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

Optimal. Leaf size=394 \[ -\frac {(e x)^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right ) (A b (b c (m-n+1)-a d (m+2 n+1))-a B (b c (m+1)-a d (m+3 n+1)))}{a^2 b^4 e (m+1) n}-\frac {d (e x)^{m+1} \left (A b \left (a^2 d^2 (m+2 n+1)-3 a b c d (m+n+1)+3 b^2 c^2 (m+1)\right )-a B \left (a^2 d^2 (m+3 n+1)-3 a b c d (m+2 n+1)+3 b^2 c^2 (m+n+1)\right )\right )}{a b^4 e (m+1) n}-\frac {d^2 x^{n+1} (e x)^m (A b (3 b c (m+n+1)-a d (m+2 n+1))-a B (3 b c (m+2 n+1)-a d (m+3 n+1)))}{a b^3 n (m+n+1)}-\frac {d^3 x^{2 n+1} (e x)^m (A b (m+2 n+1)-a B (m+3 n+1))}{a b^2 n (m+2 n+1)}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )} \]

[Out]

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

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Rubi [A]  time = 1.02, antiderivative size = 389, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {594, 570, 20, 30, 364} \[ -\frac {d (e x)^{m+1} \left (A b \left (a^2 d^2 (m+2 n+1)-3 a b c d (m+n+1)+3 b^2 c^2 (m+1)\right )-a B \left (a^2 d^2 (m+3 n+1)-3 a b c d (m+2 n+1)+3 b^2 c^2 (m+n+1)\right )\right )}{a b^4 e (m+1) n}-\frac {(e x)^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right ) (A b (b c (m-n+1)-a d (m+2 n+1))-a B (b c (m+1)-a d (m+3 n+1)))}{a^2 b^4 e (m+1) n}-\frac {d^2 x^{n+1} (e x)^m (A b (3 b c (m+n+1)-a d (m+2 n+1))-a B (3 b c (m+2 n+1)-a d (m+3 n+1)))}{a b^3 n (m+n+1)}+\frac {(e x)^{m+1} (A b-a B) \left (c+d x^n\right )^3}{a b e n \left (a+b x^n\right )}-\frac {d^3 x^{2 n+1} (e x)^m \left (A-\frac {a B (m+3 n+1)}{b (m+2 n+1)}\right )}{a b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

Rule 594

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

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Mathematica [A]  time = 0.57, size = 217, normalized size = 0.55 \[ \frac {x (e x)^m \left (\frac {d \left (3 a^2 B d^2-2 a b d (A d+3 B c)+3 b^2 c (A d+B c)\right )}{m+1}+\frac {(a B-A b) (a d-b c)^3 \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a^2 (m+1)}+\frac {b d^2 x^n (-2 a B d+A b d+3 b B c)}{m+n+1}+\frac {(b c-a d)^2 (-4 a B d+3 A b d+b B c) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a (m+1)}+\frac {b^2 B d^3 x^{2 n}}{m+2 n+1}\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((d*(3*a^2*B*d^2 + 3*b^2*c*(B*c + A*d) - 2*a*b*d*(3*B*c + A*d)))/(1 + m) + (b*d^2*(3*b*B*c + A*b*d
- 2*a*B*d)*x^n)/(1 + m + n) + (b^2*B*d^3*x^(2*n))/(1 + m + 2*n) + ((b*c - a*d)^2*(b*B*c + 3*A*b*d - 4*a*B*d)*H
ypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*(1 + m)) + ((-(A*b) + a*B)*(-(b*c) + a*d)^3*Hy
pergeometric2F1[2, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a^2*(1 + m))))/b^4

