\(\int \frac {x}{(c+a^2 c x^2)^{3/2} \arctan (a x)^{5/2}} \, dx\) [1096]

Optimal result

Integrand size = 24, antiderivative size = 129 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=-\frac {2 x}{3 a c \sqrt {c+a^2 c x^2} \arctan (a x)^{3/2}}-\frac {4}{3 a^2 c \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}}-\frac {4 \sqrt {2 \pi } \sqrt {1+a^2 x^2} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{3 a^2 c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5062, 5022, 5091, 5090, 3386, 3432} \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=-\frac {4 \sqrt {2 \pi } \sqrt {a^2 x^2+1} \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{3 a^2 c \sqrt {a^2 c x^2+c}}-\frac {2 x}{3 a c \arctan (a x)^{3/2} \sqrt {a^2 c x^2+c}}-\frac {4}{3 a^2 c \sqrt {\arctan (a x)} \sqrt {a^2 c x^2+c}} \]

[In]

Int[x/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(5/2)),x]

[Out]

(-2*x)/(3*a*c*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^(3/2)) - 4/(3*a^2*c*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]]) - (4*
Sqrt[2*Pi]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*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 5022

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

Rule 5062

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=-\frac {2 \left (a x+2 \arctan (a x)-i \sqrt {1+a^2 x^2} (-i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+i \sqrt {1+a^2 x^2} (i \arctan (a x))^{3/2} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )\right )}{3 a^2 c \sqrt {c+a^2 c x^2} \arctan (a x)^{3/2}} \]

[In]

Integrate[x/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(5/2)),x]

[Out]

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

Maple [F]

\[\int \frac {x}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}} \arctan \left (a x \right )^{\frac {5}{2}}}d x\]

[In]

int(x/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x)

[Out]

int(x/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(a^2*c*x^2+c)^(3/2)/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]

\[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\int \frac {x}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx \]

[In]

integrate(x/(a**2*c*x**2+c)**(3/2)/atan(a*x)**(5/2),x)

[Out]

Integral(x/((c*(a**2*x**2 + 1))**(3/2)*atan(a*x)**(5/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(x/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(5/2),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. \[ \int \frac {x}{\left (c+a^2 c x^2\right )^{3/2} \arctan (a x)^{5/2}} \, dx=\int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

[In]

int(x/(atan(a*x)^(5/2)*(c + a^2*c*x^2)^(3/2)),x)

[Out]

int(x/(atan(a*x)^(5/2)*(c + a^2*c*x^2)^(3/2)), x)