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

Optimal. Leaf size=96 \[ -\frac {\sqrt {\frac {\pi }{2}} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a^4 c^3}+\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{a^4 c^3}-\frac {2 x^3}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\tan ^{-1}(a x)}} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4968, 4970, 3312, 3304, 3352, 4406} \[ -\frac {\sqrt {\frac {\pi }{2}} \text {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a^4 c^3}+\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{a^4 c^3}-\frac {2 x^3}{a c^3 \left (a^2 x^2+1\right )^2 \sqrt {\tan ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3304

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 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 3352

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

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4968

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

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])

Rubi steps

\begin {align*} \int \frac {x^3}{\left (c+a^2 c x^2\right )^3 \tan ^{-1}(a x)^{3/2}} \, dx &=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\tan ^{-1}(a x)}}+\frac {6 \int \frac {x^2}{\left (c+a^2 c x^2\right )^3 \sqrt {\tan ^{-1}(a x)}} \, dx}{a}-(2 a) \int \frac {x^4}{\left (c+a^2 c x^2\right )^3 \sqrt {\tan ^{-1}(a x)}} \, dx\\ &=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {\sin ^4(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}+\frac {6 \operatorname {Subst}\left (\int \frac {\cos ^2(x) \sin ^2(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {2 \operatorname {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,\tan ^{-1}(a x)\right )}{a^4 c^3}+\frac {6 \operatorname {Subst}\left (\int \left (\frac {1}{8 \sqrt {x}}-\frac {\cos (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {\operatorname {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^4 c^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^4 c^3}+\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^4 c^3}\\ &=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {\operatorname {Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^3}-\frac {3 \operatorname {Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2 a^4 c^3}+\frac {2 \operatorname {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a^4 c^3}\\ &=-\frac {2 x^3}{a c^3 \left (1+a^2 x^2\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {\sqrt {\frac {\pi }{2}} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a^4 c^3}+\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{a^4 c^3}\\ \end {align*}

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Mathematica [C]  time = 0.37, size = 148, normalized size = 1.54 \[ \frac {\frac {-\frac {32 a^3 x^3}{\left (a^2 x^2+1\right )^2}+3 i \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-4 i \tan ^{-1}(a x)\right )-3 i \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},4 i \tan ^{-1}(a x)\right )}{\sqrt {\tan ^{-1}(a x)}}-2 \sqrt {2 \pi } C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )+16 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{16 a^4 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[2*Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]] + 16*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]
+ ((-32*a^3*x^3)/(1 + a^2*x^2)^2 + (3*I)*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*I)*ArcTan[a*x]] - (3*I)*Sqrt[I
*ArcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]])/Sqrt[ArcTan[a*x]])/(16*a^4*c^3)

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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)^3/arctan(a*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.50, size = 86, normalized size = 0.90 \[ -\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )-4 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+2 \sin \left (2 \arctan \left (a x \right )\right )-\sin \left (4 \arctan \left (a x \right )\right )}{4 a^{4} c^{3} \sqrt {\arctan \left (a x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3}}{a^{6} x^{6} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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