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

Optimal. Leaf size=143 \[ -\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}-\frac {b \sqrt {c} \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2}}+\frac {b \sqrt {c} \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2}}-\frac {b \sqrt {c} \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2}}+\frac {b \sqrt {c} \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5033, 211, 1165, 628, 1162, 617, 204} \[ -\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}-\frac {b \sqrt {c} \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2}}+\frac {b \sqrt {c} \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2}}-\frac {b \sqrt {c} \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2}}+\frac {b \sqrt {c} \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])/x^2,x]

[Out]

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

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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^2\right )}{x^2} \, dx &=-\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}+(2 b c) \int \frac {1}{1+c^2 x^4} \, dx\\ &=-\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}+(b c) \int \frac {1-c x^2}{1+c^2 x^4} \, dx+(b c) \int \frac {1+c x^2}{1+c^2 x^4} \, dx\\ &=-\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}+\frac {1}{2} b \int \frac {1}{\frac {1}{c}-\frac {\sqrt {2} x}{\sqrt {c}}+x^2} \, dx+\frac {1}{2} b \int \frac {1}{\frac {1}{c}+\frac {\sqrt {2} x}{\sqrt {c}}+x^2} \, dx-\frac {\left (b \sqrt {c}\right ) \int \frac {\frac {\sqrt {2}}{\sqrt {c}}+2 x}{-\frac {1}{c}-\frac {\sqrt {2} x}{\sqrt {c}}-x^2} \, dx}{2 \sqrt {2}}-\frac {\left (b \sqrt {c}\right ) \int \frac {\frac {\sqrt {2}}{\sqrt {c}}-2 x}{-\frac {1}{c}+\frac {\sqrt {2} x}{\sqrt {c}}-x^2} \, dx}{2 \sqrt {2}}\\ &=-\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}-\frac {b \sqrt {c} \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2}}+\frac {b \sqrt {c} \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2}}+\frac {\left (b \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2}}-\frac {\left (b \sqrt {c}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {c} x\right )}{\sqrt {2}}\\ &=-\frac {a+b \tan ^{-1}\left (c x^2\right )}{x}-\frac {b \sqrt {c} \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2}}+\frac {b \sqrt {c} \tan ^{-1}\left (1+\sqrt {2} \sqrt {c} x\right )}{\sqrt {2}}-\frac {b \sqrt {c} \log \left (1-\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2}}+\frac {b \sqrt {c} \log \left (1+\sqrt {2} \sqrt {c} x+c x^2\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 158, normalized size = 1.10 \[ -\frac {a}{x}-\frac {b \sqrt {c} \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2}}+\frac {b \sqrt {c} \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2}}-\frac {b \tan ^{-1}\left (c x^2\right )}{x}+\frac {b \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} x-\sqrt {2}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {b \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} x+\sqrt {2}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 322, normalized size = 2.25 \[ -\frac {4 \, \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} x \arctan \left (-\frac {b^{4} c^{2} + \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {3}{4}} b c x - \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {3}{4}} \sqrt {b^{2} c^{2} x^{2} + \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} b c x + \sqrt {b^{4} c^{2}}}}{b^{4} c^{2}}\right ) + 4 \, \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} x \arctan \left (\frac {b^{4} c^{2} - \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {3}{4}} b c x + \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {3}{4}} \sqrt {b^{2} c^{2} x^{2} - \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} b c x + \sqrt {b^{4} c^{2}}}}{b^{4} c^{2}}\right ) - \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} x \log \left (b^{2} c^{2} x^{2} + \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} b c x + \sqrt {b^{4} c^{2}}\right ) + \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} x \log \left (b^{2} c^{2} x^{2} - \sqrt {2} \left (b^{4} c^{2}\right )^{\frac {1}{4}} b c x + \sqrt {b^{4} c^{2}}\right ) + 4 \, b \arctan \left (c x^{2}\right ) + 4 \, a}{4 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 138, normalized size = 0.97 \[ \frac {1}{4} \, b c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} + \frac {\sqrt {2} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}}\right )} - \frac {b \arctan \left (c x^{2}\right ) + a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 125, normalized size = 0.87 \[ -\frac {a}{x}-\frac {b \arctan \left (c \,x^{2}\right )}{x}+\frac {b c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{2}+\frac {b c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{4}+\frac {b c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 132, normalized size = 0.92 \[ \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{\sqrt {c}}\right )} - \frac {4 \, \arctan \left (c x^{2}\right )}{x}\right )} b - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.20, size = 55, normalized size = 0.38 \[ -\frac {a}{x}-\frac {b\,\mathrm {atan}\left (c\,x^2\right )}{x}-{\left (-1\right )}^{1/4}\,b\,\sqrt {c}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )\,1{}\mathrm {i}-{\left (-1\right )}^{1/4}\,b\,\sqrt {c}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))/x^2,x)

[Out]

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

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sympy [A]  time = 20.44, size = 151, normalized size = 1.06 \[ \begin {cases} - \frac {a}{x} - \sqrt [4]{-1} b c \sqrt [4]{\frac {1}{c^{2}}} \log {\left (x - \sqrt [4]{-1} \sqrt [4]{\frac {1}{c^{2}}} \right )} + \frac {\sqrt [4]{-1} b c \sqrt [4]{\frac {1}{c^{2}}} \log {\left (x^{2} + i \sqrt {\frac {1}{c^{2}}} \right )}}{2} - \sqrt [4]{-1} b c \sqrt [4]{\frac {1}{c^{2}}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} x}{\sqrt [4]{\frac {1}{c^{2}}}} \right )} + \frac {\left (-1\right )^{\frac {3}{4}} b \operatorname {atan}{\left (c x^{2} \right )}}{\sqrt [4]{\frac {1}{c^{2}}}} - \frac {b \operatorname {atan}{\left (c x^{2} \right )}}{x} & \text {for}\: c \neq 0 \\- \frac {a}{x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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