Integrand size = 22, antiderivative size = 43 \[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\frac {1}{21} 2^{2/3} (-1+2 x)^{3/2} \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{3},1,\frac {5}{2},1-2 x,\frac {3}{7} (1-2 x)\right ) \] Output:
1/21*2^(2/3)*(-1+2*x)^(3/2)*AppellF1(3/2,-1/3,1,5/2,1-2*x,3/7-6/7*x)
Time = 5.72 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\frac {\sqrt [3]{x} \left (-16+32 x+16 \sqrt {1-2 x} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},2 x,-\frac {3 x}{2}\right )-29 \sqrt {1-2 x} x \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},2 x,-\frac {3 x}{2}\right )\right )}{40 \sqrt {-1+2 x}} \] Input:
Integrate[(x^(1/3)*Sqrt[-1 + 2*x])/(2 + 3*x),x]
Output:
(x^(1/3)*(-16 + 32*x + 16*Sqrt[1 - 2*x]*AppellF1[1/3, 1/2, 1, 4/3, 2*x, (- 3*x)/2] - 29*Sqrt[1 - 2*x]*x*AppellF1[4/3, 1/2, 1, 7/3, 2*x, (-3*x)/2]))/( 40*Sqrt[-1 + 2*x])
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {148, 1013, 1012}
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.
\(\displaystyle \int \frac {\sqrt [3]{x} \sqrt {2 x-1}}{3 x+2} \, dx\) |
\(\Big \downarrow \) 148 |
\(\displaystyle 3 \int \frac {x \sqrt {2 x-1}}{3 x+2}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {3 \sqrt {2 x-1} \int \frac {\sqrt {1-2 x} x}{3 x+2}d\sqrt [3]{x}}{\sqrt {1-2 x}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {3 x^{4/3} \sqrt {2 x-1} \operatorname {AppellF1}\left (\frac {4}{3},-\frac {1}{2},1,\frac {7}{3},2 x,-\frac {3 x}{2}\right )}{8 \sqrt {1-2 x}}\) |
Input:
Int[(x^(1/3)*Sqrt[-1 + 2*x])/(2 + 3*x),x]
Output:
(3*x^(4/3)*Sqrt[-1 + 2*x]*AppellF1[4/3, -1/2, 1, 7/3, 2*x, (-3*x)/2])/(8*S qrt[1 - 2*x])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {x^{\frac {1}{3}} \sqrt {-1+2 x}}{2+3 x}d x\]
Input:
int(x^(1/3)*(-1+2*x)^(1/2)/(2+3*x),x)
Output:
int(x^(1/3)*(-1+2*x)^(1/2)/(2+3*x),x)
Timed out. \[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\text {Timed out} \] Input:
integrate(x^(1/3)*(-1+2*x)^(1/2)/(2+3*x),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\int \frac {\sqrt [3]{x} \sqrt {2 x - 1}}{3 x + 2}\, dx \] Input:
integrate(x**(1/3)*(-1+2*x)**(1/2)/(2+3*x),x)
Output:
Integral(x**(1/3)*sqrt(2*x - 1)/(3*x + 2), x)
\[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\int { \frac {\sqrt {2 \, x - 1} x^{\frac {1}{3}}}{3 \, x + 2} \,d x } \] Input:
integrate(x^(1/3)*(-1+2*x)^(1/2)/(2+3*x),x, algorithm="maxima")
Output:
integrate(sqrt(2*x - 1)*x^(1/3)/(3*x + 2), x)
\[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\int { \frac {\sqrt {2 \, x - 1} x^{\frac {1}{3}}}{3 \, x + 2} \,d x } \] Input:
integrate(x^(1/3)*(-1+2*x)^(1/2)/(2+3*x),x, algorithm="giac")
Output:
integrate(sqrt(2*x - 1)*x^(1/3)/(3*x + 2), x)
Timed out. \[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=\int \frac {x^{1/3}\,\sqrt {2\,x-1}}{3\,x+2} \,d x \] Input:
int((x^(1/3)*(2*x - 1)^(1/2))/(3*x + 2),x)
Output:
int((x^(1/3)*(2*x - 1)^(1/2))/(3*x + 2), x)
\[ \int \frac {\sqrt [3]{x} \sqrt {-1+2 x}}{2+3 x} \, dx=-\frac {3 x^{\frac {1}{3}} \sqrt {2 x -1}}{7}+\frac {29 \left (\int \frac {x^{\frac {4}{3}} \sqrt {2 x -1}}{6 x^{2}+x -2}d x \right )}{7}-\frac {2 \left (\int \frac {x^{\frac {1}{3}} \sqrt {2 x -1}}{6 x^{3}+x^{2}-2 x}d x \right )}{7} \] Input:
int(x^(1/3)*(-1+2*x)^(1/2)/(2+3*x),x)
Output:
( - 3*x**(1/3)*sqrt(2*x - 1) + 29*int((x**(1/3)*sqrt(2*x - 1)*x)/(6*x**2 + x - 2),x) - 2*int((x**(1/3)*sqrt(2*x - 1))/(6*x**3 + x**2 - 2*x),x))/7