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

Optimal. Leaf size=140 \[ a x-\frac {b \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+b x \tan ^{-1}\left (c x^2\right )+\frac {b \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5027, 297, 1162, 617, 204, 1165, 628} \[ a x-\frac {b \log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+\frac {b \log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )}{2 \sqrt {2} \sqrt {c}}+b x \tan ^{-1}\left (c x^2\right )+\frac {b \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )}{\sqrt {2} \sqrt {c}}-\frac {b \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )}{\sqrt {2} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTan[c*x^2],x]

[Out]

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

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 5027

Int[ArcTan[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTan[c*x^n], x] - Dist[c*n, Int[x^n/(1 + c^2*x^(2*n)), x],
x] /; FreeQ[{c, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 107, normalized size = 0.76 \[ a x+b x \tan ^{-1}\left (c x^2\right )-\frac {b \left (\log \left (c x^2-\sqrt {2} \sqrt {c} x+1\right )-\log \left (c x^2+\sqrt {2} \sqrt {c} x+1\right )-2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {c} x\right )+2 \tan ^{-1}\left (\sqrt {2} \sqrt {c} x+1\right )\right )}{2 \sqrt {2} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTan[c*x^2],x]

[Out]

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

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fricas [B]  time = 0.44, size = 319, normalized size = 2.28 \[ b x \arctan \left (c x^{2}\right ) + a x + \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} b^{3} c x + b^{4} - \sqrt {2} \sqrt {b^{6} x^{2} + \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c x + \sqrt {\frac {b^{4}}{c^{2}}} b^{4}} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} c}{b^{4}}\right ) + \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} b^{3} c x - b^{4} - \sqrt {2} \sqrt {b^{6} x^{2} - \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c x + \sqrt {\frac {b^{4}}{c^{2}}} b^{4}} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} c}{b^{4}}\right ) + \frac {1}{4} \, \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} \log \left (b^{6} x^{2} + \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c x + \sqrt {\frac {b^{4}}{c^{2}}} b^{4}\right ) - \frac {1}{4} \, \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {1}{4}} \log \left (b^{6} x^{2} - \sqrt {2} \left (\frac {b^{4}}{c^{2}}\right )^{\frac {3}{4}} b^{3} c x + \sqrt {\frac {b^{4}}{c^{2}}} b^{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x*arctan(c*x^2) + a*x + sqrt(2)*(b^4/c^2)^(1/4)*arctan(-(sqrt(2)*(b^4/c^2)^(1/4)*b^3*c*x + b^4 - sqrt(2)*sqr
t(b^6*x^2 + sqrt(2)*(b^4/c^2)^(3/4)*b^3*c*x + sqrt(b^4/c^2)*b^4)*(b^4/c^2)^(1/4)*c)/b^4) + sqrt(2)*(b^4/c^2)^(
1/4)*arctan(-(sqrt(2)*(b^4/c^2)^(1/4)*b^3*c*x - b^4 - sqrt(2)*sqrt(b^6*x^2 - sqrt(2)*(b^4/c^2)^(3/4)*b^3*c*x +
sqrt(b^4/c^2)*b^4)*(b^4/c^2)^(1/4)*c)/b^4) + 1/4*sqrt(2)*(b^4/c^2)^(1/4)*log(b^6*x^2 + sqrt(2)*(b^4/c^2)^(3/4
)*b^3*c*x + sqrt(b^4/c^2)*b^4) - 1/4*sqrt(2)*(b^4/c^2)^(1/4)*log(b^6*x^2 - sqrt(2)*(b^4/c^2)^(3/4)*b^3*c*x + s
qrt(b^4/c^2)*b^4)

