\(\int (a+b x^{4/3})^{3/2} \, dx\) [142]

Optimal result

Integrand size = 13, antiderivative size = 287 \[ \int \left (a+b x^{4/3}\right )^{3/2} \, dx=\frac {4 a^2 \sqrt [3]{x} \sqrt {a+b x^{4/3}}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right )}+\frac {2}{5} a x \sqrt {a+b x^{4/3}}+\frac {1}{3} x \left (a+b x^{4/3}\right )^{3/2}-\frac {4 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^{4/3}}}+\frac {2 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^{4/3}}} \] Output:

4/5*a^2*x^(1/3)*(a+b*x^(4/3))^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)*x^(2/3))+2/5* 
a*x*(a+b*x^(4/3))^(1/2)+1/3*x*(a+b*x^(4/3))^(3/2)-4/5*a^(9/4)*(a^(1/2)+b^( 
1/2)*x^(2/3))*((a+b*x^(4/3))/(a^(1/2)+b^(1/2)*x^(2/3))^2)^(1/2)*EllipticE( 
sin(2*arctan(b^(1/4)*x^(1/3)/a^(1/4))),1/2*2^(1/2))/b^(3/4)/(a+b*x^(4/3))^ 
(1/2)+2/5*a^(9/4)*(a^(1/2)+b^(1/2)*x^(2/3))*((a+b*x^(4/3))/(a^(1/2)+b^(1/2 
)*x^(2/3))^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x^(1/3)/a^(1/4)),1/2* 
2^(1/2))/b^(3/4)/(a+b*x^(4/3))^(1/2)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.18 \[ \int \left (a+b x^{4/3}\right )^{3/2} \, dx=\frac {a x \sqrt {a+b x^{4/3}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^{4/3}}{a}\right )}{\sqrt {1+\frac {b x^{4/3}}{a}}} \] Input:

Integrate[(a + b*x^(4/3))^(3/2),x]
 

Output:

(a*x*Sqrt[a + b*x^(4/3)]*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^(4/3))/a 
)])/Sqrt[1 + (b*x^(4/3))/a]
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {774, 811, 811, 834, 27, 761, 1510}

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.

\(\displaystyle \int \left (a+b x^{4/3}\right )^{3/2} \, dx\)

\(\Big \downarrow \) 774

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 811

\(\displaystyle 3 \left (\frac {2}{3} a \left (\frac {2}{5} a \int \frac {x^{2/3}}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )+\frac {1}{9} x \left (a+b x^{4/3}\right )^{3/2}\right )\)

\(\Big \downarrow \) 834

\(\displaystyle 3 \left (\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^{2/3}}{\sqrt {a} \sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )+\frac {1}{9} x \left (a+b x^{4/3}\right )^{3/2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \left (\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^{2/3}}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )+\frac {1}{9} x \left (a+b x^{4/3}\right )^{3/2}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle 3 \left (\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^{4/3}}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^{2/3}}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )+\frac {1}{9} x \left (a+b x^{4/3}\right )^{3/2}\right )\)

\(\Big \downarrow \) 1510

\(\displaystyle 3 \left (\frac {2}{3} a \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^{4/3}}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^{4/3}}}-\frac {\sqrt [3]{x} \sqrt {a+b x^{4/3}}}{\sqrt {a}+\sqrt {b} x^{2/3}}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )+\frac {1}{9} x \left (a+b x^{4/3}\right )^{3/2}\right )\)

Input:

Int[(a + b*x^(4/3))^(3/2),x]
 

Output:

3*((x*(a + b*x^(4/3))^(3/2))/9 + (2*a*((x*Sqrt[a + b*x^(4/3)])/5 + (2*a*(- 
((-((x^(1/3)*Sqrt[a + b*x^(4/3)])/(Sqrt[a] + Sqrt[b]*x^(2/3))) + (a^(1/4)* 
(Sqrt[a] + Sqrt[b]*x^(2/3))*Sqrt[(a + b*x^(4/3))/(Sqrt[a] + Sqrt[b]*x^(2/3 
))^2]*EllipticE[2*ArcTan[(b^(1/4)*x^(1/3))/a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a 
 + b*x^(4/3)]))/Sqrt[b]) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^(2/3))*Sqrt[(a + 
b*x^(4/3))/(Sqrt[a] + Sqrt[b]*x^(2/3))^2]*EllipticF[2*ArcTan[(b^(1/4)*x^(1 
/3))/a^(1/4)], 1/2])/(2*b^(3/4)*Sqrt[a + b*x^(4/3)])))/5))/3)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 774

