3.22 \(\int \frac{(e x)^m (a+b x^n)^3 (A+B x^n)}{c+d x^n} \, dx\)
Optimal. Leaf size=272 \[ \frac{(e x)^{m+1} \left (-3 a^2 b d^2 (B c-A d)+a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e (m+1)}+\frac{b x^{n+1} (e x)^m \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 (m+n+1)}-\frac{b^2 x^{2 n+1} (e x)^m (-3 a B d-A b d+b B c)}{d^2 (m+2 n+1)}+\frac{(e x)^{m+1} (b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c d^4 e (m+1)}+\frac{b^3 B x^{3 n+1} (e x)^m}{d (m+3 n+1)} \]
[Out]
(b*(3*a^2*B*d^2 + b^2*c*(B*c - A*d) - 3*a*b*d*(B*c - A*d))*x^(1 + n)*(e*x)^m)/(d^3*(1 + m + n)) - (b^2*(b*B*c
- A*b*d - 3*a*B*d)*x^(1 + 2*n)*(e*x)^m)/(d^2*(1 + m + 2*n)) + (b^3*B*x^(1 + 3*n)*(e*x)^m)/(d*(1 + m + 3*n)) +
((a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 3*a*b^2*c*d*(B*c - A*d) - 3*a^2*b*d^2*(B*c - A*d))*(e*x)^(1 + m))/(d^4*e*(
1 + m)) + ((b*c - a*d)^3*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)
])/(c*d^4*e*(1 + m))
________________________________________________________________________________________
Rubi [A] time = 0.397714, antiderivative size = 272, normalized size of antiderivative = 1.,
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 =
{570, 20, 30, 364} \[ \frac{(e x)^{m+1} \left (-3 a^2 b d^2 (B c-A d)+a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e (m+1)}+\frac{b x^{n+1} (e x)^m \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 (m+n+1)}-\frac{b^2 x^{2 n+1} (e x)^m (-3 a B d-A b d+b B c)}{d^2 (m+2 n+1)}+\frac{(e x)^{m+1} (b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c d^4 e (m+1)}+\frac{b^3 B x^{3 n+1} (e x)^m}{d (m+3 n+1)} \]
Antiderivative was successfully verified.
[In]
Int[((e*x)^m*(a + b*x^n)^3*(A + B*x^n))/(c + d*x^n),x]
[Out]
(b*(3*a^2*B*d^2 + b^2*c*(B*c - A*d) - 3*a*b*d*(B*c - A*d))*x^(1 + n)*(e*x)^m)/(d^3*(1 + m + n)) - (b^2*(b*B*c
- A*b*d - 3*a*B*d)*x^(1 + 2*n)*(e*x)^m)/(d^2*(1 + m + 2*n)) + (b^3*B*x^(1 + 3*n)*(e*x)^m)/(d*(1 + m + 3*n)) +
((a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 3*a*b^2*c*d*(B*c - A*d) - 3*a^2*b*d^2*(B*c - A*d))*(e*x)^(1 + m))/(d^4*e*(
1 + m)) + ((b*c - a*d)^3*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)
])/(c*d^4*e*(1 + m))
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 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])
Rubi steps
\begin{align*} \int \frac{(e x)^m \left (a+b x^n\right )^3 \left (A+B x^n\right )}{c+d x^n} \, dx &=\int \left (\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^m}{d^4}+\frac{b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^n (e x)^m}{d^3}+\frac{b^2 (-b B c+A b d+3 a B d) x^{2 n} (e x)^m}{d^2}+\frac{b^3 B x^{3 n} (e x)^m}{d}+\frac{(-b c+a d)^3 (-B c+A d) (e x)^m}{d^4 \left (c+d x^n\right )}\right ) \, dx\\ &=\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac{\left (b^3 B\right ) \int x^{3 n} (e x)^m \, dx}{d}+\frac{\left ((b c-a d)^3 (B c-A d)\right ) \int \frac{(e x)^m}{c+d x^n} \, dx}{d^4}-\frac{\left (b^2 (b B c-A b d-3 a B d)\right ) \int x^{2 n} (e x)^m \, dx}{d^2}+\frac{\left (b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right )\right ) \int x^n (e x)^m \, dx}{d^3}\\ &=\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac{(b c-a d)^3 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{d x^n}{c}\right )}{c d^4 e (1+m)}+\frac{\left (b^3 B x^{-m} (e x)^m\right ) \int x^{m+3 n} \, dx}{d}-\frac{\left (b^2 (b B c-A b d-3 a B d) x^{-m} (e x)^m\right ) \int x^{m+2 n} \, dx}{d^2}+\frac{\left (b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^{-m} (e x)^m\right ) \int x^{m+n} \, dx}{d^3}\\ &=\frac{b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) x^{1+n} (e x)^m}{d^3 (1+m+n)}-\frac{b^2 (b B c-A b d-3 a B d) x^{1+2 n} (e x)^m}{d^2 (1+m+2 n)}+\frac{b^3 B x^{1+3 n} (e x)^m}{d (1+m+3 n)}+\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac{(b c-a d)^3 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{d x^n}{c}\right )}{c d^4 e (1+m)}\\ \end{align*}
Mathematica [A] time = 0.530397, size = 231, normalized size = 0.85 \[ \frac{x (e x)^m \left (\frac{3 a^2 b d^2 (A d-B c)+a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 c^2 (A d-B c)}{m+1}+\frac{b d x^n \left (3 a^2 B d^2+3 a b d (A d-B c)+b^2 c (B c-A d)\right )}{m+n+1}+\frac{b^2 d^2 x^{2 n} (3 a B d+A b d-b B c)}{m+2 n+1}+\frac{(b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c (m+1)}+\frac{b^3 B d^3 x^{3 n}}{m+3 n+1}\right )}{d^4} \]
Antiderivative was successfully verified.
