3.753 \(\int \frac {x \sqrt {\tan ^{-1}(a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=163 \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

1/72*FresnelC(6^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*6^(1/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)/a^2/c^2/(a^2*c*x^2+c)^(1/
2)+1/8*FresnelC(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^2/(a^2*c*x^2+c)^(
1/2)-1/3*arctan(a*x)^(1/2)/a^2/c/(a^2*c*x^2+c)^(3/2)

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Rubi [A]  time = 0.25, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4930, 4905, 4904, 3312, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[ArcTan[a*x]]/(3*a^2*c*(c + a^2*c*x^2)^(3/2)) + (Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[Ar
cTan[a*x]]])/(4*a^2*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[Pi/6]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a
*x]]])/(12*a^2*c^2*Sqrt[c + a^2*c*x^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 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 \sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx}{6 a}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\sqrt {1+a^2 x^2} \int \frac {1}{\left (1+a^2 x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx}{6 a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\cos ^3(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{6 a^2 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \left (\frac {3 \cos (x)}{4 \sqrt {x}}+\frac {\cos (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{6 a^2 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{24 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a^2 c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {\frac {\pi }{6}} \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{12 a^2 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.44, size = 167, normalized size = 1.02 \[ \frac {-48 \tan ^{-1}(a x)-i \left (a^2 x^2+1\right )^{3/2} \left (9 \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-i \tan ^{-1}(a x)\right )-9 \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},i \tan ^{-1}(a x)\right )+\sqrt {3} \left (\sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 i \tan ^{-1}(a x)\right )-\sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 i \tan ^{-1}(a x)\right )\right )\right )}{144 a^2 c \left (a^2 c x^2+c\right )^{3/2} \sqrt {\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-48*ArcTan[a*x] - I*(1 + a^2*x^2)^(3/2)*(9*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-I)*ArcTan[a*x]] - 9*Sqrt[I*Arc
Tan[a*x]]*Gamma[1/2, I*ArcTan[a*x]] + Sqrt[3]*(Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-3*I)*ArcTan[a*x]] - Sqrt[I*
ArcTan[a*x]]*Gamma[1/2, (3*I)*ArcTan[a*x]])))/(144*a^2*c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[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)^(1/2)/(a^2*c*x^2+c)^(5/2),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)^(1/2)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [F]  time = 3.57, size = 0, normalized size = 0.00 \[ \int \frac {x \sqrt {\arctan \left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(5/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*atan(a*x)^(1/2))/(c + a^2*c*x^2)^(5/2),x)

[Out]

int((x*atan(a*x)^(1/2))/(c + a^2*c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sqrt {\operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)**(1/2)/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(x*sqrt(atan(a*x))/(c*(a**2*x**2 + 1))**(5/2), x)

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