\(\int \frac {1}{\sqrt {a+b x} (c+d x)^{5/3} (3 b c-4 a d-b d x)} \, dx\) [1621]

Optimal result

Integrand size = 35, antiderivative size = 105 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/3} (3 b c-4 a d-b d x)} \, dx=-\frac {3 \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {AppellF1}\left (-\frac {2}{3},\frac {1}{2},1,\frac {1}{3},\frac {b (c+d x)}{b c-a d},\frac {b (c+d x)}{4 (b c-a d)}\right )}{8 d (b c-a d) \sqrt {a+b x} (c+d x)^{2/3}} \] Output:

-3/8*(-d*(b*x+a)/(-a*d+b*c))^(1/2)*AppellF1(-2/3,1,1/2,1/3,b*(d*x+c)/(-4*a 
*d+4*b*c),b*(d*x+c)/(-a*d+b*c))/d/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(2/3)
 

Mathematica [A] (warning: unable to verify)

Time = 21.61 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.92 \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/3} (3 b c-4 a d-b d x)} \, dx=-\frac {3 \left (-d (a+b x)+d (a+b x) \left (\frac {b (c+d x)}{d (a+b x)}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{6},\frac {2}{3},1,\frac {7}{6},\frac {-b c+a d}{d (a+b x)},\frac {3 b c-3 a d}{a d+b d x}\right )+(-b c+a d) \left (\frac {b (c+d x)}{d (a+b x)}\right )^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},\frac {2}{3},1,\frac {13}{6},\frac {-b c+a d}{d (a+b x)},\frac {3 b c-3 a d}{a d+b d x}\right )\right )}{8 d (b c-a d)^2 \sqrt {a+b x} (c+d x)^{2/3}} \] Input:

Integrate[1/(Sqrt[a + b*x]*(c + d*x)^(5/3)*(3*b*c - 4*a*d - b*d*x)),x]
 

Output:

(-3*(-(d*(a + b*x)) + d*(a + b*x)*((b*(c + d*x))/(d*(a + b*x)))^(2/3)*Appe 
llF1[1/6, 2/3, 1, 7/6, (-(b*c) + a*d)/(d*(a + b*x)), (3*b*c - 3*a*d)/(a*d 
+ b*d*x)] + (-(b*c) + a*d)*((b*(c + d*x))/(d*(a + b*x)))^(2/3)*AppellF1[7/ 
6, 2/3, 1, 13/6, (-(b*c) + a*d)/(d*(a + b*x)), (3*b*c - 3*a*d)/(a*d + b*d* 
x)]))/(8*d*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(2/3))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {149, 1013, 1012}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/3} (-4 a d+3 b c-b d x)} \, dx\)

\(\Big \downarrow \) 149

\(\displaystyle \frac {3 \int \frac {1}{(c+d x) (4 (b c-a d)-b (c+d x)) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{d}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {3 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{(c+d x) (4 (b c-a d)-b (c+d x)) \sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [3]{c+d x}}{d \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\)

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {3 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},\frac {b (c+d x)}{4 (b c-a d)},\frac {b (c+d x)}{b c-a d}\right )}{8 d (c+d x)^{2/3} (b c-a d) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\)

Input:

Int[1/(Sqrt[a + b*x]*(c + d*x)^(5/3)*(3*b*c - 4*a*d - b*d*x)),x]
 

Output:

(-3*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*AppellF1[-2/3, 1, 1/2, 1/3, (b*(c 
+ d*x))/(4*(b*c - a*d)), (b*(c + d*x))/(b*c - a*d)])/(8*d*(b*c - a*d)*(c + 
 d*x)^(2/3)*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])
 

Defintions of rubi rules used

rule 149

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) 
)^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1 
) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + 
 b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && 
IntegerQ[2*n] && IntegerQ[p]
 

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

\[\int \frac {1}{\sqrt {b x +a}\, \left (x d +c \right )^{\frac {5}{3}} \left (-b d x -4 a d +3 b c \right )}d x\]

Input:

int(1/(b*x+a)^(1/2)/(d*x+c)^(5/3)/(-b*d*x-4*a*d+3*b*c),x)
 

Output:

int(1/(b*x+a)^(1/2)/(d*x+c)^(5/3)/(-b*d*x-4*a*d+3*b*c),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/3} (3 b c-4 a d-b d x)} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(5/3)/(-b*d*x-4*a*d+3*b*c),x, algorithm= 
"fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(1/(b*x+a)**(1/2)/(d*x+c)**(5/3)/(-b*d*x-4*a*d+3*b*c),x)
 

Output:

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

Maxima [F]

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

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(5/3)/(-b*d*x-4*a*d+3*b*c),x, algorithm= 
"maxima")
 

Output:

-integrate(1/((b*d*x - 3*b*c + 4*a*d)*sqrt(b*x + a)*(d*x + c)^(5/3)), x)
 

Giac [F]

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

integrate(1/(b*x+a)^(1/2)/(d*x+c)^(5/3)/(-b*d*x-4*a*d+3*b*c),x, algorithm= 
"giac")
 

Output:

integrate(-1/((b*d*x - 3*b*c + 4*a*d)*sqrt(b*x + a)*(d*x + c)^(5/3)), x)
 

Mupad [F(-1)]

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

int(-1/((a + b*x)^(1/2)*(c + d*x)^(5/3)*(4*a*d - 3*b*c + b*d*x)),x)
 

Output:

-int(1/((a + b*x)^(1/2)*(c + d*x)^(5/3)*(4*a*d - 3*b*c + b*d*x)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/3} (3 b c-4 a d-b d x)} \, dx=-\left (\int \frac {\left (d x +c \right )^{\frac {1}{3}} \sqrt {b x +a}}{b^{2} d^{3} x^{4}+5 a b \,d^{3} x^{3}-b^{2} c \,d^{2} x^{3}+4 a^{2} d^{3} x^{2}+7 a b c \,d^{2} x^{2}-5 b^{2} c^{2} d \,x^{2}+8 a^{2} c \,d^{2} x -a b \,c^{2} d x -3 b^{2} c^{3} x +4 a^{2} c^{2} d -3 a b \,c^{3}}d x \right ) \] Input:

int(1/(b*x+a)^(1/2)/(d*x+c)^(5/3)/(-b*d*x-4*a*d+3*b*c),x)
 

Output:

 - int(((c + d*x)**(1/3)*sqrt(a + b*x))/(4*a**2*c**2*d + 8*a**2*c*d**2*x + 
 4*a**2*d**3*x**2 - 3*a*b*c**3 - a*b*c**2*d*x + 7*a*b*c*d**2*x**2 + 5*a*b* 
d**3*x**3 - 3*b**2*c**3*x - 5*b**2*c**2*d*x**2 - b**2*c*d**2*x**3 + b**2*d 
**3*x**4),x)