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

Optimal. Leaf size=124 \[ \frac {x \tan ^{-1}(a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {3 \sqrt {\tan ^{-1}(a x)}}{8 a c^2 \left (a^2 x^2+1\right )}-\frac {3 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{32 a c^2}+\frac {\tan ^{-1}(a x)^{5/2}}{5 a c^2}-\frac {3 \sqrt {\tan ^{-1}(a x)}}{16 a c^2} \]

[Out]

1/2*x*arctan(a*x)^(3/2)/c^2/(a^2*x^2+1)+1/5*arctan(a*x)^(5/2)/a/c^2-3/32*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2)
)*Pi^(1/2)/a/c^2-3/16*arctan(a*x)^(1/2)/a/c^2+3/8*arctan(a*x)^(1/2)/a/c^2/(a^2*x^2+1)

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Rubi [A]  time = 0.15, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4892, 4930, 4904, 3312, 3304, 3352} \[ \frac {x \tan ^{-1}(a x)^{3/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {3 \sqrt {\tan ^{-1}(a x)}}{8 a c^2 \left (a^2 x^2+1\right )}-\frac {3 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{32 a c^2}+\frac {\tan ^{-1}(a x)^{5/2}}{5 a c^2}-\frac {3 \sqrt {\tan ^{-1}(a x)}}{16 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[ArcTan[a*x]])/(16*a*c^2) + (3*Sqrt[ArcTan[a*x]])/(8*a*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x]^(3/2))/(2*c
^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(5*a*c^2) - (3*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(32*a*
c^2)

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 4892

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

Rule 4904

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 4930

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 90, normalized size = 0.73 \[ \frac {\frac {2 \sqrt {\tan ^{-1}(a x)} \left (-15 a^2 x^2+16 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2+40 a x \tan ^{-1}(a x)+15\right )}{a^2 x^2+1}-15 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{160 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[ArcTan[a*x]]*(15 - 15*a^2*x^2 + 40*a*x*ArcTan[a*x] + 16*(1 + a^2*x^2)*ArcTan[a*x]^2))/(1 + a^2*x^2) -
15*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(160*a*c^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(arctan(a*x)^(3/2)/(a^2*c*x^2+c)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.53, size = 75, normalized size = 0.60 \[ \frac {32 \arctan \left (a x \right )^{3}+40 \arctan \left (a x \right )^{2} \sin \left (2 \arctan \left (a x \right )\right )+30 \cos \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-15 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )}{160 a \,c^{2} \sqrt {\arctan \left (a x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160/a/c^2/arctan(a*x)^(1/2)*(32*arctan(a*x)^3+40*arctan(a*x)^2*sin(2*arctan(a*x))+30*cos(2*arctan(a*x))*arct
an(a*x)-15*arctan(a*x)^(1/2)*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(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(arctan(a*x)^(3/2)/(a^2*c*x^2+c)^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 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^{3/2}}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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