3.74 \(\int \frac {1}{(d+e x^n)^3 (a+b x^n+c x^{2 n})} \, dx\)

Optimal. Leaf size=552 \[ \frac {e^2 x \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^3}-\frac {c x \left (-3 c^2 d e \left (d \sqrt {b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt {b^2-4 a c}+a e\right )+a e \sqrt {b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt {b^2-4 a c}+b\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}-\frac {c x \left (-3 c^2 d e \left (-d \sqrt {b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt {b^2-4 a c}-a e \sqrt {b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt {b^2-4 a c}\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}+\frac {e^2 x (2 c d-b e) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (a e^2-b d e+c d^2\right )^2}+\frac {e^2 x \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^3 \left (a e^2-b d e+c d^2\right )} \]

[Out]

e^2*(3*c^2*d^2+b^2*e^2-c*e*(a*e+3*b*d))*x*hypergeom([1, 1/n],[1+1/n],-e*x^n/d)/d/(a*e^2-b*d*e+c*d^2)^3+e^2*(-b
*e+2*c*d)*x*hypergeom([2, 1/n],[1+1/n],-e*x^n/d)/d^2/(a*e^2-b*d*e+c*d^2)^2+e^2*x*hypergeom([3, 1/n],[1+1/n],-e
*x^n/d)/d^3/(a*e^2-b*d*e+c*d^2)-c*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*(2*c^3*d^3-b^2
*e^3*(b-(-4*a*c+b^2)^(1/2))-3*c^2*d*e*(b*d+2*a*e-d*(-4*a*c+b^2)^(1/2))+c*e^2*(3*b^2*d+3*a*b*e-3*b*d*(-4*a*c+b^
2)^(1/2)-(-4*a*c+b^2)^(1/2)*a*e))/(a*e^2-b*d*e+c*d^2)^3/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-c*x*hypergeom([1, 1/n
],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(2*c^3*d^3-b^2*e^3*(b+(-4*a*c+b^2)^(1/2))-3*c^2*d*e*(b*d+2*a*e+d*(-
4*a*c+b^2)^(1/2))+c*e^2*(3*b^2*d+(-4*a*c+b^2)^(1/2)*a*e+3*b*(a*e+d*(-4*a*c+b^2)^(1/2))))/(a*e^2-b*d*e+c*d^2)^3
/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))

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Rubi [A]  time = 1.02, antiderivative size = 552, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1424, 245, 1422} \[ \frac {e^2 x \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^3}-\frac {c x \left (-3 c^2 d e \left (d \sqrt {b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt {b^2-4 a c}+a e\right )+a e \sqrt {b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt {b^2-4 a c}+b\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}-\frac {c x \left (-3 c^2 d e \left (-d \sqrt {b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt {b^2-4 a c}-a e \sqrt {b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt {b^2-4 a c}\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}+\frac {e^2 x (2 c d-b e) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (a e^2-b d e+c d^2\right )^2}+\frac {e^2 x \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^3 \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x]

[Out]

-((c*(2*c^3*d^3 - b^2*(b + Sqrt[b^2 - 4*a*c])*e^3 - 3*c^2*d*e*(b*d + Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b
^2*d + a*Sqrt[b^2 - 4*a*c]*e + 3*b*(Sqrt[b^2 - 4*a*c]*d + a*e)))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-
2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^3)) - (c*(2*c^
3*d^3 - b^2*(b - Sqrt[b^2 - 4*a*c])*e^3 - 3*c^2*d*e*(b*d - Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d - 3*b
*Sqrt[b^2 - 4*a*c]*d + 3*a*b*e - a*Sqrt[b^2 - 4*a*c]*e))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)
/(b + Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^3) + (e^2*(3*c^2*d^2 +
b^2*e^2 - c*e*(3*b*d + a*e))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d*(c*d^2 - b*d*e + a*
e^2)^3) + (e^2*(2*c*d - b*e)*x*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^2*(c*d^2 - b*d*e + a
*e^2)^2) + (e^2*x*Hypergeometric2F1[3, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^3*(c*d^2 - b*d*e + a*e^2))

