3.1.20 \(\int \frac {(e x)^m (A+B x^n) (c+d x^n)^3}{(a+b x^n)^2} \, dx\) [20]

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

[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 = 0.66, 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 = {608, 584, 20, 30, 371} \begin {gather*} -\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 {(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} \end {gather*}

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 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 584

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 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*} \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 = 2.18, size = 345, normalized size = 0.88 \begin {gather*} \frac {x (e x)^m \left (\frac {b^3 c^2 (A b c-a B c-3 a A d)}{a n \left (a+b x^n\right )}+a^2 B d^3 \left (\frac {3}{1+m}+\frac {a}{a n+b n x^n}\right )+a b d^2 \left (A d \left (-\frac {2}{1+m}-\frac {a}{a n+b n x^n}\right )+B \left (-\frac {6 c}{1+m}-\frac {2 d x^n}{1+m+n}-\frac {3 a c}{a n+b n x^n}\right )\right )+b^2 d \left (A d \left (\frac {d x^n}{1+m+n}+3 c \left (\frac {1}{1+m}+\frac {a}{a n+b n x^n}\right )\right )+B \left (\frac {3 c d x^n}{1+m+n}+\frac {d^2 x^{2 n}}{1+m+2 n}+3 c^2 \left (\frac {1}{1+m}+\frac {a}{a n+b n x^n}\right )\right )\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))) \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a^2 (1+m) n}\right )}{b^4} \end {gather*}

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

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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right ) \left (c +d \,x^{n}\right )^{3}}{\left (a +b \,x^{n}\right )^{2}}\, dx\]

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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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)*(c+d*x^n)^3/(a+b*x^n)^2,x, algorithm="maxima")

[Out]

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

[Out]

integrate((B*x^n + A)*(d*x^n + c)^3*(x*e)^m/(b*x^n + a)^2, 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 (c+d\,x^n\right )}^3}{{\left (a+b\,x^n\right )}^2} \,d x \end {gather*}

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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