\(\int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{x^2} \, dx\) [484]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{x^2} \, dx=\frac {3 b (a+b x)^{4/3} \sqrt {c+d x} \operatorname {AppellF1}\left (\frac {4}{3},2,-\frac {1}{2},\frac {7}{3},1+\frac {b x}{a},-\frac {d (a+b x)}{b c-a d}\right )}{4 a^2 \sqrt {\frac {b (c+d x)}{b c-a d}}} \] Output:

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

Mathematica [A] (verified)

Time = 10.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt [3]{a+b x} \sqrt {c+d x}}{x^2} \, dx=\frac {-3 (2 b c+3 a d) \left (1+\frac {a}{b x}\right )^{2/3} \sqrt {1+\frac {c}{d x}} x \operatorname {AppellF1}\left (\frac {7}{6},\frac {2}{3},\frac {1}{2},\frac {13}{6},-\frac {a}{b x},-\frac {c}{d x}\right )-7 (c+d x) \left (3 (a+b x)-5 b x \left (\frac {d (a+b x)}{-b c+a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {b (c+d x)}{b c-a d}\right )\right )}{21 x (a+b x)^{2/3} \sqrt {c+d x}} \] Input:

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

Output:

(-3*(2*b*c + 3*a*d)*(1 + a/(b*x))^(2/3)*Sqrt[1 + c/(d*x)]*x*AppellF1[7/6, 
2/3, 1/2, 13/6, -(a/(b*x)), -(c/(d*x))] - 7*(c + d*x)*(3*(a + b*x) - 5*b*x 
*((d*(a + b*x))/(-(b*c) + a*d))^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, (b* 
(c + d*x))/(b*c - a*d)]))/(21*x*(a + b*x)^(2/3)*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.29 (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.

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

\(\Big \downarrow \) 149

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 54*sqrt(c + d*x)*(a + b*x)**(1/3)*a**2*d**2 - 108*sqrt(c + d*x)*(a + b 
*x)**(1/3)*a*b*c*d + 72*sqrt(c + d*x)*(a + b*x)**(1/3)*a*b*d**2*x - 48*sqr 
t(c + d*x)*(a + b*x)**(1/3)*b**2*c**2 + 108*sqrt(c + d*x)*(a + b*x)**(1/3) 
*b**2*c*d*x - 540*int((sqrt(c + d*x)*(a + b*x)**(1/3)*x)/(9*a**3*c*d**2 + 
9*a**3*d**3*x + 18*a**2*b*c**2*d + 27*a**2*b*c*d**2*x + 9*a**2*b*d**3*x**2 
 + 8*a*b**2*c**3 + 26*a*b**2*c**2*d*x + 18*a*b**2*c*d**2*x**2 + 8*b**3*c** 
3*x + 8*b**3*c**2*d*x**2),x)*a**3*b**2*d**5*x - 1890*int((sqrt(c + d*x)*(a 
 + b*x)**(1/3)*x)/(9*a**3*c*d**2 + 9*a**3*d**3*x + 18*a**2*b*c**2*d + 27*a 
**2*b*c*d**2*x + 9*a**2*b*d**3*x**2 + 8*a*b**2*c**3 + 26*a*b**2*c**2*d*x + 
 18*a*b**2*c*d**2*x**2 + 8*b**3*c**3*x + 8*b**3*c**2*d*x**2),x)*a**2*b**3* 
c*d**4*x - 2100*int((sqrt(c + d*x)*(a + b*x)**(1/3)*x)/(9*a**3*c*d**2 + 9* 
a**3*d**3*x + 18*a**2*b*c**2*d + 27*a**2*b*c*d**2*x + 9*a**2*b*d**3*x**2 + 
 8*a*b**2*c**3 + 26*a*b**2*c**2*d*x + 18*a*b**2*c*d**2*x**2 + 8*b**3*c**3* 
x + 8*b**3*c**2*d*x**2),x)*a*b**4*c**2*d**3*x - 720*int((sqrt(c + d*x)*(a 
+ b*x)**(1/3)*x)/(9*a**3*c*d**2 + 9*a**3*d**3*x + 18*a**2*b*c**2*d + 27*a* 
*2*b*c*d**2*x + 9*a**2*b*d**3*x**2 + 8*a*b**2*c**3 + 26*a*b**2*c**2*d*x + 
18*a*b**2*c*d**2*x**2 + 8*b**3*c**3*x + 8*b**3*c**2*d*x**2),x)*b**5*c**3*d 
**2*x + 243*int((sqrt(c + d*x)*(a + b*x)**(1/3))/(9*a**3*c*d**2*x + 9*a**3 
*d**3*x**2 + 18*a**2*b*c**2*d*x + 27*a**2*b*c*d**2*x**2 + 9*a**2*b*d**3*x* 
*3 + 8*a*b**2*c**3*x + 26*a*b**2*c**2*d*x**2 + 18*a*b**2*c*d**2*x**3 + ...