\(\int \frac {x^2}{(c+a^2 c x^2)^3 \arctan (a x)^{3/2}} \, dx\) [1000]

Optimal result

Integrand size = 24, antiderivative size = 67 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{3/2}} \, dx=-\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^3} \]

[Out]

1/2*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^3/c^3-2*x^2/a/c^3/(a^2*x^2+1)^2/arctan(a
*x)^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5088, 5090, 4491, 3386, 3432} \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{3/2}} \, dx=\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^3}-\frac {2 x^2}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

(-2*x^2)/(a*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) + (Sqrt[Pi/2]*FresnelS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(a^
3*c^3)

Rule 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[f*(x^2/d)], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 5088

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

Rule 5090

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(m
+ 1), Subst[Int[(a + b*x)^p*(Sin[x]^m/Cos[x]^(m + 2*(q + 1))), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d,
e, p}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+\frac {4 \int \frac {x}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx}{a}-(4 a) \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx \\ & = -\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+\frac {4 \text {Subst}\left (\int \frac {\cos ^3(x) \sin (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^3 c^3}-\frac {4 \text {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^3 c^3} \\ & = -\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}-\frac {4 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 \sqrt {x}}-\frac {\sin (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{a^3 c^3}+\frac {4 \text {Subst}\left (\int \left (\frac {\sin (2 x)}{4 \sqrt {x}}+\frac {\sin (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{a^3 c^3} \\ & = -\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+2 \frac {\text {Subst}\left (\int \frac {\sin (4 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{2 a^3 c^3} \\ & = -\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+2 \frac {\text {Subst}\left (\int \sin \left (4 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^3 c^3} \\ & = -\frac {2 x^2}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^3 c^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.67 \[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{3/2}} \, dx=\frac {-8 a^2 x^2-\left (1+a^2 x^2\right )^2 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )-\left (1+a^2 x^2\right )^2 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )}{4 a^3 c^3 \left (1+a^2 x^2\right )^2 \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

(-8*a^2*x^2 - (1 + a^2*x^2)^2*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*I)*ArcTan[a*x]] - (1 + a^2*x^2)^2*Sqrt[I*A
rcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]])/(4*a^3*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]])

Maple [A] (verified)

Time = 1.75 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {2 \,\operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {2}\, \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }+\cos \left (4 \arctan \left (a x \right )\right )-1}{4 c^{3} a^{3} \sqrt {\arctan \left (a x \right )}}\)	\(53\)

[In]

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

[Out]

1/4/c^3/a^3*(2*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*arctan(a*x)^(1/2)*Pi^(1/2)+cos(4*arctan(
a*x))-1)/arctan(a*x)^(1/2)

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \arctan (a x)^{3/2}} \, dx=\frac {\int \frac {x^{2}}{a^{6} x^{6} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx}{c^{3}} \]

[In]

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

[Out]

Integral(x**2/(a**6*x**6*atan(a*x)**(3/2) + 3*a**4*x**4*atan(a*x)**(3/2) + 3*a**2*x**2*atan(a*x)**(3/2) + atan
(a*x)**(3/2)), x)/c**3

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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