3.11.60 \(\int \frac {x^2}{(c+a^2 c x^2)^2 \text {ArcTan}(a x)^{5/2}} \, dx\) [1060]

Optimal. Leaf size=180 \[ -\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \text {ArcTan}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\text {ArcTan}(a x)}}+\frac {16 \sqrt {\text {ArcTan}(a x)}}{3 a^3 c^2}-\frac {32 \sqrt {\text {ArcTan}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\text {ArcTan}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {8 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{3 a^3 c^2} \]

[Out]

-2/3*x^2/a/c^2/(a^2*x^2+1)/arctan(a*x)^(3/2)+8/3*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^3/c^2-8/3*x
/a^2/c^2/(a^2*x^2+1)/arctan(a*x)^(1/2)+16/3*arctan(a*x)^(1/2)/a^3/c^2-32/3*arctan(a*x)^(1/2)/a^3/c^2/(a^2*x^2+
1)+16/3*(-a^2*x^2+1)*arctan(a*x)^(1/2)/a^3/c^2/(a^2*x^2+1)

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Rubi [A]
time = 0.17, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5062, 5052, 5050, 5024, 3393, 3385, 3433} \begin {gather*} \frac {8 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{3 a^3 c^2}+\frac {16 \sqrt {\text {ArcTan}(a x)}}{3 a^3 c^2}-\frac {2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \text {ArcTan}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt {\text {ArcTan}(a x)}}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\text {ArcTan}(a x)}}{3 a^3 c^2 \left (a^2 x^2+1\right )}-\frac {32 \sqrt {\text {ArcTan}(a x)}}{3 a^3 c^2 \left (a^2 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (8*x)/(3*a^2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) + (16*S
qrt[ArcTan[a*x]])/(3*a^3*c^2) - (32*Sqrt[ArcTan[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)) + (16*(1 - a^2*x^2)*Sqrt[ArcT
an[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)) + (8*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^3*c^2)

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 5024

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(a
+ b*x)^p/Cos[x]^(2*(q + 1)), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && ILtQ
[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 5050

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

Rule 5052

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTan
[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2))), x] + (-Dist[4/(b^2*(p + 1)*(p + 2)), Int[x*((a + b*ArcTan[c*x])^(
p + 2)/(d + e*x^2)^2), x], x] - Simp[(1 - c^2*x^2)*((a + b*ArcTan[c*x])^(p + 2)/(b^2*e*(p + 1)*(p + 2)*(d + e*
x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -2]

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]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{5/2}} \, dx &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {4 \int \frac {x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{3/2}} \, dx}{3 a}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {64 \int \frac {x \sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{3 a}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \int \frac {1}{\left (c+a^2 c x^2\right )^2 \sqrt {\tan ^{-1}(a x)}} \, dx}{3 a^2}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^3 c^2}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{3 a^3 c^2}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2}-\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {8 \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^3 c^2}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2}-\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{3 a^3 c^2}\\ &=-\frac {2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2}-\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac {8 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a^3 c^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.18, size = 162, normalized size = 0.90 \begin {gather*} \frac {-2 a x (a x+4 \text {ArcTan}(a x))+4 \sqrt {\pi } \left (1+a^2 x^2\right ) \text {ArcTan}(a x)^{3/2} \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )+\sqrt {2} \left (1+a^2 x^2\right ) (-i \text {ArcTan}(a x))^{3/2} \text {Gamma}\left (\frac {1}{2},-2 i \text {ArcTan}(a x)\right )+\sqrt {2} \left (1+a^2 x^2\right ) (i \text {ArcTan}(a x))^{3/2} \text {Gamma}\left (\frac {1}{2},2 i \text {ArcTan}(a x)\right )}{3 a^3 c^2 \left (1+a^2 x^2\right ) \text {ArcTan}(a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.35, size = 62, normalized size = 0.34

method	result	size
default	\(-\frac {-8 \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+4 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\cos \left (2 \arctan \left (a x \right )\right )+1}{3 c^{2} a^{3} \arctan \left (a x \right )^{\frac {3}{2}}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),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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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {x^{2}}{a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a**4*x**4*atan(a*x)**(5/2) + 2*a**2*x**2*atan(a*x)**(5/2) + atan(a*x)**(5/2)), x)/c**2

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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