3.2025 \(\int \frac {(a+b x) (d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=210 \[ -\frac {(d+e x)^4}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e (a+b x) (d+e x)^3}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e (a+b x) (b d-a e)^3 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^2 x (a+b x) (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e (a+b x) (d+e x)^2 (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

4*e^2*(-a*e+b*d)^2*x*(b*x+a)/b^4/((b*x+a)^2)^(1/2)+2*e*(-a*e+b*d)*(b*x+a)*(e*x+d)^2/b^3/((b*x+a)^2)^(1/2)+4/3*
e*(b*x+a)*(e*x+d)^3/b^2/((b*x+a)^2)^(1/2)-(e*x+d)^4/b/((b*x+a)^2)^(1/2)+4*e*(-a*e+b*d)^3*(b*x+a)*ln(b*x+a)/b^5
/((b*x+a)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {768, 646, 43} \[ \frac {4 e^2 x (a+b x) (b d-a e)^2}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^4}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e (a+b x) (d+e x)^3}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e (a+b x) (d+e x)^2 (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e (a+b x) (b d-a e)^3 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(4*e^2*(b*d - a*e)^2*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*e*(b*d - a*e)*(a + b*x)*(d + e*x)^2
)/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*e*(a + b*x)*(d + e*x)^3)/(3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d
+ e*x)^4/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*e*(b*d - a*e)^3*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=-\frac {(d+e x)^4}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(4 e) \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx}{b}\\ &=-\frac {(d+e x)^4}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^3}{a b+b^2 x} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^4}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 e \left (a b+b^2 x\right )\right ) \int \left (\frac {e (b d-a e)^2}{b^4}+\frac {(b d-a e)^3}{b^3 \left (a b+b^2 x\right )}+\frac {e (b d-a e) (d+e x)}{b^3}+\frac {e (d+e x)^2}{b^2}\right ) \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {4 e^2 (b d-a e)^2 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 e (b d-a e) (a+b x) (d+e x)^2}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e (a+b x) (d+e x)^3}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^4}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e (b d-a e)^3 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 170, normalized size = 0.81 \[ \frac {-3 a^4 e^4+3 a^3 b e^3 (4 d+3 e x)+6 a^2 b^2 e^2 \left (-3 d^2-4 d e x+e^2 x^2\right )-2 a b^3 e \left (-6 d^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3\right )-12 e (a+b x) (a e-b d)^3 \log (a+b x)+b^4 \left (-3 d^4+18 d^2 e^2 x^2+6 d e^3 x^3+e^4 x^4\right )}{3 b^5 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(-3*a^4*e^4 + 3*a^3*b*e^3*(4*d + 3*e*x) + 6*a^2*b^2*e^2*(-3*d^2 - 4*d*e*x + e^2*x^2) - 2*a*b^3*e*(-6*d^3 - 9*d
^2*e*x + 9*d*e^2*x^2 + e^3*x^3) + b^4*(-3*d^4 + 18*d^2*e^2*x^2 + 6*d*e^3*x^3 + e^4*x^4) - 12*e*(-(b*d) + a*e)^
3*(a + b*x)*Log[a + b*x])/(3*b^5*Sqrt[(a + b*x)^2])

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 268, normalized size = 1.28 \[ \frac {b^{4} e^{4} x^{4} - 3 \, b^{4} d^{4} + 12 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 2 \, {\left (3 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (3 \, b^{4} d^{2} e^{2} - 3 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (6 \, a b^{3} d^{2} e^{2} - 8 \, a^{2} b^{2} d e^{3} + 3 \, a^{3} b e^{4}\right )} x + 12 \, {\left (a b^{3} d^{3} e - 3 \, a^{2} b^{2} d^{2} e^{2} + 3 \, a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/3*(b^4*e^4*x^4 - 3*b^4*d^4 + 12*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 12*a^3*b*d*e^3 - 3*a^4*e^4 + 2*(3*b^4*d*e
^3 - a*b^3*e^4)*x^3 + 6*(3*b^4*d^2*e^2 - 3*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 3*(6*a*b^3*d^2*e^2 - 8*a^2*b^2*d*e
^3 + 3*a^3*b*e^4)*x + 12*(a*b^3*d^3*e - 3*a^2*b^2*d^2*e^2 + 3*a^3*b*d*e^3 - a^4*e^4 + (b^4*d^3*e - 3*a*b^3*d^2
*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)*log(b*x + a))/(b^6*x + a*b^5)

________________________________________________________________________________________

giac [A]  time = 0.33, size = 237, normalized size = 1.13 \[ \frac {1}{3} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (x {\left (\frac {x e^{4}}{b^{3}} + \frac {2 \, {\left (3 \, b^{13} d e^{3} - 2 \, a b^{12} e^{4}\right )}}{b^{16}}\right )} + \frac {18 \, b^{13} d^{2} e^{2} - 30 \, a b^{12} d e^{3} + 13 \, a^{2} b^{11} e^{4}}{b^{16}}\right )} - \frac {4 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} \log \left ({\left | -3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{2} a b - a^{3} b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{3} {\left | b \right |} - 3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )} a^{2} {\left | b \right |} \right |}\right )}{3 \, b^{4} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(x*(x*e^4/b^3 + 2*(3*b^13*d*e^3 - 2*a*b^12*e^4)/b^16) + (18*b^13*d^2*e^2 - 3
0*a*b^12*d*e^3 + 13*a^2*b^11*e^4)/b^16) - 4/3*(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*log(abs(
-3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))^2*a*b - a^3*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))^3*abs
(b) - 3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))*a^2*abs(b)))/(b^4*abs(b))

