3.601 \(\int \frac{(d x)^m}{(a+b x^n+c x^{2 n})^3} \, dx\)

Optimal. Leaf size=615 \[ -\frac{c (d x)^{m+1} \left (-8 a^2 c^2 \left (m^2+m (2-6 n)+8 n^2-6 n+1\right )+6 a b^2 c \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b (m-n+1) \sqrt{b^2-4 a c} \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )+b^4 \left (-\left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{2 a^2 d (m+1) n^2 \left (b^2-4 a c\right )^{5/2} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{c (d x)^{m+1} \left (8 a^2 c^2 \left (m^2+m (2-6 n)+8 n^2-6 n+1\right )-6 a b^2 c \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b (m-n+1) \sqrt{b^2-4 a c} \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )+b^4 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 a^2 d (m+1) n^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}+b\right )}-\frac{(d x)^{m+1} \left (4 a^2 c^2 (m-4 n+1)-b c x^n \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )-5 a b^2 c (m-3 n+1)+b^4 (m-2 n+1)\right )}{2 a^2 d n^2 \left (b^2-4 a c\right )^2 \left (a+b x^n+c x^{2 n}\right )}+\frac{(d x)^{m+1} \left (-2 a c+b^2+b c x^n\right )}{2 a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2} \]

[Out]

((d*x)^(1 + m)*(b^2 - 2*a*c + b*c*x^n))/(2*a*(b^2 - 4*a*c)*d*n*(a + b*x^n + c*x^(2*n))^2) - ((d*x)^(1 + m)*(4*
a^2*c^2*(1 + m - 4*n) - 5*a*b^2*c*(1 + m - 3*n) + b^4*(1 + m - 2*n) - b*c*(2*a*c*(2 + 2*m - 7*n) - b^2*(1 + m
- 2*n))*x^n))/(2*a^2*(b^2 - 4*a*c)^2*d*n^2*(a + b*x^n + c*x^(2*n))) - (c*(b*Sqrt[b^2 - 4*a*c]*(2*a*c*(2 + 2*m
- 7*n) - b^2*(1 + m - 2*n))*(1 + m - n) - b^4*(1 + m^2 + m*(2 - 3*n) - 3*n + 2*n^2) + 6*a*b^2*c*(1 + m^2 + m*(
2 - 4*n) - 4*n + 3*n^2) - 8*a^2*c^2*(1 + m^2 + m*(2 - 6*n) - 6*n + 8*n^2))*(d*x)^(1 + m)*Hypergeometric2F1[1,
(1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)^(5/2)*(b - Sqrt[b^2 - 4*a*
c])*d*(1 + m)*n^2) - (c*(b*Sqrt[b^2 - 4*a*c]*(2*a*c*(2 + 2*m - 7*n) - b^2*(1 + m - 2*n))*(1 + m - n) + b^4*(1
+ m^2 + m*(2 - 3*n) - 3*n + 2*n^2) - 6*a*b^2*c*(1 + m^2 + m*(2 - 4*n) - 4*n + 3*n^2) + 8*a^2*c^2*(1 + m^2 + m*
(2 - 6*n) - 6*n + 8*n^2))*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^
2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)^(5/2)*(b + Sqrt[b^2 - 4*a*c])*d*(1 + m)*n^2)

