3.1.19 \(\int \frac {(e x)^m (A+B x^n) (c+d x^n)^3}{a+b x^n} \, dx\) [19]
Optimal. Leaf size=270 \[ \frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) x^{1+n} (e x)^m}{b^3 (1+m+n)}+\frac {d^2 (3 b B c+A b d-a B d) x^{1+2 n} (e x)^m}{b^2 (1+m+2 n)}+\frac {B d^3 x^{1+3 n} (e x)^m}{b (1+m+3 n)}-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^{1+m}}{b^4 e (1+m)}+\frac {(A b-a B) (b c-a d)^3 (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a b^4 e (1+m)} \]
[Out]
d*(a^2*B*d^2+3*b^2*c*(A*d+B*c)-a*b*d*(A*d+3*B*c))*x^(1+n)*(e*x)^m/b^3/(1+m+n)+d^2*(A*b*d-B*a*d+3*B*b*c)*x^(1+2
*n)*(e*x)^m/b^2/(1+m+2*n)+B*d^3*x^(1+3*n)*(e*x)^m/b/(1+m+3*n)-(a^3*B*d^3+3*a*b^2*c*d*(A*d+B*c)-a^2*b*d^2*(A*d+
3*B*c)-b^3*c^2*(3*A*d+B*c))*(e*x)^(1+m)/b^4/e/(1+m)+(A*b-B*a)*(-a*d+b*c)^3*(e*x)^(1+m)*hypergeom([1, (1+m)/n],
[(1+m+n)/n],-b*x^n/a)/a/b^4/e/(1+m)
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Rubi [A]
time = 0.26, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {584, 20, 30,
371} \begin {gather*} \frac {d x^{n+1} (e x)^m \left (a^2 B d^2-a b d (A d+3 B c)+3 b^2 c (A d+B c)\right )}{b^3 (m+n+1)}-\frac {(e x)^{m+1} \left (a^3 B d^3-a^2 b d^2 (A d+3 B c)+3 a b^2 c d (A d+B c)+b^3 \left (-c^2\right ) (3 A d+B c)\right )}{b^4 e (m+1)}+\frac {(e x)^{m+1} (A b-a B) (b c-a d)^3 \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a b^4 e (m+1)}+\frac {d^2 x^{2 n+1} (e x)^m (-a B d+A b d+3 b B c)}{b^2 (m+2 n+1)}+\frac {B d^3 x^{3 n+1} (e x)^m}{b (m+3 n+1)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((e*x)^m*(A + B*x^n)*(c + d*x^n)^3)/(a + b*x^n),x]
[Out]
(d*(a^2*B*d^2 + 3*b^2*c*(B*c + A*d) - a*b*d*(3*B*c + A*d))*x^(1 + n)*(e*x)^m)/(b^3*(1 + m + n)) + (d^2*(3*b*B*
c + A*b*d - a*B*d)*x^(1 + 2*n)*(e*x)^m)/(b^2*(1 + m + 2*n)) + (B*d^3*x^(1 + 3*n)*(e*x)^m)/(b*(1 + m + 3*n)) -
((a^3*B*d^3 + 3*a*b^2*c*d*(B*c + A*d) - a^2*b*d^2*(3*B*c + A*d) - b^3*c^2*(B*c + 3*A*d))*(e*x)^(1 + m))/(b^4*e
*(1 + m)) + ((A*b - a*B)*(b*c - a*d)^3*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/
a)])/(a*b^4*e*(1 + m))
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]
Rubi steps
\begin {align*} \int \frac {(e x)^m \left (A+B x^n\right ) \left (c+d x^n\right )^3}{a+b x^n} \, dx &=\int \left (\frac {\left (-a^3 B d^3-3 a b^2 c d (B c+A d)+a^2 b d^2 (3 B c+A d)+b^3 c^2 (B c+3 A d)\right ) (e x)^m}{b^4}+\frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) x^n (e x)^m}{b^3}+\frac {d^2 (3 b B c+A b d-a B d) x^{2 n} (e x)^m}{b^2}+\frac {B d^3 x^{3 n} (e x)^m}{b}+\frac {(A b-a B) (b c-a d)^3 (e x)^m}{b^4 \left (a+b x^n\right )}\right ) \, dx\\ &=-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^{1+m}}{b^4 e (1+m)}+\frac {\left (B d^3\right ) \int x^{3 n} (e x)^m \, dx}{b}+\frac {\left ((A b-a B) (b c-a d)^3\right ) \int \frac {(e x)^m}{a+b x^n} \, dx}{b^4}+\frac {\left (d^2 (3 b B c+A b d-a B d)\right ) \int x^{2 n} (e x)^m \, dx}{b^2}+\frac {\left (d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right )\right ) \int x^n (e x)^m \, dx}{b^3}\\ &=-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^{1+m}}{b^4 e (1+m)}+\frac {(A b-a B) (b c-a d)^3 (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^4 e (1+m)}+\frac {\left (B d^3 x^{-m} (e x)^m\right ) \int x^{m+3 n} \, dx}{b}+\frac {\left (d^2 (3 b B c+A b d-a B d) x^{-m} (e x)^m\right ) \int x^{m+2 n} \, dx}{b^2}+\frac {\left (d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{b^3}\\ &=\frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) x^{1+n} (e x)^m}{b^3 (1+m+n)}+\frac {d^2 (3 b B c+A b d-a B d) x^{1+2 n} (e x)^m}{b^2 (1+m+2 n)}+\frac {B d^3 x^{1+3 n} (e x)^m}{b (1+m+3 n)}-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^{1+m}}{b^4 e (1+m)}+\frac {(A b-a B) (b c-a d)^3 (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^4 e (1+m)}\\ \end {align*}
