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

Optimal. Leaf size=129 \[ -\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^2 c \sqrt {a^2 c x^2+c}}-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \sqrt {a^2 c x^2+c}}+\frac {3 x \sqrt {\tan ^{-1}(a x)}}{2 a c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4930, 4905, 4904, 3296, 3305, 3351} \[ -\frac {3 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{2 a^2 c \sqrt {a^2 c x^2+c}}-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \sqrt {a^2 c x^2+c}}+\frac {3 x \sqrt {\tan ^{-1}(a x)}}{2 a c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3296

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 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 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 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 4905

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q + 1/2)*Sqrt[1
+ c^2*x^2])/Sqrt[d + e*x^2], Int[(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] && 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 {x \tan ^{-1}(a x)^{3/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \sqrt {c+a^2 c x^2}}+\frac {3 \int \frac {\sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{2 a}\\ &=-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \int \frac {\sqrt {\tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{2 a c \sqrt {c+a^2 c x^2}}\\ &=-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \sqrt {x} \cos (x) \, dx,x,\tan ^{-1}(a x)\right )}{2 a^2 c \sqrt {c+a^2 c x^2}}\\ &=\frac {3 x \sqrt {\tan ^{-1}(a x)}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \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^2 c \sqrt {c+a^2 c x^2}}\\ &=\frac {3 x \sqrt {\tan ^{-1}(a x)}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \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^2 c \sqrt {c+a^2 c x^2}}\\ &=\frac {3 x \sqrt {\tan ^{-1}(a x)}}{2 a c \sqrt {c+a^2 c x^2}}-\frac {\tan ^{-1}(a x)^{3/2}}{a^2 c \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^2 c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.18, size = 128, normalized size = 0.99 \[ \frac {3 \sqrt {a^2 x^2+1} \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-i \tan ^{-1}(a x)\right )+3 \sqrt {a^2 x^2+1} \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},i \tan ^{-1}(a x)\right )+4 \left (3 a x-2 \tan ^{-1}(a x)\right ) \tan ^{-1}(a x)}{8 a^2 c \sqrt {a^2 c x^2+c} \sqrt {\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*(3*a*x - 2*ArcTan[a*x])*ArcTan[a*x] + 3*Sqrt[1 + a^2*x^2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x
]] + 3*Sqrt[1 + a^2*x^2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, I*ArcTan[a*x]])/(8*a^2*c*Sqrt[c + a^2*c*x^2]*Sqrt[ArcT
an[a*x]])

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

[Out]

sage0*x

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maple [F]  time = 3.10, size = 0, normalized size = 0.00 \[ \int \frac {x \arctan \left (a x \right )^{\frac {3}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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