\(\int (b \sqrt [3]{x}+a x)^{3/2} \, dx\) [121]

Optimal result

Integrand size = 15, antiderivative size = 208 \[ \int \left (b \sqrt [3]{x}+a x\right )^{3/2} \, dx=-\frac {8 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {4 b^{15/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{77 a^{9/4} \sqrt {b \sqrt [3]{x}+a x}} \] Output:

-8/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^2+24/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^(1/ 
2)/a+12/55*b*x^(4/3)*(b*x^(1/3)+a*x)^(1/2)+2/5*x*(b*x^(1/3)+a*x)^(3/2)+4/7 
7*b^(15/4)*(b^(1/2)+a^(1/2)*x^(1/3))*((b+a*x^(2/3))/(b^(1/2)+a^(1/2)*x^(1/ 
3))^2)^(1/2)*x^(1/6)*InverseJacobiAM(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)),1/2 
*2^(1/2))/a^(9/4)/(b*x^(1/3)+a*x)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.51 \[ \int \left (b \sqrt [3]{x}+a x\right )^{3/2} \, dx=\frac {2 \sqrt {b \sqrt [3]{x}+a x} \left (-\left (\left (5 b-11 a x^{2/3}\right ) \left (b+a x^{2/3}\right )^2 \sqrt {1+\frac {a x^{2/3}}{b}}\right )+5 b^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {a x^{2/3}}{b}\right )\right )}{55 a^2 \sqrt {1+\frac {a x^{2/3}}{b}}} \] Input:

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

Output:

(2*Sqrt[b*x^(1/3) + a*x]*(-((5*b - 11*a*x^(2/3))*(b + a*x^(2/3))^2*Sqrt[1 
+ (a*x^(2/3))/b]) + 5*b^3*Hypergeometric2F1[-3/2, 1/4, 5/4, -((a*x^(2/3))/ 
b)]))/(55*a^2*Sqrt[1 + (a*x^(2/3))/b])
 

Rubi [A] (warning: unable to verify)

Time = 0.66 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1910, 1924, 1927, 1930, 1930, 1917, 266, 761}

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.

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

\(\Big \downarrow \) 1910

\(\displaystyle \frac {2}{5} b \int \sqrt [3]{x} \sqrt {\sqrt [3]{x} b+a x}dx+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 1924

\(\displaystyle \frac {6}{5} b \int x \sqrt {\sqrt [3]{x} b+a x}d\sqrt [3]{x}+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {6}{5} b \left (\frac {2}{11} b \int \frac {x^{4/3}}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}+\frac {2}{11} x^{4/3} \sqrt {a x+b \sqrt [3]{x}}\right )+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {6}{5} b \left (\frac {2}{11} b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \int \frac {x^{2/3}}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{7 a}\right )+\frac {2}{11} x^{4/3} \sqrt {a x+b \sqrt [3]{x}}\right )+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 1930

\(\displaystyle \frac {6}{5} b \left (\frac {2}{11} b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {b \int \frac {1}{\sqrt {\sqrt [3]{x} b+a x}}d\sqrt [3]{x}}{3 a}\right )}{7 a}\right )+\frac {2}{11} x^{4/3} \sqrt {a x+b \sqrt [3]{x}}\right )+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 1917

\(\displaystyle \frac {6}{5} b \left (\frac {2}{11} b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {b \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {x^{2/3} a+b} \sqrt [6]{x}}d\sqrt [3]{x}}{3 a \sqrt {a x+b \sqrt [3]{x}}}\right )}{7 a}\right )+\frac {2}{11} x^{4/3} \sqrt {a x+b \sqrt [3]{x}}\right )+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {6}{5} b \left (\frac {2}{11} b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {2 b \sqrt [6]{x} \sqrt {a x^{2/3}+b} \int \frac {1}{\sqrt {a x^{4/3}+b}}d\sqrt [6]{x}}{3 a \sqrt {a x+b \sqrt [3]{x}}}\right )}{7 a}\right )+\frac {2}{11} x^{4/3} \sqrt {a x+b \sqrt [3]{x}}\right )+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {6}{5} b \left (\frac {2}{11} b \left (\frac {2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{7 a}-\frac {5 b \left (\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{3 a}-\frac {b^{3/4} \sqrt [6]{x} \left (\sqrt {a} x^{2/3}+\sqrt {b}\right ) \sqrt {a x^{2/3}+b} \sqrt {\frac {a x^{4/3}+b}{\left (\sqrt {a} x^{2/3}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 a^{5/4} \sqrt {a x+b \sqrt [3]{x}} \sqrt {a x^{4/3}+b}}\right )}{7 a}\right )+\frac {2}{11} x^{4/3} \sqrt {a x+b \sqrt [3]{x}}\right )+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2}\)