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B d^{3} x^{4 \, n} + A c^{3} + {\left (3 \, B c d^{2} + A d^{3}\right )} x^{3 \, n} + 3 \, {\left (B c^{2} d + A c d^{2}\right )} x^{2 \, n} + {\left (B c^{3} + 3 \, A c^{2} d\right )} x^{n}\right )} \left (e x\right )^{m}}{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((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)*(e*x)^m/(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.84, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{n}+A \right ) \left (d \,x^{n}+c \right )^{3} \left (e x \right )^{m}}{\left (b \,x^{n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^3*b*d^3*e^m*(m + 2*n + 1) - 3*a^2*b^2*c*d^2*e^m*(m + n + 1) - b^4*c^3*e^m*(m - n + 1) + 3*a*b^3*c^2*d*e^m*
(m + 1))*A - (a^4*d^3*e^m*(m + 3*n + 1) - 3*a^3*b*c*d^2*e^m*(m + 2*n + 1) + 3*a^2*b^2*c^2*d*e^m*(m + n + 1) -
a*b^3*c^3*e^m*(m + 1))*B)*integrate(x^m/(a*b^5*n*x^n + a^2*b^4*n), x) + ((m^2*n + (n^2 + 2*n)*m + n^2 + n)*B*a
*b^3*d^3*e^m*x*e^(m*log(x) + 3*n*log(x)) + (((m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*b^4
*c^3*e^m - 3*(m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*a*b^3*c^2*d*e^m + 3*(m^3 + m^2*(4*n
+ 3) + 2*n^3 + (5*n^2 + 8*n + 3)*m + 5*n^2 + 4*n + 1)*a^2*b^2*c*d^2*e^m - (m^3 + m^2*(5*n + 3) + 4*n^3 + (8*n
^2 + 10*n + 3)*m + 8*n^2 + 5*n + 1)*a^3*b*d^3*e^m)*A - ((m^3 + 3*m^2*(n + 1) + (2*n^2 + 6*n + 3)*m + 2*n^2 + 3
*n + 1)*a*b^3*c^3*e^m - 3*(m^3 + m^2*(4*n + 3) + 2*n^3 + (5*n^2 + 8*n + 3)*m + 5*n^2 + 4*n + 1)*a^2*b^2*c^2*d*
e^m + 3*(m^3 + m^2*(5*n + 3) + 4*n^3 + (8*n^2 + 10*n + 3)*m + 8*n^2 + 5*n + 1)*a^3*b*c*d^2*e^m - (m^3 + 3*m^2*
(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*a^4*d^3*e^m)*B)*x*x^m + ((m^2*n + 2*(n^2 + n)*m
+ 2*n^2 + n)*A*a*b^3*d^3*e^m + (3*(m^2*n + 2*(n^2 + n)*m + 2*n^2 + n)*a*b^3*c*d^2*e^m - (m^2*n + (3*n^2 + 2*n)
*m + 3*n^2 + n)*a^2*b^2*d^3*e^m)*B)*x*e^(m*log(x) + 2*n*log(x)) + ((3*(m^2*n + 2*n^3 + (3*n^2 + 2*n)*m + 3*n^2
+ n)*a*b^3*c*d^2*e^m - (m^2*n + 4*n^3 + 2*(2*n^2 + n)*m + 4*n^2 + n)*a^2*b^2*d^3*e^m)*A + (3*(m^2*n + 2*n^3 +
(3*n^2 + 2*n)*m + 3*n^2 + n)*a*b^3*c^2*d*e^m - 3*(m^2*n + 4*n^3 + 2*(2*n^2 + n)*m + 4*n^2 + n)*a^2*b^2*c*d^2*
e^m + (m^2*n + 6*n^3 + (5*n^2 + 2*n)*m + 5*n^2 + n)*a^3*b*d^3*e^m)*B)*x*e^(m*log(x) + n*log(x)))/((m^3*n + 3*(
n^2 + n)*m^2 + 2*n^3 + (2*n^3 + 6*n^2 + 3*n)*m + 3*n^2 + n)*a*b^5*x^n + (m^3*n + 3*(n^2 + n)*m^2 + 2*n^3 + (2*
n^3 + 6*n^2 + 3*n)*m + 3*n^2 + n)*a^2*b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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