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rule 1424

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Int[ExpandIntegran
d[(d + e*x^n)^q/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4
*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx &=\int \left (\frac {e^2}{\left (c d^2-b d e+a e^2\right ) \left (d+e x^n\right )^3}-\frac {e^2 (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right )^2 \left (d+e x^n\right )^2}+\frac {e^2 \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )}{\left (c d^2-b d e+a e^2\right )^3 \left (d+e x^n\right )}+\frac {c^3 d^3-3 b c^2 d^2 e+3 b^2 c d e^2-3 a c^2 d e^2-b^3 e^3+2 a b c e^3-\left (3 c^3 d^2 e-3 b c^2 d e^2+b^2 c e^3-a c^2 e^3\right ) x^n}{\left (c d^2-b d e+a e^2\right )^3 \left (a+b x^n+c x^{2 n}\right )}\right ) \, dx\\ &=\frac {\int \frac {c^3 d^3-3 b c^2 d^2 e+3 b^2 c d e^2-3 a c^2 d e^2-b^3 e^3+2 a b c e^3-\left (3 c^3 d^2 e-3 b c^2 d e^2+b^2 c e^3-a c^2 e^3\right ) x^n}{a+b x^n+c x^{2 n}} \, dx}{\left (c d^2-b d e+a e^2\right )^3}+\frac {\left (e^2 (2 c d-b e)\right ) \int \frac {1}{\left (d+e x^n\right )^2} \, dx}{\left (c d^2-b d e+a e^2\right )^2}+\frac {e^2 \int \frac {1}{\left (d+e x^n\right )^3} \, dx}{c d^2-b d e+a e^2}+\frac {\left (e^2 \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) \int \frac {1}{d+e x^n} \, dx}{\left (c d^2-b d e+a e^2\right )^3}\\ &=\frac {e^2 \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^3}+\frac {e^2 (2 c d-b e) x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2}+\frac {e^2 x \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^3 \left (c d^2-b d e+a e^2\right )}-\frac {\left (c \left (2 c^3 d^3-b^2 \left (b-\sqrt {b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d-\sqrt {b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d-3 b \sqrt {b^2-4 a c} d+3 a b e-a \sqrt {b^2-4 a c} e\right )\right )\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )^3}+\frac {\left (c \left (2 c^3 d^3-b^2 \left (b+\sqrt {b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d+\sqrt {b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d+a \sqrt {b^2-4 a c} e+3 b \left (\sqrt {b^2-4 a c} d+a e\right )\right )\right )\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )^3}\\ &=\frac {c \left (2 c^3 d^3-b^2 \left (b+\sqrt {b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d+\sqrt {b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d+a \sqrt {b^2-4 a c} e+3 b \left (\sqrt {b^2-4 a c} d+a e\right )\right )\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}-\frac {c \left (2 c^3 d^3-b^2 \left (b-\sqrt {b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d-\sqrt {b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d-3 b \sqrt {b^2-4 a c} d+3 a b e-a \sqrt {b^2-4 a c} e\right )\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}+\frac {e^2 \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^3}+\frac {e^2 (2 c d-b e) x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2 \left (c d^2-b d e+a e^2\right )^2}+\frac {e^2 x \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^3 \left (c d^2-b d e+a e^2\right )}\\ \end {align*}

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Mathematica [A]  time = 1.74, size = 509, normalized size = 0.92 \[ \frac {x \left (\frac {e^2 \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d}+\frac {c \left (3 c^2 d e \left (d \sqrt {b^2-4 a c}+2 a e+b d\right )-c e^2 \left (3 b \left (d \sqrt {b^2-4 a c}+a e\right )+a e \sqrt {b^2-4 a c}+3 b^2 d\right )+b^2 e^3 \left (\sqrt {b^2-4 a c}+b\right )-2 c^3 d^3\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {c \left (3 c^2 d e \left (d \sqrt {b^2-4 a c}-2 a e-b d\right )+c e^2 \left (-3 b d \sqrt {b^2-4 a c}-a e \sqrt {b^2-4 a c}+3 a b e+3 b^2 d\right )+b^2 e^3 \left (\sqrt {b^2-4 a c}-b\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2}+\frac {e^2 (2 c d-b e) \left (e (a e-b d)+c d^2\right ) \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^2}+\frac {e^2 \left (e (a e-b d)+c d^2\right )^2 \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d^3}\right )}{\left (e (a e-b d)+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x]

