Integrand size = 21, antiderivative size = 86 \[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{16 e (1+m)}+\frac {a (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (4,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{16 e^2 (2+m)} \]
1/16*(e*x)^(1+m)*hypergeom([4, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/e/(1+m)+1/1 6*a*(e*x)^(2+m)*hypergeom([4, 1+1/2*m],[2+1/2*m],a^2*x^2)/e^2/(2+m)
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\frac {x (e x)^m \left (a (1+m) x \operatorname {Hypergeometric2F1}\left (4,1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )+(2+m) \operatorname {Hypergeometric2F1}\left (4,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )\right )}{16 (1+m) (2+m)} \]
(x*(e*x)^m*(a*(1 + m)*x*Hypergeometric2F1[4, 1 + m/2, 2 + m/2, a^2*x^2] + (2 + m)*Hypergeometric2F1[4, (1 + m)/2, (3 + m)/2, a^2*x^2]))/(16*(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.
\(\displaystyle \int \frac {(e x)^m}{(2-2 a x)^4 (a x+1)^3} \, dx\) |
\(\Big \downarrow \) 92 |
\(\displaystyle \int \frac {(e x)^m}{16 (1-a x)^4 (a x+1)^4}dx+\frac {a \int \frac {(e x)^{m+1}}{16 (1-a x)^4 (a x+1)^4}dx}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} \int \frac {(e x)^m}{(1-a x)^4 (a x+1)^4}dx+\frac {a \int \frac {(e x)^{m+1}}{(1-a x)^4 (a x+1)^4}dx}{16 e}\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \frac {1}{16} \int \frac {(e x)^m}{\left (1-a^2 x^2\right )^4}dx+\frac {a \int \frac {(e x)^{m+1}}{\left (1-a^2 x^2\right )^4}dx}{16 e}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {a (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (4,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{16 e^2 (m+2)}+\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (4,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{16 e (m+1)}\) |
((e*x)^(1 + m)*Hypergeometric2F1[4, (1 + m)/2, (3 + m)/2, a^2*x^2])/(16*e* (1 + m)) + (a*(e*x)^(2 + m)*Hypergeometric2F1[4, (2 + m)/2, (4 + m)/2, a^2 *x^2])/(16*e^2*(2 + m))
3.1.61.3.1 Defintions of rubi rules used
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])
\[\int \frac {\left (e x \right )^{m}}{\left (-2 a x +2\right )^{4} \left (a x +1\right )^{3}}d x\]
\[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\int { \frac {\left (e x\right )^{m}}{16 \, {\left (a x + 1\right )}^{3} {\left (a x - 1\right )}^{4}} \,d x } \]
integral(1/16*(e*x)^m/(a^7*x^7 - a^6*x^6 - 3*a^5*x^5 + 3*a^4*x^4 + 3*a^3*x ^3 - 3*a^2*x^2 - a*x + 1), x)
Result contains complex when optimal does not.
Time = 4.39 (sec) , antiderivative size = 5872, normalized size of antiderivative = 68.28 \[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\text {Too large to display} \]
2*a**5*e**m*m**4*x**5*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(- m)/(1536*a**6*x**5*gamma(1 - m) - 1536*a**5*x**4*gamma(1 - m) - 3072*a**4* x**3*gamma(1 - m) + 3072*a**3*x**2*gamma(1 - m) + 1536*a**2*x*gamma(1 - m) - 1536*a*gamma(1 - m)) - 15*a**5*e**m*m**3*x**5*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(1536*a**6*x**5*gamma(1 - m) - 1536*a**5*x** 4*gamma(1 - m) - 3072*a**4*x**3*gamma(1 - m) + 3072*a**3*x**2*gamma(1 - m) + 1536*a**2*x*gamma(1 - m) - 1536*a*gamma(1 - m)) + 3*a**5*e**m*m**3*x**5 *x**m*lerchphi(exp_polar(I*pi)/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(153 6*a**6*x**5*gamma(1 - m) - 1536*a**5*x**4*gamma(1 - m) - 3072*a**4*x**3*ga mma(1 - m) + 3072*a**3*x**2*gamma(1 - m) + 1536*a**2*x*gamma(1 - m) - 1536 *a*gamma(1 - m)) + 2*a**5*e**m*m**3*x**5*x**m*gamma(-m)/(1536*a**6*x**5*ga mma(1 - m) - 1536*a**5*x**4*gamma(1 - m) - 3072*a**4*x**3*gamma(1 - m) + 3 072*a**3*x**2*gamma(1 - m) + 1536*a**2*x*gamma(1 - m) - 1536*a*gamma(1 - m )) + 31*a**5*e**m*m**2*x**5*x**m*lerchphi(1/(a*x), 1, m*exp_polar(I*pi))*g amma(-m)/(1536*a**6*x**5*gamma(1 - m) - 1536*a**5*x**4*gamma(1 - m) - 3072 *a**4*x**3*gamma(1 - m) + 3072*a**3*x**2*gamma(1 - m) + 1536*a**2*x*gamma( 1 - m) - 1536*a*gamma(1 - m)) - 15*a**5*e**m*m**2*x**5*x**m*lerchphi(exp_p olar(I*pi)/(a*x), 1, m*exp_polar(I*pi))*gamma(-m)/(1536*a**6*x**5*gamma(1 - m) - 1536*a**5*x**4*gamma(1 - m) - 3072*a**4*x**3*gamma(1 - m) + 3072*a* *3*x**2*gamma(1 - m) + 1536*a**2*x*gamma(1 - m) - 1536*a*gamma(1 - m)) ...
\[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\int { \frac {\left (e x\right )^{m}}{16 \, {\left (a x + 1\right )}^{3} {\left (a x - 1\right )}^{4}} \,d x } \]
\[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\int { \frac {\left (e x\right )^{m}}{16 \, {\left (a x + 1\right )}^{3} {\left (a x - 1\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m}{(2-2 a x)^4 (1+a x)^3} \, dx=\int \frac {{\left (e\,x\right )}^m}{{\left (a\,x+1\right )}^3\,{\left (2\,a\,x-2\right )}^4} \,d x \]