\(\int \frac {x}{(c+a^2 c x^2)^{3/2} \sqrt {\arctan (a x)}} \, dx\) [962]

Optimal result

Integrand size = 24, antiderivative size = 60 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx=\frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5091, 5090, 3386, 3432} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx=\frac {\sqrt {2 \pi } \sqrt {a^2 x^2+1} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

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

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 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])

Rule 5091

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^(q + 1
/2)*(Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]), Int[x^m*(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, 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 {\sqrt {1+a^2 x^2} \int \frac {x}{\left (1+a^2 x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx}{c \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^2 c \sqrt {c+a^2 c x^2}} \\ & = \frac {\left (2 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^2 c \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a^2 c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.62 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \sqrt {\arctan (a x)}} \, dx=-\frac {\sqrt {1+a^2 x^2} \left (\sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+\sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )\right )}{2 a^2 c \sqrt {c \left (1+a^2 x^2\right )} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

-1/2*(Sqrt[1 + a^2*x^2]*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] + Sqrt[I*ArcTan[a*x]]*Gamma[1/2,
I*ArcTan[a*x]]))/(a^2*c*Sqrt[c*(1 + a^2*x^2)]*Sqrt[ArcTan[a*x]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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