\(\int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx\) [72]

Optimal result

Integrand size = 26, antiderivative size = 160 \[ \int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {e (e x)^{3/2}}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {3 e (e x)^{3/2}}{16 a^2 b^2 c^3 \left (a^2-b^2 x^2\right )}+\frac {3 e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 a^{5/2} b^{7/2} c^3}-\frac {3 e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 a^{5/2} b^{7/2} c^3} \] Output:

1/4*e*(e*x)^(3/2)/b^2/c^3/(-b^2*x^2+a^2)^2-3/16*e*(e*x)^(3/2)/a^2/b^2/c^3/ 
(-b^2*x^2+a^2)+3/32*e^(5/2)*arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/a^ 
(5/2)/b^(7/2)/c^3-3/32*e^(5/2)*arctanh(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2) 
)/a^(5/2)/b^(7/2)/c^3
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87 \[ \int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {(e x)^{5/2} \left (2 \sqrt {a} b^{3/2} x^{3/2} \left (a^2+3 b^2 x^2\right )+3 \left (a^2-b^2 x^2\right )^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-3 \left (a^2-b^2 x^2\right )^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{32 a^{5/2} b^{7/2} c^3 x^{5/2} \left (a^2-b^2 x^2\right )^2} \] Input:

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

Output:

((e*x)^(5/2)*(2*Sqrt[a]*b^(3/2)*x^(3/2)*(a^2 + 3*b^2*x^2) + 3*(a^2 - b^2*x 
^2)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] - 3*(a^2 - b^2*x^2)^2*ArcTanh[(Sqr 
t[b]*Sqrt[x])/Sqrt[a]]))/(32*a^(5/2)*b^(7/2)*c^3*x^(5/2)*(a^2 - b^2*x^2)^2 
)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {82, 252, 27, 253, 266, 27, 827, 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.

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

\(\Big \downarrow \) 82

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(e*(e*x)^(3/2))/(4*b^2*c^3*(a^2 - b^2*x^2)^2) - (3*e^2*((e*x)^(3/2)/(2*a^2 
*e*(a^2 - b^2*x^2)) + (e*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e] 
)]/(Sqrt[a]*b^(3/2)*Sqrt[e]) + ArcTanh[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e 
])]/(2*Sqrt[a]*b^(3/2)*Sqrt[e])))/(2*a^2)))/(8*b^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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

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 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 827

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

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(-\frac {\left (3 e^{2} \left (b x +a \right )^{2} \left (-b x +a \right )^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )-3 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right ) e^{2} \left (b x +a \right )^{2} \left (-b x +a \right )^{2}+b \sqrt {a e b}\, \left (-3 \left (-b x +a \right )^{2} \left (e x \right )^{\frac {3}{2}}+e x \sqrt {e x}\, \left (-3 b^{2} x^{2}-6 a b x +a^{2}\right )\right )\right ) e}{32 \sqrt {a e b}\, \left (b x +a \right )^{2} a^{2} b^{3} \left (-b x +a \right )^{2} c^{3}}\)	\(152\)
derivativedivides	\(-\frac {2 e^{5} \left (\frac {\frac {-\frac {3 b \left (e x \right )^{\frac {3}{2}}}{4 a e}+\frac {\sqrt {e x}}{4}}{\left (-b e x +a e \right )^{2}}+\frac {3 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 a e \sqrt {a e b}}}{16 a e \,b^{3}}-\frac {\frac {\frac {3 b \left (e x \right )^{\frac {3}{2}}}{4 a e}+\frac {\sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 a e \sqrt {a e b}}}{16 a e \,b^{3}}\right )}{c^{3}}\)	\(157\)
default	\(\frac {2 e^{5} \left (-\frac {\frac {-\frac {3 b \left (e x \right )^{\frac {3}{2}}}{4 a e}+\frac {\sqrt {e x}}{4}}{\left (-b e x +a e \right )^{2}}+\frac {3 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 a e \sqrt {a e b}}}{16 a e \,b^{3}}+\frac {\frac {\frac {3 b \left (e x \right )^{\frac {3}{2}}}{4 a e}+\frac {\sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {3 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 a e \sqrt {a e b}}}{16 a e \,b^{3}}\right )}{c^{3}}\)	\(157\)

