3.1583 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^9} \, dx\)

Optimal. Leaf size=149 \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{168 (d+e x)^6 (b d-a e)^3}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{28 (d+e x)^7 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{8 (d+e x)^8 (b d-a e)} \]

[Out]

1/8*(b*x+a)^5*((b*x+a)^2)^(1/2)/(-a*e+b*d)/(e*x+d)^8+1/28*b*(b*x+a)^5*((b*x+a)^2)^(1/2)/(-a*e+b*d)^2/(e*x+d)^7
+1/168*b^2*(b*x+a)^5*((b*x+a)^2)^(1/2)/(-a*e+b*d)^3/(e*x+d)^6

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Rubi [A]  time = 0.05, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {646, 45, 37} \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{168 (d+e x)^6 (b d-a e)^3}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{28 (d+e x)^7 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{8 (d+e x)^8 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^9,x]

[Out]

((a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*(b*d - a*e)*(d + e*x)^8) + (b*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(28*(b*d - a*e)^2*(d + e*x)^7) + (b^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(168*(b*d - a*e)^3*
(d + e*x)^6)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^9} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{(d+e x)^9} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (b d-a e) (d+e x)^8}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{(d+e x)^8} \, dx}{4 b^3 (b d-a e) \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (b d-a e) (d+e x)^8}+\frac {b (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{28 (b d-a e)^2 (d+e x)^7}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{(d+e x)^7} \, dx}{28 b^2 (b d-a e)^2 \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (b d-a e) (d+e x)^8}+\frac {b (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{28 (b d-a e)^2 (d+e x)^7}+\frac {b^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{168 (b d-a e)^3 (d+e x)^6}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 223, normalized size = 1.50 \[ -\frac {\sqrt {(a+b x)^2} \left (21 a^5 e^5+15 a^4 b e^4 (d+8 e x)+10 a^3 b^2 e^3 \left (d^2+8 d e x+28 e^2 x^2\right )+6 a^2 b^3 e^2 \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )+3 a b^4 e \left (d^4+8 d^3 e x+28 d^2 e^2 x^2+56 d e^3 x^3+70 e^4 x^4\right )+b^5 \left (d^5+8 d^4 e x+28 d^3 e^2 x^2+56 d^2 e^3 x^3+70 d e^4 x^4+56 e^5 x^5\right )\right )}{168 e^6 (a+b x) (d+e x)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^9,x]

[Out]

-1/168*(Sqrt[(a + b*x)^2]*(21*a^5*e^5 + 15*a^4*b*e^4*(d + 8*e*x) + 10*a^3*b^2*e^3*(d^2 + 8*d*e*x + 28*e^2*x^2)
+ 6*a^2*b^3*e^2*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x^3) + 3*a*b^4*e*(d^4 + 8*d^3*e*x + 28*d^2*e^2*x^2 +
56*d*e^3*x^3 + 70*e^4*x^4) + b^5*(d^5 + 8*d^4*e*x + 28*d^3*e^2*x^2 + 56*d^2*e^3*x^3 + 70*d*e^4*x^4 + 56*e^5*x
^5)))/(e^6*(a + b*x)*(d + e*x)^8)

