3.111 \(\int \frac {a+b \tan ^{-1}(c x^3)}{x^2} \, dx\)

Optimal. Leaf size=104 \[ -\frac {a+b \tan ^{-1}\left (c x^3\right )}{x}+\frac {1}{2} b \sqrt [3]{c} \log \left (c^{2/3} x^2+1\right )-\frac {1}{2} \sqrt {3} b \sqrt [3]{c} \tan ^{-1}\left (\frac {1-2 c^{2/3} x^2}{\sqrt {3}}\right )-\frac {1}{4} b \sqrt [3]{c} \log \left (c^{4/3} x^4-c^{2/3} x^2+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.04, 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

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fricas [A]  time = 0.44, size = 90, normalized size = 0.87 \[ \frac {2 \, \sqrt {3} b c^{\frac {1}{3}} x \arctan \left (\frac {2}{3} \, \sqrt {3} c^{\frac {2}{3}} x^{2} - \frac {1}{3} \, \sqrt {3}\right ) - b c^{\frac {1}{3}} x \log \left (c^{2} x^{4} - c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right ) + 2 \, b c^{\frac {1}{3}} x \log \left (c x^{2} + c^{\frac {1}{3}}\right ) - 4 \, b \arctan \left (c x^{3}\right ) - 4 \, a}{4 \, x} \]

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

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giac [A]  time = 3.82, size = 91, normalized size = 0.88 \[ \frac {1}{4} \, b c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )} {\left | c \right |}^{\frac {2}{3}}\right )}{{\left | c \right |}^{\frac {2}{3}}} - \frac {\log \left (x^{4} - \frac {x^{2}}{{\left | c \right |}^{\frac {2}{3}}} + \frac {1}{{\left | c \right |}^{\frac {4}{3}}}\right )}{{\left | c \right |}^{\frac {2}{3}}} + \frac {2 \, \log \left (x^{2} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{{\left | c \right |}^{\frac {2}{3}}}\right )} - \frac {b \arctan \left (c x^{3}\right ) + a}{x} \]

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

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maple [A]  time = 0.03, size = 104, normalized size = 1.00 \[ -\frac {a}{x}-\frac {b \arctan \left (c \,x^{3}\right )}{x}+\frac {b \ln \left (x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}}-\frac {b \ln \left (x^{4}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}} x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}}+\frac {b \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{2}}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}}} \]

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))

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maxima [A]  time = 0.42, size = 98, normalized size = 0.94 \[ \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} - c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right )}{c^{\frac {2}{3}}} - \frac {\log \left (c^{\frac {4}{3}} x^{4} - c^{\frac {2}{3}} x^{2} + 1\right )}{c^{\frac {2}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {2}{3}} x^{2} + 1}{c^{\frac {2}{3}}}\right )}{c^{\frac {2}{3}}}\right )} - \frac {4 \, \arctan \left (c x^{3}\right )}{x}\right )} b - \frac {a}{x} \]

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

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mupad [B]  time = 1.83, size = 99, normalized size = 0.95 \[ \frac {b\,c^{1/3}\,\ln \left (c^{2/3}\,x^2+1\right )}{2}-\frac {a}{x}-\frac {b\,\mathrm {atan}\left (c\,x^3\right )}{x}-\frac {b\,c^{1/3}\,\ln \left (-\sqrt {3}-c^{2/3}\,x^2\,2{}\mathrm {i}+1{}\mathrm {i}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4}+\frac {b\,c^{1/3}\,\ln \left (-\sqrt {3}+c^{2/3}\,x^2\,2{}\mathrm {i}-\mathrm {i}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))/x^2,x)

[Out]

(b*c^(1/3)*log(c^(2/3)*x^2 + 1))/2 - a/x - (b*atan(c*x^3))/x - (b*c^(1/3)*log(1i - c^(2/3)*x^2*2i - 3^(1/2))*(
3^(1/2)*1i + 1))/4 + (b*c^(1/3)*log(c^(2/3)*x^2*2i - 3^(1/2) - 1i)*(3^(1/2)*1i - 1))/4

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sympy [A]  time = 58.27, size = 328, normalized size = 3.15 \[ \begin {cases} - \frac {a}{x} + \left (-1\right )^{\frac {5}{6}} b c^{2} \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} \operatorname {atan}{\left (c x^{3} \right )} - \sqrt [3]{-1} b c \sqrt [3]{\frac {1}{c^{2}}} \log {\left (x - \sqrt [6]{-1} \sqrt [6]{\frac {1}{c^{2}}} \right )} + \frac {3 \sqrt [3]{-1} b c \sqrt [3]{\frac {1}{c^{2}}} \log {\left (4 x^{2} - 4 \sqrt [6]{-1} x \sqrt [6]{\frac {1}{c^{2}}} + 4 \sqrt [3]{-1} \sqrt [3]{\frac {1}{c^{2}}} \right )}}{4} - \frac {\sqrt [3]{-1} b c \sqrt [3]{\frac {1}{c^{2}}} \log {\left (4 x^{2} + 4 \sqrt [6]{-1} x \sqrt [6]{\frac {1}{c^{2}}} + 4 \sqrt [3]{-1} \sqrt [3]{\frac {1}{c^{2}}} \right )}}{4} + \frac {\sqrt [3]{-1} \sqrt {3} b c \sqrt [3]{\frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {2 \left (-1\right )^{\frac {5}{6}} \sqrt {3} x}{3 \sqrt [6]{\frac {1}{c^{2}}}} - \frac {\sqrt {3}}{3} \right )}}{2} - \frac {\sqrt [3]{-1} \sqrt {3} b c \sqrt [3]{\frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {2 \left (-1\right )^{\frac {5}{6}} \sqrt {3} x}{3 \sqrt [6]{\frac {1}{c^{2}}}} + \frac {\sqrt {3}}{3} \right )}}{2} - \frac {b \operatorname {atan}{\left (c x^{3} \right )}}{x} & \text {for}\: c \neq 0 \\- \frac {a}{x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))/x**2,x)

[Out]

Piecewise((-a/x + (-1)**(5/6)*b*c**2*(c**(-2))**(5/6)*atan(c*x**3) - (-1)**(1/3)*b*c*(c**(-2))**(1/3)*log(x -
(-1)**(1/6)*(c**(-2))**(1/6)) + 3*(-1)**(1/3)*b*c*(c**(-2))**(1/3)*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/
6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/4 - (-1)**(1/3)*b*c*(c**(-2))**(1/3)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2
))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/4 + (-1)**(1/3)*sqrt(3)*b*c*(c**(-2))**(1/3)*atan(2*(-1)**(5/6)*sq
rt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/2 - (-1)**(1/3)*sqrt(3)*b*c*(c**(-2))**(1/3)*atan(2*(-1)**(5/6)*sqrt
(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/2 - b*atan(c*x**3)/x, Ne(c, 0)), (-a/x, True))

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