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

Optimal. Leaf size=86 \[ -\frac {x \sqrt {a^2 c x^2+c}}{6 a}+\frac {\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{6 a^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4930, 195, 217, 206} \[ -\frac {x \sqrt {a^2 c x^2+c}}{6 a}+\frac {\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{6 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x],x]

[Out]

-(x*Sqrt[c + a^2*c*x^2])/(6*a) + ((c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(3*a^2*c) - (Sqrt[c]*ArcTanh[(a*Sqrt[c]*x
)/Sqrt[c + a^2*c*x^2]])/(6*a^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rubi steps

\begin {align*} \int x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx &=\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {\int \sqrt {c+a^2 c x^2} \, dx}{3 a}\\ &=-\frac {x \sqrt {c+a^2 c x^2}}{6 a}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {c \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{6 a}\\ &=-\frac {x \sqrt {c+a^2 c x^2}}{6 a}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {c \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{6 a}\\ &=-\frac {x \sqrt {c+a^2 c x^2}}{6 a}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{6 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 86, normalized size = 1.00 \[ -\frac {a x \sqrt {a^2 c x^2+c}+\sqrt {c} \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+a c x\right )-2 \left (a^2 x^2+1\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{6 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x],x]

[Out]

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

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fricas [A]  time = 0.59, size = 77, normalized size = 0.90 \[ -\frac {2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - 2 \, {\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} - \sqrt {c} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right )}{12 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)*(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-1/12*(2*sqrt(a^2*c*x^2 + c)*(a*x - 2*(a^2*x^2 + 1)*arctan(a*x)) - sqrt(c)*log(-2*a^2*c*x^2 + 2*sqrt(a^2*c*x^2
+ c)*a*sqrt(c)*x - c))/a^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)*(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.93, size = 156, normalized size = 1.81 \[ \frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (2 \arctan \left (a x \right ) x^{2} a^{2}-a x +2 \arctan \left (a x \right )\right )}{6 a^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right )}{6 a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right )}{6 a^{2} \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(a*x)*(a^2*c*x^2+c)^(1/2),x)

[Out]

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

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maxima [B]  time = 0.53, size = 260, normalized size = 3.02 \[ \frac {4 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \sqrt {c} \arctan \left (a x\right ) - 2 \, {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} {\left (a x \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt {c} + \sqrt {c} {\left (\arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) + 2, a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) + \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) - 2, -a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right )\right )}}{12 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)*(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/12*(4*(a^2*x^2 + 1)^(3/2)*sqrt(c)*arctan(a*x) - 2*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(a*x*cos(1/2*arctan2(4*a*
x, -a^2*x^2 + 3)) + 2*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) + sqrt(c)*(arctan2((a^4*x^4 + 10*a^2*x^2
+ 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) + 2, a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*
a*x, a^2*x^2 - 3))) + arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) - 2, -a*x
+ (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3)))))/a^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {atan}\left (a\,x\right )\,\sqrt {c\,a^2\,x^2+c} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atan(a*x)*(c + a^2*c*x^2)^(1/2),x)

[Out]

int(x*atan(a*x)*(c + a^2*c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)*(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x*sqrt(c*(a**2*x**2 + 1))*atan(a*x), x)

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