\(\int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx\) [70]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 159 \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {e (e x)^{7/2}}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {7 e^3 (e x)^{3/2}}{16 b^4 c^3 \left (a^2-b^2 x^2\right )}-\frac {21 e^{9/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 \sqrt {a} b^{11/2} c^3}+\frac {21 e^{9/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 \sqrt {a} b^{11/2} c^3} \] Output:

1/4*e*(e*x)^(7/2)/b^2/c^3/(-b^2*x^2+a^2)^2-7/16*e^3*(e*x)^(3/2)/b^4/c^3/(- 
b^2*x^2+a^2)-21/32*e^(9/2)*arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/a^( 
1/2)/b^(11/2)/c^3+21/32*e^(9/2)*arctanh(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2 
))/a^(1/2)/b^(11/2)/c^3
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91 \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {e^4 \sqrt {e x} \left (2 \sqrt {a} b^{3/2} x^{3/2} \left (-7 a^2+11 b^2 x^2\right )-21 \left (a^2-b^2 x^2\right )^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+21 \left (a^2-b^2 x^2\right )^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{32 \sqrt {a} b^{11/2} c^3 \sqrt {x} \left (a^2-b^2 x^2\right )^2} \] Input:

Integrate[(e*x)^(9/2)/((a + b*x)^3*(a*c - b*c*x)^3),x]
 

Output:

(e^4*Sqrt[e*x]*(2*Sqrt[a]*b^(3/2)*x^(3/2)*(-7*a^2 + 11*b^2*x^2) - 21*(a^2 
- b^2*x^2)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] + 21*(a^2 - b^2*x^2)^2*ArcT 
anh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(32*Sqrt[a]*b^(11/2)*c^3*Sqrt[x]*(a^2 - b 
^2*x^2)^2)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09, 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, 252, 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.

\(\displaystyle \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx\)

\(\Big \downarrow \) 82

\(\displaystyle \int \frac {(e x)^{9/2}}{\left (a^2 c-b^2 c x^2\right )^3}dx\)

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[(e*x)^(9/2)/((a + b*x)^3*(a*c - b*c*x)^3),x]
 

Output:

(e*(e*x)^(7/2))/(4*b^2*c^3*(a^2 - b^2*x^2)^2) - (7*e^2*((e*(e*x)^(3/2))/(2 
*b^2*(a^2 - b^2*x^2)) - (3*e^3*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*S 
qrt[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*b^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 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 = 1.67 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.79

method result size
derivativedivides \(-\frac {2 e^{5} \left (\frac {\frac {-\frac {11 b \left (e x \right )^{\frac {3}{2}}}{4}-\frac {9 a e \sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {21 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}}{16 b^{5}}-\frac {\frac {\frac {11 b \left (e x \right )^{\frac {3}{2}}}{4}-\frac {9 a e \sqrt {e x}}{4}}{\left (-b e x +a e \right )^{2}}+\frac {21 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}}{16 b^{5}}\right )}{c^{3}}\) \(125\)
default \(\frac {2 e^{5} \left (-\frac {\frac {-\frac {11 b \left (e x \right )^{\frac {3}{2}}}{4}-\frac {9 a e \sqrt {e x}}{4}}{\left (b e x +a e \right )^{2}}+\frac {21 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}}{16 b^{5}}+\frac {\frac {\frac {11 b \left (e x \right )^{\frac {3}{2}}}{4}-\frac {9 a e \sqrt {e x}}{4}}{\left (-b e x +a e \right )^{2}}+\frac {21 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 \sqrt {a e b}}}{16 b^{5}}\right )}{c^{3}}\) \(125\)
pseudoelliptic \(-\frac {e^{5} \left (-\frac {11 \left (e x \right )^{\frac {3}{2}}}{e^{2} \left (b x +a \right )^{2} b^{4}}-\frac {9 a \sqrt {e x}}{e \left (b x +a \right )^{2} b^{5}}+\frac {21 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{\sqrt {a e b}\, b^{5}}+\frac {\frac {\sqrt {e x}\, \left (-11 b x +9 a \right )}{e \left (-b x +a \right )^{2}}-\frac {21 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{\sqrt {a e b}}}{b^{5}}\right )}{32 c^{3}}\) \(128\)

