\(\int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx\) [77]

Optimal result

Integrand size = 26, antiderivative size = 181 \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {77}{48 a^6 c^3 e (e x)^{3/2}}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}+\frac {11}{16 a^4 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}+\frac {77 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 a^{15/2} c^3 e^{5/2}}+\frac {77 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 a^{15/2} c^3 e^{5/2}} \] Output:

-77/48/a^6/c^3/e/(e*x)^(3/2)+1/4/a^2/c^3/e/(e*x)^(3/2)/(-b^2*x^2+a^2)^2+11 
/16/a^4/c^3/e/(e*x)^(3/2)/(-b^2*x^2+a^2)+77/32*b^(3/2)*arctan(b^(1/2)*(e*x 
)^(1/2)/a^(1/2)/e^(1/2))/a^(15/2)/c^3/e^(5/2)+77/32*b^(3/2)*arctanh(b^(1/2 
)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/a^(15/2)/c^3/e^(5/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\frac {x \left (-\frac {2 a^{3/2} \left (32 a^4-121 a^2 b^2 x^2+77 b^4 x^4\right )}{\left (a^2-b^2 x^2\right )^2}+231 b^{3/2} x^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+231 b^{3/2} x^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{96 a^{15/2} c^3 (e x)^{5/2}} \] Input:

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

Output:

(x*((-2*a^(3/2)*(32*a^4 - 121*a^2*b^2*x^2 + 77*b^4*x^4))/(a^2 - b^2*x^2)^2 
 + 231*b^(3/2)*x^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] + 231*b^(3/2)*x^( 
3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(96*a^(15/2)*c^3*(e*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 200, 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 = {82, 253, 27, 253, 264, 266, 756, 218, 221}

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.

\(\displaystyle \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx\)

\(\Big \downarrow \) 82

\(\displaystyle \int \frac {1}{(e x)^{5/2} \left (a^2 c-b^2 c x^2\right )^3}dx\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {11 \int \frac {1}{c^2 (e x)^{5/2} \left (a^2-b^2 x^2\right )^2}dx}{8 a^2 c}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {11 \int \frac {1}{(e x)^{5/2} \left (a^2-b^2 x^2\right )^2}dx}{8 a^2 c^3}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {11 \left (\frac {7 \int \frac {1}{(e x)^{5/2} \left (a^2-b^2 x^2\right )}dx}{4 a^2}+\frac {1}{2 a^2 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}\right )}{8 a^2 c^3}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {11 \left (\frac {7 \left (\frac {b^2 \int \frac {1}{\sqrt {e x} \left (a^2-b^2 x^2\right )}dx}{a^2 e^2}-\frac {2}{3 a^2 e (e x)^{3/2}}\right )}{4 a^2}+\frac {1}{2 a^2 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}\right )}{8 a^2 c^3}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {11 \left (\frac {7 \left (\frac {2 b^2 \int \frac {1}{a^2-b^2 x^2}d\sqrt {e x}}{a^2 e^3}-\frac {2}{3 a^2 e (e x)^{3/2}}\right )}{4 a^2}+\frac {1}{2 a^2 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}\right )}{8 a^2 c^3}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {11 \left (\frac {7 \left (\frac {2 b^2 \left (\frac {e \int \frac {1}{a e-b e x}d\sqrt {e x}}{2 a}+\frac {e \int \frac {1}{a e+b x e}d\sqrt {e x}}{2 a}\right )}{a^2 e^3}-\frac {2}{3 a^2 e (e x)^{3/2}}\right )}{4 a^2}+\frac {1}{2 a^2 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}\right )}{8 a^2 c^3}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {11 \left (\frac {7 \left (\frac {2 b^2 \left (\frac {e \int \frac {1}{a e-b e x}d\sqrt {e x}}{2 a}+\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {b}}\right )}{a^2 e^3}-\frac {2}{3 a^2 e (e x)^{3/2}}\right )}{4 a^2}+\frac {1}{2 a^2 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}\right )}{8 a^2 c^3}+\frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4 a^2 c^3 e (e x)^{3/2} \left (a^2-b^2 x^2\right )^2}+\frac {11 \left (\frac {1}{2 a^2 e (e x)^{3/2} \left (a^2-b^2 x^2\right )}+\frac {7 \left (\frac {2 b^2 \left (\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{2 a^{3/2} \sqrt {b}}\right )}{a^2 e^3}-\frac {2}{3 a^2 e (e x)^{3/2}}\right )}{4 a^2}\right )}{8 a^2 c^3}\)

