3.967 \(\int \frac {x^3}{(c+a^2 c x^2)^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx\)

Optimal. Leaf size=131 \[ \frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

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

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Rubi [A]  time = 0.29, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4971, 4970, 3312, 3305, 3351} \[ \frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(2*a^4*c^2*Sqrt[c + a^2*c*x^2]) - (Sqr
t[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(2*a^4*c^2*Sqrt[c + a^2*c*x^2])

Rule 3305

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 3312

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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4970

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 4971

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*} \int \frac {x^3}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {x^3}{\left (1+a^2 x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx}{c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\sin ^3(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \left (\frac {3 \sin (x)}{4 \sqrt {x}}-\frac {\sin (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\sin (3 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(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 ) \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \sin \left (3 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 95, normalized size = 0.73 \[ \frac {\sqrt {\frac {\pi }{6}} \left (a^2 x^2+1\right )^{3/2} \left (3 \sqrt {3} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )-S\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )\right )}{2 a^4 c \left (c \left (a^2 x^2+1\right )\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi/6]*(1 + a^2*x^2)^(3/2)*(3*Sqrt[3]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] - FresnelS[Sqrt[6/Pi]*Sqrt[A
rcTan[a*x]]]))/(2*a^4*c*(c*(1 + a^2*x^2))^(3/2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 8.79, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \sqrt {\arctan \left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{\sqrt {\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \sqrt {\operatorname {atan}{\left (a x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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