\(\int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx\) [157]

Optimal result

Integrand size = 26, antiderivative size = 65 \[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=-\frac {3 \sqrt {1+\frac {b x}{a}} \operatorname {AppellF1}\left (-\frac {2}{3},\frac {1}{2},1,\frac {1}{3},-\frac {b x}{a},-\frac {b x}{4 a}\right )}{8 a e (e x)^{2/3} \sqrt {a+b x}} \] Output:

-3/8*(1+b*x/a)^(1/2)*AppellF1(-2/3,1/2,1,1/3,-b*x/a,-1/4*b*x/a)/a/e/(e*x)^ 
(2/3)/(b*x+a)^(1/2)
 

Mathematica [A] (verified)

Time = 11.06 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=-\frac {3 x \left (32 a (a+b x)+32 a b x \sqrt {1+\frac {b x}{a}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},-\frac {b x}{a},-\frac {b x}{4 a}\right )+b^2 x^2 \sqrt {1+\frac {b x}{a}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},1,\frac {7}{3},-\frac {b x}{a},-\frac {b x}{4 a}\right )\right )}{256 a^3 (e x)^{5/3} \sqrt {a+b x}} \] Input:

Integrate[1/((e*x)^(5/3)*Sqrt[a + b*x]*(4*a + b*x)),x]
 

Output:

(-3*x*(32*a*(a + b*x) + 32*a*b*x*Sqrt[1 + (b*x)/a]*AppellF1[1/3, 1/2, 1, 4 
/3, -((b*x)/a), -1/4*(b*x)/a] + b^2*x^2*Sqrt[1 + (b*x)/a]*AppellF1[4/3, 1/ 
2, 1, 7/3, -((b*x)/a), -1/4*(b*x)/a]))/(256*a^3*(e*x)^(5/3)*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.25 (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, 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.

\(\displaystyle \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx\)

\(\Big \downarrow \) 148

\(\displaystyle \frac {3 \int \frac {1}{x \sqrt {a+b x} (4 a e+b x e)}d\sqrt [3]{e x}}{e}\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \int \frac {1}{e x \sqrt {a+b x} (4 a e+b x e)}d\sqrt [3]{e x}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {3 \sqrt {\frac {b x}{a}+1} \int \frac {1}{e x \sqrt {\frac {b x}{a}+1} (4 a e+b x e)}d\sqrt [3]{e x}}{\sqrt {a+b x}}\)

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {3 \sqrt {\frac {b x}{a}+1} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},-\frac {b x}{4 a},-\frac {b x}{a}\right )}{8 a e (e x)^{2/3} \sqrt {a+b x}}\)

Input:

Int[1/((e*x)^(5/3)*Sqrt[a + b*x]*(4*a + b*x)),x]
 

Output:

(-3*Sqrt[1 + (b*x)/a]*AppellF1[-2/3, 1, 1/2, 1/3, -1/4*(b*x)/a, -((b*x)/a) 
])/(8*a*e*(e*x)^(2/3)*Sqrt[a + b*x])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 148

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]
 

rule 1012

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])
 

rule 1013

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])
 

Maple [F]

\[\int \frac {1}{\left (e x \right )^{\frac {5}{3}} \sqrt {b x +a}\, \left (b x +4 a \right )}d x\]

Input:

int(1/(e*x)^(5/3)/(b*x+a)^(1/2)/(b*x+4*a),x)
 

Output:

int(1/(e*x)^(5/3)/(b*x+a)^(1/2)/(b*x+4*a),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=\text {Timed out} \] Input:

integrate(1/(e*x)^(5/3)/(b*x+a)^(1/2)/(b*x+4*a),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int \frac {1}{\left (e x\right )^{\frac {5}{3}} \sqrt {a + b x} \left (4 a + b x\right )}\, dx \] Input:

integrate(1/(e*x)**(5/3)/(b*x+a)**(1/2)/(b*x+4*a),x)
 

Output:

Integral(1/((e*x)**(5/3)*sqrt(a + b*x)*(4*a + b*x)), x)
 

Maxima [F]

\[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int { \frac {1}{{\left (b x + 4 \, a\right )} \sqrt {b x + a} \left (e x\right )^{\frac {5}{3}}} \,d x } \] Input:

integrate(1/(e*x)^(5/3)/(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)^(5/3)), x)
 

Giac [F]

\[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int { \frac {1}{{\left (b x + 4 \, a\right )} \sqrt {b x + a} \left (e x\right )^{\frac {5}{3}}} \,d x } \] Input:

integrate(1/(e*x)^(5/3)/(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)^(5/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=\int \frac {1}{{\left (e\,x\right )}^{5/3}\,\left (4\,a+b\,x\right )\,\sqrt {a+b\,x}} \,d x \] Input:

int(1/((e*x)^(5/3)*(4*a + b*x)*(a + b*x)^(1/2)),x)
 

Output:

int(1/((e*x)^(5/3)*(4*a + b*x)*(a + b*x)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(e x)^{5/3} \sqrt {a+b x} (4 a+b x)} \, dx=\frac {-6 x^{\frac {1}{3}} \sqrt {b x +a}-8 \left (\int \frac {\sqrt {b x +a}}{4 x^{\frac {2}{3}} a^{2}+5 x^{\frac {5}{3}} a b +x^{\frac {8}{3}} b^{2}}d x \right ) a b x -\left (\int \frac {x^{\frac {1}{3}} \sqrt {b x +a}}{b^{2} x^{2}+5 a b x +4 a^{2}}d x \right ) b^{2} x}{16 e^{\frac {5}{3}} a^{2} x} \] Input:

int(1/(e*x)^(5/3)/(b*x+a)^(1/2)/(b*x+4*a),x)
 

Output:

(e**(1/3)*( - 6*x**(1/3)*sqrt(a + b*x) - 8*int(sqrt(a + b*x)/(4*x**(2/3)*a 
**2 + 5*x**(2/3)*a*b*x + x**(2/3)*b**2*x**2),x)*a*b*x - int((x**(1/3)*sqrt 
(a + b*x))/(4*a**2 + 5*a*b*x + b**2*x**2),x)*b**2*x))/(16*a**2*e**2*x)