[Out]

(x*((c*(-2*c^3*d^3 + b^2*(b + Sqrt[b^2 - 4*a*c])*e^3 + 3*c^2*d*e*(b*d + Sqrt[b^2 - 4*a*c]*d + 2*a*e) - c*e^2*(
3*b^2*d + a*Sqrt[b^2 - 4*a*c]*e + 3*b*(Sqrt[b^2 - 4*a*c]*d + a*e)))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (
2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (c*(2*c^3*d^3 + b^2*(-b + Sqrt[b^2 -
4*a*c])*e^3 + 3*c^2*d*e*(-(b*d) + Sqrt[b^2 - 4*a*c]*d - 2*a*e) + c*e^2*(3*b^2*d - 3*b*Sqrt[b^2 - 4*a*c]*d + 3
*a*b*e - a*Sqrt[b^2 - 4*a*c]*e))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])
/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) + (e^2*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*Hypergeometric2F1[1, n^(
-1), 1 + n^(-1), -((e*x^n)/d)])/d + (e^2*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*Hypergeometric2F1[2, n^(-1),
1 + n^(-1), -((e*x^n)/d)])/d^2 + (e^2*(c*d^2 + e*(-(b*d) + a*e))^2*Hypergeometric2F1[3, n^(-1), 1 + n^(-1), -
((e*x^n)/d)])/d^3))/(c*d^2 + e*(-(b*d) + a*e))^3

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fricas [F]  time = 4.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b e^{3} x^{4 \, n} + a d^{3} + {\left (3 \, b d e^{2} + a e^{3}\right )} x^{3 \, n} + {\left (c e^{3} x^{3 \, n} + 3 \, c d e^{2} x^{2 \, n} + 3 \, c d^{2} e x^{n} + c d^{3}\right )} x^{2 \, n} + 3 \, {\left (b d^{2} e + a d e^{2}\right )} x^{2 \, n} + {\left (b d^{3} + 3 \, a d^{2} e\right )} x^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

integral(1/(b*e^3*x^(4*n) + a*d^3 + (3*b*d*e^2 + a*e^3)*x^(3*n) + (c*e^3*x^(3*n) + 3*c*d*e^2*x^(2*n) + 3*c*d^2
*e*x^n + c*d^3)*x^(2*n) + 3*(b*d^2*e + a*d*e^2)*x^(2*n) + (b*d^3 + 3*a*d^2*e)*x^n), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

integrate(1/((c*x^(2*n) + b*x^n + a)*(e*x^n + d)^3), x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \,x^{n}+d \right )^{3} \left (b \,x^{n}+c \,x^{2 n}+a \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x^n+d)^3/(b*x^n+c*x^(2*n)+a),x)

[Out]

