\(\int \frac {x^3 \arctan (a x)}{(c+a^2 c x^2)^{3/2}} \, dx\) [232]

Optimal result

Integrand size = 22, antiderivative size = 107 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {x}{a^3 c \sqrt {c+a^2 c x^2}}+\frac {\arctan (a x)}{a^4 c \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{a^4 c^2}-\frac {\text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^4 c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5084, 5050, 223, 212, 197} \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{a^4 c^2}+\frac {\arctan (a x)}{a^4 c \sqrt {a^2 c x^2+c}}-\frac {\text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^4 c^{3/2}}-\frac {x}{a^3 c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

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 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 5084

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

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {-a x \sqrt {c+a^2 c x^2}+\left (2+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \arctan (a x)-\sqrt {c} \left (1+a^2 x^2\right ) \log \left (a c x+\sqrt {c} \sqrt {c+a^2 c x^2}\right )}{a^4 c^2 \left (1+a^2 x^2\right )} \]

[In]

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

[Out]

(-(a*x*Sqrt[c + a^2*c*x^2]) + (2 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] - Sqrt[c]*(1 + a^2*x^2)*Log[a*c*x
+ Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(a^4*c^2*(1 + a^2*x^2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.04

method	result	size
default	\(\frac {\left (\arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}-\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right ) a^{2} x^{2}+\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right ) a^{2} x^{2}-\sqrt {a^{2} x^{2}+1}\, a x +2 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}-\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right )+\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right )\right ) \sqrt {a^{2} x^{2}+1}\, \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{a^{4} c^{2} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(218\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {{\left (a^{2} x^{2} + 1\right )} \sqrt {c} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right ) - 2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - {\left (a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )}}{2 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3 \arctan (a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x^{3} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(x^3*arctan(a*x)/(a^2*c*x^2 + c)^(3/2), x)

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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