\(\int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx\) [71]

Optimal result

Integrand size = 26, antiderivative size = 159 \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {e (e x)^{5/2}}{4 b^2 c^3 \left (a^2-b^2 x^2\right )^2}-\frac {5 e^3 \sqrt {e x}}{16 b^4 c^3 \left (a^2-b^2 x^2\right )}+\frac {5 e^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 a^{3/2} b^{9/2} c^3}+\frac {5 e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{32 a^{3/2} b^{9/2} c^3} \] Output:

1/4*e*(e*x)^(5/2)/b^2/c^3/(-b^2*x^2+a^2)^2-5/16*e^3*(e*x)^(1/2)/b^4/c^3/(- 
b^2*x^2+a^2)+5/32*e^(7/2)*arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/a^(3 
/2)/b^(9/2)/c^3+5/32*e^(7/2)*arctanh(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/ 
a^(3/2)/b^(9/2)/c^3
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91 \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {e^3 \sqrt {e x} \left (2 a^{3/2} \sqrt {b} \sqrt {x} \left (-5 a^2+9 b^2 x^2\right )+5 \left (a^2-b^2 x^2\right )^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+5 \left (a^2-b^2 x^2\right )^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{32 a^{3/2} b^{9/2} c^3 \sqrt {x} \left (a^2-b^2 x^2\right )^2} \] Input:

Integrate[(e*x)^(7/2)/((a + b*x)^3*(a*c - b*c*x)^3),x]
 

Output:

(e^3*Sqrt[e*x]*(2*a^(3/2)*Sqrt[b]*Sqrt[x]*(-5*a^2 + 9*b^2*x^2) + 5*(a^2 - 
b^2*x^2)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] + 5*(a^2 - b^2*x^2)^2*ArcTanh 
[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(32*a^(3/2)*b^(9/2)*c^3*Sqrt[x]*(a^2 - b^2*x 
^2)^2)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {82, 252, 27, 252, 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.

\(\displaystyle \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx\)

\(\Big \downarrow \) 82

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[(e*x)^(7/2)/((a + b*x)^3*(a*c - b*c*x)^3),x]
 

Output:

(e*(e*x)^(5/2))/(4*b^2*c^3*(a^2 - b^2*x^2)^2) - (5*e^2*((e*Sqrt[e*x])/(2*b 
^2*(a^2 - b^2*x^2)) - (e*((Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqr 
t[e])])/(2*a^(3/2)*Sqrt[b]) + (Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a 
]*Sqrt[e])])/(2*a^(3/2)*Sqrt[b])))/(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 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 = 1.62 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(-\frac {e^{5} \left (\frac {7 \left (e x \right )^{\frac {3}{2}}}{e^{3} \left (b x +a \right )^{2} b^{3} a}+\frac {5 \sqrt {e x}}{e^{2} \left (b x +a \right )^{2} b^{4}}-\frac {5 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{4} \sqrt {a e b}\, a e}+\frac {\frac {\sqrt {e x}\, \left (-7 b x +5 a \right )}{e \left (-b x +a \right )^{2}}-\frac {5 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{\sqrt {a e b}}}{b^{4} a e}\right )}{32 c^{3}}\)	\(142\)
derivativedivides	\(-\frac {2 e^{5} \left (\frac {\frac {-\frac {7 \left (e x \right )^{\frac {3}{2}}}{4}+\frac {5 a e \sqrt {e x}}{4 b}}{\left (-b e x +a e \right )^{2}}-\frac {5 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 b \sqrt {a e b}}}{16 b^{3} a e}-\frac {\frac {-\frac {7 \left (e x \right )^{\frac {3}{2}}}{4}-\frac {5 a e \sqrt {e x}}{4 b}}{\left (b e x +a e \right )^{2}}+\frac {5 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 b \sqrt {a e b}}}{16 b^{3} a e}\right )}{c^{3}}\)	\(147\)
default	\(\frac {2 e^{5} \left (-\frac {\frac {-\frac {7 \left (e x \right )^{\frac {3}{2}}}{4}+\frac {5 a e \sqrt {e x}}{4 b}}{\left (-b e x +a e \right )^{2}}-\frac {5 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 b \sqrt {a e b}}}{16 b^{3} a e}+\frac {\frac {-\frac {7 \left (e x \right )^{\frac {3}{2}}}{4}-\frac {5 a e \sqrt {e x}}{4 b}}{\left (b e x +a e \right )^{2}}+\frac {5 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{4 b \sqrt {a e b}}}{16 b^{3} a e}\right )}{c^{3}}\)	\(147\)