Input:

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

Output:

-1/32*(3*e^2*(b*x+a)^2*(-b*x+a)^2*arctanh(b*(e*x)^(1/2)/(a*e*b)^(1/2))-3*a 
rctan(b*(e*x)^(1/2)/(a*e*b)^(1/2))*e^2*(b*x+a)^2*(-b*x+a)^2+b*(a*e*b)^(1/2 
)*(-3*(-b*x+a)^2*(e*x)^(3/2)+e*x*(e*x)^(1/2)*(-3*b^2*x^2-6*a*b*x+a^2)))/(a 
*e*b)^(1/2)*e/(b*x+a)^2/a^2/b^3/(-b*x+a)^2/c^3
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.75 \[ \int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\left [\frac {6 \, {\left (b^{4} e^{2} x^{4} - 2 \, a^{2} b^{2} e^{2} x^{2} + a^{4} e^{2}\right )} \sqrt {\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {e}{a b}}}{e}\right ) + 3 \, {\left (b^{4} e^{2} x^{4} - 2 \, a^{2} b^{2} e^{2} x^{2} + a^{4} e^{2}\right )} \sqrt {\frac {e}{a b}} \log \left (\frac {b e x - 2 \, \sqrt {e x} a b \sqrt {\frac {e}{a b}} + a e}{b x - a}\right ) + 4 \, {\left (3 \, b^{3} e^{2} x^{3} + a^{2} b e^{2} x\right )} \sqrt {e x}}{64 \, {\left (a^{2} b^{7} c^{3} x^{4} - 2 \, a^{4} b^{5} c^{3} x^{2} + a^{6} b^{3} c^{3}\right )}}, \frac {6 \, {\left (b^{4} e^{2} x^{4} - 2 \, a^{2} b^{2} e^{2} x^{2} + a^{4} e^{2}\right )} \sqrt {-\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {e}{a b}}}{e}\right ) + 3 \, {\left (b^{4} e^{2} x^{4} - 2 \, a^{2} b^{2} e^{2} x^{2} + a^{4} e^{2}\right )} \sqrt {-\frac {e}{a b}} \log \left (\frac {b e x + 2 \, \sqrt {e x} a b \sqrt {-\frac {e}{a b}} - a e}{b x + a}\right ) + 4 \, {\left (3 \, b^{3} e^{2} x^{3} + a^{2} b e^{2} x\right )} \sqrt {e x}}{64 \, {\left (a^{2} b^{7} c^{3} x^{4} - 2 \, a^{4} b^{5} c^{3} x^{2} + a^{6} b^{3} c^{3}\right )}}\right ] \] Input:

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

Output:

