3.1971 \(\int (a+b x) (d+e x)^4 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=219 \[ \frac {e^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7 (b d-a e)}{2 b^5}+\frac {6 e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)^2}{7 b^5}+\frac {2 e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^3}{3 b^5}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^4}{5 b^5}+\frac {e^4 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^5} \]

[Out]

1/5*(-a*e+b*d)^4*(b*x+a)^4*((b*x+a)^2)^(1/2)/b^5+2/3*e*(-a*e+b*d)^3*(b*x+a)^5*((b*x+a)^2)^(1/2)/b^5+6/7*e^2*(-
a*e+b*d)^2*(b*x+a)^6*((b*x+a)^2)^(1/2)/b^5+1/2*e^3*(-a*e+b*d)*(b*x+a)^7*((b*x+a)^2)^(1/2)/b^5+1/9*e^4*(b*x+a)^
8*((b*x+a)^2)^(1/2)/b^5

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Rubi [A]  time = 0.22, antiderivative size = 219, 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 = {770, 21, 43} \[ \frac {e^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7 (b d-a e)}{2 b^5}+\frac {6 e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)^2}{7 b^5}+\frac {2 e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)^3}{3 b^5}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^4}{5 b^5}+\frac {e^4 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^5} \]

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]

((b*d - a*e)^4*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^5) + (2*e*(b*d - a*e)^3*(a + b*x)^5*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(3*b^5) + (6*e^2*(b*d - a*e)^2*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^5) + (e^3*(
b*d - a*e)*(a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b^5) + (e^4*(a + b*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(9*b^5)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int (a+b x) (d+e x)^4 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) \left (a b+b^2 x\right )^3 (d+e x)^4 \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int (a+b x)^4 (d+e x)^4 \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac {(b d-a e)^4 (a+b x)^4}{b^4}+\frac {4 e (b d-a e)^3 (a+b x)^5}{b^4}+\frac {6 e^2 (b d-a e)^2 (a+b x)^6}{b^4}+\frac {4 e^3 (b d-a e) (a+b x)^7}{b^4}+\frac {e^4 (a+b x)^8}{b^4}\right ) \, dx}{a b+b^2 x}\\ &=\frac {(b d-a e)^4 (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^5}+\frac {2 e (b d-a e)^3 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 b^5}+\frac {6 e^2 (b d-a e)^2 (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^5}+\frac {e^3 (b d-a e) (a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{2 b^5}+\frac {e^4 (a+b x)^8 \sqrt {a^2+2 a b x+b^2 x^2}}{9 b^5}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 267, normalized size = 1.22 \[ \frac {x \sqrt {(a+b x)^2} \left (126 a^4 \left (5 d^4+10 d^3 e x+10 d^2 e^2 x^2+5 d e^3 x^3+e^4 x^4\right )+84 a^3 b x \left (15 d^4+40 d^3 e x+45 d^2 e^2 x^2+24 d e^3 x^3+5 e^4 x^4\right )+36 a^2 b^2 x^2 \left (35 d^4+105 d^3 e x+126 d^2 e^2 x^2+70 d e^3 x^3+15 e^4 x^4\right )+9 a b^3 x^3 \left (70 d^4+224 d^3 e x+280 d^2 e^2 x^2+160 d e^3 x^3+35 e^4 x^4\right )+b^4 x^4 \left (126 d^4+420 d^3 e x+540 d^2 e^2 x^2+315 d e^3 x^3+70 e^4 x^4\right )\right )}{630 (a+b x)} \]

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]

(x*Sqrt[(a + b*x)^2]*(126*a^4*(5*d^4 + 10*d^3*e*x + 10*d^2*e^2*x^2 + 5*d*e^3*x^3 + e^4*x^4) + 84*a^3*b*x*(15*d
^4 + 40*d^3*e*x + 45*d^2*e^2*x^2 + 24*d*e^3*x^3 + 5*e^4*x^4) + 36*a^2*b^2*x^2*(35*d^4 + 105*d^3*e*x + 126*d^2*
e^2*x^2 + 70*d*e^3*x^3 + 15*e^4*x^4) + 9*a*b^3*x^3*(70*d^4 + 224*d^3*e*x + 280*d^2*e^2*x^2 + 160*d*e^3*x^3 + 3
5*e^4*x^4) + b^4*x^4*(126*d^4 + 420*d^3*e*x + 540*d^2*e^2*x^2 + 315*d*e^3*x^3 + 70*e^4*x^4)))/(630*(a + b*x))

