Integrand size = 17, antiderivative size = 267 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 B \sqrt [3]{a+b x^2}}{2 b}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} A \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
3/2*B*(b*x^2+a)^(1/3)/b-3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*A*(a^(1/3)-(b*x^ 2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^(1/2) )*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*a^(1/3)-(b*x^2+ a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)),2*I-I*3^(1/2))/b/x/(-a^(1/ 3)*(a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2 )
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.25 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 B \sqrt [3]{a+b x^2}}{2 b}+\frac {A x \left (\frac {a+b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )}{\left (a+b x^2\right )^{2/3}} \] Input:
Integrate[(A + B*x)/(a + b*x^2)^(2/3),x]
Output:
(3*B*(a + b*x^2)^(1/3))/(2*b) + (A*x*((a + b*x^2)/a)^(2/3)*Hypergeometric2 F1[1/2, 2/3, 3/2, -((b*x^2)/a)])/(a + b*x^2)^(2/3)
Time = 0.26 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {455, 234, 760}
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.
\(\displaystyle \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle A \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3 B \sqrt [3]{a+b x^2}}{2 b}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {3 A \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{2 b x}+\frac {3 B \sqrt [3]{a+b x^2}}{2 b}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {3 B \sqrt [3]{a+b x^2}}{2 b}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} A \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\) |
Input:
Int[(A + B*x)/(a + b*x^2)^(2/3),x]
Output:
(3*B*(a + b*x^2)^(1/3))/(2*b) - (3^(3/4)*Sqrt[2 - Sqrt[3]]*A*(a^(1/3) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^( 2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^ 2)^(1/3))], -7 + 4*Sqrt[3]])/(b*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^( 1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)])
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
\[\int \frac {B x +A}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x\]
Input:
int((B*x+A)/(b*x^2+a)^(2/3),x)
Output:
int((B*x+A)/(b*x^2+a)^(2/3),x)
\[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {B x + A}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((B*x+A)/(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
integral((B*x + A)/(b*x^2 + a)^(2/3), x)
Time = 0.76 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.20 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {2}{3}}} + B \left (\begin {cases} \frac {x^{2}}{2 a^{\frac {2}{3}}} & \text {for}\: b = 0 \\\frac {3 \sqrt [3]{a + b x^{2}}}{2 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((B*x+A)/(b*x**2+a)**(2/3),x)
Output:
A*x*hyper((1/2, 2/3), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(2/3) + B*Piece wise((x**2/(2*a**(2/3)), Eq(b, 0)), (3*(a + b*x**2)**(1/3)/(2*b), True))
\[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {B x + A}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((B*x+A)/(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
integrate((B*x + A)/(b*x^2 + a)^(2/3), x)
\[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {B x + A}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((B*x+A)/(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
integrate((B*x + A)/(b*x^2 + a)^(2/3), x)
Time = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.20 \[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3\,B\,{\left (b\,x^2+a\right )}^{1/3}}{2\,b}+\frac {A\,x\,{\left (\frac {b\,x^2}{a}+1\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {2}{3};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{2/3}} \] Input:
int((A + B*x)/(a + b*x^2)^(2/3),x)
Output:
(3*B*(a + b*x^2)^(1/3))/(2*b) + (A*x*((b*x^2)/a + 1)^(2/3)*hypergeom([1/2, 2/3], 3/2, -(b*x^2)/a))/(a + b*x^2)^(2/3)
\[ \int \frac {A+B x}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}}}{2}+\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a \] Input:
int((B*x+A)/(b*x^2+a)^(2/3),x)
Output:
(3*(a + b*x**2)**(1/3) + 2*int(1/(a + b*x**2)**(2/3),x)*a)/2