\(\int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\) [140]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {x^3}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {x}{b^4 c^2 \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {2 \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^5 c^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {3 a^2 b x-4 b^3 x^3-6 \sqrt {a-b x} \sqrt {a+b x} \left (a^2-b^2 x^2\right ) \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{3 b^5 c^2 \sqrt {c (a-b x)} (-a+b x) (a+b x)^{3/2}} \] Input:

Integrate[x^4/((a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]
 

Output:

(3*a^2*b*x - 4*b^3*x^3 - 6*Sqrt[a - b*x]*Sqrt[a + b*x]*(a^2 - b^2*x^2)*Arc 
Tan[Sqrt[a + b*x]/Sqrt[a - b*x]])/(3*b^5*c^2*Sqrt[c*(a - b*x)]*(-a + b*x)* 
(a + b*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {109, 27, 35, 100, 27, 87, 45, 218}

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.

\(\displaystyle \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx\)

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 35

\(\displaystyle \frac {x^3}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {\int \frac {x^2}{(a+b x)^{3/2} (a c-b c x)^{3/2}}dx}{b^2 c}\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {x^3}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {\frac {\int \frac {a b^2 c x}{\sqrt {a+b x} (a c-b c x)^{3/2}}dx}{a b^3 c}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}}{b^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^3}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {\frac {\int \frac {x}{\sqrt {a+b x} (a c-b c x)^{3/2}}dx}{b}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}}{b^2 c}\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^3}{3 b^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}}-\frac {\frac {\frac {\sqrt {a+b x}}{b^2 c \sqrt {a c-b c x}}-\frac {2 \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^2 c^{3/2}}}{b}-\frac {a}{b^3 c \sqrt {a+b x} \sqrt {a c-b c x}}}{b^2 c}\)

Input:

Int[x^4/((a + b*x)^(5/2)*(a*c - b*c*x)^(5/2)),x]
 

Output:

x^3/(3*b^2*c*(a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)) - (-(a/(b^3*c*Sqrt[a + b 
*x]*Sqrt[a*c - b*c*x])) + (Sqrt[a + b*x]/(b^2*c*Sqrt[a*c - b*c*x]) - (2*Ar 
cTan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[a*c - b*c*x]])/(b^2*c^(3/2)))/b)/(b^2*c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 35

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

rule 45

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(219\) vs. \(2(88)=176\).

Time = 0.27 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.08

method	result	size
default	\(\frac {\sqrt {c \left (-b x +a \right )}\, \left (3 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {\left (b x +a \right ) c \left (-b x +a \right )}}\right ) b^{4} c \,x^{4}-6 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {\left (b x +a \right ) c \left (-b x +a \right )}}\right ) a^{2} b^{2} c \,x^{2}+4 b^{2} x^{3} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}+3 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {\left (b x +a \right ) c \left (-b x +a \right )}}\right ) a^{4} c -3 \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}\, a^{2} x \right )}{3 c^{3} \sqrt {b^{2} c}\, \left (-b x +a \right )^{2} \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, b^{4} \left (b x +a \right )^{\frac {3}{2}}}\)	\(220\)

Input:

int(x^4/(b*x+a)^(5/2)/(-b*c*x+a*c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*(c*(-b*x+a))^(1/2)/c^3*(3*arctan((b^2*c)^(1/2)*x/((b*x+a)*c*(-b*x+a))^ 
(1/2))*b^4*c*x^4-6*arctan((b^2*c)^(1/2)*x/((b*x+a)*c*(-b*x+a))^(1/2))*a^2* 
b^2*c*x^2+4*b^2*x^3*(b^2*c)^(1/2)*(c*(-b^2*x^2+a^2))^(1/2)+3*arctan((b^2*c 
)^(1/2)*x/((b*x+a)*c*(-b*x+a))^(1/2))*a^4*c-3*(c*(-b^2*x^2+a^2))^(1/2)*(b^ 
2*c)^(1/2)*a^2*x)/(b^2*c)^(1/2)/(-b*x+a)^2/(c*(-b^2*x^2+a^2))^(1/2)/b^4/(b 
*x+a)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.78 \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\left [-\frac {3 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \sqrt {-c} \log \left (2 \, b^{2} c x^{2} - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) - 2 \, {\left (4 \, b^{3} x^{3} - 3 \, a^{2} b x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{6 \, {\left (b^{9} c^{3} x^{4} - 2 \, a^{2} b^{7} c^{3} x^{2} + a^{4} b^{5} c^{3}\right )}}, -\frac {3 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) - {\left (4 \, b^{3} x^{3} - 3 \, a^{2} b x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{3 \, {\left (b^{9} c^{3} x^{4} - 2 \, a^{2} b^{7} c^{3} x^{2} + a^{4} b^{5} c^{3}\right )}}\right ] \] Input:

