3.3.41 \(\int \frac {x^4 \arctan (a x)}{(c+a^2 c x^2)^{5/2}} \, dx\) [241]

3.3.41.1 Optimal result

Integrand size = 22, antiderivative size = 308 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {1}{9 a^5 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {4}{3 a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^3 \arctan (a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x \arctan (a x)}{a^4 c^2 \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a^5 c^2 \sqrt {c+a^2 c x^2}} \]

output

1/9/a^5/c/(a^2*c*x^2+c)^(3/2)-1/3*x^3*arctan(a*x)/a^2/c/(a^2*c*x^2+c)^(3/2 
)-4/3/a^5/c^2/(a^2*c*x^2+c)^(1/2)-x*arctan(a*x)/a^4/c^2/(a^2*c*x^2+c)^(1/2 
)-2*I*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2 
)/a^5/c^2/(a^2*c*x^2+c)^(1/2)+I*polylog(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/ 
2))*(a^2*x^2+1)^(1/2)/a^5/c^2/(a^2*c*x^2+c)^(1/2)-I*polylog(2,I*(1+I*a*x)^ 
(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^5/c^2/(a^2*c*x^2+c)^(1/2)
 

3.3.41.2 Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.57 \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c \left (1+a^2 x^2\right )} \left (-\frac {45}{\sqrt {1+a^2 x^2}}-\frac {45 a x \arctan (a x)}{\sqrt {1+a^2 x^2}}+\cos (3 \arctan (a x))+36 \arctan (a x) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+36 i \left (\operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-\operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )\right )+3 \arctan (a x) \sin (3 \arctan (a x))\right )}{36 a^5 c^3 \sqrt {1+a^2 x^2}} \]

input

Integrate[(x^4*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]
 

output

(Sqrt[c*(1 + a^2*x^2)]*(-45/Sqrt[1 + a^2*x^2] - (45*a*x*ArcTan[a*x])/Sqrt[ 
1 + a^2*x^2] + Cos[3*ArcTan[a*x]] + 36*ArcTan[a*x]*(Log[1 - I*E^(I*ArcTan[ 
a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + (36*I)*(PolyLog[2, (-I)*E^(I*ArcT 
an[a*x])] - PolyLog[2, I*E^(I*ArcTan[a*x])]) + 3*ArcTan[a*x]*Sin[3*ArcTan[ 
a*x]]))/(36*a^5*c^3*Sqrt[1 + a^2*x^2])
 

3.3.41.3 Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5499, 5469, 5425, 5421, 5479, 243, 53, 2009}

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.

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

\(\Big \downarrow \) 5499

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

\(\Big \downarrow \) 5469

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

\(\Big \downarrow \) 5425

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

\(\Big \downarrow \) 5421

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

\(\Big \downarrow \) 5479

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

\(\Big \downarrow \) 243

\(\displaystyle -\frac {\frac {x^3 \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{6} a \int \frac {x^2}{\left (a^2 c x^2+c\right )^{5/2}}dx^2}{a^2}+\frac {\frac {\sqrt {a^2 x^2+1} \left (-\frac {2 i \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}\right )}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {x \arctan (a x)}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {1}{a^3 c \sqrt {a^2 c x^2+c}}}{a^2 c}\)

\(\Big \downarrow \) 53

\(\displaystyle -\frac {\frac {x^3 \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{6} a \int \left (\frac {1}{a^2 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{a^2 \left (a^2 c x^2+c\right )^{5/2}}\right )dx^2}{a^2}+\frac {\frac {\sqrt {a^2 x^2+1} \left (-\frac {2 i \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}\right )}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {x \arctan (a x)}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {1}{a^3 c \sqrt {a^2 c x^2+c}}}{a^2 c}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {x^3 \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {1}{6} a \left (\frac {2}{3 a^4 c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2}{a^4 c^2 \sqrt {a^2 c x^2+c}}\right )}{a^2}+\frac {\frac {\sqrt {a^2 x^2+1} \left (-\frac {2 i \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}\right )}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {x \arctan (a x)}{a^2 c \sqrt {a^2 c x^2+c}}-\frac {1}{a^3 c \sqrt {a^2 c x^2+c}}}{a^2 c}\)