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Rubi [A]  time = 10.7238, antiderivative size = 615, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1384, 1558, 1560, 364} \[ -\frac{c (d x)^{m+1} \left (-8 a^2 c^2 \left (m^2+m (2-6 n)+8 n^2-6 n+1\right )+6 a b^2 c \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b (m-n+1) \sqrt{b^2-4 a c} \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )+b^4 \left (-\left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{2 a^2 d (m+1) n^2 \left (b^2-4 a c\right )^{5/2} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{c (d x)^{m+1} \left (8 a^2 c^2 \left (m^2+m (2-6 n)+8 n^2-6 n+1\right )-6 a b^2 c \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b (m-n+1) \sqrt{b^2-4 a c} \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )+b^4 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 a^2 d (m+1) n^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt{b^2-4 a c}+b\right )}-\frac{(d x)^{m+1} \left (4 a^2 c^2 (m-4 n+1)-b c x^n \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )-5 a b^2 c (m-3 n+1)+b^4 (m-2 n+1)\right )}{2 a^2 d n^2 \left (b^2-4 a c\right )^2 \left (a+b x^n+c x^{2 n}\right )}+\frac{(d x)^{m+1} \left (-2 a c+b^2+b c x^n\right )}{2 a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(1 + m)*(b^2 - 2*a*c + b*c*x^n))/(2*a*(b^2 - 4*a*c)*d*n*(a + b*x^n + c*x^(2*n))^2) - ((d*x)^(1 + m)*(4*
a^2*c^2*(1 + m - 4*n) - 5*a*b^2*c*(1 + m - 3*n) + b^4*(1 + m - 2*n) - b*c*(2*a*c*(2 + 2*m - 7*n) - b^2*(1 + m
- 2*n))*x^n))/(2*a^2*(b^2 - 4*a*c)^2*d*n^2*(a + b*x^n + c*x^(2*n))) - (c*(b*Sqrt[b^2 - 4*a*c]*(2*a*c*(2 + 2*m
- 7*n) - b^2*(1 + m - 2*n))*(1 + m - n) - b^4*(1 + m^2 + m*(2 - 3*n) - 3*n + 2*n^2) + 6*a*b^2*c*(1 + m^2 + m*(
2 - 4*n) - 4*n + 3*n^2) - 8*a^2*c^2*(1 + m^2 + m*(2 - 6*n) - 6*n + 8*n^2))*(d*x)^(1 + m)*Hypergeometric2F1[1,
(1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)^(5/2)*(b - Sqrt[b^2 - 4*a*
c])*d*(1 + m)*n^2) - (c*(b*Sqrt[b^2 - 4*a*c]*(2*a*c*(2 + 2*m - 7*n) - b^2*(1 + m - 2*n))*(1 + m - n) + b^4*(1
+ m^2 + m*(2 - 3*n) - 3*n + 2*n^2) - 6*a*b^2*c*(1 + m^2 + m*(2 - 4*n) - 4*n + 3*n^2) + 8*a^2*c^2*(1 + m^2 + m*
(2 - 6*n) - 6*n + 8*n^2))*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^
2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)^(5/2)*(b + Sqrt[b^2 - 4*a*c])*d*(1 + m)*n^2)

Rule 1384

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((d*x)^(m + 1)*(b
^2 - 2*a*c + b*c*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*d*n*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(a*n*(p + 1)
*(b^2 - 4*a*c)), Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(n*(p + 1) + m + 1) - 2*a*c*(m + 2*n*(p
+ 1) + 1) + b*c*(2*n*p + 3*n + m + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b
^2 - 4*a*c, 0] && ILtQ[p + 1, 0]

Rule 1558

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
-Simp[((f*x)^(m + 1)*(a + b*x^n + c*x^(2*n))^(p + 1)*(d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^n))/(a*f*n*
(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(a*n*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*S
imp[d*(b^2*(m + n*(p + 1) + 1) - 2*a*c*(m + 2*n*(p + 1) + 1)) - a*b*e*(m + 1) + (m + n*(2*p + 3) + 1)*(b*d - 2
*a*e)*c*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p
+ 1, 0]

Rule 1560

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0])

Rule 364

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

Rubi steps

\begin{align*} \int \frac{(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx &=\frac{(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac{\int \frac{(d x)^m \left (-2 a c (1+m-4 n)+b^2 (1+m-2 n)+b c (1+m-3 n) x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{2 a \left (b^2-4 a c\right ) n}\\ &=\frac{(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac{(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}+\frac{\int \frac{(d x)^m \left (\left (2 a c (1+m-4 n)-b^2 (1+m-2 n)\right ) \left (2 a c (1+m-2 n)-b^2 (1+m-n)\right )-a b^2 c (1+m) (1+m-3 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}\\ &=\frac{(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac{(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}+\frac{\int \left (\frac{\left (-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+\frac{c \left (b^4-6 a b^2 c+8 a^2 c^2+2 b^4 m-12 a b^2 c m+16 a^2 c^2 m+b^4 m^2-6 a b^2 c m^2+8 a^2 c^2 m^2-3 b^4 n+24 a b^2 c n-48 a^2 c^2 n-3 b^4 m n+24 a b^2 c m n-48 a^2 c^2 m n+2 b^4 n^2-18 a b^2 c n^2+64 a^2 c^2 n^2\right )}{\sqrt{b^2-4 a c}}\right ) (d x)^m}{b-\sqrt{b^2-4 a c}+2 c x^n}+\frac{\left (-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-\frac{c \left (b^4-6 a b^2 c+8 a^2 c^2+2 b^4 m-12 a b^2 c m+16 a^2 c^2 m+b^4 m^2-6 a b^2 c m^2+8 a^2 c^2 m^2-3 b^4 n+24 a b^2 c n-48 a^2 c^2 n-3 b^4 m n+24 a b^2 c m n-48 a^2 c^2 m n+2 b^4 n^2-18 a b^2 c n^2+64 a^2 c^2 n^2\right )}{\sqrt{b^2-4 a c}}\right ) (d x)^m}{b+\sqrt{b^2-4 a c}+2 c x^n}\right ) \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}\\ &=\frac{(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac{(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac{\left (c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-\frac{b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(d x)^m}{b-\sqrt{b^2-4 a c}+2 c x^n} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}-\frac{\left (c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+\frac{b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(d x)^m}{b+\sqrt{b^2-4 a c}+2 c x^n} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}\\ &=\frac{(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac{(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac{c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-\frac{b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b-\sqrt{b^2-4 a c}\right ) d (1+m) n^2}-\frac{c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+\frac{b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b+\sqrt{b^2-4 a c}\right ) d (1+m) n^2}\\ \end{align*}

