\(\int \frac {x}{(c+a^2 c x^2)^{5/2} \sqrt {\arctan (a x)}} \, dx\) [969]

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx=\frac {\sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

1/12*FresnelS(6^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^2/c^2/(a^2*c*x^2+c)^(1/
2)+1/4*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^2/(a^2*c*x^2+c)^(
1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5091, 5090, 4491, 3386, 3432} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx=\frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(2*a^2*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[
Pi/6]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(2*a^2*c^2*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 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 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 )^{5/2} \sqrt {\arctan (a x)}} \, dx}{c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\cos ^2(x) \sin (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \left (\frac {\sin (x)}{4 \sqrt {x}}+\frac {\sin (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{a^2 c^2 \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 )}{4 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\sin (3 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{4 a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \sin \left (3 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{2 a^2 c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.19 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx=-\frac {\left (1+a^2 x^2\right )^{3/2} \left (3 \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+3 \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+\sqrt {3} \left (\sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+\sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )\right )\right )}{24 a^2 c \left (c \left (1+a^2 x^2\right )\right )^{3/2} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

-1/24*((1 + a^2*x^2)^(3/2)*(3*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] + 3*Sqrt[I*ArcTan[a*x]]*Gamm
a[1/2, I*ArcTan[a*x]] + Sqrt[3]*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] + Sqrt[I*ArcTan[a*x]]*G
amma[1/2, (3*I)*ArcTan[a*x]])))/(a^2*c*(c*(1 + a^2*x^2))^(3/2)*Sqrt[ArcTan[a*x]])

Maple [F]

\[\int \frac {x}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {\arctan \left (a x \right )}}d x\]

[In]

int(x/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),x)

[Out]

int(x/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(1/2),x)

Fricas [F(-2)]

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

[In]

integrate(x/(a^2*c*x^2+c)^(5/2)/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]

\[ \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\arctan (a x)}} \, dx=\int \frac {x}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \sqrt {\operatorname {atan}{\left (a x \right )}}}\, dx \]

[In]

integrate(x/(a**2*c*x**2+c)**(5/2)/atan(a*x)**(1/2),x)

[Out]

Integral(x/((c*(a**2*x**2 + 1))**(5/2)*sqrt(atan(a*x))), x)

Maxima [F(-2)]

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

[In]

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

[In]

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

[In]

int(x/(atan(a*x)^(1/2)*(c + a^2*c*x^2)^(5/2)),x)

[Out]

int(x/(atan(a*x)^(1/2)*(c + a^2*c*x^2)^(5/2)), x)