3.21.85 \(\int \frac {(a+b x) (d+e x)^{7/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [2085]

3.21.85.1 Optimal result
3.21.85.2 Mathematica [A] (verified)
3.21.85.3 Rubi [A] (verified)
3.21.85.4 Maple [A] (verified)
3.21.85.5 Fricas [B] (verification not implemented)
3.21.85.6 Sympy [F(-1)]
3.21.85.7 Maxima [F(-2)]
3.21.85.8 Giac [A] (verification not implemented)
3.21.85.9 Mupad [B] (verification not implemented)

3.21.85.1 Optimal result

Integrand size = 33, antiderivative size = 152 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {35 e^3 \sqrt {d+e x}}{64 b^4 (a+b x)}-\frac {35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x)^2}-\frac {7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^3}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}-\frac {35 e^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{9/2} \sqrt {b d-a e}} \]

output
-35/96*e^2*(e*x+d)^(3/2)/b^3/(b*x+a)^2-7/24*e*(e*x+d)^(5/2)/b^2/(b*x+a)^3- 
1/4*(e*x+d)^(7/2)/b/(b*x+a)^4-35/64*e^4*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a* 
e+b*d)^(1/2))/b^(9/2)/(-a*e+b*d)^(1/2)-35/64*e^3*(e*x+d)^(1/2)/b^4/(b*x+a)
 
3.21.85.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (105 a^3 e^3+35 a^2 b e^2 (2 d+11 e x)+7 a b^2 e \left (8 d^2+36 d e x+73 e^2 x^2\right )+b^3 \left (48 d^3+200 d^2 e x+326 d e^2 x^2+279 e^3 x^3\right )\right )}{192 b^4 (a+b x)^4}+\frac {35 e^4 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{64 b^{9/2} \sqrt {-b d+a e}} \]

input
Integrate[((a + b*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 
output
-1/192*(Sqrt[d + e*x]*(105*a^3*e^3 + 35*a^2*b*e^2*(2*d + 11*e*x) + 7*a*b^2 
*e*(8*d^2 + 36*d*e*x + 73*e^2*x^2) + b^3*(48*d^3 + 200*d^2*e*x + 326*d*e^2 
*x^2 + 279*e^3*x^3)))/(b^4*(a + b*x)^4) + (35*e^4*ArcTan[(Sqrt[b]*Sqrt[d + 
 e*x])/Sqrt[-(b*d) + a*e]])/(64*b^(9/2)*Sqrt[-(b*d) + a*e])
 
3.21.85.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {1184, 27, 51, 51, 51, 51, 73, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^6 \int \frac {(d+e x)^{7/2}}{b^6 (a+b x)^5}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {(d+e x)^{7/2}}{(a+b x)^5}dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {7 e \int \frac {(d+e x)^{5/2}}{(a+b x)^4}dx}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {7 e \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{(a+b x)^3}dx}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {7 e \left (\frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {7 e \left (\frac {5 e \left (\frac {3 e \left (\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {7 e \left (\frac {5 e \left (\frac {3 e \left (\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {7 e \left (\frac {5 e \left (\frac {3 e \left (-\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\)

input
Int[((a + b*x)*(d + e*x)^(7/2))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
 
output
-1/4*(d + e*x)^(7/2)/(b*(a + b*x)^4) + (7*e*(-1/3*(d + e*x)^(5/2)/(b*(a + 
b*x)^3) + (5*e*(-1/2*(d + e*x)^(3/2)/(b*(a + b*x)^2) + (3*e*(-(Sqrt[d + e* 
x]/(b*(a + b*x))) - (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/( 
b^(3/2)*Sqrt[b*d - a*e])))/(4*b)))/(6*b)))/(8*b)
 

3.21.85.3.1 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 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 1184
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 
3.21.85.4 Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.06

method result size
pseudoelliptic \(\frac {\frac {35 e^{4} \left (b x +a \right )^{4} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{64}-\frac {35 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, \left (\left (\frac {93}{35} e^{3} x^{3}+\frac {326}{105} d \,e^{2} x^{2}+\frac {40}{21} d^{2} e x +\frac {16}{35} d^{3}\right ) b^{3}+\frac {8 e \left (\frac {73}{8} e^{2} x^{2}+\frac {9}{2} d e x +d^{2}\right ) a \,b^{2}}{15}+\frac {2 e^{2} \left (\frac {11 e x}{2}+d \right ) a^{2} b}{3}+a^{3} e^{3}\right )}{64}}{b^{4} \left (b x +a \right )^{4} \sqrt {\left (a e -b d \right ) b}}\) \(161\)
derivativedivides \(2 e^{4} \left (\frac {-\frac {93 \left (e x +d \right )^{\frac {7}{2}}}{128 b}-\frac {511 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}{384 b^{2}}-\frac {385 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{384 b^{3}}-\frac {35 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {e x +d}}{128 b^{4}}}{\left (b \left (e x +d \right )+a e -b d \right )^{4}}+\frac {35 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{128 b^{4} \sqrt {\left (a e -b d \right ) b}}\right )\) \(178\)
default \(2 e^{4} \left (\frac {-\frac {93 \left (e x +d \right )^{\frac {7}{2}}}{128 b}-\frac {511 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}{384 b^{2}}-\frac {385 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{384 b^{3}}-\frac {35 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {e x +d}}{128 b^{4}}}{\left (b \left (e x +d \right )+a e -b d \right )^{4}}+\frac {35 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{128 b^{4} \sqrt {\left (a e -b d \right ) b}}\right )\) \(178\)