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fricas [B]  time = 1.02, size = 337, normalized size = 2.26 \[ -\frac {56 \, b^{5} e^{5} x^{5} + b^{5} d^{5} + 3 \, a b^{4} d^{4} e + 6 \, a^{2} b^{3} d^{3} e^{2} + 10 \, a^{3} b^{2} d^{2} e^{3} + 15 \, a^{4} b d e^{4} + 21 \, a^{5} e^{5} + 70 \, {\left (b^{5} d e^{4} + 3 \, a b^{4} e^{5}\right )} x^{4} + 56 \, {\left (b^{5} d^{2} e^{3} + 3 \, a b^{4} d e^{4} + 6 \, a^{2} b^{3} e^{5}\right )} x^{3} + 28 \, {\left (b^{5} d^{3} e^{2} + 3 \, a b^{4} d^{2} e^{3} + 6 \, a^{2} b^{3} d e^{4} + 10 \, a^{3} b^{2} e^{5}\right )} x^{2} + 8 \, {\left (b^{5} d^{4} e + 3 \, a b^{4} d^{3} e^{2} + 6 \, a^{2} b^{3} d^{2} e^{3} + 10 \, a^{3} b^{2} d e^{4} + 15 \, a^{4} b e^{5}\right )} x}{168 \, {\left (e^{14} x^{8} + 8 \, d e^{13} x^{7} + 28 \, d^{2} e^{12} x^{6} + 56 \, d^{3} e^{11} x^{5} + 70 \, d^{4} e^{10} x^{4} + 56 \, d^{5} e^{9} x^{3} + 28 \, d^{6} e^{8} x^{2} + 8 \, d^{7} e^{7} x + d^{8} e^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^9,x, algorithm="fricas")

[Out]

-1/168*(56*b^5*e^5*x^5 + b^5*d^5 + 3*a*b^4*d^4*e + 6*a^2*b^3*d^3*e^2 + 10*a^3*b^2*d^2*e^3 + 15*a^4*b*d*e^4 + 2
1*a^5*e^5 + 70*(b^5*d*e^4 + 3*a*b^4*e^5)*x^4 + 56*(b^5*d^2*e^3 + 3*a*b^4*d*e^4 + 6*a^2*b^3*e^5)*x^3 + 28*(b^5*
d^3*e^2 + 3*a*b^4*d^2*e^3 + 6*a^2*b^3*d*e^4 + 10*a^3*b^2*e^5)*x^2 + 8*(b^5*d^4*e + 3*a*b^4*d^3*e^2 + 6*a^2*b^3
*d^2*e^3 + 10*a^3*b^2*d*e^4 + 15*a^4*b*e^5)*x)/(e^14*x^8 + 8*d*e^13*x^7 + 28*d^2*e^12*x^6 + 56*d^3*e^11*x^5 +
70*d^4*e^10*x^4 + 56*d^5*e^9*x^3 + 28*d^6*e^8*x^2 + 8*d^7*e^7*x + d^8*e^6)

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giac [B]  time = 0.19, size = 381, normalized size = 2.56 \[ -\frac {{\left (56 \, b^{5} x^{5} e^{5} \mathrm {sgn}\left (b x + a\right ) + 70 \, b^{5} d x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 56 \, b^{5} d^{2} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, b^{5} d^{3} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, b^{5} d^{4} x e \mathrm {sgn}\left (b x + a\right ) + b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) + 210 \, a b^{4} x^{4} e^{5} \mathrm {sgn}\left (b x + a\right ) + 168 \, a b^{4} d x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 84 \, a b^{4} d^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 24 \, a b^{4} d^{3} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 336 \, a^{2} b^{3} x^{3} e^{5} \mathrm {sgn}\left (b x + a\right ) + 168 \, a^{2} b^{3} d x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 48 \, a^{2} b^{3} d^{2} x e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 280 \, a^{3} b^{2} x^{2} e^{5} \mathrm {sgn}\left (b x + a\right ) + 80 \, a^{3} b^{2} d x e^{4} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{4} b x e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) + 21 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )}}{168 \, {\left (x e + d\right )}^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^9,x, algorithm="giac")

[Out]