input

Int[(x^4*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]
 

output

-((-1/6*(a*(2/(3*a^4*c*(c + a^2*c*x^2)^(3/2)) - 2/(a^4*c^2*Sqrt[c + a^2*c* 
x^2]))) + (x^3*ArcTan[a*x])/(3*c*(c + a^2*c*x^2)^(3/2)))/a^2) + (-(1/(a^3* 
c*Sqrt[c + a^2*c*x^2])) - (x*ArcTan[a*x])/(a^2*c*Sqrt[c + a^2*c*x^2]) + (S 
qrt[1 + a^2*x^2]*(((-2*I)*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a* 
x]])/a + (I*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/a - (I*Pol 
yLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/a))/(a^2*c*Sqrt[c + a^2*c*x^ 
2]))/(a^2*c)
 

3.3.41.3.1 Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5421

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

rule 5425

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

rule 5469

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*(x_)^2*((d_) + (e_.)*(x_)^2)^(q_), x 
_Symbol] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c^3*d*(q + 1)^2)), x] + (Simp 
[x*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*c^2*d*(q + 1))), x] - Simp[1 
/(2*c^2*d*(q + 1))   Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x]) / 
; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -5/2]
 

rule 5479

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

rule 5499

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

3.3.41.4 Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\left (i+3 \arctan \left (a x \right )\right ) \left (a^{3} x^{3}-3 i a^{2} x^{2}-3 a x +i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} a^{5} c^{3}}-\frac {5 \left (\arctan \left (a x \right )+i\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 c^{3} a^{5} \left (a^{2} x^{2}+1\right )}-\frac {5 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )-i\right )}{8 c^{3} a^{5} \left (a^{2} x^{2}+1\right )}-\frac {\left (-i+3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a^{3} x^{3}+3 i a^{2} x^{2}-3 a x -i\right )}{72 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) a^{5} c^{3}}-\frac {\left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{3} a^{5}}\)	\(389\)

input

int(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

output

-1/72*(I+3*arctan(a*x))*(a^3*x^3-3*I*a^2*x^2-3*a*x+I)*(c*(a*x-I)*(I+a*x))^ 
(1/2)/(a^2*x^2+1)^2/a^5/c^3-5/8*(arctan(a*x)+I)*(a*x-I)*(c*(a*x-I)*(I+a*x) 
)^(1/2)/c^3/a^5/(a^2*x^2+1)-5/8*(c*(a*x-I)*(I+a*x))^(1/2)*(I+a*x)*(arctan( 
a*x)-I)/c^3/a^5/(a^2*x^2+1)-1/72*(-I+3*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1 
/2)*(a^3*x^3+3*I*a^2*x^2-3*a*x-I)/(a^4*x^4+2*a^2*x^2+1)/a^5/c^3-(arctan(a* 
x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x 
^2+1)^(1/2))-I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+I*dilog(1-I*(1+I*a*x 
)/(a^2*x^2+1)^(1/2)))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/c^3/a^5
 

3.3.41.5 Fricas [F]

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

input

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
 

output

integral(sqrt(a^2*c*x^2 + c)*x^4*arctan(a*x)/(a^6*c^3*x^6 + 3*a^4*c^3*x^4 
+ 3*a^2*c^3*x^2 + c^3), x)
 

3.3.41.6 Sympy [F]

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

input

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

output

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

3.3.41.7 Maxima [F]

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

input

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
 

output

integrate(x^4*arctan(a*x)/(a^2*c*x^2 + c)^(5/2), x)
 

3.3.41.8 Giac [F]

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

input

integrate(x^4*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")
 

output

sage0*x
 

3.3.41.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \arctan (a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

input

int((x^4*atan(a*x))/(c + a^2*c*x^2)^(5/2),x)
 

output

int((x^4*atan(a*x))/(c + a^2*c*x^2)^(5/2), x)