Integrand size = 26, antiderivative size = 65 \[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\frac {3 \sqrt [3]{e x} \sqrt {1+\frac {b x}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x}{a},-\frac {b x}{4 a}\right )}{4 a e \sqrt {a+b x}} \] Output:
3/4*(e*x)^(1/3)*(1+b*x/a)^(1/2)*AppellF1(1/3,1/2,1,4/3,-b*x/a,-1/4*b*x/a)/ a/e/(b*x+a)^(1/2)
Time = 11.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\frac {3 x \sqrt {\frac {a+b x}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x}{a},-\frac {b x}{4 a}\right )}{4 a (e x)^{2/3} \sqrt {a+b x}} \] Input:
Integrate[1/((e*x)^(2/3)*Sqrt[a + b*x]*(4*a + b*x)),x]
Output:
(3*x*Sqrt[(a + b*x)/a]*AppellF1[1/3, 1/2, 1, 4/3, -((b*x)/a), -1/4*(b*x)/a ])/(4*a*(e*x)^(2/3)*Sqrt[a + b*x])
Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {148, 27, 937, 936}
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.
\(\displaystyle \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {3 \int \frac {e}{\sqrt {a+b x} (4 a e+b x e)}d\sqrt [3]{e x}}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int \frac {1}{\sqrt {a+b x} (4 a e+b x e)}d\sqrt [3]{e x}\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {3 \sqrt {\frac {b x}{a}+1} \int \frac {1}{\sqrt {\frac {b x}{a}+1} (4 a e+b x e)}d\sqrt [3]{e x}}{\sqrt {a+b x}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {3 \sqrt [3]{e x} \sqrt {\frac {b x}{a}+1} \operatorname {AppellF1}\left (\frac {1}{3},1,\frac {1}{2},\frac {4}{3},-\frac {b x}{4 a},-\frac {b x}{a}\right )}{4 a e \sqrt {a+b x}}\) |
Input:
Int[1/((e*x)^(2/3)*Sqrt[a + b*x]*(4*a + b*x)),x]
Output:
(3*(e*x)^(1/3)*Sqrt[1 + (b*x)/a]*AppellF1[1/3, 1, 1/2, 4/3, -1/4*(b*x)/a, -((b*x)/a)])/(4*a*e*Sqrt[a + b*x])
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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q }, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {1}{\left (e x \right )^{\frac {2}{3}} \sqrt {b x +a}\, \left (b x +4 a \right )}d x\]
Input:
int(1/(e*x)^(2/3)/(b*x+a)^(1/2)/(b*x+4*a),x)
Output:
int(1/(e*x)^(2/3)/(b*x+a)^(1/2)/(b*x+4*a),x)
Timed out. \[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x)^(2/3)/(b*x+a)^(1/2)/(b*x+4*a),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int \frac {1}{\left (e x\right )^{\frac {2}{3}} \sqrt {a + b x} \left (4 a + b x\right )}\, dx \] Input:
integrate(1/(e*x)**(2/3)/(b*x+a)**(1/2)/(b*x+4*a),x)
Output:
Integral(1/((e*x)**(2/3)*sqrt(a + b*x)*(4*a + b*x)), x)
\[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int { \frac {1}{{\left (b x + 4 \, a\right )} \sqrt {b x + a} \left (e x\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(e*x)^(2/3)/(b*x+a)^(1/2)/(b*x+4*a),x, algorithm="maxima")
Output:
integrate(1/((b*x + 4*a)*sqrt(b*x + a)*(e*x)^(2/3)), x)
\[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int { \frac {1}{{\left (b x + 4 \, a\right )} \sqrt {b x + a} \left (e x\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(1/(e*x)^(2/3)/(b*x+a)^(1/2)/(b*x+4*a),x, algorithm="giac")
Output:
integrate(1/((b*x + 4*a)*sqrt(b*x + a)*(e*x)^(2/3)), x)
Timed out. \[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int \frac {1}{{\left (e\,x\right )}^{2/3}\,\left (4\,a+b\,x\right )\,\sqrt {a+b\,x}} \,d x \] Input:
int(1/((e*x)^(2/3)*(4*a + b*x)*(a + b*x)^(1/2)),x)
Output:
int(1/((e*x)^(2/3)*(4*a + b*x)*(a + b*x)^(1/2)), x)
\[ \int \frac {1}{(e x)^{2/3} \sqrt {a+b x} (4 a+b x)} \, dx=\frac {\int \frac {x^{\frac {1}{3}} \sqrt {b x +a}}{b^{2} x^{3}+5 a b \,x^{2}+4 a^{2} x}d x}{e^{\frac {2}{3}}} \] Input:
int(1/(e*x)^(2/3)/(b*x+a)^(1/2)/(b*x+4*a),x)
Output:
(e**(1/3)*int((x**(1/3)*sqrt(a + b*x))/(4*a**2*x + 5*a*b*x**2 + b**2*x**3) ,x))/e