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, 
Simp[k   Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre 
eQ[{a, b, p}, x] && FractionQ[n]
 

rule 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m 
, p, x]
 

rule 834

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

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.45

method	result	size
derivativedivides	\(\frac {b \,x^{\frac {7}{3}} \sqrt {a +b \,x^{\frac {4}{3}}}}{3}+\frac {11 a x \sqrt {a +b \,x^{\frac {4}{3}}}}{15}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \,x^{\frac {4}{3}}}\, \sqrt {b}}\)	\(130\)
default	\(\frac {b \,x^{\frac {7}{3}} \sqrt {a +b \,x^{\frac {4}{3}}}}{3}+\frac {11 a x \sqrt {a +b \,x^{\frac {4}{3}}}}{15}+\frac {4 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \,x^{\frac {4}{3}}}\, \sqrt {b}}\)	\(130\)

Input:

int((a+b*x^(4/3))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*b*x^(7/3)*(a+b*x^(4/3))^(1/2)+11/15*a*x*(a+b*x^(4/3))^(1/2)+4/5*I*a^(5 
/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^(2/3))^(1/2)*(1+I/a^( 
1/2)*b^(1/2)*x^(2/3))^(1/2)/(a+b*x^(4/3))^(1/2)/b^(1/2)*(EllipticF(x^(1/3) 
*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x^(1/3)*(I/a^(1/2)*b^(1/2))^(1/2), 
I))
 

Fricas [F]

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

integrate((a+b*x^(4/3))^(3/2),x, algorithm="fricas")
 

Output:

integral((b*x^(4/3) + a)^(3/2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.14 \[ \int \left (a+b x^{4/3}\right )^{3/2} \, dx=\frac {3 a^{\frac {3}{2}} x \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{\frac {4}{3}} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate((a+b*x**(4/3))**(3/2),x)
 

Output:

3*a**(3/2)*x*gamma(3/4)*hyper((-3/2, 3/4), (7/4,), b*x**(4/3)*exp_polar(I* 
pi)/a)/(4*gamma(7/4))
 

Maxima [F]

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

integrate((a+b*x^(4/3))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*x^(4/3) + a)^(3/2), x)
 

Giac [F]

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

integrate((a+b*x^(4/3))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*x^(4/3) + a)^(3/2), x)
 

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.13 \[ \int \left (a+b x^{4/3}\right )^{3/2} \, dx=\frac {x\,{\left (a+b\,x^{4/3}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {b\,x^{4/3}}{a}\right )}{{\left (\frac {b\,x^{4/3}}{a}+1\right )}^{3/2}} \] Input:

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

Output:

(x*(a + b*x^(4/3))^(3/2)*hypergeom([-3/2, 3/4], 7/4, -(b*x^(4/3))/a))/((b* 
x^(4/3))/a + 1)^(3/2)
 

Reduce [F]

\[ \int \left (a+b x^{4/3}\right )^{3/2} \, dx=\frac {x^{\frac {7}{3}} \sqrt {x^{\frac {4}{3}} b +a}\, b}{3}+\frac {11 \sqrt {x^{\frac {4}{3}} b +a}\, a x}{15}+\frac {4 \left (\int \frac {\sqrt {x^{\frac {4}{3}} b +a}}{x^{\frac {4}{3}} b +a}d x \right ) a^{2}}{15} \] Input:

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

Output:

(5*x**(1/3)*sqrt(x**(1/3)*b*x + a)*b*x**2 + 11*sqrt(x**(1/3)*b*x + a)*a*x 
+ 4*int(sqrt(x**(1/3)*b*x + a)/(x**(1/3)*b*x + a),x)*a**2)/15