\(\int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx\) [483]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx=\frac {2 b (a+b x)^{3/2} \sqrt [3]{c+d x} \operatorname {AppellF1}\left (\frac {3}{2},2,-\frac {1}{3},\frac {5}{2},1+\frac {b x}{a},-\frac {d (a+b x)}{b c-a d}\right )}{3 a^2 \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \] Output:

2/3*b*(b*x+a)^(3/2)*(d*x+c)^(1/3)*AppellF1(3/2,-1/3,2,5/2,-d*(b*x+a)/(-a*d 
+b*c),1+b*x/a)/a^2/(b*(d*x+c)/(-a*d+b*c))^(1/3)
 

Mathematica [A] (verified)

Time = 10.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx=\frac {-2 (3 b c+2 a d) \sqrt {1+\frac {a}{b x}} \left (1+\frac {c}{d x}\right )^{2/3} x \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},\frac {2}{3},\frac {13}{6},-\frac {a}{b x},-\frac {c}{d x}\right )-7 (c+d x) \left (2 (a+b x)-5 b x \sqrt {\frac {d (a+b x)}{-b c+a d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\frac {b (c+d x)}{b c-a d}\right )\right )}{14 x \sqrt {a+b x} (c+d x)^{2/3}} \] Input:

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

Output:

(-2*(3*b*c + 2*a*d)*Sqrt[1 + a/(b*x)]*(1 + c/(d*x))^(2/3)*x*AppellF1[7/6, 
1/2, 2/3, 13/6, -(a/(b*x)), -(c/(d*x))] - 7*(c + d*x)*(2*(a + b*x) - 5*b*x 
*Sqrt[(d*(a + b*x))/(-(b*c) + a*d)]*Hypergeometric2F1[1/3, 1/2, 4/3, (b*(c 
 + d*x))/(b*c - a*d)]))/(14*x*Sqrt[a + b*x]*(c + d*x)^(2/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {149, 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.

\(\displaystyle \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx\)

\(\Big \downarrow \) 149

\(\displaystyle \frac {3 \int \frac {(c+d x) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{x^2}d\sqrt [3]{c+d x}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle 3 d \int \frac {(c+d x) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}{d^2 x^2}d\sqrt [3]{c+d x}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {3 d \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \int \frac {(c+d x) \sqrt {1-\frac {b (c+d x)}{b c-a d}}}{d^2 x^2}d\sqrt [3]{c+d x}}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {3 d (c+d x)^{4/3} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}} \operatorname {AppellF1}\left (\frac {4}{3},2,-\frac {1}{2},\frac {7}{3},\frac {c+d x}{c},\frac {b (c+d x)}{b c-a d}\right )}{4 c^2 \sqrt {1-\frac {b (c+d x)}{b c-a d}}}\)

Input:

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

Output:

(3*d*(c + d*x)^(4/3)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d]*AppellF1[4/3, 2, 
-1/2, 7/3, (c + d*x)/c, (b*(c + d*x))/(b*c - a*d)])/(4*c^2*Sqrt[1 - (b*(c 
+ d*x))/(b*c - a*d)])
 

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 149

Int[((a_) + (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 - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + 
 b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && 
IntegerQ[2*n] && 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]

\[\int \frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {1}{3}}}{x^{2}}d x\]

Input:

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

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx=\int \frac {\sqrt {a + b x} \sqrt [3]{c + d x}}{x^{2}}\, dx \] Input:

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

Output:

Integral(sqrt(a + b*x)*(c + d*x)**(1/3)/x**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx=\int \frac {\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{1/3}}{x^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+b x} \sqrt [3]{c+d x}}{x^2} \, dx=\text {too large to display} \] Input:

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

Output:

( - 48*(c + d*x)**(1/3)*sqrt(a + b*x)*a**2*d**2 - 108*(c + d*x)**(1/3)*sqr 
t(a + b*x)*a*b*c*d + 108*(c + d*x)**(1/3)*sqrt(a + b*x)*a*b*d**2*x - 54*(c 
 + d*x)**(1/3)*sqrt(a + b*x)*b**2*c**2 + 72*(c + d*x)**(1/3)*sqrt(a + b*x) 
*b**2*c*d*x - 720*int(((c + d*x)**(1/3)*sqrt(a + b*x)*x)/(8*a**3*c*d**2 + 
8*a**3*d**3*x + 18*a**2*b*c**2*d + 26*a**2*b*c*d**2*x + 8*a**2*b*d**3*x**2 
 + 9*a*b**2*c**3 + 27*a*b**2*c**2*d*x + 18*a*b**2*c*d**2*x**2 + 9*b**3*c** 
3*x + 9*b**3*c**2*d*x**2),x)*a**3*b**2*d**5*x - 2100*int(((c + d*x)**(1/3) 
*sqrt(a + b*x)*x)/(8*a**3*c*d**2 + 8*a**3*d**3*x + 18*a**2*b*c**2*d + 26*a 
**2*b*c*d**2*x + 8*a**2*b*d**3*x**2 + 9*a*b**2*c**3 + 27*a*b**2*c**2*d*x + 
 18*a*b**2*c*d**2*x**2 + 9*b**3*c**3*x + 9*b**3*c**2*d*x**2),x)*a**2*b**3* 
c*d**4*x - 1890*int(((c + d*x)**(1/3)*sqrt(a + b*x)*x)/(8*a**3*c*d**2 + 8* 
a**3*d**3*x + 18*a**2*b*c**2*d + 26*a**2*b*c*d**2*x + 8*a**2*b*d**3*x**2 + 
 9*a*b**2*c**3 + 27*a*b**2*c**2*d*x + 18*a*b**2*c*d**2*x**2 + 9*b**3*c**3* 
x + 9*b**3*c**2*d*x**2),x)*a*b**4*c**2*d**3*x - 540*int(((c + d*x)**(1/3)* 
sqrt(a + b*x)*x)/(8*a**3*c*d**2 + 8*a**3*d**3*x + 18*a**2*b*c**2*d + 26*a* 
*2*b*c*d**2*x + 8*a**2*b*d**3*x**2 + 9*a*b**2*c**3 + 27*a*b**2*c**2*d*x + 
18*a*b**2*c*d**2*x**2 + 9*b**3*c**3*x + 9*b**3*c**2*d*x**2),x)*b**5*c**3*d 
**2*x + 128*int(((c + d*x)**(1/3)*sqrt(a + b*x))/(8*a**3*c*d**2*x + 8*a**3 
*d**3*x**2 + 18*a**2*b*c**2*d*x + 26*a**2*b*c*d**2*x**2 + 8*a**2*b*d**3*x* 
*3 + 9*a*b**2*c**3*x + 27*a*b**2*c**2*d*x**2 + 18*a*b**2*c*d**2*x**3 + ...