3.1.69 \(\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx\) [69]

3.1.69.1 Optimal result

Integrand size = 35, antiderivative size = 35 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\text {Int}\left (\frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+(a+b x)^2}},x\right ) \]

output

Unintegrable((b*x+a)^2*arctan(b*x+a)/(1+(b*x+a)^2)^(1/3),x)
 

3.1.69.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(181\) vs. \(2(30)=60\).

Time = 4.75 (sec) , antiderivative size = 181, normalized size of antiderivative = 5.17 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {3 \left (1+(a+b x)^2\right )^{2/3} \left (\frac {5 \sqrt [3]{2} \sqrt {\pi } \operatorname {Gamma}\left (\frac {5}{3}\right ) \, _3F_2\left (1,\frac {4}{3},\frac {4}{3};\frac {11}{6},\frac {7}{3};\frac {1}{1+(a+b x)^2}\right )}{\left (1+(a+b x)^2\right )^2}+\operatorname {Gamma}\left (\frac {11}{6}\right ) \operatorname {Gamma}\left (\frac {7}{3}\right ) \left (15+\frac {90}{1+(a+b x)^2}+\frac {24 (a+b x) \arctan (a+b x) \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {11}{6},\frac {1}{1+(a+b x)^2}\right )}{\left (1+(a+b x)^2\right )^2}+5 \arctan (a+b x) (-4 (a+b x)+6 \sin (2 \arctan (a+b x)))\right )\right )}{140 b \operatorname {Gamma}\left (\frac {11}{6}\right ) \operatorname {Gamma}\left (\frac {7}{3}\right )} \]

input

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

output

(-3*(1 + (a + b*x)^2)^(2/3)*((5*2^(1/3)*Sqrt[Pi]*Gamma[5/3]*Hypergeometric 
PFQ[{1, 4/3, 4/3}, {11/6, 7/3}, (1 + (a + b*x)^2)^(-1)])/(1 + (a + b*x)^2) 
^2 + Gamma[11/6]*Gamma[7/3]*(15 + 90/(1 + (a + b*x)^2) + (24*(a + b*x)*Arc 
Tan[a + b*x]*Hypergeometric2F1[1, 4/3, 11/6, (1 + (a + b*x)^2)^(-1)])/(1 + 
 (a + b*x)^2)^2 + 5*ArcTan[a + b*x]*(-4*(a + b*x) + 6*Sin[2*ArcTan[a + b*x 
]]))))/(140*b*Gamma[11/6]*Gamma[7/3])
 

3.1.69.3 Rubi [N/A]

Not integrable

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {5580, 5560}

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.

\(\displaystyle \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{a^2+2 a b x+b^2 x^2+1}} \, dx\)

\(\Big \downarrow \) 5580

\(\displaystyle \frac {\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{(a+b x)^2+1}}d(a+b x)}{b}\)

\(\Big \downarrow \) 5560

\(\displaystyle \frac {\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{(a+b x)^2+1}}d(a+b x)}{b}\)

input

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

output

$Aborted
 

3.1.69.3.1 Defintions of rubi rules used

rule 5560

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Unintegrab 
le[u*(a + b*ArcTan[c*x])^p, x] /; FreeQ[{a, b, c, p}, x] && (EqQ[u, 1] || M 
atchQ[u, ((d_.) + (e_.)*x)^(q_.) /; FreeQ[{d, e, q}, x]] || MatchQ[u, ((f_. 
)*x)^(m_.)*((d_.) + (e_.)*x)^(q_.) /; FreeQ[{d, e, f, m, q}, x]] || MatchQ[ 
u, ((d_.) + (e_.)*x^2)^(q_.) /; FreeQ[{d, e, q}, x]] || MatchQ[u, ((f_.)*x) 
^(m_.)*((d_.) + (e_.)*x^2)^(q_.) /; FreeQ[{d, e, f, m, q}, x]])
 

rule 5580

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/d   Subs 
t[Int[((d*e - c*f)/d + f*(x/d))^m*(C/d^2 + (C/d^2)*x^2)^q*(a + b*ArcTan[x]) 
^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, p, q}, x] & 
& EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]
 

3.1.69.4 Maple [N/A] (verified)

Not integrable

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94

\[\int \frac {\left (b x +a \right )^{2} \arctan \left (b x +a \right )}{\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {1}{3}}}d x\]

input

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

output

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

3.1.69.5 Fricas [N/A]

Not integrable

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]

input

integrate((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x, algorit 
hm="fricas")
 

output

integral((b^2*x^2 + 2*a*b*x + a^2)*arctan(b*x + a)/(b^2*x^2 + 2*a*b*x + a^ 
2 + 1)^(1/3), x)
 

3.1.69.6 Sympy [N/A]

Not integrable

Time = 3.72 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\left (a + b x\right )^{2} \operatorname {atan}{\left (a + b x \right )}}{\sqrt [3]{a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \]

input

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

output

Integral((a + b*x)**2*atan(a + b*x)/(a**2 + 2*a*b*x + b**2*x**2 + 1)**(1/3 
), x)
 

3.1.69.7 Maxima [N/A]

Not integrable

Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]

input

integrate((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x, algorit 
hm="maxima")
 

output

integrate((b*x + a)^2*arctan(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(1/3), 
 x)
 

3.1.69.8 Giac [N/A]

Not integrable

Time = 172.49 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.09 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {{\left (b x + a\right )}^{2} \arctan \left (b x + a\right )}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {1}{3}}} \,d x } \]

input

integrate((b*x+a)^2*arctan(b*x+a)/(b^2*x^2+2*a*b*x+a^2+1)^(1/3),x, algorit 
hm="giac")
 

output

sage0*x
 

3.1.69.9 Mupad [N/A]

Not integrable

Time = 0.59 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\mathrm {atan}\left (a+b\,x\right )\,{\left (a+b\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{1/3}} \,d x \]

input

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

output

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