-1/168*(56*b^5*x^5*e^5*sgn(b*x + a) + 70*b^5*d*x^4*e^4*sgn(b*x + a) + 56*b^5*d^2*x^3*e^3*sgn(b*x + a) + 28*b^5
*d^3*x^2*e^2*sgn(b*x + a) + 8*b^5*d^4*x*e*sgn(b*x + a) + b^5*d^5*sgn(b*x + a) + 210*a*b^4*x^4*e^5*sgn(b*x + a)
+ 168*a*b^4*d*x^3*e^4*sgn(b*x + a) + 84*a*b^4*d^2*x^2*e^3*sgn(b*x + a) + 24*a*b^4*d^3*x*e^2*sgn(b*x + a) + 3*
a*b^4*d^4*e*sgn(b*x + a) + 336*a^2*b^3*x^3*e^5*sgn(b*x + a) + 168*a^2*b^3*d*x^2*e^4*sgn(b*x + a) + 48*a^2*b^3*
d^2*x*e^3*sgn(b*x + a) + 6*a^2*b^3*d^3*e^2*sgn(b*x + a) + 280*a^3*b^2*x^2*e^5*sgn(b*x + a) + 80*a^3*b^2*d*x*e^
4*sgn(b*x + a) + 10*a^3*b^2*d^2*e^3*sgn(b*x + a) + 120*a^4*b*x*e^5*sgn(b*x + a) + 15*a^4*b*d*e^4*sgn(b*x + a)
+ 21*a^5*e^5*sgn(b*x + a))*e^(-6)/(x*e + d)^8

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maple [B]  time = 0.05, size = 288, normalized size = 1.93 \[ -\frac {\left (56 b^{5} e^{5} x^{5}+210 a \,b^{4} e^{5} x^{4}+70 b^{5} d \,e^{4} x^{4}+336 a^{2} b^{3} e^{5} x^{3}+168 a \,b^{4} d \,e^{4} x^{3}+56 b^{5} d^{2} e^{3} x^{3}+280 a^{3} b^{2} e^{5} x^{2}+168 a^{2} b^{3} d \,e^{4} x^{2}+84 a \,b^{4} d^{2} e^{3} x^{2}+28 b^{5} d^{3} e^{2} x^{2}+120 a^{4} b \,e^{5} x +80 a^{3} b^{2} d \,e^{4} x +48 a^{2} b^{3} d^{2} e^{3} x +24 a \,b^{4} d^{3} e^{2} x +8 b^{5} d^{4} e x +21 a^{5} e^{5}+15 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}+6 a^{2} b^{3} d^{3} e^{2}+3 a \,b^{4} d^{4} e +b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{168 \left (e x +d \right )^{8} \left (b x +a \right )^{5} e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^9,x)

[Out]

-1/168/e^6*(56*b^5*e^5*x^5+210*a*b^4*e^5*x^4+70*b^5*d*e^4*x^4+336*a^2*b^3*e^5*x^3+168*a*b^4*d*e^4*x^3+56*b^5*d
^2*e^3*x^3+280*a^3*b^2*e^5*x^2+168*a^2*b^3*d*e^4*x^2+84*a*b^4*d^2*e^3*x^2+28*b^5*d^3*e^2*x^2+120*a^4*b*e^5*x+8
0*a^3*b^2*d*e^4*x+48*a^2*b^3*d^2*e^3*x+24*a*b^4*d^3*e^2*x+8*b^5*d^4*e*x+21*a^5*e^5+15*a^4*b*d*e^4+10*a^3*b^2*d
^2*e^3+6*a^2*b^3*d^3*e^2+3*a*b^4*d^4*e+b^5*d^5)*((b*x+a)^2)^(5/2)/(e*x+d)^8/(b*x+a)^5