int(1/(e*x^n+d)^3/(b*x^n+c*x^(2*n)+a),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left ({\left (12 \, n^{2} - 7 \, n + 1\right )} c^{2} d^{4} e^{2} - 2 \, {\left (8 \, n^{2} - 6 \, n + 1\right )} b c d^{3} e^{3} + {\left (6 \, n^{2} - 5 \, n + 1\right )} b^{2} d^{2} e^{4} + {\left (2 \, n^{2} - 3 \, n + 1\right )} a^{2} e^{6} + 2 \, {\left ({\left (3 \, n^{2} - 5 \, n + 1\right )} c d^{2} e^{4} - {\left (3 \, n^{2} - 4 \, n + 1\right )} b d e^{5}\right )} a\right )} \int \frac {1}{2 \, {\left (c^{3} d^{9} n^{2} - 3 \, b c^{2} d^{8} e n^{2} + 3 \, b^{2} c d^{7} e^{2} n^{2} - b^{3} d^{6} e^{3} n^{2} + a^{3} d^{3} e^{6} n^{2} + 3 \, {\left (c d^{5} e^{4} n^{2} - b d^{4} e^{5} n^{2}\right )} a^{2} + 3 \, {\left (c^{2} d^{7} e^{2} n^{2} - 2 \, b c d^{6} e^{3} n^{2} + b^{2} d^{5} e^{4} n^{2}\right )} a + {\left (c^{3} d^{8} e n^{2} - 3 \, b c^{2} d^{7} e^{2} n^{2} + 3 \, b^{2} c d^{6} e^{3} n^{2} - b^{3} d^{5} e^{4} n^{2} + a^{3} d^{2} e^{7} n^{2} + 3 \, {\left (c d^{4} e^{5} n^{2} - b d^{3} e^{6} n^{2}\right )} a^{2} + 3 \, {\left (c^{2} d^{6} e^{3} n^{2} - 2 \, b c d^{5} e^{4} n^{2} + b^{2} d^{4} e^{5} n^{2}\right )} a\right )} x^{n}\right )}}\,{d x} + \frac {{\left (c d^{2} e^{3} {\left (6 \, n - 1\right )} - b d e^{4} {\left (4 \, n - 1\right )} + a e^{5} {\left (2 \, n - 1\right )}\right )} x x^{n} + {\left (c d^{3} e^{2} {\left (7 \, n - 1\right )} - b d^{2} e^{3} {\left (5 \, n - 1\right )} + a d e^{4} {\left (3 \, n - 1\right )}\right )} x}{2 \, {\left (c^{2} d^{8} n^{2} - 2 \, b c d^{7} e n^{2} + b^{2} d^{6} e^{2} n^{2} + a^{2} d^{4} e^{4} n^{2} + 2 \, {\left (c d^{6} e^{2} n^{2} - b d^{5} e^{3} n^{2}\right )} a + {\left (c^{2} d^{6} e^{2} n^{2} - 2 \, b c d^{5} e^{3} n^{2} + b^{2} d^{4} e^{4} n^{2} + a^{2} d^{2} e^{6} n^{2} + 2 \, {\left (c d^{4} e^{4} n^{2} - b d^{3} e^{5} n^{2}\right )} a\right )} x^{2 \, n} + 2 \, {\left (c^{2} d^{7} e n^{2} - 2 \, b c d^{6} e^{2} n^{2} + b^{2} d^{5} e^{3} n^{2} + a^{2} d^{3} e^{5} n^{2} + 2 \, {\left (c d^{5} e^{3} n^{2} - b d^{4} e^{4} n^{2}\right )} a\right )} x^{n}\right )}} + \int \frac {c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3} - {\left (3 \, c^{2} d e^{2} - 2 \, b c e^{3}\right )} a - {\left (3 \, c^{3} d^{2} e - 3 \, b c^{2} d e^{2} + b^{2} c e^{3} - a c^{2} e^{3}\right )} x^{n}}{a^{4} e^{6} + 3 \, {\left (c d^{2} e^{4} - b d e^{5}\right )} a^{3} + 3 \, {\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} a^{2} + {\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )} a + {\left (c^{4} d^{6} - 3 \, b c^{3} d^{5} e + 3 \, b^{2} c^{2} d^{4} e^{2} - b^{3} c d^{3} e^{3} + a^{3} c e^{6} + 3 \, {\left (c^{2} d^{2} e^{4} - b c d e^{5}\right )} a^{2} + 3 \, {\left (c^{3} d^{4} e^{2} - 2 \, b c^{2} d^{3} e^{3} + b^{2} c d^{2} e^{4}\right )} a\right )} x^{2 \, n} + {\left (b c^{3} d^{6} - 3 \, b^{2} c^{2} d^{5} e + 3 \, b^{3} c d^{4} e^{2} - b^{4} d^{3} e^{3} + a^{3} b e^{6} + 3 \, {\left (b c d^{2} e^{4} - b^{2} d e^{5}\right )} a^{2} + 3 \, {\left (b c^{2} d^{4} e^{2} - 2 \, b^{2} c d^{3} e^{3} + b^{3} d^{2} e^{4}\right )} a\right )} x^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