Mathematica [B]  time = 6.72233, size = 7827, normalized size = 12.73 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx \right ) ^{m}}{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*b^2*c*d^m*(5*m - 21*n + 5) - a*b^4*d^m*(m - 3*n + 1) - 4*a^3*c^2*d^m*(m - 6*n + 1))*x*x^m + (2*a*b*c
^3*d^m*(2*m - 7*n + 2) - b^3*c^2*d^m*(m - 2*n + 1))*x*e^(m*log(x) + 3*n*log(x)) + (a*b^2*c^2*d^m*(9*m - 29*n +
9) - 2*b^4*c*d^m*(m - 2*n + 1) - 4*a^2*c^3*d^m*(m - 4*n + 1))*x*e^(m*log(x) + 2*n*log(x)) - (b^5*d^m*(m - 2*n
+ 1) - 4*a*b^3*c*d^m*(m - 3*n + 1) + 2*a^2*b*c^2*d^m*n)*x*e^(m*log(x) + n*log(x)))/(a^4*b^4*n^2 - 8*a^5*b^2*c
*n^2 + 16*a^6*c^2*n^2 + (a^2*b^4*c^2*n^2 - 8*a^3*b^2*c^3*n^2 + 16*a^4*c^4*n^2)*x^(4*n) + 2*(a^2*b^5*c*n^2 - 8*
a^3*b^3*c^2*n^2 + 16*a^4*b*c^3*n^2)*x^(3*n) + (a^2*b^6*n^2 - 6*a^3*b^4*c*n^2 + 32*a^5*c^3*n^2)*x^(2*n) + 2*(a^
3*b^5*n^2 - 8*a^4*b^3*c*n^2 + 16*a^5*b*c^2*n^2)*x^n) - integrate(-1/2*(((m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*
b^4*d^m - (5*m^2 - m*(21*n - 10) + 16*n^2 - 21*n + 5)*a*b^2*c*d^m + 4*(m^2 - 2*m*(3*n - 1) + 8*n^2 - 6*n + 1)*
a^2*c^2*d^m)*x^m + ((m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*b^3*c*d^m - 2*(2*m^2 - m*(9*n - 4) + 7*n^2 - 9*n + 2
)*a*b*c^2*d^m)*e^(m*log(x) + n*log(x)))/(a^3*b^4*n^2 - 8*a^4*b^2*c*n^2 + 16*a^5*c^2*n^2 + (a^2*b^4*c*n^2 - 8*a
^3*b^2*c^2*n^2 + 16*a^4*c^3*n^2)*x^(2*n) + (a^2*b^5*n^2 - 8*a^3*b^3*c*n^2 + 16*a^4*b*c^2*n^2)*x^n), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d x\right )^{m}}{c^{3} x^{6 \, n} + b^{3} x^{3 \, n} + 3 \, a b^{2} x^{2 \, n} + 3 \, a^{2} b x^{n} + a^{3} + 3 \,{\left (b c^{2} x^{n} + a c^{2}\right )} x^{4 \, n} + 3 \,{\left (b^{2} c x^{2 \, n} + 2 \, a b c x^{n} + a^{2} c\right )} x^{2 \, n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m/(c^3*x^(6*n) + b^3*x^(3*n) + 3*a*b^2*x^(2*n) + 3*a^2*b*x^n + a^3 + 3*(b*c^2*x^n + a*c^2)*x^(4
*n) + 3*(b^2*c*x^(2*n) + 2*a*b*c*x^n + a^2*c)*x^(2*n)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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