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

Optimal. Leaf size=366 \[ -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )\right )+b^2 c^2 (m-2 n+1) (A d (m-3 n+1)-B c (m-n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^3}+\frac {b^2 (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{2 c^2 e n^2 (b c-a d)^2 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2} \]

[Out]

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

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Rubi [A]  time = 1.22, antiderivative size = 366, 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 = {595, 597, 364} \[ -\frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right ) \left (-a^2 d^2 (m-n+1) (B c (m+1)-A d (m-2 n+1))+2 a b c d \left (B c (m+1) (m-2 n+1)-A d \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )\right )+b^2 c^2 (m-2 n+1) (A d (m-3 n+1)-B c (m-n+1))\right )}{2 c^3 e (m+1) n^2 (b c-a d)^3}+\frac {b^2 (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (a d (B c (m+1)-A d (m-2 n+1))+b c (A d (m-4 n+1)-B c (m-2 n+1)))}{2 c^2 e n^2 (b c-a d)^2 \left (c+d x^n\right )}+\frac {(e x)^{m+1} (B c-A d)}{2 c e n (b c-a d) \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 201, normalized size = 0.55 \[ \frac {x (e x)^m \left (\frac {b^2 (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a}+\frac {(b c-a d)^2 (B c-A d) \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c^3}-\frac {d (A b-a B) (b c-a d) \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c^2}-\frac {b d (A b-a B) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{c}\right )}{(m+1) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((b^2*(A*b - a*B)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/a - (b*(A*b - a*B)*
d*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c - ((A*b - a*B)*d*(b*c - a*d)*Hypergeometric2
F1[2, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/c^2 + ((b*c - a*d)^2*(B*c - A*d)*Hypergeometric2F1[3, (1 + m)/n
, (1 + m + n)/n, -((d*x^n)/c)])/c^3))/((b*c - a*d)^3*(1 + m))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{b d^{3} x^{4 \, n} + a c^{3} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{3 \, n} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} x^{2 \, n} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{n}+A \right ) \left (e x \right )^{m}}{\left (b \,x^{n}+a \right ) \left (d \,x^{n}+c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: HeuristicGCDFailed

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