\(\int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx\) [78]

Optimal result

Integrand size = 29, antiderivative size = 85 \[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=-\frac {2 \sqrt {2} \sqrt {1-d x} \sqrt [3]{e+f x} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-d x),\frac {f (1-d x)}{d e+f}\right )}{d \sqrt [3]{\frac {d (e+f x)}{d e+f}}} \] Output:

-2*2^(1/2)*(-d*x+1)^(1/2)*(f*x+e)^(1/3)*AppellF1(1/2,-1/3,-1/2,3/2,f*(-d*x 
+1)/(d*e+f),-1/2*d*x+1/2)/d/(d*(f*x+e)/(d*e+f))^(1/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(494\) vs. \(2(85)=170\).

Time = 19.50 (sec) , antiderivative size = 494, normalized size of antiderivative = 5.81 \[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=\frac {\sqrt {1-d x} \left (-9 d (1+d x) (e+f x)+\frac {18 (d e+f) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right ) \left (-30 (d e+f)+(d e+4 f) (1-d x) \sqrt {2+2 d x} \left (\frac {d (e+f x)}{d e+f}\right )^{2/3} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},\frac {2}{3},\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )\right )+(d e+4 f) (-1+d x)^2 \sqrt {2+2 d x} \left (\frac {d (e+f x)}{d e+f}\right )^{2/3} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},\frac {2}{3},\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right ) \left (8 f \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},\frac {5}{3},\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )+3 (d e+f) \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{2},\frac {2}{3},\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )\right )}{18 (d e+f) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )+(1-d x) \left (8 f \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},\frac {5}{3},\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )+3 (d e+f) \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{2},\frac {2}{3},\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )\right )}\right )}{12 d^2 \sqrt {1+d x} (e+f x)^{2/3}} \] Input:

Integrate[(Sqrt[1 + d*x]*(e + f*x)^(1/3))/Sqrt[1 - d*x],x]
 

Output:

(Sqrt[1 - d*x]*(-9*d*(1 + d*x)*(e + f*x) + (18*(d*e + f)*AppellF1[1/2, 1/2 
, 2/3, 3/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)]*(-30*(d*e + f) + (d*e + 
4*f)*(1 - d*x)*Sqrt[2 + 2*d*x]*((d*(e + f*x))/(d*e + f))^(2/3)*AppellF1[3/ 
2, 1/2, 2/3, 5/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)]) + (d*e + 4*f)*(-1 
 + d*x)^2*Sqrt[2 + 2*d*x]*((d*(e + f*x))/(d*e + f))^(2/3)*AppellF1[3/2, 1/ 
2, 2/3, 5/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)]*(8*f*AppellF1[3/2, 1/2, 
 5/3, 5/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)] + 3*(d*e + f)*AppellF1[3/ 
2, 3/2, 2/3, 5/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)]))/(18*(d*e + f)*Ap 
pellF1[1/2, 1/2, 2/3, 3/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)] + (1 - d* 
x)*(8*f*AppellF1[3/2, 1/2, 5/3, 5/2, 1/2 - (d*x)/2, (f - d*f*x)/(d*e + f)] 
 + 3*(d*e + f)*AppellF1[3/2, 3/2, 2/3, 5/2, 1/2 - (d*x)/2, (f - d*f*x)/(d* 
e + f)]))))/(12*d^2*Sqrt[1 + d*x]*(e + f*x)^(2/3))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {156, 155}

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.

\(\displaystyle \int \frac {\sqrt {d x+1} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\sqrt [3]{e+f x} \int \frac {\sqrt {d x+1} \sqrt [3]{\frac {d e}{d e+f}+\frac {d f x}{d e+f}}}{\sqrt {1-d x}}dx}{\sqrt [3]{\frac {d (e+f x)}{d e+f}}}\)

\(\Big \downarrow \) 155

\(\displaystyle -\frac {2 \sqrt {2} \sqrt {1-d x} \sqrt [3]{e+f x} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-d x),\frac {f (1-d x)}{d e+f}\right )}{d \sqrt [3]{\frac {d (e+f x)}{d e+f}}}\)

Input:

Int[(Sqrt[1 + d*x]*(e + f*x)^(1/3))/Sqrt[1 - d*x],x]
 

Output:

(-2*Sqrt[2]*Sqrt[1 - d*x]*(e + f*x)^(1/3)*AppellF1[1/2, -1/2, -1/3, 3/2, ( 
1 - d*x)/2, (f*(1 - d*x))/(d*e + f)])/(d*((d*(e + f*x))/(d*e + f))^(1/3))
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

Maple [F]

\[\int \frac {\sqrt {d x +1}\, \left (f x +e \right )^{\frac {1}{3}}}{\sqrt {-d x +1}}d x\]

Input:

int((d*x+1)^(1/2)*(f*x+e)^(1/3)/(-d*x+1)^(1/2),x)
 