Input:

int((e*x)^(9/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
 

Output:

-2*e^5/c^3*(1/16/b^5*((-11/4*b*(e*x)^(3/2)-9/4*a*e*(e*x)^(1/2))/(b*e*x+a*e 
)^2+21/4/(a*e*b)^(1/2)*arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2)))-1/16/b^5*((11/ 
4*b*(e*x)^(3/2)-9/4*a*e*(e*x)^(1/2))/(-b*e*x+a*e)^2+21/4/(a*e*b)^(1/2)*arc 
tanh(b*(e*x)^(1/2)/(a*e*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.74 \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\left [-\frac {42 \, {\left (b^{4} e^{4} x^{4} - 2 \, a^{2} b^{2} e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt {\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {e}{a b}}}{e}\right ) - 21 \, {\left (b^{4} e^{4} x^{4} - 2 \, a^{2} b^{2} e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt {\frac {e}{a b}} \log \left (\frac {b e x + 2 \, \sqrt {e x} a b \sqrt {\frac {e}{a b}} + a e}{b x - a}\right ) - 4 \, {\left (11 \, b^{3} e^{4} x^{3} - 7 \, a^{2} b e^{4} x\right )} \sqrt {e x}}{64 \, {\left (b^{9} c^{3} x^{4} - 2 \, a^{2} b^{7} c^{3} x^{2} + a^{4} b^{5} c^{3}\right )}}, -\frac {42 \, {\left (b^{4} e^{4} x^{4} - 2 \, a^{2} b^{2} e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt {-\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {e}{a b}}}{e}\right ) - 21 \, {\left (b^{4} e^{4} x^{4} - 2 \, a^{2} b^{2} e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt {-\frac {e}{a b}} \log \left (\frac {b e x - 2 \, \sqrt {e x} a b \sqrt {-\frac {e}{a b}} - a e}{b x + a}\right ) - 4 \, {\left (11 \, b^{3} e^{4} x^{3} - 7 \, a^{2} b e^{4} x\right )} \sqrt {e x}}{64 \, {\left (b^{9} c^{3} x^{4} - 2 \, a^{2} b^{7} c^{3} x^{2} + a^{4} b^{5} c^{3}\right )}}\right ] \] Input:

integrate((e*x)^(9/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="fricas")
 

Output:

[-1/64*(42*(b^4*e^4*x^4 - 2*a^2*b^2*e^4*x^2 + a^4*e^4)*sqrt(e/(a*b))*arcta 
n(sqrt(e*x)*b*sqrt(e/(a*b))/e) - 21*(b^4*e^4*x^4 - 2*a^2*b^2*e^4*x^2 + a^4 
*e^4)*sqrt(e/(a*b))*log((b*e*x + 2*sqrt(e*x)*a*b*sqrt(e/(a*b)) + a*e)/(b*x 
 - a)) - 4*(11*b^3*e^4*x^3 - 7*a^2*b*e^4*x)*sqrt(e*x))/(b^9*c^3*x^4 - 2*a^ 
2*b^7*c^3*x^2 + a^4*b^5*c^3), -1/64*(42*(b^4*e^4*x^4 - 2*a^2*b^2*e^4*x^2 + 
 a^4*e^4)*sqrt(-e/(a*b))*arctan(sqrt(e*x)*b*sqrt(-e/(a*b))/e) - 21*(b^4*e^ 
4*x^4 - 2*a^2*b^2*e^4*x^2 + a^4*e^4)*sqrt(-e/(a*b))*log((b*e*x - 2*sqrt(e* 
x)*a*b*sqrt(-e/(a*b)) - a*e)/(b*x + a)) - 4*(11*b^3*e^4*x^3 - 7*a^2*b*e^4* 
x)*sqrt(e*x))/(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. \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\text {Timed out} \] Input:

