3.1096 \(\int \frac {x}{(c+a^2 c x^2)^{3/2} \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=129 \[ -\frac {4 \sqrt {2 \pi } \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{3 a^2 c \sqrt {a^2 c x^2+c}}-\frac {2 x}{3 a c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 c \sqrt {a^2 c x^2+c} \sqrt {\tan ^{-1}(a x)}} \]

[Out]

-2/3*x/a/c/arctan(a*x)^(3/2)/(a^2*c*x^2+c)^(1/2)-4/3*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)-4/3/a^2/c/(a^2*c*x^2+c)^(1/2)/arctan(a*x)^(1/2)

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Rubi [A]  time = 0.30, 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 = {4942, 4902, 4971, 4970, 3305, 3351} \[ -\frac {4 \sqrt {2 \pi } \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{3 a^2 c \sqrt {a^2 c x^2+c}}-\frac {2 x}{3 a c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^{3/2}}-\frac {4}{3 a^2 c \sqrt {a^2 c x^2+c} \sqrt {\tan ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/(3*a*c*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^(3/2)) - 4/(3*a^2*c*Sqrt[c + a^2*c*x^2]*Sqrt[ArcTan[a*x]]) - (4*
Sqrt[2*Pi]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*a^2*c*Sqrt[c + a^2*c*x^2])

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 4902

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

Rule 4942

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]

Rule 4970

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])

Rule 4971

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q +
1/2)*Sqrt[1 + c^2*x^2])/Sqrt[d + e*x^2], Int[x^m*(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] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(IntegerQ[q] || GtQ[d, 0])

Rubi steps

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

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Mathematica [C]  time = 0.19, size = 124, normalized size = 0.96 \[ -\frac {2 \left (-i \sqrt {a^2 x^2+1} \left (-i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-i \tan ^{-1}(a x)\right )+i \sqrt {a^2 x^2+1} \left (i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},i \tan ^{-1}(a x)\right )+a x+2 \tan ^{-1}(a x)\right )}{3 a^2 c \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 3.12, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}} \arctan \left (a x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(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/(a^2*c*x^2+c)^(3/2)/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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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