\(\int \frac {1}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx\) [1657]

Optimal result

Integrand size = 26, antiderivative size = 68 \[ \int \frac {1}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=-\frac {3 (c+d x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},1,\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{2 d (b c-a d) \sqrt [3]{b c+a d+2 b d x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\sqrt [3]{c+d x} (b c+a d+2 b d x)^{4/3}} \, dx=\frac {3 (c+d x)^{2/3} \left (\frac {a d+b (c+2 d x)}{-b c+a d}\right )^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{2 d (a d+b (c+2 d x))^{4/3}} \] Input:

Integrate[1/((c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(4/3)),x]
 

Output:

(3*(c + d*x)^(2/3)*((a*d + b*(c + 2*d*x))/(-(b*c) + a*d))^(4/3)*Hypergeome 
tric2F1[2/3, 4/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)])/(2*d*(a*d + b*(c + 2* 
d*x))^(4/3))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.44, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

\(\displaystyle \int \frac {1}{\sqrt [3]{c+d x} (a d+b c+2 b d x)^{4/3}} \, dx\)

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {3 (c+d x)^{2/3} \sqrt [3]{-\frac {a d+b c+2 b d x}{b c-a d}} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},\frac {2 b (c+d x)}{b c-a d}\right )}{2 d (b c-a d) \sqrt [3]{a d+b c+2 b d x}}\)

Input:

Int[1/((c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(4/3)),x]
 

Output:

(-3*(c + d*x)^(2/3)*(-((b*c + a*d + 2*b*d*x)/(b*c - a*d)))^(1/3)*Hypergeom 
etric2F1[2/3, 4/3, 5/3, (2*b*(c + d*x))/(b*c - a*d)])/(2*d*(b*c - a*d)*(b* 
c + a*d + 2*b*d*x)^(1/3))
 

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]

\[\int \frac {1}{\left (x d +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {4}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm="fricas")
 

Output:

integral((2*b*d*x + b*c + a*d)^(2/3)*(d*x + c)^(2/3)/(4*b^2*d^3*x^3 + b^2* 
c^3 + 2*a*b*c^2*d + a^2*c*d^2 + 4*(2*b^2*c*d^2 + a*b*d^3)*x^2 + (5*b^2*c^2 
*d + 6*a*b*c*d^2 + a^2*d^3)*x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(4/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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