\(\int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx\) [224]

Optimal result

Integrand size = 22, antiderivative size = 120 \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^2 c}+\frac {5 \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{6 a^4 \sqrt {c}} \]

[Out]

5/6*arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))/a^4/c^(1/2)-1/6*x*(a^2*c*x^2+c)^(1/2)/a^3/c-2/3*arctan(a*x)*(a^2*
c*x^2+c)^(1/2)/a^4/c+1/3*x^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a^2/c

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5072, 327, 223, 212, 5050} \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=\frac {x^2 \arctan (a x) \sqrt {a^2 c x^2+c}}{3 a^2 c}-\frac {2 \arctan (a x) \sqrt {a^2 c x^2+c}}{3 a^4 c}+\frac {5 \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{6 a^4 \sqrt {c}}-\frac {x \sqrt {a^2 c x^2+c}}{6 a^3 c} \]

[In]

Int[(x^3*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]

[Out]

-1/6*(x*Sqrt[c + a^2*c*x^2])/(a^3*c) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*a^4*c) + (x^2*Sqrt[c + a^2*c*x^2
]*ArcTan[a*x])/(3*a^2*c) + (5*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(6*a^4*Sqrt[c])

Rule 212

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

Rule 223

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 5050

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]

Rule 5072

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

Rubi steps \begin{align*} \text {integral}& = \frac {x^2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^2 c}-\frac {2 \int \frac {x \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^2}-\frac {\int \frac {x^2}{\sqrt {c+a^2 c x^2}} \, dx}{3 a} \\ & = -\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^2 c}+\frac {\int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{6 a^3}+\frac {2 \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^3} \\ & = -\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^2 c}+\frac {\text {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^3}+\frac {2 \text {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{3 a^3} \\ & = -\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \arctan (a x)}{3 a^2 c}+\frac {5 \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{6 a^4 \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.76 \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=\frac {-a x \sqrt {c+a^2 c x^2}+2 \left (-2+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \arctan (a x)+5 \sqrt {c} \log \left (a c x+\sqrt {c} \sqrt {c+a^2 c x^2}\right )}{6 a^4 c} \]

[In]

Integrate[(x^3*ArcTan[a*x])/Sqrt[c + a^2*c*x^2],x]

[Out]

(-(a*x*Sqrt[c + a^2*c*x^2]) + 2*(-2 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] + 5*Sqrt[c]*Log[a*c*x + Sqrt[c]
*Sqrt[c + a^2*c*x^2]])/(6*a^4*c)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {\left (2 a^{2} \arctan \left (a x \right ) x^{2}-a x -4 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 a^{4} c}-\frac {5 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, a^{4} c}+\frac {5 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, a^{4} c}\)	\(165\)

[In]

int(x^3*arctan(a*x)/(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(2*a^2*arctan(a*x)*x^2-a*x-4*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)/a^4/c-5/6*ln((1+I*a*x)/(a^2*x^2+1)^(1/
2)-I)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/a^4/c+5/6*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+I)*(c*(a*x-I)*(I+a*
x))^(1/2)/(a^2*x^2+1)^(1/2)/a^4/c

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.67 \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - 2 \, {\left (a^{2} x^{2} - 2\right )} \arctan \left (a x\right )\right )} - 5 \, \sqrt {c} \log \left (-2 \, a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right )}{12 \, a^{4} c} \]

[In]

integrate(x^3*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 - 2)*arctan(a*x)) - 5*sqrt(c)*log(-2*a^2*c*x^2 - 2*sqrt(a^2*c*x
^2 + c)*a*sqrt(c)*x - c))/(a^4*c)

Sympy [F]

\[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {x^{3} \operatorname {atan}{\left (a x \right )}}{\sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

[In]

integrate(x**3*atan(a*x)/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x**3*atan(a*x)/sqrt(c*(a**2*x**2 + 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {a {\left (\frac {\frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}}{a^{2}} - \frac {4 \, \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )} - 2 \, {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{a^{2}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{a^{4}}\right )} \arctan \left (a x\right )}{6 \, \sqrt {c}} \]

[In]

integrate(x^3*arctan(a*x)/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

-1/6*(a*((sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a*x)/a^3)/a^2 - 4*arcsinh(a*x)/a^5) - 2*(sqrt(a^2*x^2 + 1)*x^2/a^2
- 2*sqrt(a^2*x^2 + 1)/a^4)*arctan(a*x))/sqrt(c)

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^3*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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {x^3\,\mathrm {atan}\left (a\,x\right )}{\sqrt {c\,a^2\,x^2+c}} \,d x \]

[In]

int((x^3*atan(a*x))/(c + a^2*c*x^2)^(1/2),x)

[Out]

int((x^3*atan(a*x))/(c + a^2*c*x^2)^(1/2), x)