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Mathematica [A]
time = 0.59, size = 212, normalized size = 0.79 \begin {gather*} x (e x)^m \left (-\frac {(-A b+a B) (-b c+a d)^3}{a b^4 (1+m)}+\frac {A c^3}{a+a m}+\frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) x^n}{b^3 (1+m+n)}+\frac {d^2 (3 b B c+A b d-a B d) x^{2 n}}{b^2 (1+m+2 n)}+\frac {B d^3 x^{3 n}}{b+b m+3 b n}+\frac {(-A b+a B) (-b c+a d)^3 \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a b^4 (1+m)}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((e*x)^m*(A + B*x^n)*(c + d*x^n)^3)/(a + b*x^n),x]
[Out]
x*(e*x)^m*(-(((-(A*b) + a*B)*(-(b*c) + a*d)^3)/(a*b^4*(1 + m))) + (A*c^3)/(a + a*m) + (d*(a^2*B*d^2 + 3*b^2*c*
(B*c + A*d) - a*b*d*(3*B*c + A*d))*x^n)/(b^3*(1 + m + n)) + (d^2*(3*b*B*c + A*b*d - a*B*d)*x^(2*n))/(b^2*(1 +
m + 2*n)) + (B*d^3*x^(3*n))/(b + b*m + 3*b*n) + ((-(A*b) + a*B)*(-(b*c) + a*d)^3*Hypergeometric2F1[1, (1 + m)/
n, (1 + m + n)/n, -((b*x^n)/a)])/(a*b^4*(1 + m)))
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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (A +B \,x^{n}\right ) \left (c +d \,x^{n}\right )^{3}}{a +b \,x^{n}}\, dx\]
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),x)
[Out]
int((e*x)^m*(A+B*x^n)*(c+d*x^n)^3/(a+b*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)*(c+d*x^n)^3/(a+b*x^n),x, algorithm="maxima")
[Out]
((b^4*c^3*e^m - 3*a*b^3*c^2*d*e^m + 3*a^2*b^2*c*d^2*e^m - a^3*b*d^3*e^m)*A - (a*b^3*c^3*e^m - 3*a^2*b^2*c^2*d*
e^m + 3*a^3*b*c*d^2*e^m - a^4*d^3*e^m)*B)*integrate(x^m/(b^5*x^n + a*b^4), 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*b^3*d^3*x*e^(m*log(x) + 3*n*log(x)) + ((3*(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)*b^3*c^2*d - 3*(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*b^2*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^2*b*d^3)*A + ((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)*b^3*c^3 - 3*(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*b^2*c^2*d + 3*(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^2*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^3*d^3)*B)*x*x^m + ((m^3*e^m + m^2*(4*n + 3)*e^
m + (3*n^2 + 8*n + 3)*m*e^m + (3*n^2 + 4*n + 1)*e^m)*A*b^3*d^3 + (3*(m^3*e^m + m^2*(4*n + 3)*e^m + (3*n^2 + 8*
n + 3)*m*e^m + (3*n^2 + 4*n + 1)*e^m)*b^3*c*d^2 - (m^3*e^m + m^2*(4*n + 3)*e^m + (3*n^2 + 8*n + 3)*m*e^m + (3*
n^2 + 4*n + 1)*e^m)*a*b^2*d^3)*B)*x*e^(m*log(x) + 2*n*log(x)) + ((3*(m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10
*n + 3)*m*e^m + (6*n^2 + 5*n + 1)*e^m)*b^3*c*d^2 - (m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (
6*n^2 + 5*n + 1)*e^m)*a*b^2*d^3)*A + (3*(m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (6*n^2 + 5*n
+ 1)*e^m)*b^3*c^2*d - 3*(m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (6*n^2 + 5*n + 1)*e^m)*a*b^
2*c*d^2 + (m^3*e^m + m^2*(5*n + 3)*e^m + (6*n^2 + 10*n + 3)*m*e^m + (6*n^2 + 5*n + 1)*e^m)*a^2*b*d^3)*B)*x*e^(
m*log(x) + n*log(x)))/((m^4 + 2*m^3*(3*n + 2) + (11*n^2 + 18*n + 6)*m^2 + 6*n^3 + 2*(3*n^3 + 11*n^2 + 9*n + 2)
*m + 11*n^2 + 6*n + 1)*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),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*x^n + a), x)
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Sympy [C] Result contains complex when optimal does not.
