3.292 \(\int (a+b x^n)^2 (d+e x^n)^3 \, dx\)

Optimal. Leaf size=158 \[ \frac {d x^{2 n+1} \left (3 a^2 e^2+6 a b d e+b^2 d^2\right )}{2 n+1}+\frac {e x^{3 n+1} \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )}{3 n+1}+a^2 d^3 x+\frac {a d^2 x^{n+1} (3 a e+2 b d)}{n+1}+\frac {b e^2 x^{4 n+1} (2 a e+3 b d)}{4 n+1}+\frac {b^2 e^3 x^{5 n+1}}{5 n+1} \]

[Out]

a^2*d^3*x+a*d^2*(3*a*e+2*b*d)*x^(1+n)/(1+n)+d*(3*a^2*e^2+6*a*b*d*e+b^2*d^2)*x^(1+2*n)/(1+2*n)+e*(a^2*e^2+6*a*b
*d*e+3*b^2*d^2)*x^(1+3*n)/(1+3*n)+b*e^2*(2*a*e+3*b*d)*x^(1+4*n)/(1+4*n)+b^2*e^3*x^(1+5*n)/(1+5*n)

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Rubi [A]  time = 0.13, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {373} \[ \frac {d x^{2 n+1} \left (3 a^2 e^2+6 a b d e+b^2 d^2\right )}{2 n+1}+\frac {e x^{3 n+1} \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )}{3 n+1}+a^2 d^3 x+\frac {a d^2 x^{n+1} (3 a e+2 b d)}{n+1}+\frac {b e^2 x^{4 n+1} (2 a e+3 b d)}{4 n+1}+\frac {b^2 e^3 x^{5 n+1}}{5 n+1} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^2*(d + e*x^n)^3,x]

[Out]

a^2*d^3*x + (a*d^2*(2*b*d + 3*a*e)*x^(1 + n))/(1 + n) + (d*(b^2*d^2 + 6*a*b*d*e + 3*a^2*e^2)*x^(1 + 2*n))/(1 +
2*n) + (e*(3*b^2*d^2 + 6*a*b*d*e + a^2*e^2)*x^(1 + 3*n))/(1 + 3*n) + (b*e^2*(3*b*d + 2*a*e)*x^(1 + 4*n))/(1 +
4*n) + (b^2*e^3*x^(1 + 5*n))/(1 + 5*n)

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin {align*} \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^3 \, dx &=\int \left (a^2 d^3+a d^2 (2 b d+3 a e) x^n+d \left (b^2 d^2+6 a b d e+3 a^2 e^2\right ) x^{2 n}+e \left (3 b^2 d^2+6 a b d e+a^2 e^2\right ) x^{3 n}+b e^2 (3 b d+2 a e) x^{4 n}+b^2 e^3 x^{5 n}\right ) \, dx\\ &=a^2 d^3 x+\frac {a d^2 (2 b d+3 a e) x^{1+n}}{1+n}+\frac {d \left (b^2 d^2+6 a b d e+3 a^2 e^2\right ) x^{1+2 n}}{1+2 n}+\frac {e \left (3 b^2 d^2+6 a b d e+a^2 e^2\right ) x^{1+3 n}}{1+3 n}+\frac {b e^2 (3 b d+2 a e) x^{1+4 n}}{1+4 n}+\frac {b^2 e^3 x^{1+5 n}}{1+5 n}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 149, normalized size = 0.94 \[ x \left (\frac {d x^{2 n} \left (3 a^2 e^2+6 a b d e+b^2 d^2\right )}{2 n+1}+\frac {e x^{3 n} \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )}{3 n+1}+a^2 d^3+\frac {a d^2 x^n (3 a e+2 b d)}{n+1}+\frac {b e^2 x^{4 n} (2 a e+3 b d)}{4 n+1}+\frac {b^2 e^3 x^{5 n}}{5 n+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*(a^2*d^3 + (a*d^2*(2*b*d + 3*a*e)*x^n)/(1 + n) + (d*(b^2*d^2 + 6*a*b*d*e + 3*a^2*e^2)*x^(2*n))/(1 + 2*n) + (
e*(3*b^2*d^2 + 6*a*b*d*e + a^2*e^2)*x^(3*n))/(1 + 3*n) + (b*e^2*(3*b*d + 2*a*e)*x^(4*n))/(1 + 4*n) + (b^2*e^3*
x^(5*n))/(1 + 5*n))

