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

Optimal result

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

[Out]

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

Rubi [N/A]

Not integrable

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {(a+b x)^2 \arctan (a+b x)}{\sqrt [3]{1+a^2+2 a b x+b^2 x^2}} \, dx \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2 \arctan (x)}{\sqrt [3]{1+x^2}} \, dx,x,a+b x\right )}{b} \\ \end{align*}

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

[In]

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

[Out]

(-3*(1 + (a + b*x)^2)^(2/3)*((5*2^(1/3)*Sqrt[Pi]*Gamma[5/3]*HypergeometricPFQ[{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)*A
rcTan[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])

Maple [N/A] (verified)

Not integrable

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

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

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

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

[In]

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

[Out]

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)

Sympy [N/A]

Not integrable

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

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

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

[In]

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

[Out]

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

Giac [N/A]

Not integrable

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

[In]

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

[Out]

sage0*x

Mupad [N/A]

Not integrable

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

[In]

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

[Out]

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