3.9.68 \(\int \frac {\text {ArcTan}(a x)^{5/2}}{(c+a^2 c x^2)^2} \, dx\) [868]

Optimal. Leaf size=151 \[ -\frac {15 x \sqrt {\text {ArcTan}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \text {ArcTan}(a x)^{3/2}}{16 a c^2}+\frac {5 \text {ArcTan}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \text {ArcTan}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\text {ArcTan}(a x)^{7/2}}{7 a c^2}+\frac {15 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{128 a c^2} \]

[Out]

-5/16*arctan(a*x)^(3/2)/a/c^2+5/8*arctan(a*x)^(3/2)/a/c^2/(a^2*x^2+1)+1/2*x*arctan(a*x)^(5/2)/c^2/(a^2*x^2+1)+
1/7*arctan(a*x)^(7/2)/a/c^2+15/128*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a/c^2-15/32*x*arctan(a*x)^(
1/2)/c^2/(a^2*x^2+1)

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Rubi [A]
time = 0.14, antiderivative size = 151, 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 = {5012, 5050, 5090, 4491, 12, 3386, 3432} \begin {gather*} \frac {x \text {ArcTan}(a x)^{5/2}}{2 c^2 \left (a^2 x^2+1\right )}+\frac {5 \text {ArcTan}(a x)^{3/2}}{8 a c^2 \left (a^2 x^2+1\right )}-\frac {15 x \sqrt {\text {ArcTan}(a x)}}{32 c^2 \left (a^2 x^2+1\right )}+\frac {15 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{128 a c^2}+\frac {\text {ArcTan}(a x)^{7/2}}{7 a c^2}-\frac {5 \text {ArcTan}(a x)^{3/2}}{16 a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*x*Sqrt[ArcTan[a*x]])/(32*c^2*(1 + a^2*x^2)) - (5*ArcTan[a*x]^(3/2))/(16*a*c^2) + (5*ArcTan[a*x]^(3/2))/(8
*a*c^2*(1 + a^2*x^2)) + (x*ArcTan[a*x]^(5/2))/(2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7*a*c^2) + (15*Sqrt[P
i]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(128*a*c^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3386

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 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 4491

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 5012

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

Rule 5050

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 5090

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 {\tan ^{-1}(a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx &=\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}-\frac {1}{4} (5 a) \int \frac {x \tan ^{-1}(a x)^{3/2}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}-\frac {15}{16} \int \frac {\sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac {1}{64} (15 a) \int \frac {x}{\left (c+a^2 c x^2\right )^2 \sqrt {\tan ^{-1}(a x)}} \, dx\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac {15 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a c^2}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac {15 \text {Subst}\left (\int \frac {\sin (2 x)}{2 \sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{64 a c^2}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac {15 \text {Subst}\left (\int \frac {\sin (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{128 a c^2}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac {15 \text {Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{64 a c^2}\\ &=-\frac {15 x \sqrt {\tan ^{-1}(a x)}}{32 c^2 \left (1+a^2 x^2\right )}-\frac {5 \tan ^{-1}(a x)^{3/2}}{16 a c^2}+\frac {5 \tan ^{-1}(a x)^{3/2}}{8 a c^2 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^{5/2}}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^{7/2}}{7 a c^2}+\frac {15 \sqrt {\pi } S\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{128 a c^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 108, normalized size = 0.72 \begin {gather*} \frac {4 \sqrt {\text {ArcTan}(a x)} \left (-105 a x-70 \left (-1+a^2 x^2\right ) \text {ArcTan}(a x)+112 a x \text {ArcTan}(a x)^2+32 \left (1+a^2 x^2\right ) \text {ArcTan}(a x)^3\right )+105 \sqrt {\pi } \left (1+a^2 x^2\right ) S\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{896 c^2 \left (a+a^3 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[ArcTan[a*x]]*(-105*a*x - 70*(-1 + a^2*x^2)*ArcTan[a*x] + 112*a*x*ArcTan[a*x]^2 + 32*(1 + a^2*x^2)*ArcT
an[a*x]^3) + 105*Sqrt[Pi]*(1 + a^2*x^2)*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(896*c^2*(a + a^3*x^2))

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Maple [A]
time = 0.32, size = 93, normalized size = 0.62

method	result	size
default	\(\frac {128 \arctan \left (a x \right )^{\frac {7}{2}} \sqrt {\pi }+224 \arctan \left (a x \right )^{\frac {5}{2}} \sqrt {\pi }\, \sin \left (2 \arctan \left (a x \right )\right )+280 \arctan \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \cos \left (2 \arctan \left (a x \right )\right )-210 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \sin \left (2 \arctan \left (a x \right )\right )+105 \pi \,\mathrm {S}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )}{896 c^{2} a \sqrt {\pi }}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^(5/2)/(a^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)

[Out]

1/896/c^2/a/Pi^(1/2)*(128*arctan(a*x)^(7/2)*Pi^(1/2)+224*arctan(a*x)^(5/2)*Pi^(1/2)*sin(2*arctan(a*x))+280*arc
tan(a*x)^(3/2)*Pi^(1/2)*cos(2*arctan(a*x))-210*arctan(a*x)^(1/2)*Pi^(1/2)*sin(2*arctan(a*x))+105*Pi*FresnelS(2
*arctan(a*x)^(1/2)/Pi^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^(5/2)/(a^2*c*x^2+c)^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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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^(5/2)/(a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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