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fricas [B]  time = 1.00, size = 667, normalized size = 4.22 \[ \frac {{\left (24 \, b^{2} e^{3} n^{4} + 50 \, b^{2} e^{3} n^{3} + 35 \, b^{2} e^{3} n^{2} + 10 \, b^{2} e^{3} n + b^{2} e^{3}\right )} x x^{5 \, n} + {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3} + 30 \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} n^{4} + 61 \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} n^{3} + 41 \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} n^{2} + 11 \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} n\right )} x x^{4 \, n} + {\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3} + 40 \, {\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3}\right )} n^{4} + 78 \, {\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3}\right )} n^{3} + 49 \, {\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3}\right )} n^{2} + 12 \, {\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3}\right )} n\right )} x x^{3 \, n} + {\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2} + 60 \, {\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2}\right )} n^{4} + 107 \, {\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2}\right )} n^{3} + 59 \, {\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2}\right )} n^{2} + 13 \, {\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2}\right )} n\right )} x x^{2 \, n} + {\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e + 120 \, {\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e\right )} n^{4} + 154 \, {\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e\right )} n^{3} + 71 \, {\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e\right )} n^{2} + 14 \, {\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e\right )} n\right )} x x^{n} + {\left (120 \, a^{2} d^{3} n^{5} + 274 \, a^{2} d^{3} n^{4} + 225 \, a^{2} d^{3} n^{3} + 85 \, a^{2} d^{3} n^{2} + 15 \, a^{2} d^{3} n + a^{2} d^{3}\right )} x}{120 \, n^{5} + 274 \, n^{4} + 225 \, n^{3} + 85 \, n^{2} + 15 \, n + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((24*b^2*e^3*n^4 + 50*b^2*e^3*n^3 + 35*b^2*e^3*n^2 + 10*b^2*e^3*n + b^2*e^3)*x*x^(5*n) + (3*b^2*d*e^2 + 2*a*b*
e^3 + 30*(3*b^2*d*e^2 + 2*a*b*e^3)*n^4 + 61*(3*b^2*d*e^2 + 2*a*b*e^3)*n^3 + 41*(3*b^2*d*e^2 + 2*a*b*e^3)*n^2 +
11*(3*b^2*d*e^2 + 2*a*b*e^3)*n)*x*x^(4*n) + (3*b^2*d^2*e + 6*a*b*d*e^2 + a^2*e^3 + 40*(3*b^2*d^2*e + 6*a*b*d*
e^2 + a^2*e^3)*n^4 + 78*(3*b^2*d^2*e + 6*a*b*d*e^2 + a^2*e^3)*n^3 + 49*(3*b^2*d^2*e + 6*a*b*d*e^2 + a^2*e^3)*n
^2 + 12*(3*b^2*d^2*e + 6*a*b*d*e^2 + a^2*e^3)*n)*x*x^(3*n) + (b^2*d^3 + 6*a*b*d^2*e + 3*a^2*d*e^2 + 60*(b^2*d^
3 + 6*a*b*d^2*e + 3*a^2*d*e^2)*n^4 + 107*(b^2*d^3 + 6*a*b*d^2*e + 3*a^2*d*e^2)*n^3 + 59*(b^2*d^3 + 6*a*b*d^2*e
+ 3*a^2*d*e^2)*n^2 + 13*(b^2*d^3 + 6*a*b*d^2*e + 3*a^2*d*e^2)*n)*x*x^(2*n) + (2*a*b*d^3 + 3*a^2*d^2*e + 120*(
2*a*b*d^3 + 3*a^2*d^2*e)*n^4 + 154*(2*a*b*d^3 + 3*a^2*d^2*e)*n^3 + 71*(2*a*b*d^3 + 3*a^2*d^2*e)*n^2 + 14*(2*a*
b*d^3 + 3*a^2*d^2*e)*n)*x*x^n + (120*a^2*d^3*n^5 + 274*a^2*d^3*n^4 + 225*a^2*d^3*n^3 + 85*a^2*d^3*n^2 + 15*a^2
*d^3*n + a^2*d^3)*x)/(120*n^5 + 274*n^4 + 225*n^3 + 85*n^2 + 15*n + 1)

