3.111 \(\int \frac{a+b \tan ^{-1}(c x^3)}{x^2} \, dx\)
Optimal. Leaf size=104 \[ -\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac{1}{2} b \sqrt [3]{c} \log \left (c^{2/3} x^2+1\right )-\frac{1}{4} b \sqrt [3]{c} \log \left (c^{4/3} x^4-c^{2/3} x^2+1\right )-\frac{1}{2} \sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\frac{1-2 c^{2/3} x^2}{\sqrt{3}}\right ) \]
[Out]
-((a + b*ArcTan[c*x^3])/x) - (Sqrt[3]*b*c^(1/3)*ArcTan[(1 - 2*c^(2/3)*x^2)/Sqrt[3]])/2 + (b*c^(1/3)*Log[1 + c^
(2/3)*x^2])/2 - (b*c^(1/3)*Log[1 - c^(2/3)*x^2 + c^(4/3)*x^4])/4
________________________________________________________________________________________
Rubi [A] time = 0.0851211, antiderivative size = 104, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used =
{5033, 275, 200, 31, 634, 617, 204, 628} \[ -\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac{1}{2} b \sqrt [3]{c} \log \left (c^{2/3} x^2+1\right )-\frac{1}{4} b \sqrt [3]{c} \log \left (c^{4/3} x^4-c^{2/3} x^2+1\right )-\frac{1}{2} \sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\frac{1-2 c^{2/3} x^2}{\sqrt{3}}\right ) \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c*x^3])/x^2,x]
[Out]
-((a + b*ArcTan[c*x^3])/x) - (Sqrt[3]*b*c^(1/3)*ArcTan[(1 - 2*c^(2/3)*x^2)/Sqrt[3]])/2 + (b*c^(1/3)*Log[1 + c^
(2/3)*x^2])/2 - (b*c^(1/3)*Log[1 - c^(2/3)*x^2 + c^(4/3)*x^4])/4
Rule 5033
Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Rule 275
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 200
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 634
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && !NiceSqrtQ[b^2 - 4*a*c]
Rule 617
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 628
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]
Rubi steps
\begin{align*} \int \frac{a+b \tan ^{-1}\left (c x^3\right )}{x^2} \, dx &=-\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+(3 b c) \int \frac{x}{1+c^2 x^6} \, dx\\ &=-\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac{1}{2} (3 b c) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x^3} \, dx,x,x^2\right )\\ &=-\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{1}{1+c^{2/3} x} \, dx,x,x^2\right )+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{2-c^{2/3} x}{1-c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=-\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac{1}{2} b \sqrt [3]{c} \log \left (1+c^{2/3} x^2\right )-\frac{1}{4} \left (b \sqrt [3]{c}\right ) \operatorname{Subst}\left (\int \frac{-c^{2/3}+2 c^{4/3} x}{1-c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )+\frac{1}{4} (3 b c) \operatorname{Subst}\left (\int \frac{1}{1-c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )\\ &=-\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac{1}{2} b \sqrt [3]{c} \log \left (1+c^{2/3} x^2\right )-\frac{1}{4} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2+c^{4/3} x^4\right )+\frac{1}{2} \left (3 b \sqrt [3]{c}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-2 c^{2/3} x^2\right )\\ &=-\frac{a+b \tan ^{-1}\left (c x^3\right )}{x}-\frac{1}{2} \sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\frac{1-2 c^{2/3} x^2}{\sqrt{3}}\right )+\frac{1}{2} b \sqrt [3]{c} \log \left (1+c^{2/3} x^2\right )-\frac{1}{4} b \sqrt [3]{c} \log \left (1-c^{2/3} x^2+c^{4/3} x^4\right )\\ \end{align*}
Mathematica [A] time = 0.0369877, size = 170, normalized size = 1.63 \[ -\frac{a}{x}+\frac{1}{2} b \sqrt [3]{c} \log \left (c^{2/3} x^2+1\right )-\frac{1}{4} b \sqrt [3]{c} \log \left (c^{2/3} x^2-\sqrt{3} \sqrt [3]{c} x+1\right )-\frac{1}{4} b \sqrt [3]{c} \log \left (c^{2/3} x^2+\sqrt{3} \sqrt [3]{c} x+1\right )-\frac{b \tan ^{-1}\left (c x^3\right )}{x}-\frac{1}{2} \sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (\sqrt{3}-2 \sqrt [3]{c} x\right )-\frac{1}{2} \sqrt{3} b \sqrt [3]{c} \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt{3}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcTan[c*x^3])/x^2,x]
[Out]
