\(\int \frac {x^3 \sqrt {\arctan (a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\) [751]

Optimal result

Integrand size = 26, antiderivative size = 215 \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 \sqrt {\arctan (a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}} \]

[Out]

-1/72*FresnelC(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^4/c^2/(a^2*c*x^2+c)^(1
/2)+3/8*FresnelC(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^4/c^2/(a^2*c*x^2+c)^
(1/2)-3/4*arctan(a*x)^(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)+1/12*cos(3*arctan(a*x))*(a^2*x^2+1)^(1/2)*arctan(a*x)^
(1/2)/a^4/c^2/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5091, 5090, 3393, 3377, 3385, 3433} \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{12 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {3 \sqrt {\arctan (a x)}}{4 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{12 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

(-3*Sqrt[ArcTan[a*x]])/(4*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[1 + a^2*x^2]*Sqrt[ArcTan[a*x]]*Cos[3*ArcTan[a*x
]])/(12*a^4*c^2*Sqrt[c + a^2*c*x^2]) + (3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])
/(4*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(12*a
^4*c^2*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 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[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^3 \sqrt {\arctan (a x)}}{\left (1+a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \sqrt {x} \sin ^3(x) \, dx,x,\arctan (a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \left (\frac {3}{4} \sqrt {x} \sin (x)-\frac {1}{4} \sqrt {x} \sin (3 x)\right ) \, dx,x,\arctan (a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \sqrt {x} \sin (3 x) \, dx,x,\arctan (a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \sqrt {x} \sin (x) \, dx,x,\arctan (a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {3 \sqrt {\arctan (a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{24 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{8 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {3 \sqrt {\arctan (a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {3 \sqrt {\arctan (a x)}}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \sqrt {\arctan (a x)} \cos (3 \arctan (a x))}{12 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} \operatorname {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\arctan (a x)}\right )}{12 a^4 c^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.51 \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {-96 \arctan (a x)-144 a^2 x^2 \arctan (a x)-27 i \left (1+a^2 x^2\right )^{3/2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-i \arctan (a x)\right )+27 i \left (1+a^2 x^2\right )^{3/2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},i \arctan (a x)\right )+i \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )+i a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},-3 i \arctan (a x)\right )-i \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )-i a^2 x^2 \sqrt {3+3 a^2 x^2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},3 i \arctan (a x)\right )}{144 a^4 c^2 \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sqrt {\arctan (a x)}} \]

[In]

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

[Out]

(-96*ArcTan[a*x] - 144*a^2*x^2*ArcTan[a*x] - (27*I)*(1 + a^2*x^2)^(3/2)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)
*ArcTan[a*x]] + (27*I)*(1 + a^2*x^2)^(3/2)*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]] + I*Sqrt[3 + 3*a^2*x^
2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] + I*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]
]*Gamma[1/2, (-3*I)*ArcTan[a*x]] - I*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]] - I
*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])/(144*a^4*c^2*(1 + a^2*x^2)*Sqr
t[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]])

Maple [F]

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

[In]

int(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

[Out]

int(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

Fricas [F(-2)]

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

[In]

integrate(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

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

[In]

integrate(x^3*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(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. \[ \int \frac {x^3 \sqrt {\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\sqrt {\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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