\(\int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx\) [103]

Optimal result

Integrand size = 26, antiderivative size = 177 \[ \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx=-\frac {3 a^6 c x \sqrt {a+b x} \sqrt {a c-b c x}}{128 b^4}-\frac {a^4 c x^3 \sqrt {a+b x} \sqrt {a c-b c x}}{64 b^2}+\frac {1}{16} a^2 c x^5 \sqrt {a+b x} \sqrt {a c-b c x}+\frac {1}{8} x^5 (a+b x)^{3/2} (a c-b c x)^{3/2}+\frac {3 a^8 c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{64 b^5} \] Output:

-3/128*a^6*c*x*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2)/b^4-1/64*a^4*c*x^3*(b*x+a) 
^(1/2)*(-b*c*x+a*c)^(1/2)/b^2+1/16*a^2*c*x^5*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1 
/2)+1/8*x^5*(b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2)+3/64*a^8*c^(3/2)*arctan(c^(1/ 
2)*(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2))/b^5
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.65 \[ \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx=\frac {(c (a-b x))^{3/2} \left (-b x \sqrt {a-b x} \sqrt {a+b x} \left (3 a^6+2 a^4 b^2 x^2-24 a^2 b^4 x^4+16 b^6 x^6\right )+6 a^8 \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )\right )}{128 b^5 (a-b x)^{3/2}} \] Input:

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

Output:

((c*(a - b*x))^(3/2)*(-(b*x*Sqrt[a - b*x]*Sqrt[a + b*x]*(3*a^6 + 2*a^4*b^2 
*x^2 - 24*a^2*b^4*x^4 + 16*b^6*x^6)) + 6*a^8*ArcTan[Sqrt[a + b*x]/Sqrt[a - 
 b*x]]))/(128*b^5*(a - b*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {111, 27, 101, 25, 27, 40, 40, 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.

\(\displaystyle \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx\)

\(\Big \downarrow \) 111

\(\displaystyle -\frac {\int -3 a^2 c x^2 (a+b x)^{3/2} (a c-b c x)^{3/2}dx}{8 b^2 c}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 a^2 \int x^2 (a+b x)^{3/2} (a c-b c x)^{3/2}dx}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 101

\(\displaystyle \frac {3 a^2 \left (-\frac {\int -a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}dx}{6 b^2 c}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 a^2 \left (\frac {\int a^2 c (a+b x)^{3/2} (a c-b c x)^{3/2}dx}{6 b^2 c}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 a^2 \left (\frac {a^2 \int (a+b x)^{3/2} (a c-b c x)^{3/2}dx}{6 b^2}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 40

\(\displaystyle \frac {3 a^2 \left (\frac {a^2 \left (\frac {3}{4} a^2 c \int \sqrt {a+b x} \sqrt {a c-b c x}dx+\frac {1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}\right )}{6 b^2}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 40

\(\displaystyle \frac {3 a^2 \left (\frac {a^2 \left (\frac {3}{4} a^2 c \left (\frac {1}{2} a^2 c \int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}}dx+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}\right )+\frac {1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}\right )}{6 b^2}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 45

\(\displaystyle \frac {3 a^2 \left (\frac {a^2 \left (\frac {3}{4} a^2 c \left (a^2 c \int \frac {1}{\frac {c (a+b x) b}{a c-b c x}+b}d\frac {\sqrt {a+b x}}{\sqrt {a c-b c x}}+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}\right )+\frac {1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}\right )}{6 b^2}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {3 a^2 \left (\frac {a^2 \left (\frac {3}{4} a^2 c \left (\frac {a^2 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b}+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}\right )+\frac {1}{4} x (a+b x)^{3/2} (a c-b c x)^{3/2}\right )}{6 b^2}-\frac {x (a+b x)^{5/2} (a c-b c x)^{5/2}}{6 b^2 c}\right )}{8 b^2}-\frac {x^3 (a+b x)^{5/2} (a c-b c x)^{5/2}}{8 b^2 c}\)

