3.19.15 \(\int x^2 \sqrt [4]{b x^3+a x^4} \, dx\) [1815]

3.19.15.1 Optimal result
3.19.15.2 Mathematica [A] (verified)
3.19.15.3 Rubi [A] (verified)
3.19.15.4 Maple [A] (verified)
3.19.15.5 Fricas [C] (verification not implemented)
3.19.15.6 Sympy [F]
3.19.15.7 Maxima [F]
3.19.15.8 Giac [B] (verification not implemented)
3.19.15.9 Mupad [F(-1)]

3.19.15.1 Optimal result

Integrand size = 19, antiderivative size = 123 \[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=\frac {\left (77 b^3-44 a b^2 x+32 a^2 b x^2+384 a^3 x^3\right ) \sqrt [4]{b x^3+a x^4}}{1536 a^3}+\frac {77 b^4 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{1024 a^{15/4}}-\frac {77 b^4 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^3+a x^4}}\right )}{1024 a^{15/4}} \]

output
1/1536*(384*a^3*x^3+32*a^2*b*x^2-44*a*b^2*x+77*b^3)*(a*x^4+b*x^3)^(1/4)/a^ 
3+77/1024*b^4*arctan(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(15/4)-77/1024*b^4*a 
rctanh(a^(1/4)*x/(a*x^4+b*x^3)^(1/4))/a^(15/4)
 
3.19.15.2 Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.14 \[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=\frac {x^{9/4} (b+a x)^{3/4} \left (2 a^{3/4} x^{3/4} \sqrt [4]{b+a x} \left (77 b^3-44 a b^2 x+32 a^2 b x^2+384 a^3 x^3\right )+231 b^4 \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )-231 b^4 \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{b+a x}}\right )\right )}{3072 a^{15/4} \left (x^3 (b+a x)\right )^{3/4}} \]

input
Integrate[x^2*(b*x^3 + a*x^4)^(1/4),x]
 
output
(x^(9/4)*(b + a*x)^(3/4)*(2*a^(3/4)*x^(3/4)*(b + a*x)^(1/4)*(77*b^3 - 44*a 
*b^2*x + 32*a^2*b*x^2 + 384*a^3*x^3) + 231*b^4*ArcTan[(a^(1/4)*x^(1/4))/(b 
 + a*x)^(1/4)] - 231*b^4*ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]))/(307 
2*a^(15/4)*(x^3*(b + a*x))^(3/4))
 
3.19.15.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.71, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1927, 1930, 1930, 1930, 1938, 73, 854, 827, 216, 219}

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.

\(\displaystyle \int x^2 \sqrt [4]{a x^4+b x^3} \, dx\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {1}{16} b \int \frac {x^5}{\left (a x^4+b x^3\right )^{3/4}}dx+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \int \frac {x^4}{\left (a x^4+b x^3\right )^{3/4}}dx}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \int \frac {x^3}{\left (a x^4+b x^3\right )^{3/4}}dx}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b \int \frac {x^2}{\left (a x^4+b x^3\right )^{3/4}}dx}{4 a}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \int \frac {1}{\sqrt [4]{x} (b+a x)^{3/4}}dx}{4 a \left (a x^4+b x^3\right )^{3/4}}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \int \frac {\sqrt {x}}{(b+a x)^{3/4}}d\sqrt [4]{x}}{a \left (a x^4+b x^3\right )^{3/4}}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \int \frac {\sqrt {x}}{1-a x}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{a \left (a x^4+b x^3\right )^{3/4}}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}\right )}{a \left (a x^4+b x^3\right )^{3/4}}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{a \left (a x^4+b x^3\right )^{3/4}}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{16} b \left (\frac {x^2 \sqrt [4]{a x^4+b x^3}}{3 a}-\frac {11 b \left (\frac {x \sqrt [4]{a x^4+b x^3}}{2 a}-\frac {7 b \left (\frac {\sqrt [4]{a x^4+b x^3}}{a}-\frac {3 b x^{9/4} (a x+b)^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{a \left (a x^4+b x^3\right )^{3/4}}\right )}{8 a}\right )}{12 a}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b x^3}\)

input
Int[x^2*(b*x^3 + a*x^4)^(1/4),x]
 
output
(x^3*(b*x^3 + a*x^4)^(1/4))/4 + (b*((x^2*(b*x^3 + a*x^4)^(1/4))/(3*a) - (1 
1*b*((x*(b*x^3 + a*x^4)^(1/4))/(2*a) - (7*b*((b*x^3 + a*x^4)^(1/4)/a - (3* 
b*x^(9/4)*(b + a*x)^(3/4)*(-1/2*ArcTan[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]/ 
a^(3/4) + ArcTanh[(a^(1/4)*x^(1/4))/(b + a*x)^(1/4)]/(2*a^(3/4))))/(a*(b*x 
^3 + a*x^4)^(3/4))))/(8*a)))/(12*a)))/16
 

