\(\int \frac {a+b \arctan (c x^3)}{x^4} \, dx\) [101]

Optimal result

Integrand size = 14, antiderivative size = 39 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=-\frac {a+b \arctan \left (c x^3\right )}{3 x^3}+b c \log (x)-\frac {1}{6} b c \log \left (1+c^2 x^6\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4946, 272, 36, 29, 31} \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=-\frac {a+b \arctan \left (c x^3\right )}{3 x^3}-\frac {1}{6} b c \log \left (c^2 x^6+1\right )+b c \log (x) \]

[In]

Int[(a + b*ArcTan[c*x^3])/x^4,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

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

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^3\right )}{3 x^3}+(b c) \int \frac {1}{x \left (1+c^2 x^6\right )} \, dx \\ & = -\frac {a+b \arctan \left (c x^3\right )}{3 x^3}+\frac {1}{6} (b c) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^6\right ) \\ & = -\frac {a+b \arctan \left (c x^3\right )}{3 x^3}+\frac {1}{6} (b c) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^6\right )-\frac {1}{6} \left (b c^3\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^6\right ) \\ & = -\frac {a+b \arctan \left (c x^3\right )}{3 x^3}+b c \log (x)-\frac {1}{6} b c \log \left (1+c^2 x^6\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b \arctan \left (c x^3\right )}{3 x^3}+b c \log (x)-\frac {1}{6} b c \log \left (1+c^2 x^6\right ) \]

[In]

Integrate[(a + b*ArcTan[c*x^3])/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {a}{3 x^{3}}+b \left (-\frac {\arctan \left (c \,x^{3}\right )}{3 x^{3}}+c \left (\ln \left (x \right )-\frac {\ln \left (c^{2} x^{6}+1\right )}{6}\right )\right )\)	\(39\)
parts	\(-\frac {a}{3 x^{3}}+b \left (-\frac {\arctan \left (c \,x^{3}\right )}{3 x^{3}}+c \left (\ln \left (x \right )-\frac {\ln \left (c^{2} x^{6}+1\right )}{6}\right )\right )\)	\(39\)
parallelrisch	\(\frac {6 b c \ln \left (x \right ) x^{3}-b c \ln \left (c^{2} x^{6}+1\right ) x^{3}-2 b \arctan \left (c \,x^{3}\right )-2 a}{6 x^{3}}\)	\(45\)
risch	\(\frac {i b \ln \left (i c \,x^{3}+1\right )}{6 x^{3}}-\frac {-6 b c \ln \left (x \right ) x^{3}+b c \ln \left (-c^{2} x^{6}-1\right ) x^{3}+i b \ln \left (-i c \,x^{3}+1\right )+2 a}{6 x^{3}}\)	\(68\)

[In]

int((a+b*arctan(c*x^3))/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=-\frac {b c x^{3} \log \left (c^{2} x^{6} + 1\right ) - 6 \, b c x^{3} \log \left (x\right ) + 2 \, b \arctan \left (c x^{3}\right ) + 2 \, a}{6 \, x^{3}} \]

[In]

integrate((a+b*arctan(c*x^3))/x^4,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (39) = 78\).

Time = 39.53 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.82 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=\begin {cases} - \frac {a}{3 x^{3}} + b c \log {\left (x \right )} - \frac {b c \log {\left (x - \sqrt [6]{- \frac {1}{c^{2}}} \right )}}{3} - \frac {b c \log {\left (4 x^{2} + 4 x \sqrt [6]{- \frac {1}{c^{2}}} + 4 \sqrt [3]{- \frac {1}{c^{2}}} \right )}}{3} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{3 \sqrt {- \frac {1}{c^{2}}}} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\- \frac {a}{3 x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*atan(c*x**3))/x**4,x)

[Out]

Piecewise((-a/(3*x**3) + b*c*log(x) - b*c*log(x - (-1/c**2)**(1/6))/3 - b*c*log(4*x**2 + 4*x*(-1/c**2)**(1/6)
+ 4*(-1/c**2)**(1/3))/3 - b*atan(c*x**3)/(3*sqrt(-1/c**2)) - b*atan(c*x**3)/(3*x**3), Ne(c, 0)), (-a/(3*x**3),
True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=-\frac {1}{6} \, {\left (c {\left (\log \left (c^{2} x^{6} + 1\right ) - \log \left (x^{6}\right )\right )} + \frac {2 \, \arctan \left (c x^{3}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \]

[In]

integrate((a+b*arctan(c*x^3))/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.54 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=-\frac {b c^{3} x^{3} \log \left (c^{2} x^{6} + 1\right ) - 2 \, b c^{3} x^{3} \log \left (c x^{3}\right ) + 2 \, b c^{2} \arctan \left (c x^{3}\right ) + 2 \, a c^{2}}{6 \, c^{2} x^{3}} \]

[In]

integrate((a+b*arctan(c*x^3))/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \arctan \left (c x^3\right )}{x^4} \, dx=b\,c\,\ln \left (x\right )-\frac {a}{3\,x^3}-\frac {b\,\mathrm {atan}\left (c\,x^3\right )}{3\,x^3}-\frac {b\,c\,\ln \left (c^2\,x^6+1\right )}{6} \]

[In]

int((a + b*atan(c*x^3))/x^4,x)

[Out]

b*c*log(x) - a/(3*x^3) - (b*atan(c*x^3))/(3*x^3) - (b*c*log(c^2*x^6 + 1))/6