Input:

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

Output:

1/(4*a^2*c^3*e*(e*x)^(3/2)*(a^2 - b^2*x^2)^2) + (11*(1/(2*a^2*e*(e*x)^(3/2 
)*(a^2 - b^2*x^2)) + (7*(-2/(3*a^2*e*(e*x)^(3/2)) + (2*b^2*((Sqrt[e]*ArcTa 
n[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(2*a^(3/2)*Sqrt[b]) + (Sqrt[e]*A 
rcTanh[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(2*a^(3/2)*Sqrt[b])))/(a^2* 
e^3)))/(4*a^2)))/(8*a^2*c^3)
 

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 82

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

rule 221

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

rule 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\frac {231 b^{2} \left (e x \right )^{\frac {3}{2}} \left (b x +a \right )^{2} \left (-b x +a \right )^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{64}-\frac {231 b^{2} \left (e x \right )^{\frac {3}{2}} \left (b x +a \right )^{2} \left (-b x +a \right )^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{64}+a \left (\frac {77}{32} b^{4} x^{4}-\frac {121}{32} a^{2} b^{2} x^{2}+a^{4}\right ) e \sqrt {a e b}\right )}{3 e^{2} \left (e x \right )^{\frac {3}{2}} \sqrt {a e b}\, \left (b x +a \right )^{2} a^{7} \left (-b x +a \right )^{2} c^{3}}\)	\(149\)
derivativedivides	\(-\frac {2 e^{5} \left (-\frac {b^{2} \left (\frac {-\frac {17 b \left (e x \right )^{\frac {3}{2}}}{4}+\frac {19 a e \sqrt {e x}}{4}}{\left (-b e x +a e \right )^{2}}+\frac {77 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}\right )}{16 a^{7} e^{7}}+\frac {1}{3 a^{6} e^{6} \left (e x \right )^{\frac {3}{2}}}-\frac {b^{2} \left (\frac {\frac {17 b \left (e x \right )^{\frac {3}{2}}}{4}+\frac {19 a e \sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {77 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}\right )}{16 a^{7} e^{7}}\right )}{c^{3}}\)	\(150\)
default	\(\frac {2 e^{5} \left (\frac {b^{2} \left (\frac {-\frac {17 b \left (e x \right )^{\frac {3}{2}}}{4}+\frac {19 a e \sqrt {e x}}{4}}{\left (-b e x +a e \right )^{2}}+\frac {77 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}\right )}{16 a^{7} e^{7}}-\frac {1}{3 a^{6} e^{6} \left (e x \right )^{\frac {3}{2}}}+\frac {b^{2} \left (\frac {\frac {17 b \left (e x \right )^{\frac {3}{2}}}{4}+\frac {19 a e \sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {77 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}\right )}{16 a^{7} e^{7}}\right )}{c^{3}}\)	\(150\)
risch	\(-\frac {2}{3 a^{6} x \sqrt {e x}\, e^{2} c^{3}}-\frac {2 b^{2} \left (-\frac {\frac {\frac {17 b \left (e x \right )^{\frac {3}{2}}}{4}+\frac {19 a e \sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {77 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}}{16 a e}+\frac {\frac {\frac {17 b \left (e x \right )^{\frac {3}{2}}}{4}-\frac {19 a e \sqrt {e x}}{4}}{\left (b e x -a e \right )^{2}}-\frac {77 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}}{16 a e}\right )}{a^{6} e \,c^{3}}\)	\(157\)

Input:

int(1/(e*x)^(5/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
 

Output:

