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

Optimal. Leaf size=155 \[ \frac {15 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{4 c \sqrt {a^2 c x^2+c}} \]

[Out]

5/2*arctan(a*x)^(3/2)/a/c/(a^2*c*x^2+c)^(1/2)+x*arctan(a*x)^(5/2)/c/(a^2*c*x^2+c)^(1/2)+15/8*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/c/(a^2*c*x^2+c)^(1/2)-15/4*x*arctan(a*x)^(1/2
)/c/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4898, 4905, 4904, 3296, 3305, 3351} \[ \frac {15 \sqrt {\frac {\pi }{2}} \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a c \sqrt {a^2 c x^2+c}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {a^2 c x^2+c}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {a^2 c x^2+c}}-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{4 c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*x*Sqrt[ArcTan[a*x]])/(4*c*Sqrt[c + a^2*c*x^2]) + (5*ArcTan[a*x]^(3/2))/(2*a*c*Sqrt[c + a^2*c*x^2]) + (x*A
rcTan[a*x]^(5/2))/(c*Sqrt[c + a^2*c*x^2]) + (15*Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcTan[a
*x]]])/(4*a*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 4898

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(b*p*(a + b*ArcTan[
c*x])^(p - 1))/(c*d*Sqrt[d + e*x^2]), x] + (-Dist[b^2*p*(p - 1), Int[(a + b*ArcTan[c*x])^(p - 2)/(d + e*x^2)^(
3/2), x], x] + Simp[(x*(a + b*ArcTan[c*x])^p)/(d*Sqrt[d + e*x^2]), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e,
c^2*d] && GtQ[p, 1]

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])

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}-\frac {15}{4} \int \frac {\sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}-\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \int \frac {\sqrt {\tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{4 c \sqrt {c+a^2 c x^2}}\\ &=\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}-\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \sqrt {x} \cos (x) \, dx,x,\tan ^{-1}(a x)\right )}{4 a c \sqrt {c+a^2 c x^2}}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{4 c \sqrt {c+a^2 c x^2}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}+\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a c \sqrt {c+a^2 c x^2}}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{4 c \sqrt {c+a^2 c x^2}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}+\frac {\left (15 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{4 a c \sqrt {c+a^2 c x^2}}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{4 c \sqrt {c+a^2 c x^2}}+\frac {5 \tan ^{-1}(a x)^{3/2}}{2 a c \sqrt {c+a^2 c x^2}}+\frac {x \tan ^{-1}(a x)^{5/2}}{c \sqrt {c+a^2 c x^2}}+\frac {15 \sqrt {\frac {\pi }{2}} \sqrt {1+a^2 x^2} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{4 a c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 97, normalized size = 0.63 \[ \frac {15 \sqrt {2 \pi } \sqrt {a^2 x^2+1} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )+2 \sqrt {\tan ^{-1}(a x)} \left (-15 a x+4 a x \tan ^{-1}(a x)^2+10 \tan ^{-1}(a x)\right )}{8 a c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[ArcTan[a*x]]*(-15*a*x + 10*ArcTan[a*x] + 4*a*x*ArcTan[a*x]^2) + 15*Sqrt[2*Pi]*Sqrt[1 + a^2*x^2]*Fresne
lS[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(8*a*c*Sqrt[c + a^2*c*x^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)^(5/2)/(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(arctan(a*x)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [F]  time = 1.36, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x \right )^{\frac {5}{2}}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(arctan(a*x)^(5/2)/(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(arctan(a*x)^(5/2)/(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 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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