3.19.15.3.1 Defintions of rubi rules used

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 827
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 854
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

rule 1927
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1930
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

rule 1938
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 
3.19.15.4 Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.24

method result size
pseudoelliptic \(\frac {1536 \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}} a^{\frac {15}{4}} x^{3}+128 a^{\frac {11}{4}} b \,x^{2} \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}-176 a^{\frac {7}{4}} b^{2} x \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}+308 b^{3} \left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}} a^{\frac {3}{4}}-231 \ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}\right ) b^{4}-462 \arctan \left (\frac {\left (x^{3} \left (a x +b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{4}}{6144 a^{\frac {15}{4}}}\) \(153\)

input
int(x^2*(a*x^4+b*x^3)^(1/4),x,method=_RETURNVERBOSE)
 
output
1/6144*(1536*(x^3*(a*x+b))^(1/4)*a^(15/4)*x^3+128*a^(11/4)*b*x^2*(x^3*(a*x 
+b))^(1/4)-176*a^(7/4)*b^2*x*(x^3*(a*x+b))^(1/4)+308*b^3*(x^3*(a*x+b))^(1/ 
4)*a^(3/4)-231*ln((a^(1/4)*x+(x^3*(a*x+b))^(1/4))/(-a^(1/4)*x+(x^3*(a*x+b) 
)^(1/4)))*b^4-462*arctan(1/a^(1/4)/x*(x^3*(a*x+b))^(1/4))*b^4)/a^(15/4)
 
3.19.15.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.15 \[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=-\frac {231 \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \log \left (\frac {77 \, {\left (a^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{4}\right )}}{x}\right ) - 231 \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-\frac {77 \, {\left (a^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{4}\right )}}{x}\right ) - 231 i \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-\frac {77 \, {\left (i \, a^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{4}\right )}}{x}\right ) + 231 i \, a^{3} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} \log \left (-\frac {77 \, {\left (-i \, a^{4} \left (\frac {b^{16}}{a^{15}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} b^{4}\right )}}{x}\right ) - 4 \, {\left (384 \, a^{3} x^{3} + 32 \, a^{2} b x^{2} - 44 \, a b^{2} x + 77 \, b^{3}\right )} {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}}}{6144 \, a^{3}} \]

input
integrate(x^2*(a*x^4+b*x^3)^(1/4),x, algorithm="fricas")
 
output
-1/6144*(231*a^3*(b^16/a^15)^(1/4)*log(77*(a^4*(b^16/a^15)^(1/4)*x + (a*x^ 
4 + b*x^3)^(1/4)*b^4)/x) - 231*a^3*(b^16/a^15)^(1/4)*log(-77*(a^4*(b^16/a^ 
15)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^4)/x) - 231*I*a^3*(b^16/a^15)^(1/4)* 
log(-77*(I*a^4*(b^16/a^15)^(1/4)*x - (a*x^4 + b*x^3)^(1/4)*b^4)/x) + 231*I 
*a^3*(b^16/a^15)^(1/4)*log(-77*(-I*a^4*(b^16/a^15)^(1/4)*x - (a*x^4 + b*x^ 
3)^(1/4)*b^4)/x) - 4*(384*a^3*x^3 + 32*a^2*b*x^2 - 44*a*b^2*x + 77*b^3)*(a 
*x^4 + b*x^3)^(1/4))/a^3
 
3.19.15.6 Sympy [F]

\[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=\int x^{2} \sqrt [4]{x^{3} \left (a x + b\right )}\, dx \]

input
integrate(x**2*(a*x**4+b*x**3)**(1/4),x)
 
output
Integral(x**2*(x**3*(a*x + b))**(1/4), x)
 
3.19.15.7 Maxima [F]

\[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=\int { {\left (a x^{4} + b x^{3}\right )}^{\frac {1}{4}} x^{2} \,d x } \]

input
integrate(x^2*(a*x^4+b*x^3)^(1/4),x, algorithm="maxima")
 
output
integrate((a*x^4 + b*x^3)^(1/4)*x^2, x)
 
3.19.15.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (103) = 206\).

Time = 0.30 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.26 \[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=\frac {\frac {462 \, \sqrt {2} b^{5} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{3}} + \frac {462 \, \sqrt {2} b^{5} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{3}} + \frac {231 \, \sqrt {2} b^{5} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{3}} + \frac {231 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{5} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x}}\right )}{a^{4}} + \frac {8 \, {\left (77 \, {\left (a + \frac {b}{x}\right )}^{\frac {13}{4}} b^{5} - 275 \, {\left (a + \frac {b}{x}\right )}^{\frac {9}{4}} a b^{5} + 351 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{4}} a^{2} b^{5} + 231 \, {\left (a + \frac {b}{x}\right )}^{\frac {1}{4}} a^{3} b^{5}\right )} x^{4}}{a^{3} b^{4}}}{12288 \, b} \]

input
integrate(x^2*(a*x^4+b*x^3)^(1/4),x, algorithm="giac")
 
output
1/12288*(462*sqrt(2)*b^5*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b 
/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a^3) + 462*sqrt(2)*b^5*arctan(-1/2*sqrt 
(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a^3) 
+ 231*sqrt(2)*b^5*log(sqrt(2)*(-a)^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt 
(a + b/x))/((-a)^(3/4)*a^3) + 231*sqrt(2)*(-a)^(1/4)*b^5*log(-sqrt(2)*(-a) 
^(1/4)*(a + b/x)^(1/4) + sqrt(-a) + sqrt(a + b/x))/a^4 + 8*(77*(a + b/x)^( 
13/4)*b^5 - 275*(a + b/x)^(9/4)*a*b^5 + 351*(a + b/x)^(5/4)*a^2*b^5 + 231* 
(a + b/x)^(1/4)*a^3*b^5)*x^4/(a^3*b^4))/b
 
3.19.15.9 Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt [4]{b x^3+a x^4} \, dx=\int x^2\,{\left (a\,x^4+b\,x^3\right )}^{1/4} \,d x \]

input
int(x^2*(a*x^4 + b*x^3)^(1/4),x)
 
output
int(x^2*(a*x^4 + b*x^3)^(1/4), x)