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giac [B]  time = 0.24, size = 947, normalized size = 5.99 \[ \frac {120 \, a^{2} d^{3} n^{5} x + 60 \, b^{2} d^{3} n^{4} x x^{2 \, n} + 240 \, a b d^{3} n^{4} x x^{n} + 120 \, b^{2} d^{2} n^{4} x x^{3 \, n} e + 360 \, a b d^{2} n^{4} x x^{2 \, n} e + 360 \, a^{2} d^{2} n^{4} x x^{n} e + 274 \, a^{2} d^{3} n^{4} x + 107 \, b^{2} d^{3} n^{3} x x^{2 \, n} + 308 \, a b d^{3} n^{3} x x^{n} + 90 \, b^{2} d n^{4} x x^{4 \, n} e^{2} + 240 \, a b d n^{4} x x^{3 \, n} e^{2} + 180 \, a^{2} d n^{4} x x^{2 \, n} e^{2} + 234 \, b^{2} d^{2} n^{3} x x^{3 \, n} e + 642 \, a b d^{2} n^{3} x x^{2 \, n} e + 462 \, a^{2} d^{2} n^{3} x x^{n} e + 225 \, a^{2} d^{3} n^{3} x + 59 \, b^{2} d^{3} n^{2} x x^{2 \, n} + 142 \, a b d^{3} n^{2} x x^{n} + 24 \, b^{2} n^{4} x x^{5 \, n} e^{3} + 60 \, a b n^{4} x x^{4 \, n} e^{3} + 40 \, a^{2} n^{4} x x^{3 \, n} e^{3} + 183 \, b^{2} d n^{3} x x^{4 \, n} e^{2} + 468 \, a b d n^{3} x x^{3 \, n} e^{2} + 321 \, a^{2} d n^{3} x x^{2 \, n} e^{2} + 147 \, b^{2} d^{2} n^{2} x x^{3 \, n} e + 354 \, a b d^{2} n^{2} x x^{2 \, n} e + 213 \, a^{2} d^{2} n^{2} x x^{n} e + 85 \, a^{2} d^{3} n^{2} x + 13 \, b^{2} d^{3} n x x^{2 \, n} + 28 \, a b d^{3} n x x^{n} + 50 \, b^{2} n^{3} x x^{5 \, n} e^{3} + 122 \, a b n^{3} x x^{4 \, n} e^{3} + 78 \, a^{2} n^{3} x x^{3 \, n} e^{3} + 123 \, b^{2} d n^{2} x x^{4 \, n} e^{2} + 294 \, a b d n^{2} x x^{3 \, n} e^{2} + 177 \, a^{2} d n^{2} x x^{2 \, n} e^{2} + 36 \, b^{2} d^{2} n x x^{3 \, n} e + 78 \, a b d^{2} n x x^{2 \, n} e + 42 \, a^{2} d^{2} n x x^{n} e + 15 \, a^{2} d^{3} n x + b^{2} d^{3} x x^{2 \, n} + 2 \, a b d^{3} x x^{n} + 35 \, b^{2} n^{2} x x^{5 \, n} e^{3} + 82 \, a b n^{2} x x^{4 \, n} e^{3} + 49 \, a^{2} n^{2} x x^{3 \, n} e^{3} + 33 \, b^{2} d n x x^{4 \, n} e^{2} + 72 \, a b d n x x^{3 \, n} e^{2} + 39 \, a^{2} d n x x^{2 \, n} e^{2} + 3 \, b^{2} d^{2} x x^{3 \, n} e + 6 \, a b d^{2} x x^{2 \, n} e + 3 \, a^{2} d^{2} x x^{n} e + a^{2} d^{3} x + 10 \, b^{2} n x x^{5 \, n} e^{3} + 22 \, a b n x x^{4 \, n} e^{3} + 12 \, a^{2} n x x^{3 \, n} e^{3} + 3 \, b^{2} d x x^{4 \, n} e^{2} + 6 \, a b d x x^{3 \, n} e^{2} + 3 \, a^{2} d x x^{2 \, n} e^{2} + b^{2} x x^{5 \, n} e^{3} + 2 \, a b x x^{4 \, n} e^{3} + a^{2} x x^{3 \, n} e^{3}}{120 \, n^{5} + 274 \, n^{4} + 225 \, n^{3} + 85 \, n^{2} + 15 \, n + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(120*a^2*d^3*n^5*x + 60*b^2*d^3*n^4*x*x^(2*n) + 240*a*b*d^3*n^4*x*x^n + 120*b^2*d^2*n^4*x*x^(3*n)*e + 360*a*b*
d^2*n^4*x*x^(2*n)*e + 360*a^2*d^2*n^4*x*x^n*e + 274*a^2*d^3*n^4*x + 107*b^2*d^3*n^3*x*x^(2*n) + 308*a*b*d^3*n^
3*x*x^n + 90*b^2*d*n^4*x*x^(4*n)*e^2 + 240*a*b*d*n^4*x*x^(3*n)*e^2 + 180*a^2*d*n^4*x*x^(2*n)*e^2 + 234*b^2*d^2
*n^3*x*x^(3*n)*e + 642*a*b*d^2*n^3*x*x^(2*n)*e + 462*a^2*d^2*n^3*x*x^n*e + 225*a^2*d^3*n^3*x + 59*b^2*d^3*n^2*
x*x^(2*n) + 142*a*b*d^3*n^2*x*x^n + 24*b^2*n^4*x*x^(5*n)*e^3 + 60*a*b*n^4*x*x^(4*n)*e^3 + 40*a^2*n^4*x*x^(3*n)
*e^3 + 183*b^2*d*n^3*x*x^(4*n)*e^2 + 468*a*b*d*n^3*x*x^(3*n)*e^2 + 321*a^2*d*n^3*x*x^(2*n)*e^2 + 147*b^2*d^2*n
^2*x*x^(3*n)*e + 354*a*b*d^2*n^2*x*x^(2*n)*e + 213*a^2*d^2*n^2*x*x^n*e + 85*a^2*d^3*n^2*x + 13*b^2*d^3*n*x*x^(
2*n) + 28*a*b*d^3*n*x*x^n + 50*b^2*n^3*x*x^(5*n)*e^3 + 122*a*b*n^3*x*x^(4*n)*e^3 + 78*a^2*n^3*x*x^(3*n)*e^3 +
123*b^2*d*n^2*x*x^(4*n)*e^2 + 294*a*b*d*n^2*x*x^(3*n)*e^2 + 177*a^2*d*n^2*x*x^(2*n)*e^2 + 36*b^2*d^2*n*x*x^(3*
n)*e + 78*a*b*d^2*n*x*x^(2*n)*e + 42*a^2*d^2*n*x*x^n*e + 15*a^2*d^3*n*x + b^2*d^3*x*x^(2*n) + 2*a*b*d^3*x*x^n
+ 35*b^2*n^2*x*x^(5*n)*e^3 + 82*a*b*n^2*x*x^(4*n)*e^3 + 49*a^2*n^2*x*x^(3*n)*e^3 + 33*b^2*d*n*x*x^(4*n)*e^2 +
72*a*b*d*n*x*x^(3*n)*e^2 + 39*a^2*d*n*x*x^(2*n)*e^2 + 3*b^2*d^2*x*x^(3*n)*e + 6*a*b*d^2*x*x^(2*n)*e + 3*a^2*d^
2*x*x^n*e + a^2*d^3*x + 10*b^2*n*x*x^(5*n)*e^3 + 22*a*b*n*x*x^(4*n)*e^3 + 12*a^2*n*x*x^(3*n)*e^3 + 3*b^2*d*x*x
^(4*n)*e^2 + 6*a*b*d*x*x^(3*n)*e^2 + 3*a^2*d*x*x^(2*n)*e^2 + b^2*x*x^(5*n)*e^3 + 2*a*b*x*x^(4*n)*e^3 + a^2*x*x
^(3*n)*e^3)/(120*n^5 + 274*n^4 + 225*n^3 + 85*n^2 + 15*n + 1)