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

Output:

[-1/6*(3*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*sqrt(-c)*log(2*b^2*c*x^2 - 2*sqrt 
(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(-c)*x - a^2*c) - 2*(4*b^3*x^3 - 3*a^2* 
b*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(b^9*c^3*x^4 - 2*a^2*b^7*c^3*x^2 + 
a^4*b^5*c^3), -1/3*(3*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*sqrt(c)*arctan(sqrt( 
-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - a^2*c)) - (4*b^3*x^3 
- 3*a^2*b*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(b^9*c^3*x^4 - 2*a^2*b^7*c^ 
3*x^2 + a^4*b^5*c^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.95 \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{2} c} - \frac {2 \, a^{2}}{{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} b^{4} c}\right )} - \frac {x}{3 \, \sqrt {-b^{2} c x^{2} + a^{2} c} b^{4} c^{2}} + \frac {\arcsin \left (\frac {b x}{a}\right )}{b^{5} c^{\frac {5}{2}}} \] Input:

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

Output:

1/3*x*(3*x^2/((-b^2*c*x^2 + a^2*c)^(3/2)*b^2*c) - 2*a^2/((-b^2*c*x^2 + a^2 
*c)^(3/2)*b^4*c)) - 1/3*x/(sqrt(-b^2*c*x^2 + a^2*c)*b^4*c^2) + arcsin(b*x/ 
a)/(b^5*c^(5/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (88) = 176\).

Time = 0.21 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.36 \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {\frac {\sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a} {\left (\frac {8 \, {\left (b x + a\right )}}{b^{4} c} - \frac {15 \, a}{b^{4} c}\right )}}{{\left ({\left (b x + a\right )} c - 2 \, a c\right )}^{2}} - \frac {12 \, \log \left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}\right )}{b^{4} \sqrt {-c} c^{2}} + \frac {4 \, {\left (9 \, a {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} - 30 \, a^{2} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} c + 32 \, a^{3} c^{2}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} - 2 \, a c\right )}^{3} b^{4} \sqrt {-c} c}}{12 \, b} \] Input:

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

Output:

1/12*(sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)*(8*(b*x + a)/(b^4*c) - 15*a 
/(b^4*c))/((b*x + a)*c - 2*a*c)^2 - 12*log((sqrt(b*x + a)*sqrt(-c) - sqrt( 
-(b*x + a)*c + 2*a*c))^2)/(b^4*sqrt(-c)*c^2) + 4*(9*a*(sqrt(b*x + a)*sqrt( 
-c) - sqrt(-(b*x + a)*c + 2*a*c))^4 - 30*a^2*(sqrt(b*x + a)*sqrt(-c) - sqr 
t(-(b*x + a)*c + 2*a*c))^2*c + 32*a^3*c^2)/(((sqrt(b*x + a)*sqrt(-c) - sqr 
t(-(b*x + a)*c + 2*a*c))^2 - 2*a*c)^3*b^4*sqrt(-c)*c))/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.23 \[ \int \frac {x^4}{(a+b x)^{5/2} (a c-b c x)^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-6 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{2}+6 \sqrt {b x +a}\, \sqrt {-b x +a}\, \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) b^{2} x^{2}-3 a^{2} b x +4 b^{3} x^{3}\right )}{3 \sqrt {b x +a}\, \sqrt {-b x +a}\, b^{5} c^{3} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:

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

Output:

(sqrt(c)*( - 6*sqrt(a + b*x)*sqrt(a - b*x)*asin(sqrt(a - b*x)/(sqrt(a)*sqr 
t(2)))*a**2 + 6*sqrt(a + b*x)*sqrt(a - b*x)*asin(sqrt(a - b*x)/(sqrt(a)*sq 
rt(2)))*b**2*x**2 - 3*a**2*b*x + 4*b**3*x**3))/(3*sqrt(a + b*x)*sqrt(a - b 
*x)*b**5*c**3*(a**2 - b**2*x**2))