\(\int \frac {\sqrt {\arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [746]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {\sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {x \sqrt {\arctan (a x)}}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5025, 5024, 3377, 3386, 3432} \[ \int \frac {\sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {x \sqrt {\arctan (a x)}}{c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{a c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

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

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 5024

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

Rule 5025

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^(q + 1/2)*(Sqrt[1
+ c^2*x^2]/Sqrt[d + e*x^2]), Int[(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] && ILtQ[2*(q + 1), 0] &&  !(IntegerQ[q] || GtQ[d, 0])

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {4 a x \arctan (a x)+\sqrt {1+a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+\sqrt {1+a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )}{4 a c \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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