-(a/x) - (b*ArcTan[c*x^3])/x - (Sqrt[3]*b*c^(1/3)*ArcTan[Sqrt[3] - 2*c^(1/3)*x])/2 - (Sqrt[3]*b*c^(1/3)*ArcTan
[Sqrt[3] + 2*c^(1/3)*x])/2 + (b*c^(1/3)*Log[1 + c^(2/3)*x^2])/2 - (b*c^(1/3)*Log[1 - Sqrt[3]*c^(1/3)*x + c^(2/
3)*x^2])/4 - (b*c^(1/3)*Log[1 + Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/4
________________________________________________________________________________________
Maple [A] time = 0.026, size = 104, normalized size = 1. \begin{align*} -{\frac{a}{x}}-{\frac{b\arctan \left ( c{x}^{3} \right ) }{x}}+{\frac{b}{2\,c}\ln \left ({x}^{2}+\sqrt [3]{{c}^{-2}} \right ) \left ({c}^{-2} \right ) ^{-{\frac{2}{3}}}}-{\frac{b}{4\,c}\ln \left ({x}^{4}-\sqrt [3]{{c}^{-2}}{x}^{2}+ \left ({c}^{-2} \right ) ^{{\frac{2}{3}}} \right ) \left ({c}^{-2} \right ) ^{-{\frac{2}{3}}}}+{\frac{b\sqrt{3}}{2\,c}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{{x}^{2}}{\sqrt [3]{{c}^{-2}}}}-1 \right ) } \right ) \left ({c}^{-2} \right ) ^{-{\frac{2}{3}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(c*x^3))/x^2,x)
[Out]
-a/x-b/x*arctan(c*x^3)+1/2*b/c/(1/c^2)^(2/3)*ln(x^2+(1/c^2)^(1/3))-1/4*b/c/(1/c^2)^(2/3)*ln(x^4-(1/c^2)^(1/3)*
x^2+(1/c^2)^(2/3))+1/2*b/c/(1/c^2)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c^2)^(1/3)*x^2-1))
________________________________________________________________________________________
Maxima [A] time = 1.54006, size = 151, normalized size = 1.45 \begin{align*} \frac{1}{4} \,{\left (c{\left (\frac{2 \, \sqrt{3}{\left (c^{2}\right )}^{\frac{2}{3}} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (c^{2}\right )}^{\frac{1}{3}}{\left (2 \, x^{2} - \frac{1}{c^{2}}^{\frac{1}{3}}\right )}\right )}{c^{2}} - \frac{{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (x^{4} - \frac{1}{c^{2}}^{\frac{1}{3}} x^{2} + \frac{1}{c^{2}}^{\frac{2}{3}}\right )}{c^{2}} + \frac{2 \,{\left (c^{2}\right )}^{\frac{2}{3}} \log \left (x^{2} + \frac{1}{c^{2}}^{\frac{1}{3}}\right )}{c^{2}}\right )} - \frac{4 \, \arctan \left (c x^{3}\right )}{x}\right )} b - \frac{a}{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x^3))/x^2,x, algorithm="maxima")
[Out]
1/4*(c*(2*sqrt(3)*(c^2)^(2/3)*arctan(1/3*sqrt(3)*(c^2)^(1/3)*(2*x^2 - (c^(-2))^(1/3)))/c^2 - (c^2)^(2/3)*log(x
^4 - (c^(-2))^(1/3)*x^2 + (c^(-2))^(2/3))/c^2 + 2*(c^2)^(2/3)*log(x^2 + (c^(-2))^(1/3))/c^2) - 4*arctan(c*x^3)
/x)*b - a/x
________________________________________________________________________________________
Fricas [A] time = 2.65988, size = 258, normalized size = 2.48 \begin{align*} \frac{2 \, \sqrt{3} b c^{\frac{1}{3}} x \arctan \left (\frac{2}{3} \, \sqrt{3} c^{\frac{2}{3}} x^{2} - \frac{1}{3} \, \sqrt{3}\right ) - b c^{\frac{1}{3}} x \log \left (c^{2} x^{4} - c^{\frac{4}{3}} x^{2} + c^{\frac{2}{3}}\right ) + 2 \, b c^{\frac{1}{3}} x \log \left (c x^{2} + c^{\frac{1}{3}}\right ) - 4 \, b \arctan \left (c x^{3}\right ) - 4 \, a}{4 \, x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x^3))/x^2,x, algorithm="fricas")
[Out]
1/4*(2*sqrt(3)*b*c^(1/3)*x*arctan(2/3*sqrt(3)*c^(2/3)*x^2 - 1/3*sqrt(3)) - b*c^(1/3)*x*log(c^2*x^4 - c^(4/3)*x
^2 + c^(2/3)) + 2*b*c^(1/3)*x*log(c*x^2 + c^(1/3)) - 4*b*arctan(c*x^3) - 4*a)/x
________________________________________________________________________________________
Sympy [A] time = 92.5375, size = 1975, normalized size = 18.99 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(c*x**3))/x**2,x)
[Out]
Piecewise((-(a - oo*I*b)/x, Eq(c, -I/x**3)), (-(a + oo*I*b)/x, Eq(c, I/x**3)), (-(a - b*atan((-sqrt(3)/2 - I/2
)**(-3)))/x, Eq(c, -1/(x**3*(-sqrt(3)/2 - I/2)**3))), (-(a - b*atan((-sqrt(3)/2 + I/2)**(-3)))/x, Eq(c, -1/(x*
*3*(-sqrt(3)/2 + I/2)**3))), (-(a - b*atan((sqrt(3)/2 - I/2)**(-3)))/x, Eq(c, -1/(x**3*(sqrt(3)/2 - I/2)**3)))
, (-(a - b*atan((sqrt(3)/2 + I/2)**(-3)))/x, Eq(c, -1/(x**3*(sqrt(3)/2 + I/2)**3))), (-a/x, Eq(c, 0)), (-28*(-
1)**(5/6)*a*c**42*x**6*(c**(-2))**(161/6)/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40
*x*(c**(-2))**(161/6)) - 28*(-1)**(5/6)*a*c**40*(c**(-2))**(161/6)/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/