-2/3/e^2*(-231/64*b^2*(e*x)^(3/2)*(b*x+a)^2*(-b*x+a)^2*arctanh(b*(e*x)^(1/ 
2)/(a*e*b)^(1/2))-231/64*b^2*(e*x)^(3/2)*(b*x+a)^2*(-b*x+a)^2*arctan(b*(e* 
x)^(1/2)/(a*e*b)^(1/2))+a*(77/32*b^4*x^4-121/32*a^2*b^2*x^2+a^4)*e*(a*e*b) 
^(1/2))/(e*x)^(3/2)/(a*e*b)^(1/2)/(b*x+a)^2/a^7/(-b*x+a)^2/c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.46 \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\left [\frac {462 \, {\left (b^{5} e x^{6} - 2 \, a^{2} b^{3} e x^{4} + a^{4} b e x^{2}\right )} \sqrt {\frac {b}{a e}} \arctan \left (\sqrt {e x} \sqrt {\frac {b}{a e}}\right ) + 231 \, {\left (b^{5} e x^{6} - 2 \, a^{2} b^{3} e x^{4} + a^{4} b e x^{2}\right )} \sqrt {\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {\frac {b}{a e}} + a}{b x - a}\right ) - 4 \, {\left (77 \, a b^{4} x^{4} - 121 \, a^{3} b^{2} x^{2} + 32 \, a^{5}\right )} \sqrt {e x}}{192 \, {\left (a^{7} b^{4} c^{3} e^{3} x^{6} - 2 \, a^{9} b^{2} c^{3} e^{3} x^{4} + a^{11} c^{3} e^{3} x^{2}\right )}}, -\frac {462 \, {\left (b^{5} e x^{6} - 2 \, a^{2} b^{3} e x^{4} + a^{4} b e x^{2}\right )} \sqrt {-\frac {b}{a e}} \arctan \left (\sqrt {e x} \sqrt {-\frac {b}{a e}}\right ) - 231 \, {\left (b^{5} e x^{6} - 2 \, a^{2} b^{3} e x^{4} + a^{4} b e x^{2}\right )} \sqrt {-\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {-\frac {b}{a e}} - a}{b x + a}\right ) + 4 \, {\left (77 \, a b^{4} x^{4} - 121 \, a^{3} b^{2} x^{2} + 32 \, a^{5}\right )} \sqrt {e x}}{192 \, {\left (a^{7} b^{4} c^{3} e^{3} x^{6} - 2 \, a^{9} b^{2} c^{3} e^{3} x^{4} + a^{11} c^{3} e^{3} x^{2}\right )}}\right ] \] Input:

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

Output:

[1/192*(462*(b^5*e*x^6 - 2*a^2*b^3*e*x^4 + a^4*b*e*x^2)*sqrt(b/(a*e))*arct 
an(sqrt(e*x)*sqrt(b/(a*e))) + 231*(b^5*e*x^6 - 2*a^2*b^3*e*x^4 + a^4*b*e*x 
^2)*sqrt(b/(a*e))*log((b*x + 2*sqrt(e*x)*a*sqrt(b/(a*e)) + a)/(b*x - a)) - 
 4*(77*a*b^4*x^4 - 121*a^3*b^2*x^2 + 32*a^5)*sqrt(e*x))/(a^7*b^4*c^3*e^3*x 
^6 - 2*a^9*b^2*c^3*e^3*x^4 + a^11*c^3*e^3*x^2), -1/192*(462*(b^5*e*x^6 - 2 
*a^2*b^3*e*x^4 + a^4*b*e*x^2)*sqrt(-b/(a*e))*arctan(sqrt(e*x)*sqrt(-b/(a*e 
))) - 231*(b^5*e*x^6 - 2*a^2*b^3*e*x^4 + a^4*b*e*x^2)*sqrt(-b/(a*e))*log(( 
b*x + 2*sqrt(e*x)*a*sqrt(-b/(a*e)) - a)/(b*x + a)) + 4*(77*a*b^4*x^4 - 121 
*a^3*b^2*x^2 + 32*a^5)*sqrt(e*x))/(a^7*b^4*c^3*e^3*x^6 - 2*a^9*b^2*c^3*e^3 
*x^4 + a^11*c^3*e^3*x^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\frac {77 \, b^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{32 \, \sqrt {a b e} a^{7} c^{3} e^{2}} - \frac {77 \, b^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{32 \, \sqrt {-a b e} a^{7} c^{3} e^{2}} - \frac {15 \, \sqrt {e x} b^{4} e^{2} x^{2} - 19 \, \sqrt {e x} a^{2} b^{2} e^{2}}{16 \, {\left (b^{2} e^{2} x^{2} - a^{2} e^{2}\right )}^{2} a^{6} c^{3} e} - \frac {2}{3 \, \sqrt {e x} a^{6} c^{3} e^{2} x} \] Input:

