\(\int \frac {x}{(c+a^2 c x^2)^2 \arctan (a x)^{5/2}} \, dx\) [1061]

Optimal result

Integrand size = 22, antiderivative size = 101 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^2 c^2} \]

[Out]

-2/3*x/a/c^2/(a^2*x^2+1)/arctan(a*x)^(3/2)-8/3*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2/c^2-4/3*(-a
^2*x^2+1)/a^2/c^2/(a^2*x^2+1)/arctan(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5052, 5090, 4491, 12, 3386, 3432} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^2 c^2}-\frac {2 x}{3 a c^2 \left (a^2 x^2+1\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}} \]

[In]

Int[x/((c + a^2*c*x^2)^2*ArcTan[a*x]^(5/2)),x]

[Out]

(-2*x)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (4*(1 - a^2*x^2))/(3*a^2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]
]) - (8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^2*c^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5052

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTan
[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2))), x] + (-Dist[4/(b^2*(p + 1)*(p + 2)), Int[x*((a + b*ArcTan[c*x])^(
p + 2)/(d + e*x^2)^2), x], x] - Simp[(1 - c^2*x^2)*((a + b*ArcTan[c*x])^(p + 2)/(b^2*e*(p + 1)*(p + 2)*(d + e*
x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -2]

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}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {16}{3} \int \frac {x}{\left (c+a^2 c x^2\right )^2 \sqrt {\arctan (a x)}} \, dx \\ & = -\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {16 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{3 a^2 c^2} \\ & = -\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {16 \text {Subst}\left (\int \frac {\sin (2 x)}{2 \sqrt {x}} \, dx,x,\arctan (a x)\right )}{3 a^2 c^2} \\ & = -\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {8 \text {Subst}\left (\int \frac {\sin (2 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{3 a^2 c^2} \\ & = -\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {16 \text {Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{3 a^2 c^2} \\ & = -\frac {2 x}{3 a c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}}-\frac {4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\arctan (a x)}}-\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{3 a^2 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=-\frac {2 \left (a x+\left (2-2 a^2 x^2\right ) \arctan (a x)+4 \sqrt {\pi } \left (1+a^2 x^2\right ) \arctan (a x)^{3/2} \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \arctan (a x)^{3/2}} \]

[In]

Integrate[x/((c + a^2*c*x^2)^2*ArcTan[a*x]^(5/2)),x]

[Out]

(-2*(a*x + (2 - 2*a^2*x^2)*ArcTan[a*x] + 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)*FresnelS[(2*Sqrt[ArcTan[a*
x]])/Sqrt[Pi]]))/(3*a^2*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2))

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.58

method	result	size
default	\(-\frac {8 \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+4 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+\sin \left (2 \arctan \left (a x \right )\right )}{3 c^{2} a^{2} \arctan \left (a x \right )^{\frac {3}{2}}}\)	\(59\)

[In]

int(x/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/c^2/a^2*(8*Pi^(1/2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+4*cos(2*arctan(a*x))*arctan(
a*x)+sin(2*arctan(a*x)))/arctan(a*x)^(3/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\frac {\int \frac {x}{a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{2}} \]

[In]

integrate(x/(a**2*c*x**2+c)**2/atan(a*x)**(5/2),x)

[Out]

Integral(x/(a**4*x**4*atan(a*x)**(5/2) + 2*a**2*x**2*atan(a*x)**(5/2) + atan(a*x)**(5/2)), x)/c**2

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(x/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(x/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^2 \arctan (a x)^{5/2}} \, dx=\int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

[In]

int(x/(atan(a*x)^(5/2)*(c + a^2*c*x^2)^2),x)

[Out]

int(x/(atan(a*x)^(5/2)*(c + a^2*c*x^2)^2), x)