Integrand size = 17, antiderivative size = 287 \[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\frac {3}{5} A x \sqrt [3]{a+b x^2}+\frac {3 B \left (a+b x^2\right )^{4/3}}{8 b}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a A \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
3/5*A*x*(b*x^2+a)^(1/3)+3/8*B*(b*x^2+a)^(4/3)/b-2/5*3^(3/4)*(1/2*6^(1/2)-1 /2*2^(1/2))*a*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)*Elliptic F(((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 = 4.60 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.29 \[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\frac {3 \left (a+b x^2\right ) (5 a B+b x (8 A+5 B x))+16 a A b x \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )}{40 b \left (a+b x^2\right )^{2/3}} \] Input:
Integrate[(A + B*x)*(a + b*x^2)^(1/3),x]
Output:
(3*(a + b*x^2)*(5*a*B + b*x*(8*A + 5*B*x)) + 16*a*A*b*x*(1 + (b*x^2)/a)^(2 /3)*Hypergeometric2F1[1/2, 2/3, 3/2, -((b*x^2)/a)])/(40*b*(a + b*x^2)^(2/3 ))
Time = 0.28 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {455, 211, 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.
\(\displaystyle \int \sqrt [3]{a+b x^2} (A+B x) \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle A \int \sqrt [3]{b x^2+a}dx+\frac {3 B \left (a+b x^2\right )^{4/3}}{8 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {3 B \left (a+b x^2\right )^{4/3}}{8 b}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle A \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )+\frac {3 B \left (a+b x^2\right )^{4/3}}{8 b}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle A \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )+\frac {3 B \left (a+b x^2\right )^{4/3}}{8 b}\) |
Input:
Int[(A + B*x)*(a + b*x^2)^(1/3),x]
Output:
(3*B*(a + b*x^2)^(4/3))/(8*b) + A*((3*x*(a + b*x^2)^(1/3))/5 - (2*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]])/(5*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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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]
\[\int \left (B x +A \right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}d x\]
Input:
int((B*x+A)*(b*x^2+a)^(1/3),x)
Output:
int((B*x+A)*(b*x^2+a)^(1/3),x)
\[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (B x + A\right )} \,d x } \] Input:
integrate((B*x+A)*(b*x^2+a)^(1/3),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/3)*(B*x + A), x)
Time = 0.84 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.19 \[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=A \sqrt [3]{a} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + B \left (\begin {cases} \frac {\sqrt [3]{a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {3 \left (a + b x^{2}\right )^{\frac {4}{3}}}{8 b} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((B*x+A)*(b*x**2+a)**(1/3),x)
Output:
A*a**(1/3)*x*hyper((-1/3, 1/2), (3/2,), b*x**2*exp_polar(I*pi)/a) + B*Piec ewise((a**(1/3)*x**2/2, Eq(b, 0)), (3*(a + b*x**2)**(4/3)/(8*b), True))
\[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (B x + A\right )} \,d x } \] Input:
integrate((B*x+A)*(b*x^2+a)^(1/3),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(1/3)*(B*x + A), x)
\[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{3}} {\left (B x + A\right )} \,d x } \] Input:
integrate((B*x+A)*(b*x^2+a)^(1/3),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(1/3)*(B*x + A), x)
Time = 0.45 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.19 \[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\frac {3\,B\,{\left (b\,x^2+a\right )}^{4/3}}{8\,b}+\frac {A\,x\,{\left (b\,x^2+a\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{1/3}} \] Input:
int((a + b*x^2)^(1/3)*(A + B*x),x)
Output:
(3*B*(a + b*x^2)^(4/3))/(8*b) + (A*x*(a + b*x^2)^(1/3)*hypergeom([-1/3, 1/ 2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(1/3)
\[ \int (A+B x) \sqrt [3]{a+b x^2} \, dx=\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a x}{5}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a}{8}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}}{8}+\frac {2 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{2}}{5} \] Input:
int((B*x+A)*(b*x^2+a)^(1/3),x)
Output:
(24*(a + b*x**2)**(1/3)*a*x + 15*(a + b*x**2)**(1/3)*a + 15*(a + b*x**2)** (1/3)*b*x**2 + 16*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**2)/40