integrate((e*x)**(9/2)/(b*x+a)**3/(-b*c*x+a*c)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x)^(9/2)/(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 = 0.14 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.79 \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=-\frac {1}{32} \, e^{4} {\left (\frac {21 \, e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b^{5} c^{3}} + \frac {21 \, e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} b^{5} c^{3}} - \frac {2 \, {\left (11 \, \sqrt {e x} b^{2} e^{4} x^{3} - 7 \, \sqrt {e x} a^{2} e^{4} x\right )}}{{\left (b^{2} e^{2} x^{2} - a^{2} e^{2}\right )}^{2} b^{4} c^{3}}\right )} \] Input:

integrate((e*x)^(9/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="giac")
 

Output:

-1/32*e^4*(21*e*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*b^5*c^3) + 21 
*e*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*e)*b^5*c^3) - 2*(11*sqrt(e* 
x)*b^2*e^4*x^3 - 7*sqrt(e*x)*a^2*e^4*x)/((b^2*e^2*x^2 - a^2*e^2)^2*b^4*c^3 
))
 

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85 \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\frac {11\,e^5\,{\left (e\,x\right )}^{7/2}}{16\,b^2}-\frac {7\,a^2\,e^7\,{\left (e\,x\right )}^{3/2}}{16\,b^4}}{a^4\,c^3\,e^4-2\,a^2\,b^2\,c^3\,e^4\,x^2+b^4\,c^3\,e^4\,x^4}-\frac {21\,e^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,\sqrt {a}\,b^{11/2}\,c^3}+\frac {21\,e^{9/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,\sqrt {a}\,b^{11/2}\,c^3} \] Input:

int((e*x)^(9/2)/((a*c - b*c*x)^3*(a + b*x)^3),x)
 

Output:

((11*e^5*(e*x)^(7/2))/(16*b^2) - (7*a^2*e^7*(e*x)^(3/2))/(16*b^4))/(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) - (21*e^(9/2)*atan((b^(1/ 
2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(1/2)*b^(11/2)*c^3) + (21*e^(9/2 
)*atanh((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(1/2)*b^(11/2)*c^3 
)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.70 \[ \int \frac {(e x)^{9/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\sqrt {e}\, e^{4} \left (-42 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}+84 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}-42 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{4}-21 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4}+42 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{2} x^{2}-21 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{4} x^{4}+21 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4}-42 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{2} x^{2}+21 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{4} x^{4}-28 \sqrt {x}\, a^{3} b^{2} x +44 \sqrt {x}\, a \,b^{4} x^{3}\right )}{64 a \,b^{6} c^{3} \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:

int((e*x)^(9/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x)
 

Output:

(sqrt(e)*e**4*( - 42*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a 
**4 + 84*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b**2*x** 
2 - 42*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**4*x**4 - 21* 
sqrt(b)*sqrt(a)*log( - sqrt(a) + sqrt(x)*sqrt(b))*a**4 + 42*sqrt(b)*sqrt(a 
)*log( - sqrt(a) + sqrt(x)*sqrt(b))*a**2*b**2*x**2 - 21*sqrt(b)*sqrt(a)*lo 
g( - sqrt(a) + sqrt(x)*sqrt(b))*b**4*x**4 + 21*sqrt(b)*sqrt(a)*log(sqrt(a) 
 + sqrt(x)*sqrt(b))*a**4 - 42*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt(b 
))*a**2*b**2*x**2 + 21*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt(b))*b**4 
*x**4 - 28*sqrt(x)*a**3*b**2*x + 44*sqrt(x)*a*b**4*x**3))/(64*a*b**6*c**3* 
(a**4 - 2*a**2*b**2*x**2 + b**4*x**4))