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 0.74, size = 687, normalized size = 4.61 \[ \frac {\left (\frac {4\,b^5\,d-5\,a\,b^4\,e}{4\,e^6}+\frac {b^5\,d}{4\,e^6}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {5\,a^4\,b\,e^4-10\,a^3\,b^2\,d\,e^3+10\,a^2\,b^3\,d^2\,e^2-5\,a\,b^4\,d^3\,e+b^5\,d^4}{7\,e^6}+\frac {d\,\left (\frac {-10\,a^3\,b^2\,e^4+10\,a^2\,b^3\,d\,e^3-5\,a\,b^4\,d^2\,e^2+b^5\,d^3\,e}{7\,e^6}+\frac {d\,\left (\frac {d\,\left (\frac {b^5\,d}{7\,e^3}-\frac {b^4\,\left (5\,a\,e-b\,d\right )}{7\,e^3}\right )}{e}+\frac {b^3\,\left (10\,a^2\,e^2-5\,a\,b\,d\,e+b^2\,d^2\right )}{7\,e^4}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^7}-\frac {\left (\frac {10\,a^2\,b^3\,e^2-15\,a\,b^4\,d\,e+6\,b^5\,d^2}{5\,e^6}+\frac {d\,\left (\frac {b^5\,d}{5\,e^5}-\frac {b^4\,\left (5\,a\,e-3\,b\,d\right )}{5\,e^5}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {a^5}{8\,e}-\frac {d\,\left (\frac {5\,a^4\,b}{8\,e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {5\,a\,b^4}{8\,e}-\frac {b^5\,d}{8\,e^2}\right )}{e}-\frac {5\,a^2\,b^3}{4\,e}\right )}{e}+\frac {5\,a^3\,b^2}{4\,e}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^8}+\frac {\left (\frac {-10\,a^3\,b^2\,e^3+20\,a^2\,b^3\,d\,e^2-15\,a\,b^4\,d^2\,e+4\,b^5\,d^3}{6\,e^6}+\frac {d\,\left (\frac {d\,\left (\frac {b^5\,d}{6\,e^4}-\frac {b^4\,\left (5\,a\,e-2\,b\,d\right )}{6\,e^4}\right )}{e}+\frac {b^3\,\left (10\,a^2\,e^2-10\,a\,b\,d\,e+3\,b^2\,d^2\right )}{6\,e^5}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,e^6\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/(d + e*x)^9,x)

[Out]

(((4*b^5*d - 5*a*b^4*e)/(4*e^6) + (b^5*d)/(4*e^6))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^4) -
(((b^5*d^4 + 5*a^4*b*e^4 - 10*a^3*b^2*d*e^3 + 10*a^2*b^3*d^2*e^2 - 5*a*b^4*d^3*e)/(7*e^6) + (d*((b^5*d^3*e - 1
0*a^3*b^2*e^4 - 5*a*b^4*d^2*e^2 + 10*a^2*b^3*d*e^3)/(7*e^6) + (d*((d*((b^5*d)/(7*e^3) - (b^4*(5*a*e - b*d))/(7
*e^3)))/e + (b^3*(10*a^2*e^2 + b^2*d^2 - 5*a*b*d*e))/(7*e^4)))/e))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b
*x)*(d + e*x)^7) - (((6*b^5*d^2 + 10*a^2*b^3*e^2 - 15*a*b^4*d*e)/(5*e^6) + (d*((b^5*d)/(5*e^5) - (b^4*(5*a*e -
3*b*d))/(5*e^5)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^5) - ((a^5/(8*e) - (d*((5*a^4*b)/(
8*e) - (d*((d*((d*((5*a*b^4)/(8*e) - (b^5*d)/(8*e^2)))/e - (5*a^2*b^3)/(4*e)))/e + (5*a^3*b^2)/(4*e)))/e))/e)*
(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^8) + (((4*b^5*d^3 - 10*a^3*b^2*e^3 + 20*a^2*b^3*d*e^2 -
15*a*b^4*d^2*e)/(6*e^6) + (d*((d*((b^5*d)/(6*e^4) - (b^4*(5*a*e - 2*b*d))/(6*e^4)))/e + (b^3*(10*a^2*e^2 + 3*b
^2*d^2 - 10*a*b*d*e))/(6*e^5)))/e)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/((a + b*x)*(d + e*x)^6) - (b^5*(a^2 + b^2*
x^2 + 2*a*b*x)^(1/2))/(3*e^6*(a + b*x)*(d + e*x)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/(e*x+d)**9,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/(d + e*x)**9, x)

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