\(\int \frac {x^4}{(c+a^2 c x^2)^3 \sqrt {\arctan (a x)}} \, dx\) [938]

Optimal result

Integrand size = 24, antiderivative size = 89 \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\frac {3 \sqrt {\arctan (a x)}}{4 a^5 c^3}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a^5 c^3}-\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{2 a^5 c^3} \]

[Out]

1/16*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^5/c^3-1/2*FresnelC(2*arctan(a*x)^(1/2)/
Pi^(1/2))*Pi^(1/2)/a^5/c^3+3/4*arctan(a*x)^(1/2)/a^5/c^3

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5090, 3393, 3385, 3433} \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a^5 c^3}-\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{2 a^5 c^3}+\frac {3 \sqrt {\arctan (a x)}}{4 a^5 c^3} \]

[In]

Int[x^4/((c + a^2*c*x^2)^3*Sqrt[ArcTan[a*x]]),x]

[Out]

(3*Sqrt[ArcTan[a*x]])/(4*a^5*c^3) + (Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(8*a^5*c^3) - (Sqrt[
Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(2*a^5*c^3)

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sin ^4(x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{a^5 c^3} \\ & = \frac {\text {Subst}\left (\int \left (\frac {3}{8 \sqrt {x}}-\frac {\cos (2 x)}{2 \sqrt {x}}+\frac {\cos (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\arctan (a x)\right )}{a^5 c^3} \\ & = \frac {3 \sqrt {\arctan (a x)}}{4 a^5 c^3}+\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{8 a^5 c^3}-\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\arctan (a x)\right )}{2 a^5 c^3} \\ & = \frac {3 \sqrt {\arctan (a x)}}{4 a^5 c^3}+\frac {\text {Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{4 a^5 c^3}-\frac {\text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\arctan (a x)}\right )}{a^5 c^3} \\ & = \frac {3 \sqrt {\arctan (a x)}}{4 a^5 c^3}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )}{8 a^5 c^3}-\frac {\sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{2 a^5 c^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.58 \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\frac {10 \sqrt {2 \pi } \sqrt {\arctan (a x)^2} \operatorname {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\arctan (a x)}\right )-80 \sqrt {\pi } \sqrt {\arctan (a x)^2} \operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )+3 \sqrt {\arctan (a x)} \left (64 \sqrt {\arctan (a x)^2}+4 \sqrt {2} \sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},-2 i \arctan (a x)\right )+4 \sqrt {2} \sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},2 i \arctan (a x)\right )-\sqrt {i \arctan (a x)} \Gamma \left (\frac {1}{2},-4 i \arctan (a x)\right )-\sqrt {-i \arctan (a x)} \Gamma \left (\frac {1}{2},4 i \arctan (a x)\right )\right )}{256 a^5 c^3 \sqrt {\arctan (a x)^2}} \]

[In]

Integrate[x^4/((c + a^2*c*x^2)^3*Sqrt[ArcTan[a*x]]),x]

[Out]

(10*Sqrt[2*Pi]*Sqrt[ArcTan[a*x]^2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] - 80*Sqrt[Pi]*Sqrt[ArcTan[a*x]^2]*
FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + 3*Sqrt[ArcTan[a*x]]*(64*Sqrt[ArcTan[a*x]^2] + 4*Sqrt[2]*Sqrt[I*ArcT
an[a*x]]*Gamma[1/2, (-2*I)*ArcTan[a*x]] + 4*Sqrt[2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (2*I)*ArcTan[a*x]] - Sqr
t[I*ArcTan[a*x]]*Gamma[1/2, (-4*I)*ArcTan[a*x]] - Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]]))/(256*
a^5*c^3*Sqrt[ArcTan[a*x]^2])

Maple [A] (verified)

Time = 1.44 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {\pi \sqrt {2}\, \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-8 \pi \,\operatorname {FresnelC}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+12 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }}{16 c^{3} a^{5} \sqrt {\pi }}\)	\(59\)

[In]

int(x^4/(a^2*c*x^2+c)^3/arctan(a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/c^3/a^5*(Pi*2^(1/2)*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))-8*Pi*FresnelC(2*arctan(a*x)^(1/2)/Pi^(
1/2))+12*arctan(a*x)^(1/2)*Pi^(1/2))/Pi^(1/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^4/(a^2*c*x^2+c)^3/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]

\[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\frac {\int \frac {x^{4}}{a^{6} x^{6} \sqrt {\operatorname {atan}{\left (a x \right )}} + 3 a^{4} x^{4} \sqrt {\operatorname {atan}{\left (a x \right )}} + 3 a^{2} x^{2} \sqrt {\operatorname {atan}{\left (a x \right )}} + \sqrt {\operatorname {atan}{\left (a x \right )}}}\, dx}{c^{3}} \]

[In]

integrate(x**4/(a**2*c*x**2+c)**3/atan(a*x)**(1/2),x)

[Out]

Integral(x**4/(a**6*x**6*sqrt(atan(a*x)) + 3*a**4*x**4*sqrt(atan(a*x)) + 3*a**2*x**2*sqrt(atan(a*x)) + sqrt(at
an(a*x))), x)/c**3

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(x^4/(a^2*c*x^2+c)^3/arctan(a*x)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\int { \frac {x^{4}}{{\left (a^{2} c x^{2} + c\right )}^{3} \sqrt {\arctan \left (a x\right )}} \,d x } \]

[In]

integrate(x^4/(a^2*c*x^2+c)^3/arctan(a*x)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\arctan (a x)}} \, dx=\int \frac {x^4}{\sqrt {\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]

[In]

int(x^4/(atan(a*x)^(1/2)*(c + a^2*c*x^2)^3),x)

[Out]

int(x^4/(atan(a*x)^(1/2)*(c + a^2*c*x^2)^3), x)