[In]
Integrate[((e*x)^m*(a + b*x^n)^3*(A + B*x^n))/(c + d*x^n),x]
[Out]
(x*(e*x)^m*((a^3*B*d^3 + 3*a*b^2*c*d*(B*c - A*d) + b^3*c^2*(-(B*c) + A*d) + 3*a^2*b*d^2*(-(B*c) + A*d))/(1 + m
) + (b*d*(3*a^2*B*d^2 + b^2*c*(B*c - A*d) + 3*a*b*d*(-(B*c) + A*d))*x^n)/(1 + m + n) + (b^2*d^2*(-(b*B*c) + A*
b*d + 3*a*B*d)*x^(2*n))/(1 + m + 2*n) + (b^3*B*d^3*x^(3*n))/(1 + m + 3*n) + ((b*c - a*d)^3*(B*c - A*d)*Hyperge
ometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(1 + m))))/d^4
________________________________________________________________________________________
Maple [F] time = 0.493, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( a+b{x}^{n} \right ) ^{3} \left ( A+B{x}^{n} \right ) }{c+d{x}^{n}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x)
[Out]
int((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="maxima")
[Out]
-((b^3*c^3*d*e^m - 3*a*b^2*c^2*d^2*e^m + 3*a^2*b*c*d^3*e^m - a^3*d^4*e^m)*A - (b^3*c^4*e^m - 3*a*b^2*c^3*d*e^m
+ 3*a^2*b*c^2*d^2*e^m - a^3*c*d^3*e^m)*B)*integrate(x^m/(d^5*x^n + c*d^4), x) + ((m^3 + 3*m^2*(n + 1) + (2*n^
2 + 6*n + 3)*m + 2*n^2 + 3*n + 1)*B*b^3*d^3*e^m*x*e^(m*log(x) + 3*n*log(x)) + (((m^3 + 3*m^2*(2*n + 1) + 6*n^3
+ (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*b^3*c^2*d*e^m - 3*(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*b^2*c*d^2*e^m + 3*(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^2*b*d^3*e^m)*A - ((m^3 + 3*m^2*(2*n + 1) + 6*n^3 + (11*n^2 + 12*n + 3)*m + 11*n^2 + 6*n + 1)*b^
3*c^3*e^m - 3*(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*b^2*c^2*d*e^m + 3*(
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^2*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^3*d^3*e^m)*B)*x*x^m + ((m^3 + m^2*(4*n + 3) + (3*n^
2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*A*b^3*d^3*e^m - ((m^3 + m^2*(4*n + 3) + (3*n^2 + 8*n + 3)*m + 3*n^2 + 4*n +
1)*b^3*c*d^2*e^m - 3*(m^3 + m^2*(4*n + 3) + (3*n^2 + 8*n + 3)*m + 3*n^2 + 4*n + 1)*a*b^2*d^3*e^m)*B)*x*e^(m*lo
g(x) + 2*n*log(x)) - (((m^3 + m^2*(5*n + 3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*b^3*c*d^2*e^m - 3*(m^3 +
m^2*(5*n + 3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*a*b^2*d^3*e^m)*A - ((m^3 + m^2*(5*n + 3) + (6*n^2 + 1
0*n + 3)*m + 6*n^2 + 5*n + 1)*b^3*c^2*d*e^m - 3*(m^3 + m^2*(5*n + 3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)
*a*b^2*c*d^2*e^m + 3*(m^3 + m^2*(5*n + 3) + (6*n^2 + 10*n + 3)*m + 6*n^2 + 5*n + 1)*a^2*b*d^3*e^m)*B)*x*e^(m*l
og(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)*d^4)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b^{3} x^{4 \, n} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} x^{3 \, n} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} x^{2 \, n} +{\left (B a^{3} + 3 \, A a^{2} b\right )} x^{n}\right )} \left (e x\right )^{m}}{d x^{n} + c}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="fricas")
[Out]
integral((B*b^3*x^(4*n) + A*a^3 + (3*B*a*b^2 + A*b^3)*x^(3*n) + 3*(B*a^2*b + A*a*b^2)*x^(2*n) + (B*a^3 + 3*A*a
^2*b)*x^n)*(e*x)^m/(d*x^n + c), x)
________________________________________________________________________________________
Sympy [C] time = 56.4896, size = 1503, normalized size = 5.53 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)**m*(a+b*x**n)**3*(A+B*x**n)/(c+d*x**n),x)
