Integrand size = 21, antiderivative size = 86 \[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{8 e (1+m)}+\frac {a (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{8 e^2 (2+m)} \] Output:
1/8*(e*x)^(1+m)*hypergeom([3, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/e/(1+m)+1/8* a*(e*x)^(2+m)*hypergeom([3, 1+1/2*m],[2+1/2*m],a^2*x^2)/e^2/(2+m)
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\frac {x (e x)^m \left (a (1+m) x \operatorname {Hypergeometric2F1}\left (3,1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )+(2+m) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )\right )}{8 (1+m) (2+m)} \] Input:
Integrate[(e*x)^m/((2 - 2*a*x)^3*(1 + a*x)^2),x]
Output:
(x*(e*x)^m*(a*(1 + m)*x*Hypergeometric2F1[3, 1 + m/2, 2 + m/2, a^2*x^2] + (2 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, a^2*x^2]))/(8*(1 + m)*( 2 + m))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {92, 27, 82, 278}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(e x)^m}{(2-2 a x)^3 (a x+1)^2} \, dx\) |
| \(\Big \downarrow \) 92 |
| \(\displaystyle \int \frac {(e x)^m}{8 (1-a x)^3 (a x+1)^3}dx+\frac {a \int \frac {(e x)^{m+1}}{8 (1-a x)^3 (a x+1)^3}dx}{e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \frac {(e x)^m}{(1-a x)^3 (a x+1)^3}dx+\frac {a \int \frac {(e x)^{m+1}}{(1-a x)^3 (a x+1)^3}dx}{8 e}\) |
| \(\Big \downarrow \) 82 |
| \(\displaystyle \frac {1}{8} \int \frac {(e x)^m}{\left (1-a^2 x^2\right )^3}dx+\frac {a \int \frac {(e x)^{m+1}}{\left (1-a^2 x^2\right )^3}dx}{8 e}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {a (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (3,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{8 e^2 (m+2)}+\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{8 e (m+1)}\) |
Input:
Int[(e*x)^m/((2 - 2*a*x)^3*(1 + a*x)^2),x]
Output:
((e*x)^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, a^2*x^2])/(8*e*( 1 + m)) + (a*(e*x)^(2 + m)*Hypergeometric2F1[3, (2 + m)/2, (4 + m)/2, a^2* x^2])/(8*e^2*(2 + m))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Simp[a Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f In t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] && !RationalQ[p] && !IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
\[\int \frac {\left (e x \right )^{m}}{\left (-2 a x +2\right )^{3} \left (a x +1\right )^{2}}d x\]
Input:
int((e*x)^m/(-2*a*x+2)^3/(a*x+1)^2,x)
Output:
int((e*x)^m/(-2*a*x+2)^3/(a*x+1)^2,x)
\[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\int { -\frac {\left (e x\right )^{m}}{8 \, {\left (a x + 1\right )}^{2} {\left (a x - 1\right )}^{3}} \,d x } \] Input:
integrate((e*x)^m/(-2*a*x+2)^3/(a*x+1)^2,x, algorithm="fricas")
Output:
integral(-1/8*(e*x)^m/(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1 ), x)
Result contains complex when optimal does not.
Time = 2.23 (sec) , antiderivative size = 1972, normalized size of antiderivative = 22.93 \[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\text {Too large to display} \] Input:
integrate((e*x)**m/(-2*a*x+2)**3/(a*x+1)**2,x)
Output:
-2*a**3*e**m*m**3*x**3*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma( -m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128*a**2*x* gamma(1 - m) + 128*a*gamma(1 - m)) + 6*a**3*e**m*m**2*x**3*x**m*lerchphi(1 /(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128* a**3*x**2*gamma(1 - m) - 128*a**2*x*gamma(1 - m) + 128*a*gamma(1 - m)) - 2 *a**3*e**m*m**2*x**3*x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*exp_polar(I *pi))*gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128*a**2*x*gamma(1 - m) + 128*a*gamma(1 - m)) - 2*a**3*e**m*m**2*x**3*x** m*gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128 *a**2*x*gamma(1 - m) + 128*a*gamma(1 - m)) - 3*a**3*e**m*m*x**3*x**m*lerch phi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128*a**2*x*gamma(1 - m) + 128*a*gamma(1 - m) ) + 3*a**3*e**m*m*x**3*x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*exp_polar (I*pi))*gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128*a**2*x*gamma(1 - m) + 128*a*gamma(1 - m)) + 4*a**3*e**m*m*x**3*x**m *gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128* a**2*x*gamma(1 - m) + 128*a*gamma(1 - m)) + 2*a**2*e**m*m**3*x**2*x**m*ler chphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(128*a**4*x**3*gamma(1 - m) - 128*a**3*x**2*gamma(1 - m) - 128*a**2*x*gamma(1 - m) + 128*a*gamma(1 - m)) - 6*a**2*e**m*m**2*x**2*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi)...
\[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\int { -\frac {\left (e x\right )^{m}}{8 \, {\left (a x + 1\right )}^{2} {\left (a x - 1\right )}^{3}} \,d x } \] Input:
integrate((e*x)^m/(-2*a*x+2)^3/(a*x+1)^2,x, algorithm="maxima")
Output:
-1/8*integrate((e*x)^m/((a*x + 1)^2*(a*x - 1)^3), x)
\[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\int { -\frac {\left (e x\right )^{m}}{8 \, {\left (a x + 1\right )}^{2} {\left (a x - 1\right )}^{3}} \,d x } \] Input:
integrate((e*x)^m/(-2*a*x+2)^3/(a*x+1)^2,x, algorithm="giac")
Output:
integrate(-1/8*(e*x)^m/((a*x + 1)^2*(a*x - 1)^3), x)
Timed out. \[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=\int -\frac {{\left (e\,x\right )}^m}{{\left (a\,x+1\right )}^2\,{\left (2\,a\,x-2\right )}^3} \,d x \] Input:
int(-(e*x)^m/((a*x + 1)^2*(2*a*x - 2)^3),x)
Output:
int(-(e*x)^m/((a*x + 1)^2*(2*a*x - 2)^3), x)
\[ \int \frac {(e x)^m}{(2-2 a x)^3 (1+a x)^2} \, dx=-\frac {e^{m} \left (\int \frac {x^{m}}{a^{5} x^{5}-a^{4} x^{4}-2 a^{3} x^{3}+2 a^{2} x^{2}+a x -1}d x \right )}{8} \] Input:
int((e*x)^m/(-2*a*x+2)^3/(a*x+1)^2,x)
Output:
( - e**m*int(x**m/(a**5*x**5 - a**4*x**4 - 2*a**3*x**3 + 2*a**2*x**2 + a*x - 1),x))/8