Output:

int((d*x+1)^(1/2)*(f*x+e)^(1/3)/(-d*x+1)^(1/2),x)
 

Fricas [F]

\[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=\int { \frac {\sqrt {d x + 1} {\left (f x + e\right )}^{\frac {1}{3}}}{\sqrt {-d x + 1}} \,d x } \] Input:

integrate((d*x+1)^(1/2)*(f*x+e)^(1/3)/(-d*x+1)^(1/2),x, algorithm="fricas" 
)
 

Output:

integral(-sqrt(d*x + 1)*sqrt(-d*x + 1)*(f*x + e)^(1/3)/(d*x - 1), x)
 

Sympy [F]

\[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=\int \frac {\sqrt [3]{e + f x} \sqrt {d x + 1}}{\sqrt {- d x + 1}}\, dx \] Input:

integrate((d*x+1)**(1/2)*(f*x+e)**(1/3)/(-d*x+1)**(1/2),x)
 

Output:

Integral((e + f*x)**(1/3)*sqrt(d*x + 1)/sqrt(-d*x + 1), x)
 

Maxima [F]

\[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=\int { \frac {\sqrt {d x + 1} {\left (f x + e\right )}^{\frac {1}{3}}}{\sqrt {-d x + 1}} \,d x } \] Input:

integrate((d*x+1)^(1/2)*(f*x+e)^(1/3)/(-d*x+1)^(1/2),x, algorithm="maxima" 
)
 

Output:

integrate(sqrt(d*x + 1)*(f*x + e)^(1/3)/sqrt(-d*x + 1), x)
 

Giac [F]

\[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=\int { \frac {\sqrt {d x + 1} {\left (f x + e\right )}^{\frac {1}{3}}}{\sqrt {-d x + 1}} \,d x } \] Input:

integrate((d*x+1)^(1/2)*(f*x+e)^(1/3)/(-d*x+1)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(d*x + 1)*(f*x + e)^(1/3)/sqrt(-d*x + 1), x)
 

Mupad [F(-1)]

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

int(((e + f*x)^(1/3)*(d*x + 1)^(1/2))/(1 - d*x)^(1/2),x)
 

Output:

int(((e + f*x)^(1/3)*(d*x + 1)^(1/2))/(1 - d*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {1+d x} \sqrt [3]{e+f x}}{\sqrt {1-d x}} \, dx=\frac {-3 \left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}\, d e -3 \left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}\, f +\left (\int \frac {\left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}\, x^{2}}{d^{2} f \,x^{3}+d^{2} e \,x^{2}-f x -e}d x \right ) d^{3} e f +4 \left (\int \frac {\left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}\, x^{2}}{d^{2} f \,x^{3}+d^{2} e \,x^{2}-f x -e}d x \right ) d^{2} f^{2}-3 \left (\int \frac {\left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}}{d^{2} f \,x^{3}+d^{2} e \,x^{2}-f x -e}d x \right ) d^{2} e^{2}-\left (\int \frac {\left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}}{d^{2} f \,x^{3}+d^{2} e \,x^{2}-f x -e}d x \right ) d e f -\left (\int \frac {\left (f x +e \right )^{\frac {1}{3}} \sqrt {d x +1}\, \sqrt {-d x +1}}{d^{2} f \,x^{3}+d^{2} e \,x^{2}-f x -e}d x \right ) f^{2}}{3 d^{2} e} \] Input:

int((d*x+1)^(1/2)*(f*x+e)^(1/3)/(-d*x+1)^(1/2),x)
 

Output:

( - 3*(e + f*x)**(1/3)*sqrt(d*x + 1)*sqrt( - d*x + 1)*d*e - 3*(e + f*x)**( 
1/3)*sqrt(d*x + 1)*sqrt( - d*x + 1)*f + int(((e + f*x)**(1/3)*sqrt(d*x + 1 
)*sqrt( - d*x + 1)*x**2)/(d**2*e*x**2 + d**2*f*x**3 - e - f*x),x)*d**3*e*f 
 + 4*int(((e + f*x)**(1/3)*sqrt(d*x + 1)*sqrt( - d*x + 1)*x**2)/(d**2*e*x* 
*2 + d**2*f*x**3 - e - f*x),x)*d**2*f**2 - 3*int(((e + f*x)**(1/3)*sqrt(d* 
x + 1)*sqrt( - d*x + 1))/(d**2*e*x**2 + d**2*f*x**3 - e - f*x),x)*d**2*e** 
2 - int(((e + f*x)**(1/3)*sqrt(d*x + 1)*sqrt( - d*x + 1))/(d**2*e*x**2 + d 
**2*f*x**3 - e - f*x),x)*d*e*f - int(((e + f*x)**(1/3)*sqrt(d*x + 1)*sqrt( 
 - d*x + 1))/(d**2*e*x**2 + d**2*f*x**3 - e - f*x),x)*f**2)/(3*d**2*e)