[Out]
A*a**3*e**m*m*x*x**m*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(c*n**2*gamma(m/n + 1 +
1/n)) + A*a**3*e**m*x*x**m*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(c*n**2*gamma(m/
n + 1 + 1/n)) + 3*A*a**2*b*e**m*m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n +
1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + 3*A*a**2*b*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/
n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n*gamma(m/n + 2 + 1/n)) + 3*A*a**2*b*e**m*x*x**m*x**n*lerchphi(d*x**n*exp
_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + 3*A*a*b**2*e**m*m*x*x**
m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1
/n)) + 6*A*a*b**2*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/
n)/(c*n*gamma(m/n + 3 + 1/n)) + 3*A*a*b**2*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2
+ 1/n)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1/n)) + A*b**3*e**m*m*x*x**m*x**(3*n)*lerchphi(d*x**n*exp_
polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n**2*gamma(m/n + 4 + 1/n)) + 3*A*b**3*e**m*x*x**m*x**
(3*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n*gamma(m/n + 4 + 1/n)) + A
*b**3*e**m*x*x**m*x**(3*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n**2*g
amma(m/n + 4 + 1/n)) + B*a**3*e**m*m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/
n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + B*a**3*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n
+ 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n*gamma(m/n + 2 + 1/n)) + B*a**3*e**m*x*x**m*x**n*lerchphi(d*x**n*exp_pola
r(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(c*n**2*gamma(m/n + 2 + 1/n)) + 3*B*a**2*b*e**m*m*x*x**m*x**
(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1/n))
+ 6*B*a**2*b*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(c
*n*gamma(m/n + 3 + 1/n)) + 3*B*a**2*b*e**m*x*x**m*x**(2*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n
)*gamma(m/n + 2 + 1/n)/(c*n**2*gamma(m/n + 3 + 1/n)) + 3*B*a*b**2*e**m*m*x*x**m*x**(3*n)*lerchphi(d*x**n*exp_p
olar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n**2*gamma(m/n + 4 + 1/n)) + 9*B*a*b**2*e**m*x*x**m*x*
*(3*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n*gamma(m/n + 4 + 1/n)) +
3*B*a*b**2*e**m*x*x**m*x**(3*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 3 + 1/n)*gamma(m/n + 3 + 1/n)/(c*n
**2*gamma(m/n + 4 + 1/n)) + B*b**3*e**m*m*x*x**m*x**(4*n)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 4 + 1/n)
*gamma(m/n + 4 + 1/n)/(c*n**2*gamma(m/n + 5 + 1/n)) + 4*B*b**3*e**m*x*x**m*x**(4*n)*lerchphi(d*x**n*exp_polar(
I*pi)/c, 1, m/n + 4 + 1/n)*gamma(m/n + 4 + 1/n)/(c*n*gamma(m/n + 5 + 1/n)) + B*b**3*e**m*x*x**m*x**(4*n)*lerch
phi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 4 + 1/n)*gamma(m/n + 4 + 1/n)/(c*n**2*gamma(m/n + 5 + 1/n))
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{n} + A\right )}{\left (b x^{n} + a\right )}^{3} \left (e x\right )^{m}}{d x^{n} + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)^m*(a+b*x^n)^3*(A+B*x^n)/(c+d*x^n),x, algorithm="giac")
[Out]
integrate((B*x^n + A)*(b*x^n + a)^3*(e*x)^m/(d*x^n + c), x)