x^{2}\right ) + a}{4 \, x^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 12.69 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=- \frac {a}{4 x^{4}} - \frac {b c^{2} \operatorname {atan}{\left (c x^{2} \right )}}{4} - \frac {b c}{4 x^{2}} - \frac {b \operatorname {atan}{\left (c x^{2} \right )}}{4 x^{4}} \]

[In]

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

[Out]

-a/(4*x**4) - b*c**2*atan(c*x**2)/4 - b*c/(4*x**2) - b*atan(c*x**2)/(4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=-\frac {1}{4} \, {\left ({\left (c \arctan \left (c x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {\arctan \left (c x^{2}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]

[In]

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

[Out]

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.76 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=\frac {i \, b c^{5} x^{4} \log \left (i \, c x^{2} + 1\right ) - i \, b c^{5} x^{4} \log \left (-i \, c x^{2} + 1\right ) - 2 \, b c^{4} x^{2} - 2 \, b c^{3} \arctan \left (c x^{2}\right ) - 2 \, a c^{3}}{8 \, c^{3} x^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x^5} \, dx=-\frac {\frac {b\,c\,x^2}{2}+\frac {a}{2}}{2\,x^4}-\frac {b\,c^2\,\mathrm {atan}\left (c\,x^2\right )}{4}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{4\,x^4} \]

[In]

int((a + b*atan(c*x^2))/x^5,x)

[Out]

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