________________________________________________________________________________________

maple [B]  time = 0.07, size = 321, normalized size = 1.53 \[ -\frac {\left (-b^{4} e^{4} x^{4}+2 a \,b^{3} e^{4} x^{3}-6 b^{4} d \,e^{3} x^{3}+12 a^{3} b \,e^{4} x \ln \left (b x +a \right )-36 a^{2} b^{2} d \,e^{3} x \ln \left (b x +a \right )-6 a^{2} b^{2} e^{4} x^{2}+36 a \,b^{3} d^{2} e^{2} x \ln \left (b x +a \right )+18 a \,b^{3} d \,e^{3} x^{2}-12 b^{4} d^{3} e x \ln \left (b x +a \right )-18 b^{4} d^{2} e^{2} x^{2}+12 a^{4} e^{4} \ln \left (b x +a \right )-36 a^{3} b d \,e^{3} \ln \left (b x +a \right )-9 a^{3} b \,e^{4} x +36 a^{2} b^{2} d^{2} e^{2} \ln \left (b x +a \right )+24 a^{2} b^{2} d \,e^{3} x -12 a \,b^{3} d^{3} e \ln \left (b x +a \right )-18 a \,b^{3} d^{2} e^{2} x +3 a^{4} e^{4}-12 a^{3} b d \,e^{3}+18 a^{2} b^{2} d^{2} e^{2}-12 a \,b^{3} d^{3} e +3 b^{4} d^{4}\right ) \left (b x +a \right )^{2}}{3 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

-1/3*(-b^4*e^4*x^4+2*a*b^3*e^4*x^3-6*b^4*d*e^3*x^3+12*ln(b*x+a)*x*a^3*b*e^4-36*ln(b*x+a)*x*a^2*b^2*d*e^3+36*ln
(b*x+a)*x*a*b^3*d^2*e^2-12*ln(b*x+a)*x*b^4*d^3*e-6*a^2*b^2*e^4*x^2+18*a*b^3*d*e^3*x^2-18*b^4*d^2*e^2*x^2+12*ln
(b*x+a)*a^4*e^4-36*ln(b*x+a)*a^3*b*d*e^3+36*ln(b*x+a)*a^2*b^2*d^2*e^2-12*ln(b*x+a)*a*b^3*d^3*e-9*a^3*b*e^4*x+2
4*a^2*b^2*d*e^3*x-18*a*b^3*d^2*e^2*x+3*a^4*e^4-12*a^3*b*d*e^3+18*a^2*b^2*d^2*e^2-12*a*b^3*d^3*e+3*b^4*d^4)*(b*
x+a)^2/b^5/((b*x+a)^2)^(3/2)

________________________________________________________________________________________

maxima [B]  time = 0.64, size = 754, normalized size = 3.59 \[ \frac {e^{4} x^{4}}{3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b} - \frac {7 \, a e^{4} x^{3}}{6 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {9 \, a^{2} e^{4} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} - \frac {10 \, a^{3} e^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {9 \, a^{4} e^{4}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {{\left (4 \, b d e^{3} + a e^{4}\right )} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {20 \, a^{4} e^{4} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {5 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {2 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {a d^{4}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {39 \, a^{5} e^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {6 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {2 \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {5 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {4 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {b d^{4} + 4 \, a d^{3} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {12 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {12 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {4 \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, {\left (4 \, b d e^{3} + a e^{4}\right )} a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} a^{3}}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} a^{2}}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (b d^{4} + 4 \, a d^{3} e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/3*e^4*x^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b) - 7/6*a*e^4*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 9/2*a^2*e^
4*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) - 10*a^3*e^4*log(x + a/b)/b^5 + 9*a^4*e^4/(sqrt(b^2*x^2 + 2*a*b*x +
a^2)*b^5) + 1/2*(4*b*d*e^3 + a*e^4)*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 20*a^4*e^4*x/(b^6*(x + a/b)^2) -
5/2*(4*b*d*e^3 + a*e^4)*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 2*(3*b*d^2*e^2 + 2*a*d*e^3)*x^2/(sqrt(b^2
*x^2 + 2*a*b*x + a^2)*b^2) - 1/2*a*d^4/(b^3*(x + a/b)^2) - 39/2*a^5*e^4/(b^7*(x + a/b)^2) + 6*(4*b*d*e^3 + a*e
^4)*a^2*log(x + a/b)/b^5 - 6*(3*b*d^2*e^2 + 2*a*d*e^3)*a*log(x + a/b)/b^4 + 2*(2*b*d^3*e + 3*a*d^2*e^2)*log(x
+ a/b)/b^3 - 5*(4*b*d*e^3 + a*e^4)*a^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + 4*(3*b*d^2*e^2 + 2*a*d*e^3)*a^2/(
sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (b*d^4 + 4*a*d^3*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 12*(4*b*d*e^3 +
a*e^4)*a^3*x/(b^6*(x + a/b)^2) - 12*(3*b*d^2*e^2 + 2*a*d*e^3)*a^2*x/(b^5*(x + a/b)^2) + 4*(2*b*d^3*e + 3*a*d^
2*e^2)*a*x/(b^4*(x + a/b)^2) + 23/2*(4*b*d*e^3 + a*e^4)*a^4/(b^7*(x + a/b)^2) - 11*(3*b*d^2*e^2 + 2*a*d*e^3)*a
^3/(b^6*(x + a/b)^2) + 3*(2*b*d^3*e + 3*a*d^2*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(b*d^4 + 4*a*d^3*e)*a/(b^4*(x +
a/b)^2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(d + e*x)^4)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(((a + b*x)*(d + e*x)^4)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right ) \left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((a + b*x)*(d + e*x)**4/((a + b*x)**2)**(3/2), x)

________________________________________________________________________________________