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maple [A]  time = 0.06, size = 164, normalized size = 1.04 \[ \frac {b^{2} e^{3} x \,{\mathrm e}^{5 n \ln \relax (x )}}{5 n +1}+a^{2} d^{3} x +\frac {\left (3 a e +2 b d \right ) a \,d^{2} x \,{\mathrm e}^{n \ln \relax (x )}}{n +1}+\frac {\left (2 a e +3 b d \right ) b \,e^{2} x \,{\mathrm e}^{4 n \ln \relax (x )}}{4 n +1}+\frac {\left (3 a^{2} e^{2}+6 a b d e +b^{2} d^{2}\right ) d x \,{\mathrm e}^{2 n \ln \relax (x )}}{2 n +1}+\frac {\left (a^{2} e^{2}+6 a b d e +3 b^{2} d^{2}\right ) e x \,{\mathrm e}^{3 n \ln \relax (x )}}{3 n +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*d^3*x+b^2*e^3/(5*n+1)*x*exp(n*ln(x))^5+d*(3*a^2*e^2+6*a*b*d*e+b^2*d^2)/(2*n+1)*x*exp(n*ln(x))^2+e*(a^2*e^2
+6*a*b*d*e+3*b^2*d^2)/(3*n+1)*x*exp(n*ln(x))^3+a*d^2*(3*a*e+2*b*d)/(n+1)*x*exp(n*ln(x))+b*e^2*(2*a*e+3*b*d)/(4
*n+1)*x*exp(n*ln(x))^4