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fricas [A]  time = 1.28, size = 285, normalized size = 1.30 \[ \frac {1}{9} \, b^{4} e^{4} x^{9} + a^{4} d^{4} x + \frac {1}{2} \, {\left (b^{4} d e^{3} + a b^{3} e^{4}\right )} x^{8} + \frac {2}{7} \, {\left (3 \, b^{4} d^{2} e^{2} + 8 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{7} + \frac {2}{3} \, {\left (b^{4} d^{3} e + 6 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x^{6} + \frac {1}{5} \, {\left (b^{4} d^{4} + 16 \, a b^{3} d^{3} e + 36 \, a^{2} b^{2} d^{2} e^{2} + 16 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} x^{5} + {\left (a b^{3} d^{4} + 6 \, a^{2} b^{2} d^{3} e + 6 \, a^{3} b d^{2} e^{2} + a^{4} d e^{3}\right )} x^{4} + \frac {2}{3} \, {\left (3 \, a^{2} b^{2} d^{4} + 8 \, a^{3} b d^{3} e + 3 \, a^{4} d^{2} e^{2}\right )} x^{3} + 2 \, {\left (a^{3} b d^{4} + a^{4} d^{3} e\right )} x^{2} \]

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="fricas")

[Out]

1/9*b^4*e^4*x^9 + a^4*d^4*x + 1/2*(b^4*d*e^3 + a*b^3*e^4)*x^8 + 2/7*(3*b^4*d^2*e^2 + 8*a*b^3*d*e^3 + 3*a^2*b^2
*e^4)*x^7 + 2/3*(b^4*d^3*e + 6*a*b^3*d^2*e^2 + 6*a^2*b^2*d*e^3 + a^3*b*e^4)*x^6 + 1/5*(b^4*d^4 + 16*a*b^3*d^3*
e + 36*a^2*b^2*d^2*e^2 + 16*a^3*b*d*e^3 + a^4*e^4)*x^5 + (a*b^3*d^4 + 6*a^2*b^2*d^3*e + 6*a^3*b*d^2*e^2 + a^4*
d*e^3)*x^4 + 2/3*(3*a^2*b^2*d^4 + 8*a^3*b*d^3*e + 3*a^4*d^2*e^2)*x^3 + 2*(a^3*b*d^4 + a^4*d^3*e)*x^2

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giac [B]  time = 0.19, size = 461, normalized size = 2.11 \[ \frac {1}{9} \, b^{4} x^{9} e^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b^{4} d x^{8} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, b^{4} d^{2} x^{7} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, b^{4} d^{3} x^{6} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, b^{4} d^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a b^{3} x^{8} e^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {16}{7} \, a b^{3} d x^{7} e^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, a b^{3} d^{2} x^{6} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {16}{5} \, a b^{3} d^{3} x^{5} e \mathrm {sgn}\left (b x + a\right ) + a b^{3} d^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, a^{2} b^{2} x^{7} e^{4} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{2} b^{2} d x^{6} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {36}{5} \, a^{2} b^{2} d^{2} x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{3} x^{4} e \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{2} d^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, a^{3} b x^{6} e^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {16}{5} \, a^{3} b d x^{5} e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{3} b d^{2} x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {16}{3} \, a^{3} b d^{3} x^{3} e \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b d^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, a^{4} x^{5} e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{4} d x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{4} d^{2} x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{4} d^{3} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{4} d^{4} x \mathrm {sgn}\left (b x + a\right ) \]

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]

1/9*b^4*x^9*e^4*sgn(b*x + a) + 1/2*b^4*d*x^8*e^3*sgn(b*x + a) + 6/7*b^4*d^2*x^7*e^2*sgn(b*x + a) + 2/3*b^4*d^3
*x^6*e*sgn(b*x + a) + 1/5*b^4*d^4*x^5*sgn(b*x + a) + 1/2*a*b^3*x^8*e^4*sgn(b*x + a) + 16/7*a*b^3*d*x^7*e^3*sgn
(b*x + a) + 4*a*b^3*d^2*x^6*e^2*sgn(b*x + a) + 16/5*a*b^3*d^3*x^5*e*sgn(b*x + a) + a*b^3*d^4*x^4*sgn(b*x + a)
+ 6/7*a^2*b^2*x^7*e^4*sgn(b*x + a) + 4*a^2*b^2*d*x^6*e^3*sgn(b*x + a) + 36/5*a^2*b^2*d^2*x^5*e^2*sgn(b*x + a)
+ 6*a^2*b^2*d^3*x^4*e*sgn(b*x + a) + 2*a^2*b^2*d^4*x^3*sgn(b*x + a) + 2/3*a^3*b*x^6*e^4*sgn(b*x + a) + 16/5*a^
3*b*d*x^5*e^3*sgn(b*x + a) + 6*a^3*b*d^2*x^4*e^2*sgn(b*x + a) + 16/3*a^3*b*d^3*x^3*e*sgn(b*x + a) + 2*a^3*b*d^
4*x^2*sgn(b*x + a) + 1/5*a^4*x^5*e^4*sgn(b*x + a) + a^4*d*x^4*e^3*sgn(b*x + a) + 2*a^4*d^2*x^3*e^2*sgn(b*x + a
) + 2*a^4*d^3*x^2*e*sgn(b*x + a) + a^4*d^4*x*sgn(b*x + a)