Input:

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

Output:

-1/8*(x^3*(a + b*x)^(5/2)*(a*c - b*c*x)^(5/2))/(b^2*c) + (3*a^2*(-1/6*(x*( 
a + b*x)^(5/2)*(a*c - b*c*x)^(5/2))/(b^2*c) + (a^2*((x*(a + b*x)^(3/2)*(a* 
c - b*c*x)^(3/2))/4 + (3*a^2*c*((x*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/2 + (a 
^2*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[a*c - b*c*x]])/b))/4))/(6*b 
^2)))/(8*b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* 
(a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1))   Int[(a 
 + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ 
b*c + a*d, 0] && IGtQ[m + 1/2, 0]
 

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 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 111

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

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 [A] (verified)

Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {x \left (16 b^{6} x^{6}-24 a^{2} x^{4} b^{4}+2 a^{4} x^{2} b^{2}+3 a^{6}\right ) \left (-b x +a \right ) \sqrt {b x +a}\, c^{2}}{128 b^{4} \sqrt {-c \left (b x -a \right )}}+\frac {3 a^{8} \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right ) \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}\, c^{2}}{128 b^{4} \sqrt {b^{2} c}\, \sqrt {b x +a}\, \sqrt {-c \left (b x -a \right )}}\)	\(153\)
default	\(\frac {\sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}\, c \left (-16 b^{6} x^{7} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}+24 a^{2} b^{4} x^{5} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}-2 a^{4} b^{2} x^{3} \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}+3 \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{8} c -3 \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, a^{6} x \right )}{128 b^{4} \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}}\)	\(208\)

Input:

int(x^4*(b*x+a)^(3/2)*(-b*c*x+a*c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/128*x*(16*b^6*x^6-24*a^2*b^4*x^4+2*a^4*b^2*x^2+3*a^6)/b^4*(-b*x+a)*(b*x 
+a)^(1/2)/(-c*(b*x-a))^(1/2)*c^2+3/128*a^8/b^4/(b^2*c)^(1/2)*arctan((b^2*c 
)^(1/2)*x/(-b^2*c*x^2+a^2*c)^(1/2))*(-(b*x+a)*c*(b*x-a))^(1/2)/(b*x+a)^(1/ 
2)/(-c*(b*x-a))^(1/2)*c^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.36 \[ \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx=\left [\frac {3 \, a^{8} \sqrt {-c} c \log \left (2 \, b^{2} c x^{2} + 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) - 2 \, {\left (16 \, b^{7} c x^{7} - 24 \, a^{2} b^{5} c x^{5} + 2 \, a^{4} b^{3} c x^{3} + 3 \, a^{6} b c x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{256 \, b^{5}}, -\frac {3 \, a^{8} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (16 \, b^{7} c x^{7} - 24 \, a^{2} b^{5} c x^{5} + 2 \, a^{4} b^{3} c x^{3} + 3 \, a^{6} b c x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{128 \, b^{5}}\right ] \] Input:

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

Output:

[1/256*(3*a^8*sqrt(-c)*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*(16*b^7*c*x^7 - 24*a^2*b^5*c*x^5 + 2*a^4*b^3 
*c*x^3 + 3*a^6*b*c*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b^5, -1/128*(3*a^8 
*c^(3/2)*arctan(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - 
a^2*c)) + (16*b^7*c*x^7 - 24*a^2*b^5*c*x^5 + 2*a^4*b^3*c*x^3 + 3*a^6*b*c*x 
)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b^5]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.73 \[ \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx=\frac {3 \, a^{8} c^{\frac {3}{2}} \arcsin \left (\frac {b x}{a}\right )}{128 \, b^{5}} + \frac {3 \, \sqrt {-b^{2} c x^{2} + a^{2} c} a^{6} c x}{128 \, b^{4}} + \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} a^{4} x}{64 \, b^{4}} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {5}{2}} x^{3}}{8 \, b^{2} c} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {5}{2}} a^{2} x}{16 \, b^{4} c} \] Input:

