3.1.28 \(\int \frac {(e x)^m (A+B x^n)}{(a+b x^n)^3 (c+d x^n)} \, dx\) [28]

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

[Out]

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

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Rubi [A]
time = 0.81, antiderivative size = 407, normalized size of antiderivative = 1.00, 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 = {609, 611, 371} \begin {gather*} \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 {(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 {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} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 609

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 611

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]

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*}

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Mathematica [A]
time = 1.21, size = 402, normalized size = 0.99 \begin {gather*} \frac {x (e x)^m \left (a c (b c-a d) (1+m) \left (a^3 B d (1+m-3 n)+A b^3 c (1+m-2 n) x^n+a b^2 \left (A c (1+m-3 n)-B c (1+m) x^n-A d (1+m-4 n) x^n\right )-a^2 b \left (A d (1+m-5 n)+B c (1+m-n)-B d (1+m-2 n) x^n\right )\right )-c \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 ) \left (a+b x^n\right )^2 \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )+2 a^3 d^2 (-B c+A d) n^2 \left (a+b x^n\right )^2 \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )\right )}{2 a^3 c (-b c+a d)^3 (1+m) n^2 \left (a+b x^n\right )^2} \end {gather*}

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

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right )}{\left (a +b \,x^{n}\right )^{3} \left (c +d \,x^{n}\right )}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*e^m - m*(3*n - 2)*e^m + (2*n^2 - 3*n + 1)*e^m)*b^3*c^2 - 2*(m^2*e^m - 2*m*(2*n - 1)*e^m + (3*n^2 - 4*n
+ 1)*e^m)*a*b^2*c*d + (m^2*e^m - m*(5*n - 2)*e^m + (6*n^2 - 5*n + 1)*e^m)*a^2*b*d^2)*A - ((m^2*e^m - m*(n - 2
)*e^m - (n - 1)*e^m)*a*b^2*c^2 - 2*(m^2*e^m - 2*m*(n - 1)*e^m - (2*n - 1)*e^m)*a^2*b*c*d + (m^2*e^m - m*(3*n -
2)*e^m + (2*n^2 - 3*n + 1)*e^m)*a^3*d^2)*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*((((m*e^m - (3*n - 1)*e^m)*a*b^2*c - (m*e^m
- (5*n - 1)*e^m)*a^2*b*d)*A - ((m*e^m - (n - 1)*e^m)*a^2*b*c - (m*e^m - (3*n - 1)*e^m)*a^3*d)*B)*x*x^m + (((m
*e^m - (2*n - 1)*e^m)*b^3*c - (m*e^m - (4*n - 1)*e^m)*a*b^2*d)*A - ((m*e^m + e^m)*a*b^2*c - (m*e^m - (2*n - 1)
*e^m)*a^2*b*d)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(x*e)^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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{{\left (a+b\,x^n\right )}^3\,\left (c+d\,x^n\right )} \,d x \end {gather*}

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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