input
int((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
 
output
35/64/((a*e-b*d)*b)^(1/2)*(e^4*(b*x+a)^4*arctan(b*(e*x+d)^(1/2)/((a*e-b*d) 
*b)^(1/2))-((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*((93/35*e^3*x^3+326/105*d*e^2 
*x^2+40/21*d^2*e*x+16/35*d^3)*b^3+8/15*e*(73/8*e^2*x^2+9/2*d*e*x+d^2)*a*b^ 
2+2/3*e^2*(11/2*e*x+d)*a^2*b+a^3*e^3))/b^4/(b*x+a)^4
 
3.21.85.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (124) = 248\).

Time = 0.55 (sec) , antiderivative size = 765, normalized size of antiderivative = 5.03 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\left [\frac {105 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (48 \, b^{5} d^{4} + 8 \, a b^{4} d^{3} e + 14 \, a^{2} b^{3} d^{2} e^{2} + 35 \, a^{3} b^{2} d e^{3} - 105 \, a^{4} b e^{4} + 279 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (326 \, b^{5} d^{2} e^{2} + 185 \, a b^{4} d e^{3} - 511 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (200 \, b^{5} d^{3} e + 52 \, a b^{4} d^{2} e^{2} + 133 \, a^{2} b^{3} d e^{3} - 385 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{384 \, {\left (a^{4} b^{6} d - a^{5} b^{5} e + {\left (b^{10} d - a b^{9} e\right )} x^{4} + 4 \, {\left (a b^{9} d - a^{2} b^{8} e\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d - a^{3} b^{7} e\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d - a^{4} b^{6} e\right )} x\right )}}, \frac {105 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (48 \, b^{5} d^{4} + 8 \, a b^{4} d^{3} e + 14 \, a^{2} b^{3} d^{2} e^{2} + 35 \, a^{3} b^{2} d e^{3} - 105 \, a^{4} b e^{4} + 279 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (326 \, b^{5} d^{2} e^{2} + 185 \, a b^{4} d e^{3} - 511 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (200 \, b^{5} d^{3} e + 52 \, a b^{4} d^{2} e^{2} + 133 \, a^{2} b^{3} d e^{3} - 385 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{192 \, {\left (a^{4} b^{6} d - a^{5} b^{5} e + {\left (b^{10} d - a b^{9} e\right )} x^{4} + 4 \, {\left (a b^{9} d - a^{2} b^{8} e\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d - a^{3} b^{7} e\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d - a^{4} b^{6} e\right )} x\right )}}\right ] \]

input
integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fric 
as")
 
output
[1/384*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e 
^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2* 
d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(48*b^5*d^4 + 8*a*b^4*d^3*e + 14* 
a^2*b^3*d^2*e^2 + 35*a^3*b^2*d*e^3 - 105*a^4*b*e^4 + 279*(b^5*d*e^3 - a*b^ 
4*e^4)*x^3 + (326*b^5*d^2*e^2 + 185*a*b^4*d*e^3 - 511*a^2*b^3*e^4)*x^2 + ( 
200*b^5*d^3*e + 52*a*b^4*d^2*e^2 + 133*a^2*b^3*d*e^3 - 385*a^3*b^2*e^4)*x) 
*sqrt(e*x + d))/(a^4*b^6*d - a^5*b^5*e + (b^10*d - a*b^9*e)*x^4 + 4*(a*b^9 
*d - a^2*b^8*e)*x^3 + 6*(a^2*b^8*d - a^3*b^7*e)*x^2 + 4*(a^3*b^7*d - a^4*b 
^6*e)*x), 1/192*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 
4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)* 
sqrt(e*x + d)/(b*e*x + b*d)) - (48*b^5*d^4 + 8*a*b^4*d^3*e + 14*a^2*b^3*d^ 
2*e^2 + 35*a^3*b^2*d*e^3 - 105*a^4*b*e^4 + 279*(b^5*d*e^3 - a*b^4*e^4)*x^3 
 + (326*b^5*d^2*e^2 + 185*a*b^4*d*e^3 - 511*a^2*b^3*e^4)*x^2 + (200*b^5*d^ 
3*e + 52*a*b^4*d^2*e^2 + 133*a^2*b^3*d*e^3 - 385*a^3*b^2*e^4)*x)*sqrt(e*x 
+ d))/(a^4*b^6*d - a^5*b^5*e + (b^10*d - a*b^9*e)*x^4 + 4*(a*b^9*d - a^2*b 
^8*e)*x^3 + 6*(a^2*b^8*d - a^3*b^7*e)*x^2 + 4*(a^3*b^7*d - a^4*b^6*e)*x)]
 
