3.9.66 \(\int \frac {x^2 \arctan (a x)^{5/2}}{(c+a^2 c x^2)^2} \, dx\) [866]

3.9.66.1 Optimal result

Integrand size = 24, antiderivative size = 157 \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {15 x \sqrt {\arctan (a x)}}{32 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {5 \arctan (a x)^{3/2}}{16 a^3 c^2}-\frac {5 \arctan (a x)^{3/2}}{8 a^3 c^2 \left (1+a^2 x^2\right )}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {15 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{128 a^3 c^2} \]

output

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

3.9.66.2 Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.71 \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {4 \sqrt {\arctan (a x)} \left (105 a x+70 \left (-1+a^2 x^2\right ) \arctan (a x)-112 a x \arctan (a x)^2+32 \left (1+a^2 x^2\right ) \arctan (a x)^3\right )-105 \sqrt {\pi } \left (1+a^2 x^2\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{896 a^3 c^2 \left (1+a^2 x^2\right )} \]

input

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

output

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

3.9.66.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5471, 27, 5465, 5427, 5505, 4906, 27, 3042, 3786, 3832}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2 \arctan (a x)^{5/2}}{\left (a^2 c x^2+c\right )^2} \, dx\)

\(\Big \downarrow \) 5471

\(\displaystyle \frac {5 \int \frac {x \arctan (a x)^{3/2}}{c^2 \left (a^2 x^2+1\right )^2}dx}{4 a}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \int \frac {x \arctan (a x)^{3/2}}{\left (a^2 x^2+1\right )^2}dx}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 5465

\(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {\arctan (a x)}}{\left (a^2 x^2+1\right )^2}dx}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 5427

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {1}{4} a \int \frac {x}{\left (a^2 x^2+1\right )^2 \sqrt {\arctan (a x)}}dx+\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 5505

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {a x}{\left (a^2 x^2+1\right ) \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a}+\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {\sin (2 \arctan (a x))}{2 \sqrt {\arctan (a x)}}d\arctan (a x)}{4 a}+\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {\sin (2 \arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)}{8 a}+\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {\sin (2 \arctan (a x))}{\sqrt {\arctan (a x)}}d\arctan (a x)}{8 a}+\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 3786

\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \sin (2 \arctan (a x))d\sqrt {\arctan (a x)}}{4 a}+\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}+\frac {\arctan (a x)^{7/2}}{7 a^3 c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

\(\Big \downarrow \) 3832

\(\displaystyle \frac {\arctan (a x)^{7/2}}{7 a^3 c^2}+\frac {5 \left (\frac {3 \left (\frac {x \sqrt {\arctan (a x)}}{2 \left (a^2 x^2+1\right )}-\frac {\sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arctan (a x)}}{\sqrt {\pi }}\right )}{8 a}+\frac {\arctan (a x)^{3/2}}{3 a}\right )}{4 a}-\frac {\arctan (a x)^{3/2}}{2 a^2 \left (a^2 x^2+1\right )}\right )}{4 a c^2}-\frac {x \arctan (a x)^{5/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}\)

input

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

output

-1/2*(x*ArcTan[a*x]^(5/2))/(a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(7/2)/(7* 
a^3*c^2) + (5*(-1/2*ArcTan[a*x]^(3/2)/(a^2*(1 + a^2*x^2)) + (3*((x*Sqrt[Ar 
cTan[a*x]])/(2*(1 + a^2*x^2)) + ArcTan[a*x]^(3/2)/(3*a) - (Sqrt[Pi]*Fresne 
lS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(8*a)))/(4*a)))/(4*a*c^2)
 

3.9.66.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3786

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[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 3832

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 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

rule 5427

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

rule 5465

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] - Simp[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 5471

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2) 
^2, x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] 
 + (-Simp[x*((a + b*ArcTan[c*x])^p/(2*c^2*d*(d + e*x^2))), x] + Simp[b*(p/( 
2*c))   Int[x*((a + b*ArcTan[c*x])^(p - 1)/(d + e*x^2)^2), x], x]) /; FreeQ 
[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
 

rule 5505

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(q_), x_Symbol] :> Simp[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])
 

3.9.66.4 Maple [A] (verified)

Time = 25.51 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.59

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

input

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

output

1/896/c^2/a^3*(128*arctan(a*x)^(7/2)*Pi^(1/2)-224*arctan(a*x)^(5/2)*Pi^(1/ 
2)*sin(2*arctan(a*x))-280*arctan(a*x)^(3/2)*Pi^(1/2)*cos(2*arctan(a*x))+21 
0*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)))/Pi^(1/2)
 

3.9.66.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.9.66.6 Sympy [F]

\[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\frac {\int \frac {x^{2} \operatorname {atan}^{\frac {5}{2}}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

input

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

output

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

3.9.66.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\text {Exception raised: RuntimeError} \]

input

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

output

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

3.9.66.8 Giac [F]

\[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\int { \frac {x^{2} \arctan \left (a x\right )^{\frac {5}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{2}} \,d x } \]

input

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

output

sage0*x
 

3.9.66.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \arctan (a x)^{5/2}}{\left (c+a^2 c x^2\right )^2} \, dx=\int \frac {x^2\,{\mathrm {atan}\left (a\,x\right )}^{5/2}}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

input

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

output

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