Input:

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

Output:

(2*x*(b*x^(1/3) + a*x)^(3/2))/5 + (6*b*((2*x^(4/3)*Sqrt[b*x^(1/3) + a*x])/ 
11 + (2*b*((2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(7*a) - (5*b*((2*Sqrt[b*x^(1/ 
3) + a*x])/(3*a) - (b^(3/4)*(Sqrt[b] + Sqrt[a]*x^(2/3))*Sqrt[b + a*x^(2/3) 
]*x^(1/6)*Sqrt[(b + a*x^(4/3))/(Sqrt[b] + Sqrt[a]*x^(2/3))^2]*EllipticF[2* 
ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(3*a^(5/4)*Sqrt[b*x^(1/3) + a*x]* 
Sqrt[b + a*x^(4/3)])))/(7*a)))/11))/5
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

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 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1924

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp 
[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x 
], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j 
] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1 
]
 

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]
 

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\frac {262 x^{\frac {5}{3}} a^{3} b^{2}}{385}+\frac {56 x^{\frac {7}{3}} a^{4} b}{55}+\frac {4 b^{4} \sqrt {-a b}\, \sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}} a -\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x^{\frac {1}{3}} a +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77}-\frac {16 a^{2} b^{3} x}{385}+\frac {2 a^{5} x^{3}}{5}-\frac {8 x^{\frac {1}{3}} a \,b^{4}}{77}}{a^{3} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}}\)	\(163\)
derivativedivides	\(\frac {2 a \,x^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}{5}+\frac {34 b \,x^{\frac {4}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{55}+\frac {24 b^{2} x^{\frac {2}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{385 a}-\frac {8 b^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}{77 a^{2}}+\frac {4 b^{4} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77 a^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(196\)

Input:

int((b*x^(1/3)+a*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/385*(131*x^(5/3)*a^3*b^2+196*x^(7/3)*a^4*b+10*b^4*(-a*b)^(1/2)*((x^(1/3) 
*a+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(x^(1/3)*a-(-a*b)^(1/2))/(-a*b)^( 
1/2))^(1/2)*(-x^(1/3)/(-a*b)^(1/2)*a)^(1/2)*EllipticF(((x^(1/3)*a+(-a*b)^( 
1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))-8*a^2*b^3*x+77*a^5*x^3-20*x^(1/3)*a 
*b^4)/a^3/(x^(1/3)*(b+a*x^(2/3)))^(1/2)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.19 \[ \int \left (b \sqrt [3]{x}+a x\right )^{3/2} \, dx=\frac {2\,x\,{\left (a\,x+b\,x^{1/3}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {9}{4};\ \frac {13}{4};\ -\frac {a\,x^{2/3}}{b}\right )}{3\,{\left (\frac {a\,x^{2/3}}{b}+1\right )}^{3/2}} \] Input:

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

Output:

(2*x*(a*x + b*x^(1/3))^(3/2)*hypergeom([-3/2, 9/4], 13/4, -(a*x^(2/3))/b)) 
/(3*((a*x^(2/3))/b + 1)^(3/2))
 

Reduce [F]

\[ \int \left (b \sqrt [3]{x}+a x\right )^{3/2} \, dx=\frac {\frac {24 x^{\frac {5}{6}} \sqrt {x^{\frac {2}{3}} a +b}\, a \,b^{2}}{385}+\frac {34 \sqrt {x}\, \sqrt {x^{\frac {2}{3}} a +b}\, a^{2} b x}{55}+\frac {2 x^{\frac {13}{6}} \sqrt {x^{\frac {2}{3}} a +b}\, a^{3}}{5}-\frac {8 x^{\frac {1}{6}} \sqrt {x^{\frac {2}{3}} a +b}\, b^{3}}{77}+\frac {4 \left (\int \frac {\sqrt {x^{\frac {2}{3}} a +b}}{x^{\frac {5}{6}} b +\sqrt {x}\, a x}d x \right ) b^{4}}{231}}{a^{2}} \] Input:

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

Output:

(2*(36*x**(5/6)*sqrt(x**(2/3)*a + b)*a*b**2 + 357*sqrt(x)*sqrt(x**(2/3)*a 
+ b)*a**2*b*x + 231*x**(1/6)*sqrt(x**(2/3)*a + b)*a**3*x**2 - 60*x**(1/6)* 
sqrt(x**(2/3)*a + b)*b**3 + 10*int(sqrt(x**(2/3)*a + b)/(x**(5/6)*b + sqrt 
(x)*a*x),x)*b**4))/(1155*a**2)