3.28 \(\int \frac{(e x)^m (A+B x^n)}{(a+b x^n)^3 (c+d x^n)} \, dx\)

Optimal. Leaf size=407 \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2+m (2-5 n)+6 n^2-5 n+1\right )-2 a b c d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b^2 c^2 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )+a B \left (-a^2 d^2 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )+2 a b c d (m+1) (m-2 n+1)-b^2 c^2 (m+1) (m-n+1)\right )\right )}{2 a^3 e (m+1) n^2 (b c-a d)^3}+\frac{(e x)^{m+1} (A b (a d (m-4 n+1)-b c (m-2 n+1))+a B (b c (m+1)-a d (m-2 n+1)))}{2 a^2 e n^2 (b c-a d)^2 \left (a+b x^n\right )}+\frac{d^2 (e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (A b-a B)}{2 a e n (b c-a d) \left (a+b x^n\right )^2} \]

[Out]

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

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Rubi [A]  time = 1.24718, antiderivative size = 407, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {595, 597, 364} \[ \frac{(e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2+m (2-5 n)+6 n^2-5 n+1\right )-2 a b c d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b^2 c^2 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )+a B \left (-a^2 d^2 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )+2 a b c d (m+1) (m-2 n+1)-b^2 c^2 (m+1) (m-n+1)\right )\right )}{2 a^3 e (m+1) n^2 (b c-a d)^3}+\frac{(e x)^{m+1} (A b (a d (m-4 n+1)-b c (m-2 n+1))+a B (b c (m+1)-a d (m-2 n+1)))}{2 a^2 e n^2 (b c-a d)^2 \left (a+b x^n\right )}+\frac{d^2 (e x)^{m+1} (B c-A d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c e (m+1) (b c-a d)^3}+\frac{(e x)^{m+1} (A b-a B)}{2 a e n (b c-a d) \left (a+b x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 595

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, n, q}, x] && LtQ[p, -1]

Rule 597

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, n, p}, x]

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{(e x)^m \left (A+B x^n\right )}{\left (a+b x^n\right )^3 \left (c+d x^n\right )} \, dx &=\frac{(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2}-\frac{\int \frac{(e x)^m \left (-a B c (1+m)+A b c (1+m-2 n)+2 a A d n+(A b-a B) d (1+m-2 n) x^n\right )}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx}{2 a (b c-a d) n}\\ &=\frac{(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2}+\frac{(A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right )}+\frac{\int \frac{(e x)^m \left (-c (1+m) (A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n)))+(b c-a d) n (a B c (1+m)-A b c (1+m-2 n)-2 a A d n)-d (A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n))) (1+m-n) x^n\right )}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{2 a^2 (b c-a d)^2 n^2}\\ &=\frac{(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2}+\frac{(A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right )}+\frac{\int \left (\frac{\left (a B \left (2 a b c d (1+m) (1+m-2 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (a+b x^n\right )}+\frac{2 a^2 d^2 (-B c+A d) n^2 (e x)^m}{(-b c+a d) \left (c+d x^n\right )}\right ) \, dx}{2 a^2 (b c-a d)^2 n^2}\\ &=\frac{(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2}+\frac{(A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right )}+\frac{\left (d^2 (B c-A d)\right ) \int \frac{(e x)^m}{c+d x^n} \, dx}{(b c-a d)^3}+\frac{\left (a B \left (2 a b c d (1+m) (1+m-2 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )\right ) \int \frac{(e x)^m}{a+b x^n} \, dx}{2 a^2 (b c-a d)^3 n^2}\\ &=\frac{(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e n \left (a+b x^n\right )^2}+\frac{(A b (a d (1+m-4 n)-b c (1+m-2 n))+a B (b c (1+m)-a d (1+m-2 n))) (e x)^{1+m}}{2 a^2 (b c-a d)^2 e n^2 \left (a+b x^n\right )}+\frac{\left (a B \left (2 a b c d (1+m) (1+m-2 n)-b^2 c^2 (1+m) (1+m-n)-a^2 d^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )\right )+A b \left (b^2 c^2 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-2 a b c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+a^2 d^2 \left (1+m^2+m (2-5 n)-5 n+6 n^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{b x^n}{a}\right )}{2 a^3 (b c-a d)^3 e (1+m) n^2}+\frac{d^2 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{d x^n}{c}\right )}{c (b c-a d)^3 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.264498, size = 199, normalized size = 0.49 \[ \frac{x (e x)^m \left (\frac{b (b c-a d) (B c-A d) \, _2F_1\left (2,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^2}+\frac{(A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^3}+\frac{b d (A d-B c) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a}+\frac{d^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c}\right )}{(m+1) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((b*d*(-(B*c) + A*d)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/a + (d^2*(B*c -
A*d)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c + (b*(b*c - a*d)*(B*c - A*d)*Hypergeometr
ic2F1[2, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/a^2 + ((A*b - a*B)*(b*c - a*d)^2*Hypergeometric2F1[3, (1 + m
)/n, (1 + m + n)/n, -((b*x^n)/a)])/a^3))/((b*c - a*d)^3*(1 + m))

