3.9.74 \(\int \frac {x^2 \text {ArcTan}(a x)^{5/2}}{(c+a^2 c x^2)^3} \, dx\) [874]

Optimal. Leaf size=133 \[ \frac {\text {ArcTan}(a x)^{7/2}}{28 a^3 c^3}-\frac {5 \text {ArcTan}(a x)^{3/2} \cos (4 \text {ArcTan}(a x))}{256 a^3 c^3}-\frac {15 \sqrt {\frac {\pi }{2}} S\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\text {ArcTan}(a x)}\right )}{4096 a^3 c^3}+\frac {15 \sqrt {\text {ArcTan}(a x)} \sin (4 \text {ArcTan}(a x))}{2048 a^3 c^3}-\frac {\text {ArcTan}(a x)^{5/2} \sin (4 \text {ArcTan}(a x))}{32 a^3 c^3} \]

[Out]

1/28*arctan(a*x)^(7/2)/a^3/c^3-5/256*arctan(a*x)^(3/2)*cos(4*arctan(a*x))/a^3/c^3-1/32*arctan(a*x)^(5/2)*sin(4
*arctan(a*x))/a^3/c^3-15/8192*FresnelS(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^3/c^3+15/2048*
sin(4*arctan(a*x))*arctan(a*x)^(1/2)/a^3/c^3

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Rubi [A]
time = 0.14, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5090, 4491, 3377, 3386, 3432} \begin {gather*} -\frac {15 \sqrt {\frac {\pi }{2}} S\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\text {ArcTan}(a x)}\right )}{4096 a^3 c^3}+\frac {\text {ArcTan}(a x)^{7/2}}{28 a^3 c^3}-\frac {\text {ArcTan}(a x)^{5/2} \sin (4 \text {ArcTan}(a x))}{32 a^3 c^3}+\frac {15 \sqrt {\text {ArcTan}(a x)} \sin (4 \text {ArcTan}(a x))}{2048 a^3 c^3}-\frac {5 \text {ArcTan}(a x)^{3/2} \cos (4 \text {ArcTan}(a x))}{256 a^3 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[a*x]^(7/2)/(28*a^3*c^3) - (5*ArcTan[a*x]^(3/2)*Cos[4*ArcTan[a*x]])/(256*a^3*c^3) - (15*Sqrt[Pi/2]*Fresn
elS[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(4096*a^3*c^3) + (15*Sqrt[ArcTan[a*x]]*Sin[4*ArcTan[a*x]])/(2048*a^3*c^3)
- (ArcTan[a*x]^(5/2)*Sin[4*ArcTan[a*x]])/(32*a^3*c^3)

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 3386

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 3432

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

Rule 4491

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 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*} \int \frac {x^2 \tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac {\text {Subst}\left (\int x^{5/2} \cos ^2(x) \sin ^2(x) \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac {\text {Subst}\left (\int \left (\frac {x^{5/2}}{8}-\frac {1}{8} x^{5/2} \cos (4 x)\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^3}\\ &=\frac {\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac {\text {Subst}\left (\int x^{5/2} \cos (4 x) \, dx,x,\tan ^{-1}(a x)\right )}{8 a^3 c^3}\\ &=\frac {\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac {\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}+\frac {5 \text {Subst}\left (\int x^{3/2} \sin (4 x) \, dx,x,\tan ^{-1}(a x)\right )}{64 a^3 c^3}\\ &=\frac {\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac {5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}-\frac {\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}+\frac {15 \text {Subst}\left (\int \sqrt {x} \cos (4 x) \, dx,x,\tan ^{-1}(a x)\right )}{512 a^3 c^3}\\ &=\frac {\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac {5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}+\frac {15 \sqrt {\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac {\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}-\frac {15 \text {Subst}\left (\int \frac {\sin (4 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{4096 a^3 c^3}\\ &=\frac {\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac {5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}+\frac {15 \sqrt {\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac {\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}-\frac {15 \text {Subst}\left (\int \sin \left (4 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{2048 a^3 c^3}\\ &=\frac {\tan ^{-1}(a x)^{7/2}}{28 a^3 c^3}-\frac {5 \tan ^{-1}(a x)^{3/2} \cos \left (4 \tan ^{-1}(a x)\right )}{256 a^3 c^3}-\frac {15 \sqrt {\frac {\pi }{2}} S\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4096 a^3 c^3}+\frac {15 \sqrt {\tan ^{-1}(a x)} \sin \left (4 \tan ^{-1}(a x)\right )}{2048 a^3 c^3}-\frac {\tan ^{-1}(a x)^{5/2} \sin \left (4 \tan ^{-1}(a x)\right )}{32 a^3 c^3}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.18, size = 185, normalized size = 1.39 \begin {gather*} \frac {32 \text {ArcTan}(a x) \left (-105 a x \left (-1+a^2 x^2\right )-70 \left (1-6 a^2 x^2+a^4 x^4\right ) \text {ArcTan}(a x)+448 a x \left (-1+a^2 x^2\right ) \text {ArcTan}(a x)^2+128 \left (1+a^2 x^2\right )^2 \text {ArcTan}(a x)^3\right )+105 \left (1+a^2 x^2\right )^2 \sqrt {-i \text {ArcTan}(a x)} \text {Gamma}\left (\frac {1}{2},-4 i \text {ArcTan}(a x)\right )+105 \left (1+a^2 x^2\right )^2 \sqrt {i \text {ArcTan}(a x)} \text {Gamma}\left (\frac {1}{2},4 i \text {ArcTan}(a x)\right )}{114688 a^3 c^3 \left (1+a^2 x^2\right )^2 \sqrt {\text {ArcTan}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*ArcTan[a*x]*(-105*a*x*(-1 + a^2*x^2) - 70*(1 - 6*a^2*x^2 + a^4*x^4)*ArcTan[a*x] + 448*a*x*(-1 + a^2*x^2)*A
rcTan[a*x]^2 + 128*(1 + a^2*x^2)^2*ArcTan[a*x]^3) + 105*(1 + a^2*x^2)^2*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-4*
I)*ArcTan[a*x]] + 105*(1 + a^2*x^2)^2*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (4*I)*ArcTan[a*x]])/(114688*a^3*c^3*(1 +
a^2*x^2)^2*Sqrt[ArcTan[a*x]])

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Maple [A]
time = 0.34, size = 96, normalized size = 0.72

method	result	size
default	\(\frac {2048 \arctan \left (a x \right )^{4}-1792 \arctan \left (a x \right )^{3} \sin \left (4 \arctan \left (a x \right )\right )-105 \,\mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {2}\, \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }-1120 \arctan \left (a x \right )^{2} \cos \left (4 \arctan \left (a x \right )\right )+420 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )}{57344 c^{3} a^{3} \sqrt {\arctan \left (a x \right )}}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/57344/c^3/a^3*(2048*arctan(a*x)^4-1792*arctan(a*x)^3*sin(4*arctan(a*x))-105*FresnelS(2*2^(1/2)/Pi^(1/2)*arct
an(a*x)^(1/2))*2^(1/2)*arctan(a*x)^(1/2)*Pi^(1/2)-1120*arctan(a*x)^2*cos(4*arctan(a*x))+420*sin(4*arctan(a*x))
*arctan(a*x))/arctan(a*x)^(1/2)

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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*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,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*arctan(a*x)^(5/2)/(a^2*c*x^2+c)^3,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} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*atan(a*x)**(5/2)/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x)/c**3

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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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 )}^3} \,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)^3,x)

[Out]

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

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