Input:

int((e*x)^(7/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.74 \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\left [\frac {10 \, {\left (b^{4} e^{3} x^{4} - 2 \, a^{2} b^{2} e^{3} x^{2} + a^{4} e^{3}\right )} \sqrt {\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {e}{a b}}}{e}\right ) + 5 \, {\left (b^{4} e^{3} x^{4} - 2 \, a^{2} b^{2} e^{3} x^{2} + a^{4} e^{3}\right )} \sqrt {\frac {e}{a b}} \log \left (\frac {b e x + 2 \, \sqrt {e x} a b \sqrt {\frac {e}{a b}} + a e}{b x - a}\right ) + 4 \, {\left (9 \, a b^{2} e^{3} x^{2} - 5 \, a^{3} e^{3}\right )} \sqrt {e x}}{64 \, {\left (a b^{8} c^{3} x^{4} - 2 \, a^{3} b^{6} c^{3} x^{2} + a^{5} b^{4} c^{3}\right )}}, -\frac {10 \, {\left (b^{4} e^{3} x^{4} - 2 \, a^{2} b^{2} e^{3} x^{2} + a^{4} e^{3}\right )} \sqrt {-\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {e}{a b}}}{e}\right ) - 5 \, {\left (b^{4} e^{3} x^{4} - 2 \, a^{2} b^{2} e^{3} x^{2} + a^{4} e^{3}\right )} \sqrt {-\frac {e}{a b}} \log \left (\frac {b e x + 2 \, \sqrt {e x} a b \sqrt {-\frac {e}{a b}} - a e}{b x + a}\right ) - 4 \, {\left (9 \, a b^{2} e^{3} x^{2} - 5 \, a^{3} e^{3}\right )} \sqrt {e x}}{64 \, {\left (a b^{8} c^{3} x^{4} - 2 \, a^{3} b^{6} c^{3} x^{2} + a^{5} b^{4} c^{3}\right )}}\right ] \] Input:

integrate((e*x)^(7/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="fricas")
 

Output:

[1/64*(10*(b^4*e^3*x^4 - 2*a^2*b^2*e^3*x^2 + a^4*e^3)*sqrt(e/(a*b))*arctan 
(sqrt(e*x)*b*sqrt(e/(a*b))/e) + 5*(b^4*e^3*x^4 - 2*a^2*b^2*e^3*x^2 + a^4*e 
^3)*sqrt(e/(a*b))*log((b*e*x + 2*sqrt(e*x)*a*b*sqrt(e/(a*b)) + a*e)/(b*x - 
 a)) + 4*(9*a*b^2*e^3*x^2 - 5*a^3*e^3)*sqrt(e*x))/(a*b^8*c^3*x^4 - 2*a^3*b 
^6*c^3*x^2 + a^5*b^4*c^3), -1/64*(10*(b^4*e^3*x^4 - 2*a^2*b^2*e^3*x^2 + a^ 
4*e^3)*sqrt(-e/(a*b))*arctan(sqrt(e*x)*b*sqrt(-e/(a*b))/e) - 5*(b^4*e^3*x^ 
4 - 2*a^2*b^2*e^3*x^2 + a^4*e^3)*sqrt(-e/(a*b))*log((b*e*x + 2*sqrt(e*x)*a 
*b*sqrt(-e/(a*b)) - a*e)/(b*x + a)) - 4*(9*a*b^2*e^3*x^2 - 5*a^3*e^3)*sqrt 
(e*x))/(a*b^8*c^3*x^4 - 2*a^3*b^6*c^3*x^2 + a^5*b^4*c^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\text {Timed out} \] Input:

