3.1062 \(\int \frac {1}{(c+a^2 c x^2)^2 \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ \frac {8 x}{3 c^2 \left (a^2 x^2+1\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (a^2 x^2+1\right )}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (a^2 x^2+1\right )}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a c^2}-\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a c^2} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4902, 4932, 4930, 4904, 3312, 3304, 3352} \[ \frac {8 x}{3 c^2 \left (a^2 x^2+1\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (a^2 x^2+1\right )}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (a^2 x^2+1\right )}-\frac {2}{3 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a c^2}-\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^(5/2)),x]

[Out]

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

Rule 4932

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTan
[c*x])^(p + 1))/(b*c*d*(p + 1)*(d + e*x^2)), x] + (-Dist[4/(b^2*(p + 1)*(p + 2)), Int[(x*(a + b*ArcTan[c*x])^(
p + 2))/(d + e*x^2)^2, x], x] - Simp[((1 - c^2*x^2)*(a + b*ArcTan[c*x])^(p + 2))/(b^2*e*(p + 1)*(p + 2)*(d + e
*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -2]

Rubi steps

\begin {align*} \int \frac {1}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{5/2}} \, dx &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac {1}{3} (4 a) \int \frac {x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {1}{3} (64 a) \int \frac {x \sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16}{3} \int \frac {1}{\left (c+a^2 c x^2\right )^2 \sqrt {\tan ^{-1}(a x)}} \, dx\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \operatorname {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a c^2}\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \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 )}{3 a c^2}\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a c^2}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {8 \operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a c^2}\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a c^2}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \operatorname {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{3 a c^2}\\ &=-\frac {2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac {8 x}{3 c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}-\frac {16 \sqrt {\tan ^{-1}(a x)}}{3 a c^2}+\frac {32 \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {16 \left (1-a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}{3 a c^2 \left (1+a^2 x^2\right )}-\frac {8 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a c^2}\\ \end {align*}

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Mathematica [C]  time = 0.44, size = 170, normalized size = 0.98 \[ \frac {-4 \sqrt {\pi } \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2} C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )+\frac {\sqrt {2} \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2 \Gamma \left (\frac {1}{2},2 i \tan ^{-1}(a x)\right )}{\sqrt {i \tan ^{-1}(a x)}}+\sqrt {2} \left (a^2 x^2+1\right ) \sqrt {i \tan ^{-1}(a x)} \sqrt {\tan ^{-1}(a x)^2} \Gamma \left (\frac {1}{2},-2 i \tan ^{-1}(a x)\right )+8 a x \tan ^{-1}(a x)-2}{3 c^2 \left (a^3 x^2+a\right ) \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((c + a^2*c*x^2)^2*ArcTan[a*x]^(5/2)),x]

[Out]

(-2 + 8*a*x*ArcTan[a*x] - 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]
+ Sqrt[2]*(1 + a^2*x^2)*Sqrt[I*ArcTan[a*x]]*Sqrt[ArcTan[a*x]^2]*Gamma[1/2, (-2*I)*ArcTan[a*x]] + (Sqrt[2]*(1 +
a^2*x^2)*ArcTan[a*x]^2*Gamma[1/2, (2*I)*ArcTan[a*x]])/Sqrt[I*ArcTan[a*x]])/(3*c^2*(a + a^3*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(1/(a^2*c*x^2+c)^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(-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(1/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.47, size = 62, normalized size = 0.36 \[ \frac {-8 \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+4 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-\cos \left (2 \arctan \left (a x \right )\right )-1}{3 a \,c^{2} \arctan \left (a x \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a/c^2*(-8*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+4*sin(2*arctan(a*x))*arctan(a*
x)-cos(2*arctan(a*x))-1)/arctan(a*x)^(3/2)

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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(1/(a^2*c*x^2+c)^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 \[ \int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a**4*x**4*atan(a*x)**(5/2) + 2*a**2*x**2*atan(a*x)**(5/2) + atan(a*x)**(5/2)), x)/c**2

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