[1/64*(6*(b^4*e^2*x^4 - 2*a^2*b^2*e^2*x^2 + a^4*e^2)*sqrt(e/(a*b))*arctan( 
sqrt(e*x)*b*sqrt(e/(a*b))/e) + 3*(b^4*e^2*x^4 - 2*a^2*b^2*e^2*x^2 + a^4*e^ 
2)*sqrt(e/(a*b))*log((b*e*x - 2*sqrt(e*x)*a*b*sqrt(e/(a*b)) + a*e)/(b*x - 
a)) + 4*(3*b^3*e^2*x^3 + a^2*b*e^2*x)*sqrt(e*x))/(a^2*b^7*c^3*x^4 - 2*a^4* 
b^5*c^3*x^2 + a^6*b^3*c^3), 1/64*(6*(b^4*e^2*x^4 - 2*a^2*b^2*e^2*x^2 + a^4 
*e^2)*sqrt(-e/(a*b))*arctan(sqrt(e*x)*b*sqrt(-e/(a*b))/e) + 3*(b^4*e^2*x^4 
 - 2*a^2*b^2*e^2*x^2 + a^4*e^2)*sqrt(-e/(a*b))*log((b*e*x + 2*sqrt(e*x)*a* 
b*sqrt(-e/(a*b)) - a*e)/(b*x + a)) + 4*(3*b^3*e^2*x^3 + a^2*b*e^2*x)*sqrt( 
e*x))/(a^2*b^7*c^3*x^4 - 2*a^4*b^5*c^3*x^2 + a^6*b^3*c^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((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 = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84 \[ \int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {1}{32} \, e^{2} {\left (\frac {3 \, e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a^{2} b^{3} c^{3}} + \frac {3 \, e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a^{2} b^{3} c^{3}} + \frac {2 \, {\left (3 \, \sqrt {e x} b^{2} e^{4} x^{3} + \sqrt {e x} a^{2} e^{4} x\right )}}{{\left (b^{2} e^{2} x^{2} - a^{2} e^{2}\right )}^{2} a^{2} b^{2} c^{3}}\right )} \] Input:

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

Output:

1/32*e^2*(3*e*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*a^2*b^3*c^3) + 
3*e*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*e)*a^2*b^3*c^3) + 2*(3*sqr 
t(e*x)*b^2*e^4*x^3 + sqrt(e*x)*a^2*e^4*x)/((b^2*e^2*x^2 - a^2*e^2)^2*a^2*b 
^2*c^3))
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.82 \[ \int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\frac {3\,e^3\,{\left (e\,x\right )}^{7/2}}{16\,a^2}+\frac {e^5\,{\left (e\,x\right )}^{3/2}}{16\,b^2}}{a^4\,c^3\,e^4-2\,a^2\,b^2\,c^3\,e^4\,x^2+b^4\,c^3\,e^4\,x^4}+\frac {3\,e^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,a^{5/2}\,b^{7/2}\,c^3}-\frac {3\,e^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,a^{5/2}\,b^{7/2}\,c^3} \] Input:

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

Output:

((3*e^3*(e*x)^(7/2))/(16*a^2) + (e^5*(e*x)^(3/2))/(16*b^2))/(a^4*c^3*e^4 + 
 b^4*c^3*e^4*x^4 - 2*a^2*b^2*c^3*e^4*x^2) + (3*e^(5/2)*atan((b^(1/2)*(e*x) 
^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(5/2)*b^(7/2)*c^3) - (3*e^(5/2)*atanh((b 
^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(5/2)*b^(7/2)*c^3)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.69 \[ \int \frac {(e x)^{5/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\sqrt {e}\, e^{2} \left (6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}-12 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{4}+3 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4}-6 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{2} x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{4} x^{4}-3 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4}+6 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{2} x^{2}-3 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{4} x^{4}+4 \sqrt {x}\, a^{3} b^{2} x +12 \sqrt {x}\, a \,b^{4} x^{3}\right )}{64 a^{3} b^{4} c^{3} \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:

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

Output:

(sqrt(e)*e**2*(6*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**4 
- 12*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**2 + 
6*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**4*x**4 + 3*sqrt(b 
)*sqrt(a)*log( - sqrt(a) + sqrt(x)*sqrt(b))*a**4 - 6*sqrt(b)*sqrt(a)*log( 
- sqrt(a) + sqrt(x)*sqrt(b))*a**2*b**2*x**2 + 3*sqrt(b)*sqrt(a)*log( - sqr 
t(a) + sqrt(x)*sqrt(b))*b**4*x**4 - 3*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x 
)*sqrt(b))*a**4 + 6*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt(b))*a**2*b* 
*2*x**2 - 3*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt(b))*b**4*x**4 + 4*s 
qrt(x)*a**3*b**2*x + 12*sqrt(x)*a*b**4*x**3))/(64*a**3*b**4*c**3*(a**4 - 2 
*a**2*b**2*x**2 + b**4*x**4))