Integrand size = 13, antiderivative size = 266 \[ \int \sqrt {a+b x^{4/3}} \, dx=\frac {6 a \sqrt [3]{x} \sqrt {a+b x^{4/3}}}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right )}+\frac {3}{5} x \sqrt {a+b x^{4/3}}-\frac {6 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^{4/3}}}+\frac {3 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a+b x^{4/3}}} \] Output:
6/5*a*x^(1/3)*(a+b*x^(4/3))^(1/2)/b^(1/2)/(a^(1/2)+b^(1/2)*x^(2/3))+3/5*x* (a+b*x^(4/3))^(1/2)-6/5*a^(5/4)*(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)+3/5*a^(5/4)*(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)*InverseJac obiAM(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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.71 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.20 \[ \int \sqrt {a+b x^{4/3}} \, dx=\frac {x \sqrt {a+b x^{4/3}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^{4/3}}{a}\right )}{\sqrt {1+\frac {b x^{4/3}}{a}}} \] Input:
Integrate[Sqrt[a + b*x^(4/3)],x]
Output:
(x*Sqrt[a + b*x^(4/3)]*Hypergeometric2F1[-1/2, 3/4, 7/4, -((b*x^(4/3))/a)] )/Sqrt[1 + (b*x^(4/3))/a]
Time = 0.48 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {774, 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.
\(\displaystyle \int \sqrt {a+b x^{4/3}} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 3 \int x^{2/3} \sqrt {b x^{4/3}+a}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle 3 \left (\frac {2}{5} a \int \frac {x^{2/3}}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )\) |
\(\Big \downarrow \) 834 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^{2/3}}{\sqrt {a} \sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^{2/3}}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^{4/3}}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^{2/3}}{\sqrt {b x^{4/3}+a}}d\sqrt [3]{x}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle 3 \left (\frac {2}{5} a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^{4/3}}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^{2/3}\right ) \sqrt {\frac {a+b x^{4/3}}{\left (\sqrt {a}+\sqrt {b} x^{2/3}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [3]{x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^{4/3}}}-\frac {\sqrt [3]{x} \sqrt {a+b x^{4/3}}}{\sqrt {a}+\sqrt {b} x^{2/3}}}{\sqrt {b}}\right )+\frac {1}{5} x \sqrt {a+b x^{4/3}}\right )\) |
Input:
Int[Sqrt[a + b*x^(4/3)],x]
Output:
3*((x*Sqrt[a + b*x^(4/3)])/5 + (2*a*(-((-((x^(1/3)*Sqrt[a + b*x^(4/3)])/(S qrt[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)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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]
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]
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.43
method | result | size |
derivativedivides | \(\frac {3 x \sqrt {a +b \,x^{\frac {4}{3}}}}{5}+\frac {6 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \,x^{\frac {4}{3}}}\, \sqrt {b}}\) | \(114\) |
default | \(\frac {3 x \sqrt {a +b \,x^{\frac {4}{3}}}}{5}+\frac {6 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{\frac {2}{3}}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x^{\frac {1}{3}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +b \,x^{\frac {4}{3}}}\, \sqrt {b}}\) | \(114\) |
Input:
int((a+b*x^(4/3))^(1/2),x,method=_RETURNVERBOSE)
Output:
3/5*x*(a+b*x^(4/3))^(1/2)+6/5*I*a^(3/2)/(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)-Elliptic E(x^(1/3)*(I/a^(1/2)*b^(1/2))^(1/2),I))
\[ \int \sqrt {a+b x^{4/3}} \, dx=\int { \sqrt {b x^{\frac {4}{3}} + a} \,d x } \] Input:
integrate((a+b*x^(4/3))^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(b*x^(4/3) + a), x)
Result contains complex when optimal does not.
Time = 0.49 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.15 \[ \int \sqrt {a+b x^{4/3}} \, dx=\frac {3 \sqrt {a} x \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{\frac {4}{3}} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((a+b*x**(4/3))**(1/2),x)
Output:
3*sqrt(a)*x*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**(4/3)*exp_polar(I*p i)/a)/(4*gamma(7/4))
\[ \int \sqrt {a+b x^{4/3}} \, dx=\int { \sqrt {b x^{\frac {4}{3}} + a} \,d x } \] Input:
integrate((a+b*x^(4/3))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^(4/3) + a), x)
\[ \int \sqrt {a+b x^{4/3}} \, dx=\int { \sqrt {b x^{\frac {4}{3}} + a} \,d x } \] Input:
integrate((a+b*x^(4/3))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*x^(4/3) + a), x)
Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.14 \[ \int \sqrt {a+b x^{4/3}} \, dx=\frac {x\,\sqrt {a+b\,x^{4/3}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {b\,x^{4/3}}{a}\right )}{\sqrt {\frac {b\,x^{4/3}}{a}+1}} \] Input:
int((a + b*x^(4/3))^(1/2),x)
Output:
(x*(a + b*x^(4/3))^(1/2)*hypergeom([-1/2, 3/4], 7/4, -(b*x^(4/3))/a))/((b* x^(4/3))/a + 1)^(1/2)
\[ \int \sqrt {a+b x^{4/3}} \, dx=\frac {3 \sqrt {x^{\frac {4}{3}} b +a}\, x}{5}+\frac {2 \left (\int \frac {\sqrt {x^{\frac {4}{3}} b +a}}{x^{\frac {4}{3}} b +a}d x \right ) a}{5} \] Input:
int((a+b*x^(4/3))^(1/2),x)
Output:
(3*sqrt(x**(1/3)*b*x + a)*x + 2*int(sqrt(x**(1/3)*b*x + a)/(x**(1/3)*b*x + a),x)*a)/5