time = 23.38, size = 1503, normalized size = 5.57 \begin {gather*} \text {Too large to display} \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),x)
[Out]
A*c**3*e**m*m*x*x**m*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*gamma(m/n + 1/n)/(a*n**2*gamma(m/n + 1 +
1/n)) + A*c**3*e**m*x*x**m*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1/n)*gamma(m/n + 1/n)/(a*n**2*gamma(m/
n + 1 + 1/n)) + 3*A*c**2*d*e**m*m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n +
1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + 3*A*c**2*d*e**m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/
n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n*gamma(m/n + 2 + 1/n)) + 3*A*c**2*d*e**m*x*x**m*x**n*lerchphi(b*x**n*exp
_polar(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + 3*A*c*d**2*e**m*m*x*x**
m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1
/n)) + 6*A*c*d**2*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/
n)/(a*n*gamma(m/n + 3 + 1/n)) + 3*A*c*d**2*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2
+ 1/n)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1/n)) + A*d**3*e**m*m*x*x**m*x**(3*n)*lerchphi(b*x**n*exp_
polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n**2*gamma(m/n + 4 + 1/n)) + 3*A*d**3*e**m*x*x**m*x**
(3*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n*gamma(m/n + 4 + 1/n)) + A
*d**3*e**m*x*x**m*x**(3*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n**2*g
amma(m/n + 4 + 1/n)) + B*c**3*e**m*m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/
n + 1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + B*c**3*e**m*x*x**m*x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n
+ 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n*gamma(m/n + 2 + 1/n)) + B*c**3*e**m*x*x**m*x**n*lerchphi(b*x**n*exp_pola
r(I*pi)/a, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(a*n**2*gamma(m/n + 2 + 1/n)) + 3*B*c**2*d*e**m*m*x*x**m*x**
(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1/n))
+ 6*B*c**2*d*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(a
*n*gamma(m/n + 3 + 1/n)) + 3*B*c**2*d*e**m*x*x**m*x**(2*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 2 + 1/n
)*gamma(m/n + 2 + 1/n)/(a*n**2*gamma(m/n + 3 + 1/n)) + 3*B*c*d**2*e**m*m*x*x**m*x**(3*n)*lerchphi(b*x**n*exp_p
olar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n**2*gamma(m/n + 4 + 1/n)) + 9*B*c*d**2*e**m*x*x**m*x*
*(3*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n*gamma(m/n + 4 + 1/n)) +
3*B*c*d**2*e**m*x*x**m*x**(3*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(a*n
**2*gamma(m/n + 4 + 1/n)) + B*d**3*e**m*m*x*x**m*x**(4*n)*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 4 + 1/n)
*gamma(m/n + 4 + 1/n)/(a*n**2*gamma(m/n + 5 + 1/n)) + 4*B*d**3*e**m*x*x**m*x**(4*n)*lerchphi(b*x**n*exp_polar(
I*pi)/a, 1, m/n + 4 + 1/n)*gamma(m/n + 4 + 1/n)/(a*n*gamma(m/n + 5 + 1/n)) + B*d**3*e**m*x*x**m*x**(4*n)*lerch
phi(b*x**n*exp_polar(I*pi)/a, 1, m/n + 4 + 1/n)*gamma(m/n + 4 + 1/n)/(a*n**2*gamma(m/n + 5 + 1/n))
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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),x, algorithm="giac")
[Out]
integrate((B*x^n + A)*(d*x^n + c)^3*(x*e)^m/(b*x^n + a), 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}{a+b\,x^n} \,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),x)
[Out]
int(((e*x)^m*(A + B*x^n)*(c + d*x^n)^3)/(a + b*x^n), x)
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