3.21.85.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Timed out} \]

input
integrate((b*x+a)*(e*x+d)**(7/2)/(b**2*x**2+2*a*b*x+a**2)**3,x)
 
output
Timed out
 
3.21.85.7 Maxima [F(-2)]

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

input
integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxi 
ma")
 
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(a*e-b*d>0)', see `assume?` for m 
ore detail
 
3.21.85.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {35 \, e^{4} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, \sqrt {-b^{2} d + a b e} b^{4}} - \frac {279 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{3} e^{4} - 511 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{3} d e^{4} + 385 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} - 105 \, \sqrt {e x + d} b^{3} d^{3} e^{4} + 511 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{2} e^{5} - 770 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} + 315 \, \sqrt {e x + d} a b^{2} d^{2} e^{5} + 385 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b e^{6} - 315 \, \sqrt {e x + d} a^{2} b d e^{6} + 105 \, \sqrt {e x + d} a^{3} e^{7}}{192 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{4} b^{4}} \]

input
integrate((b*x+a)*(e*x+d)^(7/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac 
")
 
output
35/64*e^4*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b* 
e)*b^4) - 1/192*(279*(e*x + d)^(7/2)*b^3*e^4 - 511*(e*x + d)^(5/2)*b^3*d*e 
^4 + 385*(e*x + d)^(3/2)*b^3*d^2*e^4 - 105*sqrt(e*x + d)*b^3*d^3*e^4 + 511 
*(e*x + d)^(5/2)*a*b^2*e^5 - 770*(e*x + d)^(3/2)*a*b^2*d*e^5 + 315*sqrt(e* 
x + d)*a*b^2*d^2*e^5 + 385*(e*x + d)^(3/2)*a^2*b*e^6 - 315*sqrt(e*x + d)*a 
^2*b*d*e^6 + 105*sqrt(e*x + d)*a^3*e^7)/(((e*x + d)*b - b*d + a*e)^4*b^4)
 
3.21.85.9 Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.22 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {35\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{64\,b^{9/2}\,\sqrt {a\,e-b\,d}}-\frac {\frac {93\,e^4\,{\left (d+e\,x\right )}^{7/2}}{64\,b}+\frac {385\,e^4\,{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{192\,b^3}+\frac {35\,e^4\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{64\,b^4}+\frac {511\,e^4\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{5/2}}{192\,b^2}}{b^4\,{\left (d+e\,x\right )}^4-\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,{\left (d+e\,x\right )}^3-\left (d+e\,x\right )\,\left (-4\,a^3\,b\,e^3+12\,a^2\,b^2\,d\,e^2-12\,a\,b^3\,d^2\,e+4\,b^4\,d^3\right )+a^4\,e^4+b^4\,d^4+{\left (d+e\,x\right )}^2\,\left (6\,a^2\,b^2\,e^2-12\,a\,b^3\,d\,e+6\,b^4\,d^2\right )+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e-4\,a^3\,b\,d\,e^3} \]

input
int(((a + b*x)*(d + e*x)^(7/2))/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
 
output
(35*e^4*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(64*b^(9/2)*(a* 
e - b*d)^(1/2)) - ((93*e^4*(d + e*x)^(7/2))/(64*b) + (385*e^4*(d + e*x)^(3 
/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e))/(192*b^3) + (35*e^4*(d + e*x)^(1/2)*( 
a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/(64*b^4) + (511*e^4*(a 
*e - b*d)*(d + e*x)^(5/2))/(192*b^2))/(b^4*(d + e*x)^4 - (4*b^4*d - 4*a*b^ 
3*e)*(d + e*x)^3 - (d + e*x)*(4*b^4*d^3 - 4*a^3*b*e^3 + 12*a^2*b^2*d*e^2 - 
 12*a*b^3*d^2*e) + a^4*e^4 + b^4*d^4 + (d + e*x)^2*(6*b^4*d^2 + 6*a^2*b^2* 
e^2 - 12*a*b^3*d*e) + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3)