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

Output:

77/32*b^2*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*a^7*c^3*e^2) - 77/3 
2*b^2*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*e)*a^7*c^3*e^2) - 1/16*( 
15*sqrt(e*x)*b^4*e^2*x^2 - 19*sqrt(e*x)*a^2*b^2*e^2)/((b^2*e^2*x^2 - a^2*e 
^2)^2*a^6*c^3*e) - 2/3/(sqrt(e*x)*a^6*c^3*e^2*x)
 

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\frac {77\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,a^{15/2}\,c^3\,e^{5/2}}-\frac {\frac {2\,e^3}{3\,a^2}-\frac {121\,b^2\,e^3\,x^2}{48\,a^4}+\frac {77\,b^4\,e^3\,x^4}{48\,a^6}}{b^4\,c^3\,{\left (e\,x\right )}^{11/2}+a^4\,c^3\,e^4\,{\left (e\,x\right )}^{3/2}-2\,a^2\,b^2\,c^3\,e^2\,{\left (e\,x\right )}^{7/2}}+\frac {77\,b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,a^{15/2}\,c^3\,e^{5/2}} \] Input:

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

Output:

(77*b^(3/2)*atan((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(15/2)*c^ 
3*e^(5/2)) - ((2*e^3)/(3*a^2) - (121*b^2*e^3*x^2)/(48*a^4) + (77*b^4*e^3*x 
^4)/(48*a^6))/(b^4*c^3*(e*x)^(11/2) + a^4*c^3*e^4*(e*x)^(3/2) - 2*a^2*b^2* 
c^3*e^2*(e*x)^(7/2)) + (77*b^(3/2)*atanh((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^ 
(1/2))))/(32*a^(15/2)*c^3*e^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(e x)^{5/2} (a+b x)^3 (a c-b c x)^3} \, dx=\frac {\sqrt {e}\, \left (462 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} b x -924 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{3} x^{3}+462 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{5} x^{5}-231 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4} b x +462 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{3} x^{3}-231 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{5} x^{5}+231 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4} b x -462 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{3} x^{3}+231 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{5} x^{5}-128 a^{6}+484 a^{4} b^{2} x^{2}-308 a^{2} b^{4} x^{4}\right )}{192 \sqrt {x}\, a^{8} c^{3} e^{3} x \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:

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

Output:

(sqrt(e)*(462*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))* 
a**4*b*x - 924*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a))) 
*a**2*b**3*x**3 + 462*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sq 
rt(a)))*b**5*x**5 - 231*sqrt(x)*sqrt(b)*sqrt(a)*log( - sqrt(a) + sqrt(x)*s 
qrt(b))*a**4*b*x + 462*sqrt(x)*sqrt(b)*sqrt(a)*log( - sqrt(a) + sqrt(x)*sq 
rt(b))*a**2*b**3*x**3 - 231*sqrt(x)*sqrt(b)*sqrt(a)*log( - sqrt(a) + sqrt( 
x)*sqrt(b))*b**5*x**5 + 231*sqrt(x)*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)* 
sqrt(b))*a**4*b*x - 462*sqrt(x)*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt 
(b))*a**2*b**3*x**3 + 231*sqrt(x)*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sq 
rt(b))*b**5*x**5 - 128*a**6 + 484*a**4*b**2*x**2 - 308*a**2*b**4*x**4))/(1 
92*sqrt(x)*a**8*c**3*e**3*x*(a**4 - 2*a**2*b**2*x**2 + b**4*x**4))