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

Output:

3/128*a^8*c^(3/2)*arcsin(b*x/a)/b^5 + 3/128*sqrt(-b^2*c*x^2 + a^2*c)*a^6*c 
*x/b^4 + 1/64*(-b^2*c*x^2 + a^2*c)^(3/2)*a^4*x/b^4 - 1/8*(-b^2*c*x^2 + a^2 
*c)^(5/2)*x^3/(b^2*c) - 1/16*(-b^2*c*x^2 + a^2*c)^(5/2)*a^2*x/(b^4*c)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 562 vs. \(2 (143) = 286\).

Time = 0.49 (sec) , antiderivative size = 562, normalized size of antiderivative = 3.18 \[ \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/13440*(112*(90*a^5*c*log(abs(-sqrt(b*x + a)*sqrt(-c) + sqrt(-(b*x + a)* 
c + 2*a*c)))/sqrt(-c) - (195*a^4 - (295*a^3 - 2*(3*(4*b*x - 17*a)*(b*x + a 
) + 133*a^2)*(b*x + a))*(b*x + a))*sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a 
))*a^3*c - 56*(150*a^6*c*log(abs(-sqrt(b*x + a)*sqrt(-c) + sqrt(-(b*x + a) 
*c + 2*a*c)))/sqrt(-c) - (405*a^5 - (745*a^4 - 2*(451*a^3 - (4*(5*b*x - 26 
*a)*(b*x + a) + 321*a^2)*(b*x + a))*(b*x + a))*(b*x + a))*sqrt(-(b*x + a)* 
c + 2*a*c)*sqrt(b*x + a))*a^2*c - 8*(1050*a^7*c*log(abs(-sqrt(b*x + a)*sqr 
t(-c) + sqrt(-(b*x + a)*c + 2*a*c)))/sqrt(-c) - (2835*a^6 - (6335*a^5 - 2* 
(4781*a^4 - (4551*a^3 - 4*(5*(6*b*x - 37*a)*(b*x + a) + 661*a^2)*(b*x + a) 
)*(b*x + a))*(b*x + a))*(b*x + a))*sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a 
))*a*c + (7350*a^8*c*log(abs(-sqrt(b*x + a)*sqrt(-c) + sqrt(-(b*x + a)*c + 
 2*a*c)))/sqrt(-c) - (23205*a^7 - (59465*a^6 - 2*(53963*a^5 - (64233*a^4 - 
 4*(12463*a^3 - 5*(6*(7*b*x - 50*a)*(b*x + a) + 1219*a^2)*(b*x + a))*(b*x 
+ a))*(b*x + a))*(b*x + a))*(b*x + a))*sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x 
 + a))*c)/b^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.68 \[ \int x^4 (a+b x)^{3/2} (a c-b c x)^{3/2} \, dx=\frac {\sqrt {c}\, c \left (-6 \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{8}-3 \sqrt {b x +a}\, \sqrt {-b x +a}\, a^{6} b x -2 \sqrt {b x +a}\, \sqrt {-b x +a}\, a^{4} b^{3} x^{3}+24 \sqrt {b x +a}\, \sqrt {-b x +a}\, a^{2} b^{5} x^{5}-16 \sqrt {b x +a}\, \sqrt {-b x +a}\, b^{7} x^{7}\right )}{128 b^{5}} \] Input:

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

Output:

(sqrt(c)*c*( - 6*asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))*a**8 - 3*sqrt(a + b 
*x)*sqrt(a - b*x)*a**6*b*x - 2*sqrt(a + b*x)*sqrt(a - b*x)*a**4*b**3*x**3 
+ 24*sqrt(a + b*x)*sqrt(a - b*x)*a**2*b**5*x**5 - 16*sqrt(a + b*x)*sqrt(a 
- b*x)*b**7*x**7))/(128*b**5)