3.1106 \(\int \frac {1}{(c+a^2 c x^2)^{5/2} \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac {\sqrt {2 \pi } \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {6 \pi } \sqrt {a^2 x^2+1} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {4 x}{c \left (a^2 c x^2+c\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {2}{3 a c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.62, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4902, 4968, 4971, 4970, 4406, 3304, 3352, 4905, 4904, 3312} \[ -\frac {\sqrt {2 \pi } \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}-\frac {\sqrt {6 \pi } \sqrt {a^2 x^2+1} \text {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {4 x}{c \left (a^2 c x^2+c\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {2}{3 a c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(3*a*c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(3/2)) + (4*x)/(c*(c + a^2*c*x^2)^(3/2)*Sqrt[ArcTan[a*x]]) - (Sqrt
[2*Pi]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcTan[a*x]]])/(a*c^2*Sqrt[c + a^2*c*x^2]) - (Sqrt[6*Pi]*Sqr
t[1 + a^2*x^2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcTan[a*x]]])/(a*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 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 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 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])

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 {1}{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^{5/2}} \, dx &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}-(2 a) \int \frac {x}{\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-4 \int \frac {1}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx+\left (8 a^2\right ) \int \frac {x^2}{\left (c+a^2 c x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \int \frac {1}{\left (1+a^2 x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (8 a^2 \sqrt {1+a^2 x^2}\right ) \int \frac {x^2}{\left (1+a^2 x^2\right )^{5/2} \sqrt {\tan ^{-1}(a x)}} \, dx}{c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos ^3(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (8 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \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 )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (8 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{4 \sqrt {x}}-\frac {\cos (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {2}{3 a c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}}+\frac {4 x}{c \left (c+a^2 c x^2\right )^{3/2} \sqrt {\tan ^{-1}(a x)}}-\frac {\sqrt {2 \pi } \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {6 \pi } \sqrt {1+a^2 x^2} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\tan ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.53, size = 300, normalized size = 1.64 \[ \frac {-3 a^2 x^2 \sqrt {3 a^2 x^2+3} \left (-i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-3 i \tan ^{-1}(a x)\right )-3 a^2 x^2 \sqrt {3 a^2 x^2+3} \left (i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},3 i \tan ^{-1}(a x)\right )-3 \left (a^2 x^2+1\right )^{3/2} \left (-i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-i \tan ^{-1}(a x)\right )-3 \left (a^2 x^2+1\right )^{3/2} \left (i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},i \tan ^{-1}(a x)\right )-3 \sqrt {3 a^2 x^2+3} \left (-i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-3 i \tan ^{-1}(a x)\right )-3 \sqrt {3 a^2 x^2+3} \left (i \tan ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},3 i \tan ^{-1}(a x)\right )+24 a x \tan ^{-1}(a x)-4}{6 c^2 \left (a^3 x^2+a\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4 + 24*a*x*ArcTan[a*x] - 3*(1 + a^2*x^2)^(3/2)*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-I)*ArcTan[a*x]] - 3*(1
+ a^2*x^2)^(3/2)*(I*ArcTan[a*x])^(3/2)*Gamma[1/2, I*ArcTan[a*x]] - 3*Sqrt[3 + 3*a^2*x^2]*((-I)*ArcTan[a*x])^(3
/2)*Gamma[1/2, (-3*I)*ArcTan[a*x]] - 3*a^2*x^2*Sqrt[3 + 3*a^2*x^2]*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-3*I)*
ArcTan[a*x]] - 3*Sqrt[3 + 3*a^2*x^2]*(I*ArcTan[a*x])^(3/2)*Gamma[1/2, (3*I)*ArcTan[a*x]] - 3*a^2*x^2*Sqrt[3 +
3*a^2*x^2]*(I*ArcTan[a*x])^(3/2)*Gamma[1/2, (3*I)*ArcTan[a*x]])/(6*c^2*(a + a^3*x^2)*Sqrt[c + a^2*c*x^2]*ArcTa
n[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)^(5/2)/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]  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(1/(a^2*c*x^2+c)^(5/2)/arctan(a*x)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [F]  time = 1.43, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a^{2} c \,x^{2}+c \right )^{\frac {5}{2}} \arctan \left (a x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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