6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) - 72*(-1)**(1/3)*b*c**47*(c**(-2))**(91/3)/(28*(-1)**(5/6)*c**
42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) - 21*(-1)**(1/6)*b*c**45*x**7*(c**(-2)
)**(169/6)*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(28*(-1)**(5/6)*c**
42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 7*(-1)**(1/6)*b*c**45*x**7*(c**(-2))
**(169/6)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(28*(-1)**(5/6)*c**4
2*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) - 14*(-1)**(1/6)*sqrt(3)*b*c**45*x**7*(
c**(-2))**(169/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/(28*(-1)**(5/6)*c**42*x**7*(c
**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 14*(-1)**(1/6)*sqrt(3)*b*c**45*x**7*(c**(-2))*
*(169/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**
(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 28*(-1)**(1/6)*b*c**45*x**7*(c**(-2))**(169/6)*log(2)/(
28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) - 21*(-1)**(1/6)*b*c
**43*x*(c**(-2))**(169/6)*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(28*
(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 7*(-1)**(1/6)*b*c**43
*x*(c**(-2))**(169/6)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(28*(-1)
**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) - 14*(-1)**(1/6)*sqrt(3)*b*
c**43*x*(c**(-2))**(169/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/(28*(-1)**(5/6)*c**4
2*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 14*(-1)**(1/6)*sqrt(3)*b*c**43*x*(c**
(-2))**(169/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(28*(-1)**(5/6)*c**42*x**7*(c**(
-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 28*(-1)**(1/6)*b*c**43*x*(c**(-2))**(169/6)*log(2
)/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) - 28*(-1)**(5/6)*
b*c**42*x**6*(c**(-2))**(161/6)*atan(c*x**3)/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c*
*40*x*(c**(-2))**(161/6)) - 28*(-1)**(5/6)*b*c**40*(c**(-2))**(161/6)*atan(c*x**3)/(28*(-1)**(5/6)*c**42*x**7*
(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 28*(-1)**(1/6)*b*c**37*x**7*(c**(-2))**(145/
6)*log(x - (-1)**(1/6)*(c**(-2))**(1/6))/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*
x*(c**(-2))**(161/6)) + 28*(-1)**(1/6)*b*c**35*x*(c**(-2))**(145/6)*log(x - (-1)**(1/6)*(c**(-2))**(1/6))/(28*
(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6)) + 72*(-1)**(1/3)*b*c**3
5*(c**(-2))**(73/3)/(28*(-1)**(5/6)*c**42*x**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**40*x*(c**(-2))**(161/6))
- 28*(-1)**(2/3)*b*x**7*(c**(-2))**(8/3)*atan(c*x**3)/(28*(-1)**(5/6)*c**48*x**7*(c**(-2))**(161/6) + 28*(-1)
**(5/6)*c**46*x*(c**(-2))**(161/6)) - 28*(-1)**(2/3)*b*x*(c**(-2))**(8/3)*atan(c*x**3)/(28*(-1)**(5/6)*c**50*x
**7*(c**(-2))**(161/6) + 28*(-1)**(5/6)*c**48*x*(c**(-2))**(161/6)), True))
________________________________________________________________________________________
Giac [A] time = 1.23004, size = 123, normalized size = 1.18 \begin{align*} \frac{1}{4} \, b c{\left (\frac{2 \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - \frac{1}{{\left | c \right |}^{\frac{2}{3}}}\right )}{\left | c \right |}^{\frac{2}{3}}\right )}{{\left | c \right |}^{\frac{2}{3}}} - \frac{\log \left (x^{4} - \frac{x^{2}}{{\left | c \right |}^{\frac{2}{3}}} + \frac{1}{{\left | c \right |}^{\frac{4}{3}}}\right )}{{\left | c \right |}^{\frac{2}{3}}} + \frac{2 \, \log \left (x^{2} + \frac{1}{{\left | c \right |}^{\frac{2}{3}}}\right )}{{\left | c \right |}^{\frac{2}{3}}}\right )} - \frac{b \arctan \left (c x^{3}\right ) + a}{x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x^3))/x^2,x, algorithm="giac")
[Out]
1/4*b*c*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1/abs(c)^(2/3))*abs(c)^(2/3))/abs(c)^(2/3) - log(x^4 - x^2/abs(
c)^(2/3) + 1/abs(c)^(4/3))/abs(c)^(2/3) + 2*log(x^2 + 1/abs(c)^(2/3))/abs(c)^(2/3)) - (b*arctan(c*x^3) + a)/x