\(\int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx\) [486]

Optimal result

Integrand size = 19, antiderivative size = 58 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=-\frac {2 (c+d x)^{4/5} \operatorname {Hypergeometric2F1}\left (-\frac {7}{10},1,-\frac {1}{2},-\frac {d (a+b x)}{b c-a d}\right )}{3 (b c-a d) (a+b x)^{3/2}} \] Output:

-2/3*(d*x+c)^(4/5)*hypergeom([-7/10, 1],[-1/2],-d*(b*x+a)/(-a*d+b*c))/(-a* 
d+b*c)/(b*x+a)^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=-\frac {2 \sqrt [5]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{5},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} \sqrt [5]{c+d x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {80, 79}

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.

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {2 \sqrt [5]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{5},-\frac {1}{2},-\frac {d (a+b x)}{b c-a d}\right )}{3 b (a+b x)^{3/2} \sqrt [5]{c+d x}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(b*x + a)*(d*x + c)^(4/5)/(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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