\(\int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx\) [480]

Optimal result
Mathematica [B] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 78 \[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=-\frac {3 \sqrt {a+b x} \sqrt {c+d x} \operatorname {AppellF1}\left (-\frac {2}{3},-\frac {1}{2},-\frac {1}{2},\frac {1}{3},-\frac {b x}{a},-\frac {d x}{c}\right )}{2 x^{2/3} \sqrt {1+\frac {b x}{a}} \sqrt {1+\frac {d x}{c}}} \] Output:

-3/2*(b*x+a)^(1/2)*(d*x+c)^(1/2)*AppellF1(-2/3,-1/2,-1/2,1/3,-b*x/a,-d*x/c 
)/x^(2/3)/(1+b*x/a)^(1/2)/(1+d*x/c)^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(159\) vs. \(2(78)=156\).

Time = 3.33 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.04 \[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\frac {-12 (a+b x) (c+d x)+18 (b c+a d) x \sqrt {1+\frac {b x}{a}} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {b x}{a},-\frac {d x}{c}\right )+9 b d x^2 \sqrt {1+\frac {b x}{a}} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},-\frac {b x}{a},-\frac {d x}{c}\right )}{8 x^{2/3} \sqrt {a+b x} \sqrt {c+d x}} \] Input:

Integrate[(Sqrt[a + b*x]*Sqrt[c + d*x])/x^(5/3),x]
 

Output:

(-12*(a + b*x)*(c + d*x) + 18*(b*c + a*d)*x*Sqrt[1 + (b*x)/a]*Sqrt[1 + (d* 
x)/c]*AppellF1[1/3, 1/2, 1/2, 4/3, -((b*x)/a), -((d*x)/c)] + 9*b*d*x^2*Sqr 
t[1 + (b*x)/a]*Sqrt[1 + (d*x)/c]*AppellF1[4/3, 1/2, 1/2, 7/3, -((b*x)/a), 
-((d*x)/c)])/(8*x^(2/3)*Sqrt[a + b*x]*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {152, 152, 150}

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.

\(\displaystyle \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {\sqrt {a+b x} \int \frac {\sqrt {\frac {b x}{a}+1} \sqrt {c+d x}}{x^{5/3}}dx}{\sqrt {\frac {b x}{a}+1}}\)

\(\Big \downarrow \) 152

\(\displaystyle \frac {\sqrt {a+b x} \sqrt {c+d x} \int \frac {\sqrt {\frac {b x}{a}+1} \sqrt {\frac {d x}{c}+1}}{x^{5/3}}dx}{\sqrt {\frac {b x}{a}+1} \sqrt {\frac {d x}{c}+1}}\)

\(\Big \downarrow \) 150

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

Input:

Int[(Sqrt[a + b*x]*Sqrt[c + d*x])/x^(5/3),x]
 

Output:

(-3*Sqrt[a + b*x]*Sqrt[c + d*x]*AppellF1[-2/3, -1/2, -1/2, 1/3, -((b*x)/a) 
, -((d*x)/c)])/(2*x^(2/3)*Sqrt[1 + (b*x)/a]*Sqrt[1 + (d*x)/c])
 

Defintions of rubi rules used

rule 150
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 152
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) 
Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, 
 n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]
 
Maple [F]

\[\int \frac {\sqrt {b x +a}\, \sqrt {x d +c}}{x^{\frac {5}{3}}}d x\]

Input:

int((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^(5/3),x)
 

Output:

int((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^(5/3),x)
 

Fricas [F]

\[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\int { \frac {\sqrt {b x + a} \sqrt {d x + c}}{x^{\frac {5}{3}}} \,d x } \] Input:

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

Output:

integral(sqrt(b*x + a)*sqrt(d*x + c)/x^(5/3), x)
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\int \frac {\sqrt {a + b x} \sqrt {c + d x}}{x^{\frac {5}{3}}}\, dx \] Input:

integrate((b*x+a)**(1/2)*(d*x+c)**(1/2)/x**(5/3),x)
 

Output:

Integral(sqrt(a + b*x)*sqrt(c + d*x)/x**(5/3), x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\int { \frac {\sqrt {b x + a} \sqrt {d x + c}}{x^{\frac {5}{3}}} \,d x } \] Input:

integrate((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^(5/3),x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x + a)*sqrt(d*x + c)/x^(5/3), x)
 

Giac [F]

\[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\int { \frac {\sqrt {b x + a} \sqrt {d x + c}}{x^{\frac {5}{3}}} \,d x } \] Input:

integrate((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^(5/3),x, algorithm="giac")
 

Output:

integrate(sqrt(b*x + a)*sqrt(d*x + c)/x^(5/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\int \frac {\sqrt {a+b\,x}\,\sqrt {c+d\,x}}{x^{5/3}} \,d x \] Input:

int(((a + b*x)^(1/2)*(c + d*x)^(1/2))/x^(5/3),x)
 

Output:

int(((a + b*x)^(1/2)*(c + d*x)^(1/2))/x^(5/3), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x} \sqrt {c+d x}}{x^{5/3}} \, dx=\frac {-6 \sqrt {d x +c}\, \sqrt {b x +a}+3 x^{\frac {2}{3}} \left (\int \frac {x^{\frac {1}{3}} \sqrt {d x +c}\, \sqrt {b x +a}}{b d \,x^{2}+a d x +b c x +a c}d x \right ) b d -3 x^{\frac {2}{3}} \left (\int \frac {\sqrt {d x +c}\, \sqrt {b x +a}}{x^{\frac {5}{3}} a c +x^{\frac {8}{3}} a d +x^{\frac {8}{3}} b c +x^{\frac {11}{3}} b d}d x \right ) a c}{x^{\frac {2}{3}}} \] Input:

int((b*x+a)^(1/2)*(d*x+c)^(1/2)/x^(5/3),x)
 

Output:

(3*( - 2*sqrt(c + d*x)*sqrt(a + b*x) + x**(2/3)*int((x**(1/3)*sqrt(c + d*x 
)*sqrt(a + b*x))/(a*c + a*d*x + b*c*x + b*d*x**2),x)*b*d - x**(2/3)*int((s 
qrt(c + d*x)*sqrt(a + b*x))/(x**(2/3)*a*c*x + x**(2/3)*a*d*x**2 + x**(2/3) 
*b*c*x**2 + x**(2/3)*b*d*x**3),x)*a*c))/x**(2/3)