\(\int \frac {\sqrt [3]{c+d x}}{\sqrt {a+b x}} \, dx\) [417]

Optimal result

Integrand size = 19, antiderivative size = 381 \[ \int \frac {\sqrt [3]{c+d x}}{\sqrt {a+b x}} \, dx=\frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 b}-\frac {4\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d) \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{5 b^{4/3} d \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

6/5*(b*x+a)^(1/2)*(d*x+c)^(1/3)/b-4/5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*(- 
a*d+b*c)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))*(((-a*d+b*c)^(2/3)+b^(1/ 
3)*(-a*d+b*c)^(1/3)*(d*x+c)^(1/3)+b^(2/3)*(d*x+c)^(2/3))/((1-3^(1/2))*(-a* 
d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*(-a*d+ 
b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3)*(d 
*x+c)^(1/3)),2*I-I*3^(1/2))/b^(4/3)/d/(b*x+a)^(1/2)/(-(-a*d+b*c)^(1/3)*((- 
a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/((1-3^(1/2))*(-a*d+b*c)^(1/3)-b^(1/3 
)*(d*x+c)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt [3]{c+d x}}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} \sqrt [3]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \] Input:

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

Output:

(2*Sqrt[a + b*x]*(c + d*x)^(1/3)*Hypergeometric2F1[-1/3, 1/2, 3/2, (d*(a + 
 b*x))/(-(b*c) + a*d)])/(b*((b*(c + d*x))/(b*c - a*d))^(1/3))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {60, 73, 760}

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.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}}dx}{5 b}+\frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {6 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{5 b d}+\frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 b}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 b}-\frac {4\ 3^{3/4} \sqrt {2-\sqrt {3}} (b c-a d) \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{5 b^{4/3} d \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\)

Input:

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

Output:

(6*Sqrt[a + b*x]*(c + d*x)^(1/3))/(5*b) - (4*3^(3/4)*Sqrt[2 - Sqrt[3]]*(b* 
c - a*d)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))*Sqrt[((b*c - a*d)^( 
2/3) + b^(1/3)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3) 
)/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF 
[ArcSin[((1 + Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - 
Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))], -7 + 4*Sqrt[3]])/( 
5*b^(4/3)*d*Sqrt[-(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d* 
x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)] 
*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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