\(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx\) [510]

Optimal result

Integrand size = 19, antiderivative size = 410 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx=-\frac {2 \sqrt [6]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {16 d \sqrt [6]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}+\frac {16 d \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{9 \sqrt [4]{3} (b c-a d)^{7/3} \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

-2/3*(d*x+c)^(1/6)/(-a*d+b*c)/(b*x+a)^(3/2)+16/9*d*(d*x+c)^(1/6)/(-a*d+b*c 
)^2/(b*x+a)^(1/2)+16/27*d*(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))*3^(3/4)/(-a*d+b*c)^(7/3)/(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 = 73, normalized size of antiderivative = 0.18 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx=-\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{6},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} (c+d x)^{5/6}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {61, 61, 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.

\(\displaystyle \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {8 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/6}}dx}{9 (b c-a d)}-\frac {2 \sqrt [6]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {8 d \left (-\frac {2 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}}dx}{3 (b c-a d)}-\frac {2 \sqrt [6]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [6]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {8 d \left (-\frac {4 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{b c-a d}-\frac {2 \sqrt [6]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [6]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 766

\(\displaystyle -\frac {8 d \left (-\frac {2 \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} (b c-a d)^{4/3} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [6]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [6]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

Input:

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

Output:

(-2*(c + d*x)^(1/6))/(3*(b*c - a*d)*(a + b*x)^(3/2)) - (8*d*((-2*(c + d*x) 
^(1/6))/((b*c - a*d)*Sqrt[a + b*x]) - (2*(c + d*x)^(1/6)*((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))/((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])/(3^(1/4)*(b*c - a*d)^(4/3) 
*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])))/(9*(b*c - a*d))
 

Defintions of rubi rules used

rule 61

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

\[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (x d +c \right )^{\frac {5}{6}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(b*x + a)*(d*x + c)^(1/6)/(b^3*d*x^4 + a^3*c + (b^3*c + 3*a*b 
^2*d)*x^3 + 3*(a*b^2*c + a^2*b*d)*x^2 + (3*a^2*b*c + a^3*d)*x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/6}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.51 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/6}} \, dx=\frac {2 \sqrt {d x +c}\, \sqrt {b x +a}\, \left (-8 b^{2} d^{2} x^{2}-12 a b \,d^{2} x -4 b^{2} c d x -3 a^{2} d^{2}-6 a b c d +b^{2} c^{2}\right )}{3 \left (d x +c \right )^{\frac {1}{3}} \left (a^{3} b^{2} d^{3} x^{2}-3 a^{2} b^{3} c \,d^{2} x^{2}+3 a \,b^{4} c^{2} d \,x^{2}-b^{5} c^{3} x^{2}+2 a^{4} b \,d^{3} x -6 a^{3} b^{2} c \,d^{2} x +6 a^{2} b^{3} c^{2} d x -2 a \,b^{4} c^{3} x +a^{5} d^{3}-3 a^{4} b c \,d^{2}+3 a^{3} b^{2} c^{2} d -a^{2} b^{3} c^{3}\right )} \] Input:

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

Output:

(2*sqrt(c + d*x)*sqrt(a + b*x)*( - 3*a**2*d**2 - 6*a*b*c*d - 12*a*b*d**2*x 
 + b**2*c**2 - 4*b**2*c*d*x - 8*b**2*d**2*x**2))/(3*(c + d*x)**(1/3)*(a**5 
*d**3 - 3*a**4*b*c*d**2 + 2*a**4*b*d**3*x + 3*a**3*b**2*c**2*d - 6*a**3*b* 
*2*c*d**2*x + a**3*b**2*d**3*x**2 - a**2*b**3*c**3 + 6*a**2*b**3*c**2*d*x 
- 3*a**2*b**3*c*d**2*x**2 - 2*a*b**4*c**3*x + 3*a*b**4*c**2*d*x**2 - b**5* 
c**3*x**2))