\(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx\) [430]

Optimal result

Integrand size = 19, antiderivative size = 421 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx=-\frac {2 \sqrt [3]{c+d x}}{3 (b c-a d) (a+b x)^{3/2}}+\frac {14 d \sqrt [3]{c+d x}}{9 (b c-a d)^2 \sqrt {a+b x}}-\frac {14 \sqrt {2-\sqrt {3}} d \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} \sqrt [3]{b} (b c-a d)^2 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

-2/3*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(3/2)+14/9*d*(d*x+c)^(1/3)/(-a*d+b*c 
)^2/(b*x+a)^(1/2)-14/27*(1/2*6^(1/2)-1/2*2^(1/2))*d*((-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))*3^( 
3/4)/b^(1/3)/(-a*d+b*c)^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 = 0.17 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx=-\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2}{3},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} (c+d x)^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 445, 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, 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.

\(\displaystyle \int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{2/3}}dx}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 d \left (-\frac {d \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}}dx}{3 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {7 d \left (-\frac {\int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{b c-a d}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 760

\(\displaystyle -\frac {7 d \left (\frac {2 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} (b c-a d) \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\)

Input:

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

Output:

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

\[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (x d +c \right )^{\frac {2}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x, algorithm="fricas")
 

Output:

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

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

integrate(1/(b*x+a)**(5/2)/(d*x+c)**(2/3),x)
 

Output:

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(2/3)), x)
 

Maxima [F]

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

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(2/3)), x)
 

Giac [F]

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

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(2/3),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(2/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{2/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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