Integrand size = 24, antiderivative size = 98 \[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )}{a^5 d^3 e (1+m)}+\frac {b (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (3,\frac {2+m}{2},\frac {4+m}{2},\frac {b^2 x^2}{a^2}\right )}{a^6 d^3 e^2 (2+m)} \]
(e*x)^(1+m)*hypergeom([3, 1/2+1/2*m],[3/2+1/2*m],b^2*x^2/a^2)/a^5/d^3/e/(1 +m)+b*(e*x)^(2+m)*hypergeom([3, 1+1/2*m],[2+1/2*m],b^2*x^2/a^2)/a^6/d^3/e^ 2/(2+m)
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89 \[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\frac {x (e x)^m \left (b (1+m) x \operatorname {Hypergeometric2F1}\left (3,1+\frac {m}{2},2+\frac {m}{2},\frac {b^2 x^2}{a^2}\right )+a (2+m) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},\frac {b^2 x^2}{a^2}\right )\right )}{a^6 d^3 (1+m) (2+m)} \]
(x*(e*x)^m*(b*(1 + m)*x*Hypergeometric2F1[3, 1 + m/2, 2 + m/2, (b^2*x^2)/a ^2] + a*(2 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2]) )/(a^6*d^3*(1 + m)*(2 + m))
Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, 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.
\(\displaystyle \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx\) |
\(\Big \downarrow \) 92 |
\(\displaystyle a \int \frac {(e x)^m}{d^3 (a-b x)^3 (a+b x)^3}dx+\frac {b \int \frac {(e x)^{m+1}}{d^3 (a-b x)^3 (a+b x)^3}dx}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \int \frac {(e x)^m}{(a-b x)^3 (a+b x)^3}dx}{d^3}+\frac {b \int \frac {(e x)^{m+1}}{(a-b x)^3 (a+b x)^3}dx}{d^3 e}\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \frac {a \int \frac {(e x)^m}{\left (a^2-b^2 x^2\right )^3}dx}{d^3}+\frac {b \int \frac {(e x)^{m+1}}{\left (a^2-b^2 x^2\right )^3}dx}{d^3 e}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {b (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (3,\frac {m+2}{2},\frac {m+4}{2},\frac {b^2 x^2}{a^2}\right )}{a^6 d^3 e^2 (m+2)}+\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (3,\frac {m+1}{2},\frac {m+3}{2},\frac {b^2 x^2}{a^2}\right )}{a^5 d^3 e (m+1)}\) |
((e*x)^(1 + m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, (b^2*x^2)/a^2])/ (a^5*d^3*e*(1 + m)) + (b*(e*x)^(2 + m)*Hypergeometric2F1[3, (2 + m)/2, (4 + m)/2, (b^2*x^2)/a^2])/(a^6*d^3*e^2*(2 + m))
3.1.66.3.1 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])
\[\int \frac {\left (e x \right )^{m}}{\left (b x +a \right )^{2} \left (-b d x +a d \right )^{3}}d x\]
\[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\int { -\frac {\left (e x\right )^{m}}{{\left (b d x - a d\right )}^{3} {\left (b x + a\right )}^{2}} \,d x } \]
integral(-(e*x)^m/(b^5*d^3*x^5 - a*b^4*d^3*x^4 - 2*a^2*b^3*d^3*x^3 + 2*a^3 *b^2*d^3*x^2 + a^4*b*d^3*x - a^5*d^3), x)
Result contains complex when optimal does not.
Time = 2.95 (sec) , antiderivative size = 2717, normalized size of antiderivative = 27.72 \[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\text {Too large to display} \]
-2*a**3*e**m*m**3*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/( 16*a**7*b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5*b **3*d**3*x**2*gamma(1 - m) + 16*a**4*b**4*d**3*x**3*gamma(1 - m)) + 6*a**3 *e**m*m**2*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a**7 *b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5*b**3*d** 3*x**2*gamma(1 - m) + 16*a**4*b**4*d**3*x**3*gamma(1 - m)) - 2*a**3*e**m*m **2*x**m*lerchphi(a*exp_polar(I*pi)/(b*x), 1, m*exp_polar(I*pi))*gamma(-m) /(16*a**7*b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5 *b**3*d**3*x**2*gamma(1 - m) + 16*a**4*b**4*d**3*x**3*gamma(1 - m)) - 3*a* *3*e**m*m*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a**7* b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5*b**3*d**3 *x**2*gamma(1 - m) + 16*a**4*b**4*d**3*x**3*gamma(1 - m)) + 3*a**3*e**m*m* x**m*lerchphi(a*exp_polar(I*pi)/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16 *a**7*b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5*b** 3*d**3*x**2*gamma(1 - m) + 16*a**4*b**4*d**3*x**3*gamma(1 - m)) + 2*a**2*b *e**m*m**3*x*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a* *7*b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5*b**3*d **3*x**2*gamma(1 - m) + 16*a**4*b**4*d**3*x**3*gamma(1 - m)) - 6*a**2*b*e* *m*m**2*x*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(16*a**7* b*d**3*gamma(1 - m) - 16*a**6*b**2*d**3*x*gamma(1 - m) - 16*a**5*b**3*d...
\[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\int { -\frac {\left (e x\right )^{m}}{{\left (b d x - a d\right )}^{3} {\left (b x + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\int { -\frac {\left (e x\right )^{m}}{{\left (b d x - a d\right )}^{3} {\left (b x + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m}{(a+b x)^2 (a d-b d x)^3} \, dx=\int \frac {{\left (e\,x\right )}^m}{{\left (a\,d-b\,d\,x\right )}^3\,{\left (a+b\,x\right )}^2} \,d x \]