Integrand size = 24, antiderivative size = 63 \[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\frac {2 (e x)^{3/2} \sqrt [3]{1+\frac {d x}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},1,\frac {5}{2},-\frac {d x}{c},-\frac {b x}{a}\right )}{3 a e \sqrt [3]{c+d x}} \] Output:
2/3*(e*x)^(3/2)*(1+d*x/c)^(1/3)*AppellF1(3/2,1,1/3,5/2,-b*x/a,-d*x/c)/a/e/ (d*x+c)^(1/3)
Time = 11.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\frac {2 x \sqrt {e x} \sqrt [3]{\frac {c+d x}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},1,\frac {5}{2},-\frac {d x}{c},-\frac {b x}{a}\right )}{3 a \sqrt [3]{c+d x}} \] Input:
Integrate[Sqrt[e*x]/((a + b*x)*(c + d*x)^(1/3)),x]
Output:
(2*x*Sqrt[e*x]*((c + d*x)/c)^(1/3)*AppellF1[3/2, 1/3, 1, 5/2, -((d*x)/c), -((b*x)/a)])/(3*a*(c + d*x)^(1/3))
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {148, 27, 395, 394}
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.
\(\displaystyle \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {2 \int \frac {e^2 x}{\sqrt [3]{c+d x} (a e+b x e)}d\sqrt {e x}}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e x}{\sqrt [3]{c+d x} (a e+b x e)}d\sqrt {e x}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {2 \sqrt [3]{\frac {d x}{c}+1} \int \frac {e x}{\sqrt [3]{\frac {d x}{c}+1} (a e+b x e)}d\sqrt {e x}}{\sqrt [3]{c+d x}}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {2 (e x)^{3/2} \sqrt [3]{\frac {d x}{c}+1} \operatorname {AppellF1}\left (\frac {3}{2},1,\frac {1}{3},\frac {5}{2},-\frac {b x}{a},-\frac {d x}{c}\right )}{3 a e \sqrt [3]{c+d x}}\) |
Input:
Int[Sqrt[e*x]/((a + b*x)*(c + d*x)^(1/3)),x]
Output:
(2*(e*x)^(3/2)*(1 + (d*x)/c)^(1/3)*AppellF1[3/2, 1, 1/3, 5/2, -((b*x)/a), -((d*x)/c)])/(3*a*e*(c + d*x)^(1/3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {\sqrt {e x}}{\left (b x +a \right ) \left (x d +c \right )^{\frac {1}{3}}}d x\]
Input:
int((e*x)^(1/2)/(b*x+a)/(d*x+c)^(1/3),x)
Output:
int((e*x)^(1/2)/(b*x+a)/(d*x+c)^(1/3),x)
Timed out. \[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\text {Timed out} \] Input:
integrate((e*x)^(1/2)/(b*x+a)/(d*x+c)^(1/3),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\int \frac {\sqrt {e x}}{\left (a + b x\right ) \sqrt [3]{c + d x}}\, dx \] Input:
integrate((e*x)**(1/2)/(b*x+a)/(d*x+c)**(1/3),x)
Output:
Integral(sqrt(e*x)/((a + b*x)*(c + d*x)**(1/3)), x)
\[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\int { \frac {\sqrt {e x}}{{\left (b x + a\right )} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((e*x)^(1/2)/(b*x+a)/(d*x+c)^(1/3),x, algorithm="maxima")
Output:
integrate(sqrt(e*x)/((b*x + a)*(d*x + c)^(1/3)), x)
\[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\int { \frac {\sqrt {e x}}{{\left (b x + a\right )} {\left (d x + c\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((e*x)^(1/2)/(b*x+a)/(d*x+c)^(1/3),x, algorithm="giac")
Output:
integrate(sqrt(e*x)/((b*x + a)*(d*x + c)^(1/3)), x)
Timed out. \[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\int \frac {\sqrt {e\,x}}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{1/3}} \,d x \] Input:
int((e*x)^(1/2)/((a + b*x)*(c + d*x)^(1/3)),x)
Output:
int((e*x)^(1/2)/((a + b*x)*(c + d*x)^(1/3)), x)
\[ \int \frac {\sqrt {e x}}{(a+b x) \sqrt [3]{c+d x}} \, dx=\sqrt {e}\, \left (\int \frac {\sqrt {x}}{\left (d x +c \right )^{\frac {1}{3}} a +\left (d x +c \right )^{\frac {1}{3}} b x}d x \right ) \] Input:
int((e*x)^(1/2)/(b*x+a)/(d*x+c)^(1/3),x)
Output:
sqrt(e)*int(sqrt(x)/((c + d*x)**(1/3)*a + (c + d*x)**(1/3)*b*x),x)