\(\int \sqrt [4]{b x^5+a x^8} \, dx\) [1258]

Optimal result

Integrand size = 15, antiderivative size = 91 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {1}{3} x \sqrt [4]{b x^5+a x^8}+\frac {b \arctan \left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}}+\frac {b \text {arctanh}\left (\frac {\sqrt [4]{b x^5+a x^8}}{\sqrt [4]{a} x^2}\right )}{6 a^{3/4}} \] Output:

1/3*x*(a*x^8+b*x^5)^(1/4)+1/6*b*arctan((a*x^8+b*x^5)^(1/4)/a^(1/4)/x^2)/a^ 
(3/4)+1/6*b*arctanh((a*x^8+b*x^5)^(1/4)/a^(1/4)/x^2)/a^(3/4)
 

Mathematica [A] (verified)

Time = 6.79 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.26 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {\sqrt [4]{x^5 \left (b+a x^3\right )} \left (2 a^{3/4} x^{9/4} \sqrt [4]{b+a x^3}-b \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )+b \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{b+a x^3}}\right )\right )}{6 a^{3/4} x^{5/4} \sqrt [4]{b+a x^3}} \] Input:

Integrate[(b*x^5 + a*x^8)^(1/4),x]
 

Output:

((x^5*(b + a*x^3))^(1/4)*(2*a^(3/4)*x^(9/4)*(b + a*x^3)^(1/4) - b*ArcTan[( 
a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)] + b*ArcTanh[(a^(1/4)*x^(3/4))/(b + a*x 
^3)^(1/4)]))/(6*a^(3/4)*x^(5/4)*(b + a*x^3)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1910, 1938, 851, 807, 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.

\(\displaystyle \int \sqrt [4]{a x^8+b x^5} \, dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {1}{4} b \int \frac {x^5}{\left (a x^8+b x^5\right )^{3/4}}dx+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {x^{5/4}}{\left (a x^3+b\right )^{3/4}}dx}{4 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {x^2}{\left (a x^3+b\right )^{3/4}}d\sqrt [4]{x}}{\left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {\sqrt {x}}{(b+a x)^{3/4}}dx^{3/4}}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \int \frac {\sqrt {x}}{1-a x}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}\right )}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{b+a x}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {b x^{15/4} \left (a x^3+b\right )^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x+b}}\right )}{2 a^{3/4}}\right )}{3 \left (a x^8+b x^5\right )^{3/4}}+\frac {1}{3} x \sqrt [4]{a x^8+b x^5}\)

Input:

Int[(b*x^5 + a*x^8)^(1/4),x]
 

Output:

(x*(b*x^5 + a*x^8)^(1/4))/3 + (b*x^(15/4)*(b + a*x^3)^(3/4)*(-1/2*ArcTan[( 
a^(1/4)*x^(3/4))/(b + a*x)^(1/4)]/a^(3/4) + ArcTanh[(a^(1/4)*x^(3/4))/(b + 
 a*x)^(1/4)]/(2*a^(3/4))))/(3*(b*x^5 + a*x^8)^(3/4))
 

Defintions of rubi rules used

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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

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 851

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

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 1910

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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13

method	result	size
pseudoelliptic	\(\frac {4 \left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}} x \,a^{\frac {3}{4}}+\ln \left (\frac {-a^{\frac {1}{4}} x^{2}-\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x^{2}-\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (x^{5} \left (a \,x^{3}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x^{2}}\right ) b}{12 a^{\frac {3}{4}}}\)	\(103\)

Input:

int((a*x^8+b*x^5)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/12*(4*(x^5*(a*x^3+b))^(1/4)*x*a^(3/4)+ln((-a^(1/4)*x^2-(x^5*(a*x^3+b))^( 
1/4))/(a^(1/4)*x^2-(x^5*(a*x^3+b))^(1/4)))*b+2*arctan((x^5*(a*x^3+b))^(1/4 
)/a^(1/4)/x^2)*b)/a^(3/4)
 

Fricas [F(-1)]

Timed out. \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\text {Timed out} \] Input:

integrate((a*x^8+b*x^5)^(1/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\int \sqrt [4]{a x^{8} + b x^{5}}\, dx \] Input:

integrate((a*x**8+b*x**5)**(1/4),x)
 

Output:

Integral((a*x**8 + b*x**5)**(1/4), x)
 

Maxima [F]

\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\int { {\left (a x^{8} + b x^{5}\right )}^{\frac {1}{4}} \,d x } \] Input:

integrate((a*x^8+b*x^5)^(1/4),x, algorithm="maxima")
 

Output:

integrate((a*x^8 + b*x^5)^(1/4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (71) = 142\).

Time = 0.32 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.25 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {1}{24} \, {\left (\frac {8 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} x^{3}}{b} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{a} - \frac {\sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{3}}}\right )}{a}\right )} b \] Input:

integrate((a*x^8+b*x^5)^(1/4),x, algorithm="giac")
 

Output:

1/24*(8*(a + b/x^3)^(1/4)*x^3/b + 2*sqrt(2)*(-a)^(1/4)*arctan(1/2*sqrt(2)* 
(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^3)^(1/4))/(-a)^(1/4))/a + 2*sqrt(2)*(-a)^ 
(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^3)^(1/4))/(-a)^ 
(1/4))/a + sqrt(2)*(-a)^(1/4)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + s 
qrt(-a) + sqrt(a + b/x^3))/a - sqrt(2)*(-a)^(1/4)*log(-sqrt(2)*(-a)^(1/4)* 
(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3))/a)*b
 

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {4\,x\,{\left (a\,x^8+b\,x^5\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {a\,x^3}{b}\right )}{9\,{\left (\frac {a\,x^3}{b}+1\right )}^{1/4}} \] Input:

int((a*x^8 + b*x^5)^(1/4),x)
 

Output:

(4*x*(a*x^8 + b*x^5)^(1/4)*hypergeom([-1/4, 3/4], 7/4, -(a*x^3)/b))/(9*((a 
*x^3)/b + 1)^(1/4))
 

Reduce [F]

\[ \int \sqrt [4]{b x^5+a x^8} \, dx=\frac {x^{\frac {9}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}}}{3}+\frac {\left (\int \frac {x^{\frac {5}{4}}}{\left (a \,x^{3}+b \right )^{\frac {3}{4}}}d x \right ) b}{4} \] Input:

int((a*x^8+b*x^5)^(1/4),x)
 

Output:

(4*x**(1/4)*(a*x**3 + b)**(1/4)*x**2 + 3*int((x**(1/4)*(a*x**3 + b)**(1/4) 
*x)/(a*x**3 + b),x)*b)/12