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maxima [A]  time = 0.62, size = 242, normalized size = 1.53 \[ a^{2} d^{3} x + \frac {b^{2} e^{3} x^{5 \, n + 1}}{5 \, n + 1} + \frac {3 \, b^{2} d e^{2} x^{4 \, n + 1}}{4 \, n + 1} + \frac {2 \, a b e^{3} x^{4 \, n + 1}}{4 \, n + 1} + \frac {3 \, b^{2} d^{2} e x^{3 \, n + 1}}{3 \, n + 1} + \frac {6 \, a b d e^{2} x^{3 \, n + 1}}{3 \, n + 1} + \frac {a^{2} e^{3} x^{3 \, n + 1}}{3 \, n + 1} + \frac {b^{2} d^{3} x^{2 \, n + 1}}{2 \, n + 1} + \frac {6 \, a b d^{2} e x^{2 \, n + 1}}{2 \, n + 1} + \frac {3 \, a^{2} d e^{2} x^{2 \, n + 1}}{2 \, n + 1} + \frac {2 \, a b d^{3} x^{n + 1}}{n + 1} + \frac {3 \, a^{2} d^{2} e x^{n + 1}}{n + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*d^3*x + b^2*e^3*x^(5*n + 1)/(5*n + 1) + 3*b^2*d*e^2*x^(4*n + 1)/(4*n + 1) + 2*a*b*e^3*x^(4*n + 1)/(4*n + 1
) + 3*b^2*d^2*e*x^(3*n + 1)/(3*n + 1) + 6*a*b*d*e^2*x^(3*n + 1)/(3*n + 1) + a^2*e^3*x^(3*n + 1)/(3*n + 1) + b^
2*d^3*x^(2*n + 1)/(2*n + 1) + 6*a*b*d^2*e*x^(2*n + 1)/(2*n + 1) + 3*a^2*d*e^2*x^(2*n + 1)/(2*n + 1) + 2*a*b*d^
3*x^(n + 1)/(n + 1) + 3*a^2*d^2*e*x^(n + 1)/(n + 1)

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mupad [B]  time = 1.71, size = 157, normalized size = 0.99 \[ a^2\,d^3\,x+\frac {x\,x^{2\,n}\,\left (3\,a^2\,d\,e^2+6\,a\,b\,d^2\,e+b^2\,d^3\right )}{2\,n+1}+\frac {x\,x^{3\,n}\,\left (a^2\,e^3+6\,a\,b\,d\,e^2+3\,b^2\,d^2\,e\right )}{3\,n+1}+\frac {b^2\,e^3\,x\,x^{5\,n}}{5\,n+1}+\frac {a\,d^2\,x\,x^n\,\left (3\,a\,e+2\,b\,d\right )}{n+1}+\frac {b\,e^2\,x\,x^{4\,n}\,\left (2\,a\,e+3\,b\,d\right )}{4\,n+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^n)^2*(d + e*x^n)^3,x)

[Out]

a^2*d^3*x + (x*x^(2*n)*(b^2*d^3 + 3*a^2*d*e^2 + 6*a*b*d^2*e))/(2*n + 1) + (x*x^(3*n)*(a^2*e^3 + 3*b^2*d^2*e +
6*a*b*d*e^2))/(3*n + 1) + (b^2*e^3*x*x^(5*n))/(5*n + 1) + (a*d^2*x*x^n*(3*a*e + 2*b*d))/(n + 1) + (b*e^2*x*x^(
4*n)*(2*a*e + 3*b*d))/(4*n + 1)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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