\(\int \frac {(e x)^m (A+B x^n) (c+d x^n)}{(a+b x^n)^3} \, dx\) [7]

Optimal result

Integrand size = 29, antiderivative size = 228 \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )}{\left (a+b x^n\right )^3} \, dx=-\frac {(A b (b c (1+m-2 n)-a d (1+m-n))-a B (b c (1+m)-a d (1+m+n))) (e x)^{1+m}}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )}{2 a b e n \left (a+b x^n\right )^2}-\frac {(b c (a B (1+m)-A b (1+m-2 n)) (1+m-n)+a d (1+m) (A b (1+m-n)-a B (1+m+n))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )}{2 a^3 b^2 e (1+m) n^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {608, 468, 371} \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )}{\left (a+b x^n\right )^3} \, dx=-\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {b x^n}{a}\right ) (b c (m-n+1) (a B (m+1)-A b (m-2 n+1))+a d (m+1) (A b (m-n+1)-a B (m+n+1)))}{2 a^3 b^2 e (m+1) n^2}+\frac {(e x)^{m+1} (b c (a B (m+1)-A b (m-2 n+1))+a d (A b (m-n+1)-a B (m+n+1)))}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )}{2 a b e n \left (a+b x^n\right )^2} \]

[In]

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

[Out]

((b*c*(a*B*(1 + m) - A*b*(1 + m - 2*n)) + a*d*(A*b*(1 + m - n) - a*B*(1 + m + n)))*(e*x)^(1 + m))/(2*a^2*b^2*e
*n^2*(a + b*x^n)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^n))/(2*a*b*e*n*(a + b*x^n)^2) - ((b*c*(a*B*(1 + m) - A
*b*(1 + m - 2*n))*(1 + m - n) + a*d*(1 + m)*(A*b*(1 + m - n) - a*B*(1 + m + n)))*(e*x)^(1 + m)*Hypergeometric2
F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(2*a^3*b^2*e*(1 + m)*n^2)

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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 608

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/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n},
x] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rubi steps \begin{align*} \text {integral}& = \frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )}{2 a b e n \left (a+b x^n\right )^2}-\frac {\int \frac {(e x)^m \left (-c (a B (1+m)-A b (1+m-2 n))+d (A b (1+m-n)-a B (1+m+n)) x^n\right )}{\left (a+b x^n\right )^2} \, dx}{2 a b n} \\ & = \frac {(b c (a B (1+m)-A b (1+m-2 n))+a d (A b (1+m-n)-a B (1+m+n))) (e x)^{1+m}}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )}{2 a b e n \left (a+b x^n\right )^2}-\frac {(b c (a B (1+m)-A b (1+m-2 n)) (1+m-n)+a d (1+m) (A b (1+m-n)-a B (1+m+n))) \int \frac {(e x)^m}{a+b x^n} \, dx}{2 a^2 b^2 n^2} \\ & = \frac {(b c (a B (1+m)-A b (1+m-2 n))+a d (A b (1+m-n)-a B (1+m+n))) (e x)^{1+m}}{2 a^2 b^2 e n^2 \left (a+b x^n\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^n\right )}{2 a b e n \left (a+b x^n\right )^2}-\frac {(b c (a B (1+m)-A b (1+m-2 n)) (1+m-n)+a d (1+m) (A b (1+m-n)-a B (1+m+n))) (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^2 e (1+m) n^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.60 \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )}{\left (a+b x^n\right )^3} \, dx=\frac {x (e x)^m \left (a^2 B d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )+a (b B c+A b d-2 a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )+(A b-a B) (b c-a d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {b x^n}{a}\right )\right )}{a^3 b^2 (1+m)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )}{\left (a+b x^n\right )^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )}{\left (a+b x^n\right )^3} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,\left (c+d\,x^n\right )}{{\left (a+b\,x^n\right )}^3} \,d x \]

[In]

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

[Out]

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