integrate((e*x)**(7/2)/(b*x+a)**3/(-b*c*x+a*c)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x)^(7/2)/(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 = 135, normalized size of antiderivative = 0.85 \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\frac {5 \, e^{5} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a b^{4} c^{3}} - \frac {5 \, e^{5} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a b^{4} c^{3}} + \frac {2 \, {\left (9 \, \sqrt {e x} b^{2} e^{8} x^{2} - 5 \, \sqrt {e x} a^{2} e^{8}\right )}}{{\left (b^{2} e^{2} x^{2} - a^{2} e^{2}\right )}^{2} b^{4} c^{3}}}{32 \, e} \] Input:

integrate((e*x)^(7/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="giac")
 

Output:

1/32*(5*e^5*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*a*b^4*c^3) - 5*e^ 
5*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*e)*a*b^4*c^3) + 2*(9*sqrt(e* 
x)*b^2*e^8*x^2 - 5*sqrt(e*x)*a^2*e^8)/((b^2*e^2*x^2 - a^2*e^2)^2*b^4*c^3)) 
/e
 

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85 \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\frac {9\,e^5\,{\left (e\,x\right )}^{5/2}}{16\,b^2}-\frac {5\,a^2\,e^7\,\sqrt {e\,x}}{16\,b^4}}{a^4\,c^3\,e^4-2\,a^2\,b^2\,c^3\,e^4\,x^2+b^4\,c^3\,e^4\,x^4}+\frac {5\,e^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,a^{3/2}\,b^{9/2}\,c^3}+\frac {5\,e^{7/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{32\,a^{3/2}\,b^{9/2}\,c^3} \] Input:

int((e*x)^(7/2)/((a*c - b*c*x)^3*(a + b*x)^3),x)
 

Output:

((9*e^5*(e*x)^(5/2))/(16*b^2) - (5*a^2*e^7*(e*x)^(1/2))/(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) + (5*e^(7/2)*atan((b^(1/2) 
*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(3/2)*b^(9/2)*c^3) + (5*e^(7/2)*at 
anh((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(32*a^(3/2)*b^(9/2)*c^3)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.69 \[ \int \frac {(e x)^{7/2}}{(a+b x)^3 (a c-b c x)^3} \, dx=\frac {\sqrt {e}\, e^{3} \left (10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}-20 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}+10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{4}-5 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4}+10 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{2} x^{2}-5 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{4} x^{4}+5 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{4}-10 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) a^{2} b^{2} x^{2}+5 \sqrt {b}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}+\sqrt {x}\, \sqrt {b}\right ) b^{4} x^{4}-20 \sqrt {x}\, a^{4} b +36 \sqrt {x}\, a^{2} b^{3} x^{2}\right )}{64 a^{2} b^{5} c^{3} \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:

int((e*x)^(7/2)/(b*x+a)^3/(-b*c*x+a*c)^3,x)
 

Output:

(sqrt(e)*e**3*(10*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**4 
 - 20*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**2 + 
 10*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**4*x**4 - 5*sqrt 
(b)*sqrt(a)*log( - sqrt(a) + sqrt(x)*sqrt(b))*a**4 + 10*sqrt(b)*sqrt(a)*lo 
g( - sqrt(a) + sqrt(x)*sqrt(b))*a**2*b**2*x**2 - 5*sqrt(b)*sqrt(a)*log( - 
sqrt(a) + sqrt(x)*sqrt(b))*b**4*x**4 + 5*sqrt(b)*sqrt(a)*log(sqrt(a) + sqr 
t(x)*sqrt(b))*a**4 - 10*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt(b))*a** 
2*b**2*x**2 + 5*sqrt(b)*sqrt(a)*log(sqrt(a) + sqrt(x)*sqrt(b))*b**4*x**4 - 
 20*sqrt(x)*a**4*b + 36*sqrt(x)*a**2*b**3*x**2))/(64*a**2*b**5*c**3*(a**4 
- 2*a**2*b**2*x**2 + b**4*x**4))