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Maple [F]  time = 0.677, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( A+B{x}^{n} \right ) }{ \left ( a+b{x}^{n} \right ) ^{3} \left ( c+d{x}^{n} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -{\left ({\left ({\left (m^{2} - m{\left (3 \, n - 2\right )} + 2 \, n^{2} - 3 \, n + 1\right )} b^{3} c^{2} e^{m} - 2 \,{\left (m^{2} - 2 \, m{\left (2 \, n - 1\right )} + 3 \, n^{2} - 4 \, n + 1\right )} a b^{2} c d e^{m} +{\left (m^{2} - m{\left (5 \, n - 2\right )} + 6 \, n^{2} - 5 \, n + 1\right )} a^{2} b d^{2} e^{m}\right )} A -{\left ({\left (m^{2} - m{\left (n - 2\right )} - n + 1\right )} a b^{2} c^{2} e^{m} - 2 \,{\left (m^{2} - 2 \, m{\left (n - 1\right )} - 2 \, n + 1\right )} a^{2} b c d e^{m} +{\left (m^{2} - m{\left (3 \, n - 2\right )} + 2 \, n^{2} - 3 \, n + 1\right )} a^{3} d^{2} e^{m}\right )} B\right )} \int -\frac{x^{m}}{2 \,{\left (a^{3} b^{3} c^{3} n^{2} - 3 \, a^{4} b^{2} c^{2} d n^{2} + 3 \, a^{5} b c d^{2} n^{2} - a^{6} d^{3} n^{2} +{\left (a^{2} b^{4} c^{3} n^{2} - 3 \, a^{3} b^{3} c^{2} d n^{2} + 3 \, a^{4} b^{2} c d^{2} n^{2} - a^{5} b d^{3} n^{2}\right )} x^{n}\right )}}\,{d x} -{\left (B c d^{2} e^{m} - A d^{3} e^{m}\right )} \int -\frac{x^{m}}{b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3} +{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} x^{n}}\,{d x} - \frac{{\left ({\left (a b^{2} c e^{m}{\left (m - 3 \, n + 1\right )} - a^{2} b d e^{m}{\left (m - 5 \, n + 1\right )}\right )} A -{\left (a^{2} b c e^{m}{\left (m - n + 1\right )} - a^{3} d e^{m}{\left (m - 3 \, n + 1\right )}\right )} B\right )} x x^{m} +{\left ({\left (b^{3} c e^{m}{\left (m - 2 \, n + 1\right )} - a b^{2} d e^{m}{\left (m - 4 \, n + 1\right )}\right )} A +{\left (a^{2} b d e^{m}{\left (m - 2 \, n + 1\right )} - a b^{2} c e^{m}{\left (m + 1\right )}\right )} B\right )} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{2 \,{\left (a^{4} b^{2} c^{2} n^{2} - 2 \, a^{5} b c d n^{2} + a^{6} d^{2} n^{2} +{\left (a^{2} b^{4} c^{2} n^{2} - 2 \, a^{3} b^{3} c d n^{2} + a^{4} b^{2} d^{2} n^{2}\right )} x^{2 \, n} + 2 \,{\left (a^{3} b^{3} c^{2} n^{2} - 2 \, a^{4} b^{2} c d n^{2} + a^{5} b d^{2} n^{2}\right )} x^{n}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*b^3*c^2*e^m - 2*(m^2 - 2*m*(2*n - 1) + 3*n^2 - 4*n + 1)*a*b^2*c*d*e^m
+ (m^2 - m*(5*n - 2) + 6*n^2 - 5*n + 1)*a^2*b*d^2*e^m)*A - ((m^2 - m*(n - 2) - n + 1)*a*b^2*c^2*e^m - 2*(m^2
- 2*m*(n - 1) - 2*n + 1)*a^2*b*c*d*e^m + (m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*a^3*d^2*e^m)*B)*integrate(-1/2*
x^m/(a^3*b^3*c^3*n^2 - 3*a^4*b^2*c^2*d*n^2 + 3*a^5*b*c*d^2*n^2 - a^6*d^3*n^2 + (a^2*b^4*c^3*n^2 - 3*a^3*b^3*c^
2*d*n^2 + 3*a^4*b^2*c*d^2*n^2 - a^5*b*d^3*n^2)*x^n), x) - (B*c*d^2*e^m - A*d^3*e^m)*integrate(-x^m/(b^3*c^4 -
3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3 + (b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*x^n), x)
- 1/2*(((a*b^2*c*e^m*(m - 3*n + 1) - a^2*b*d*e^m*(m - 5*n + 1))*A - (a^2*b*c*e^m*(m - n + 1) - a^3*d*e^m*(m -
3*n + 1))*B)*x*x^m + ((b^3*c*e^m*(m - 2*n + 1) - a*b^2*d*e^m*(m - 4*n + 1))*A + (a^2*b*d*e^m*(m - 2*n + 1) -
a*b^2*c*e^m*(m + 1))*B)*x*e^(m*log(x) + n*log(x)))/(a^4*b^2*c^2*n^2 - 2*a^5*b*c*d*n^2 + a^6*d^2*n^2 + (a^2*b^4
*c^2*n^2 - 2*a^3*b^3*c*d*n^2 + a^4*b^2*d^2*n^2)*x^(2*n) + 2*(a^3*b^3*c^2*n^2 - 2*a^4*b^2*c*d*n^2 + a^5*b*d^2*n
^2)*x^n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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