((12*n^2 - 7*n + 1)*c^2*d^4*e^2 - 2*(8*n^2 - 6*n + 1)*b*c*d^3*e^3 + (6*n^2 - 5*n + 1)*b^2*d^2*e^4 + (2*n^2 - 3
*n + 1)*a^2*e^6 + 2*((3*n^2 - 5*n + 1)*c*d^2*e^4 - (3*n^2 - 4*n + 1)*b*d*e^5)*a)*integrate(1/2/(c^3*d^9*n^2 -
3*b*c^2*d^8*e*n^2 + 3*b^2*c*d^7*e^2*n^2 - b^3*d^6*e^3*n^2 + a^3*d^3*e^6*n^2 + 3*(c*d^5*e^4*n^2 - b*d^4*e^5*n^2
)*a^2 + 3*(c^2*d^7*e^2*n^2 - 2*b*c*d^6*e^3*n^2 + b^2*d^5*e^4*n^2)*a + (c^3*d^8*e*n^2 - 3*b*c^2*d^7*e^2*n^2 + 3
*b^2*c*d^6*e^3*n^2 - b^3*d^5*e^4*n^2 + a^3*d^2*e^7*n^2 + 3*(c*d^4*e^5*n^2 - b*d^3*e^6*n^2)*a^2 + 3*(c^2*d^6*e^
3*n^2 - 2*b*c*d^5*e^4*n^2 + b^2*d^4*e^5*n^2)*a)*x^n), x) + 1/2*((c*d^2*e^3*(6*n - 1) - b*d*e^4*(4*n - 1) + a*e
^5*(2*n - 1))*x*x^n + (c*d^3*e^2*(7*n - 1) - b*d^2*e^3*(5*n - 1) + a*d*e^4*(3*n - 1))*x)/(c^2*d^8*n^2 - 2*b*c*
d^7*e*n^2 + b^2*d^6*e^2*n^2 + a^2*d^4*e^4*n^2 + 2*(c*d^6*e^2*n^2 - b*d^5*e^3*n^2)*a + (c^2*d^6*e^2*n^2 - 2*b*c
*d^5*e^3*n^2 + b^2*d^4*e^4*n^2 + a^2*d^2*e^6*n^2 + 2*(c*d^4*e^4*n^2 - b*d^3*e^5*n^2)*a)*x^(2*n) + 2*(c^2*d^7*e
*n^2 - 2*b*c*d^6*e^2*n^2 + b^2*d^5*e^3*n^2 + a^2*d^3*e^5*n^2 + 2*(c*d^5*e^3*n^2 - b*d^4*e^4*n^2)*a)*x^n) + int
egrate((c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3 - (3*c^2*d*e^2 - 2*b*c*e^3)*a - (3*c^3*d^2*e - 3*b*c
^2*d*e^2 + b^2*c*e^3 - a*c^2*e^3)*x^n)/(a^4*e^6 + 3*(c*d^2*e^4 - b*d*e^5)*a^3 + 3*(c^2*d^4*e^2 - 2*b*c*d^3*e^3
+ b^2*d^2*e^4)*a^2 + (c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*a + (c^4*d^6 - 3*b*c^3*d^5*e +
3*b^2*c^2*d^4*e^2 - b^3*c*d^3*e^3 + a^3*c*e^6 + 3*(c^2*d^2*e^4 - b*c*d*e^5)*a^2 + 3*(c^3*d^4*e^2 - 2*b*c^2*d^
3*e^3 + b^2*c*d^2*e^4)*a)*x^(2*n) + (b*c^3*d^6 - 3*b^2*c^2*d^5*e + 3*b^3*c*d^4*e^2 - b^4*d^3*e^3 + a^3*b*e^6 +
3*(b*c*d^2*e^4 - b^2*d*e^5)*a^2 + 3*(b*c^2*d^4*e^2 - 2*b^2*c*d^3*e^3 + b^3*d^2*e^4)*a)*x^n), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x)

[Out]

int(1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x**n)**3/(a+b*x**n+c*x**(2*n)),x)

[Out]

Exception raised: HeuristicGCDFailed

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