\(\int \frac {\sqrt {a+b x}}{(c+d x)^{2/3}} \, dx\) [427]

Optimal result

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

6/5*(b*x+a)^(1/2)*(d*x+c)^(1/3)/d+6/5*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^(1/3)/d^2/(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 = 73, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt {a+b x}}{(c+d x)^{2/3}} \, dx=\frac {2 (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {3}{2},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (c+d x)^{2/3}} \] Input:

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

Output:

(2*(a + b*x)^(3/2)*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometric2F1[2/3 
, 3/2, 5/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(c + d*x)^(2/3))
 

Rubi [A] (verified)

Time = 0.30 (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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {6 \sqrt {a+b x} \sqrt [3]{c+d x}}{5 d}-\frac {9 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{5 d^2}\)

\(\Big \downarrow \) 760

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

Input:

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

Output:

(6*Sqrt[a + b*x]*(c + d*x)^(1/3))/(5*d) + (6*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^(1/3)*d^2*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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