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maple [B]  time = 0.05, size = 339, normalized size = 1.55 \[ \frac {\left (70 b^{4} e^{4} x^{8}+315 x^{7} a \,b^{3} e^{4}+315 x^{7} b^{4} d \,e^{3}+540 x^{6} a^{2} b^{2} e^{4}+1440 x^{6} a \,b^{3} d \,e^{3}+540 x^{6} b^{4} d^{2} e^{2}+420 x^{5} a^{3} b \,e^{4}+2520 x^{5} a^{2} b^{2} d \,e^{3}+2520 x^{5} a \,b^{3} d^{2} e^{2}+420 x^{5} b^{4} d^{3} e +126 x^{4} a^{4} e^{4}+2016 x^{4} a^{3} b d \,e^{3}+4536 x^{4} a^{2} b^{2} d^{2} e^{2}+2016 x^{4} a \,b^{3} d^{3} e +126 x^{4} b^{4} d^{4}+630 a^{4} d \,e^{3} x^{3}+3780 a^{3} b \,d^{2} e^{2} x^{3}+3780 a^{2} b^{2} d^{3} e \,x^{3}+630 a \,b^{3} d^{4} x^{3}+1260 x^{2} a^{4} d^{2} e^{2}+3360 x^{2} a^{3} b \,d^{3} e +1260 x^{2} a^{2} b^{2} d^{4}+1260 a^{4} d^{3} e x +1260 a^{3} b \,d^{4} x +630 a^{4} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{630 \left (b x +a \right )^{3}} \]

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]

1/630*x*(70*b^4*e^4*x^8+315*a*b^3*e^4*x^7+315*b^4*d*e^3*x^7+540*a^2*b^2*e^4*x^6+1440*a*b^3*d*e^3*x^6+540*b^4*d
^2*e^2*x^6+420*a^3*b*e^4*x^5+2520*a^2*b^2*d*e^3*x^5+2520*a*b^3*d^2*e^2*x^5+420*b^4*d^3*e*x^5+126*a^4*e^4*x^4+2
016*a^3*b*d*e^3*x^4+4536*a^2*b^2*d^2*e^2*x^4+2016*a*b^3*d^3*e*x^4+126*b^4*d^4*x^4+630*a^4*d*e^3*x^3+3780*a^3*b
*d^2*e^2*x^3+3780*a^2*b^2*d^3*e*x^3+630*a*b^3*d^4*x^3+1260*a^4*d^2*e^2*x^2+3360*a^3*b*d^3*e*x^2+1260*a^2*b^2*d
^4*x^2+1260*a^4*d^3*e*x+1260*a^3*b*d^4*x+630*a^4*d^4)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.74, size = 998, normalized size = 4.56 \[ \text {result too large to display} \]

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]

1/9*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*e^4*x^4/b - 13/72*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*e^4*x^3/b^2 + 1/4*(b^2
*x^2 + 2*a*b*x + a^2)^(3/2)*a*d^4*x - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^5*e^4*x/b^4 + 37/168*(b^2*x^2 + 2*
a*b*x + a^2)^(5/2)*a^2*e^4*x^2/b^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*d^4/b - 1/4*(b^2*x^2 + 2*a*b*x +
a^2)^(3/2)*a^6*e^4/b^5 - 121/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3*e^4*x/b^4 + 125/504*(b^2*x^2 + 2*a*b*x +
a^2)^(5/2)*a^4*e^4/b^5 + 1/8*(4*b*d*e^3 + a*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^3/b^2 + 1/4*(4*b*d*e^3 + a*
e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4*x/b^4 - 1/2*(3*b*d^2*e^2 + 2*a*d*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)
*a^3*x/b^3 + 1/2*(2*b*d^3*e + 3*a*d^2*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*x/b^2 - 1/4*(b*d^4 + 4*a*d^3*e)
*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*x/b - 11/56*(4*b*d*e^3 + a*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x^2/b^3 +
2/7*(3*b*d^2*e^2 + 2*a*d*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^2/b^2 + 1/4*(4*b*d*e^3 + a*e^4)*(b^2*x^2 + 2*
a*b*x + a^2)^(3/2)*a^5/b^5 - 1/2*(3*b*d^2*e^2 + 2*a*d*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4/b^4 + 1/2*(2*b*
d^3*e + 3*a*d^2*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3/b^3 - 1/4*(b*d^4 + 4*a*d^3*e)*(b^2*x^2 + 2*a*b*x + a^
2)^(3/2)*a^2/b^2 + 13/56*(4*b*d*e^3 + a*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2*x/b^4 - 3/7*(3*b*d^2*e^2 + 2*
a*d*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x/b^3 + 1/3*(2*b*d^3*e + 3*a*d^2*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(5/
2)*x/b^2 - 69/280*(4*b*d*e^3 + a*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3/b^5 + 17/35*(3*b*d^2*e^2 + 2*a*d*e^3
)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2/b^4 - 7/15*(2*b*d^3*e + 3*a*d^2*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a/b
^3 + 1/5*(b*d^4 + 4*a*d^3*e)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/b^2

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

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)

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

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)

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