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giac [A]  time = 0.18, size = 149, normalized size = 1.06 \[ -\frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | c \right |}} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{2}} + \frac {2 \, \sqrt {2} \sqrt {{\left | c \right |}} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \frac {\sqrt {2}}{\sqrt {{\left | c \right |}}}\right )} \sqrt {{\left | c \right |}}\right )}{c^{2}} - \frac {\sqrt {2} \sqrt {{\left | c \right |}} \log \left (x^{2} + \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{2}} + \frac {\sqrt {2} \sqrt {{\left | c \right |}} \log \left (x^{2} - \frac {\sqrt {2} x}{\sqrt {{\left | c \right |}}} + \frac {1}{{\left | c \right |}}\right )}{c^{2}}\right )} - 4 \, x \arctan \left (c x^{2}\right )\right )} b + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 125, normalized size = 0.89 \[ a x +b x \arctan \left (c \,x^{2}\right )-\frac {b \sqrt {2}\, \ln \left (\frac {x^{2}-\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}{x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {1}{c^{2}}}}\right )}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-\frac {b \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x+b*x*arctan(c*x^2)-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)-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 = 0.43, size = 127, normalized size = 0.91 \[ -\frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x + \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, c x - \sqrt {2} \sqrt {c}\right )}}{2 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (c x^{2} + \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (c x^{2} - \sqrt {2} \sqrt {c} x + 1\right )}{c^{\frac {3}{2}}}\right )} - 4 \, x \arctan \left (c x^{2}\right )\right )} b + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.39, size = 49, normalized size = 0.35 \[ a\,x+b\,x\,\mathrm {atan}\left (c\,x^2\right )-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\right )}{\sqrt {c}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {c}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*atan(c*x^2),x)

[Out]

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

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sympy [A]  time = 10.23, size = 930, normalized size = 6.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*Piecewise((-x*atan((-sqrt(2)/2 - sqrt(2)*I/2)**(-2)), Eq(c, -1/(x**2*(-sqrt(2)/2 - sqrt(2)*I/2)**2))),
(-x*atan((-sqrt(2)/2 + sqrt(2)*I/2)**(-2)), Eq(c, -1/(x**2*(-sqrt(2)/2 + sqrt(2)*I/2)**2))), (-x*atan((sqrt(2
)/2 - sqrt(2)*I/2)**(-2)), Eq(c, -1/(x**2*(sqrt(2)/2 - sqrt(2)*I/2)**2))), (-x*atan((sqrt(2)/2 + sqrt(2)*I/2)*
*(-2)), Eq(c, -1/(x**2*(sqrt(2)/2 + sqrt(2)*I/2)**2))), (0, Eq(c, 0)), (-2*(-1)**(3/4)*c**5*x**5*(c**(-2))**(7
/4)*atan(c*x**2)/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2))**(7/4)) + 2*I*c**4*
x**4*(c**(-2))**(3/2)*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1
)**(3/4)*c**3*(c**(-2))**(7/4)) - I*c**4*x**4*(c**(-2))**(3/2)*log(x**2 + I*sqrt(c**(-2)))/(-2*(-1)**(3/4)*c**
5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2))**(7/4)) - 2*I*c**4*x**4*(c**(-2))**(3/2)*atan((-1)**(3/
4)*x/(c**(-2))**(1/4))/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2))**(7/4)) - 2*(
-1)**(3/4)*c**3*x*(c**(-2))**(7/4)*atan(c*x**2)/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**
3*(c**(-2))**(7/4)) + 2*I*c**2*(c**(-2))**(3/2)*log(x - (-1)**(1/4)*(c**(-2))**(1/4))/(-2*(-1)**(3/4)*c**5*x**
4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2))**(7/4)) - I*c**2*(c**(-2))**(3/2)*log(x**2 + I*sqrt(c**(-2))
)/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2))**(7/4)) - 2*I*c**2*(c**(-2))**(3/2
)*atan((-1)**(3/4)*x/(c**(-2))**(1/4))/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2
))**(7/4)) + 2*c*x**4*atan(c*x**2)/(-2*(-1)**(3/4)*c**5*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**3*(c**(-2))**
(7/4)) + 2*atan(c*x**2)/(-2*(-1)**(3/4)*c**6*x**4*(c**(-2))**(7/4) - 2*(-1)**(3/4)*c**4*(c**(-2))**(7/4)), Tru
e))

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