\(\int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx\) [508]

Optimal result

Integrand size = 19, antiderivative size = 343 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx=\frac {3^{3/4} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{d \sqrt [3]{b c-a d} \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

3^(3/4)*(d*x+c)^(1/6)*((-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))/((-a 
*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))^2)^(1/2)*InverseJacobiAM( 
arccos(((-a*d+b*c)^(1/3)-(1-3^(1/2))*b^(1/3)*(d*x+c)^(1/3))/((-a*d+b*c)^(1 
/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))/d/(-a*d+b 
*c)^(1/3)/(b*x+a)^(1/2)/(-b^(1/3)*(d*x+c)^(1/3)*((-a*d+b*c)^(1/3)-b^(1/3)* 
(d*x+c)^(1/3))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*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 = 0.21 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx=\frac {2 \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b (c+d x)^{5/6}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {73, 766}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 766

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

Input:

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

Output:

(3^(3/4)*(c + d*x)^(1/6)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))*Sqr 
t[((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))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/ 
3))^2]*EllipticF[ArcCos[((b*c - a*d)^(1/3) - (1 - Sqrt[3])*b^(1/3)*(c + d* 
x)^(1/3))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))], (2 
 + Sqrt[3])/4])/(d*(b*c - a*d)^(1/3)*Sqrt[-((b^(1/3)*(c + d*x)^(1/3)*((b*c 
 - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((b*c - a*d)^(1/3) - (1 + Sqrt[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 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 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

Maple [F]

\[\int \frac {1}{\sqrt {b x +a}\, \left (x d +c \right )^{\frac {5}{6}}}d x\]

Input:

int(1/(b*x+a)^(1/2)/(d*x+c)^(5/6),x)
 

Output:

int(1/(b*x+a)^(1/2)/(d*x+c)^(5/6),x)
 

Fricas [F]

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

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(5/6),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x + a)*(d*x + c)^(1/6)/(b*d*x^2 + a*c + (b*c + a*d)*x), x)
 

Sympy [F]

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

integrate(1/(b*x+a)**(1/2)/(d*x+c)**(5/6),x)
 

Output:

Integral(1/(sqrt(a + b*x)*(c + d*x)**(5/6)), x)
 

Maxima [F]

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

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(5/6),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(b*x + a)*(d*x + c)^(5/6)), x)
 

Giac [F]

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

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(5/6),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(b*x + a)*(d*x + c)^(5/6)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx=\int \frac {1}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/6}} \,d x \] Input:

int(1/((a + b*x)^(1/2)*(c + d*x)^(5/6)),x)
 

Output:

int(1/((a + b*x)^(1/2)*(c + d*x)^(5/6)), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.09 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx=-\frac {2 \sqrt {d x +c}\, \sqrt {b x +a}}{\left (d x +c \right )^{\frac {1}{3}} \left (a d -b c \right )} \] Input:

int(1/(b*x+a)^(1/2)/(d*x+c)^(5/6),x)
 

Output:

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