3.1071 \(\int \frac {1}{(c+a^2 c x^2)^3 \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac {16 x}{3 c^3 \left (a^2 x^2+1\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac {4 \sqrt {2 \pi } C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{3 a c^3}-\frac {8 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a c^3} \]

[Out]

-2/3/a/c^3/(a^2*x^2+1)^2/arctan(a*x)^(3/2)-8/3*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a/c^3-4/3*Fresn
elC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a/c^3+16/3*x/c^3/(a^2*x^2+1)^2/arctan(a*x)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {4902, 4968, 4970, 4406, 3304, 3352, 4904, 3312} \[ \frac {16 x}{3 c^3 \left (a^2 x^2+1\right )^2 \sqrt {\tan ^{-1}(a x)}}-\frac {2}{3 a c^3 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^{3/2}}-\frac {4 \sqrt {2 \pi } \text {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{3 a c^3}-\frac {8 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{3 a c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a^2*c*x^2)^3*ArcTan[a*x]^(5/2)),x]

[Out]

-2/(3*a*c^3*(1 + a^2*x^2)^2*ArcTan[a*x]^(3/2)) + (16*x)/(3*c^3*(1 + a^2*x^2)^2*Sqrt[ArcTan[a*x]]) - (4*Sqrt[2*
Pi]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(3*a*c^3) - (8*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]
)/(3*a*c^3)

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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

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 4968

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(x^m*(d
+ e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^(p + 1))/(b*c*d*(p + 1)), x] + (-Dist[(c*(m + 2*q + 2))/(b*(p + 1)), Int[
x^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] - Dist[m/(b*c*(p + 1)), Int[x^(m - 1)*(d + e*x^2)^
q*(a + b*ArcTan[c*x])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[e, c^2*d] && IntegerQ[m] && LtQ[
q, -1] && LtQ[p, -1] && NeQ[m + 2*q + 2, 0]

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])

Rubi steps

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

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Mathematica [C]  time = 0.72, size = 186, normalized size = 1.49 \[ \frac {2 \left (-\frac {1}{a \left (a^2 x^2+1\right )^2}+\frac {8 x \tan ^{-1}(a x)}{\left (a^2 x^2+1\right )^2}+\frac {\sqrt {2} \tan ^{-1}(a x)^2 \Gamma \left (\frac {1}{2},2 i \tan ^{-1}(a x)\right )}{a \sqrt {i \tan ^{-1}(a x)}}+\frac {\tan ^{-1}(a x)^2 \Gamma \left (\frac {1}{2},4 i \tan ^{-1}(a x)\right )}{a \sqrt {i \tan ^{-1}(a x)}}-\frac {\sqrt {2} \left (-i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-2 i \tan ^{-1}(a x)\right )}{a}-\frac {\left (-i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-4 i \tan ^{-1}(a x)\right )}{a}\right )}{3 c^3 \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((c + a^2*c*x^2)^3*ArcTan[a*x]^(5/2)),x]

[Out]

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

[Out]

Timed out

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maple [A]  time = 0.70, size = 113, normalized size = 0.90 \[ \frac {-16 \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}-32 \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \arctan \left (a x \right )^{\frac {3}{2}}+16 \sin \left (2 \arctan \left (a x \right )\right ) \arctan \left (a x \right )+8 \sin \left (4 \arctan \left (a x \right )\right ) \arctan \left (a x \right )-4 \cos \left (2 \arctan \left (a x \right )\right )-\cos \left (4 \arctan \left (a x \right )\right )-3}{12 a \,c^{3} \arctan \left (a x \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2*c*x^2+c)^3/arctan(a*x)^(5/2),x)

[Out]

1/12/a/c^3*(-16*2^(1/2)*Pi^(1/2)*FresnelC(2*2^(1/2)/Pi^(1/2)*arctan(a*x)^(1/2))*arctan(a*x)^(3/2)-32*Pi^(1/2)*
FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+16*sin(2*arctan(a*x))*arctan(a*x)+8*sin(4*arctan(a*x)
)*arctan(a*x)-4*cos(2*arctan(a*x))-cos(4*arctan(a*x))-3)/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)^3/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 \[ \int \frac {1}{{\mathrm {atan}\left (a\,x\right )}^{5/2}\,{\left (c\,a^2\,x^2+c\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{a^{6} x^{6} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 3 a^{4} x^{4} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a**6*x**6*atan(a*x)**(5/2) + 3*a**4*x**4*atan(a*x)**(5/2) + 3*a**2*x**2*atan(a*x)**(5/2) + atan(a*
x)**(5/2)), x)/c**3

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