Integrand size = 18, antiderivative size = 206 \[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=-\frac {d x^{1+m}}{2 c (b c-a d) (c+d x)^2}+\frac {d (a d (1-m)-b c (3-m)) x^{1+m}}{2 c^2 (b c-a d)^2 (c+d x)}+\frac {b^3 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{a (b c-a d)^3 (1+m)}+\frac {d \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (2-3 m+m^2\right )\right ) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{2 c^3 (b c-a d)^3 (1+m)} \] Output:
-1/2*d*x^(1+m)/c/(-a*d+b*c)/(d*x+c)^2+1/2*d*(a*d*(1-m)-b*c*(3-m))*x^(1+m)/ c^2/(-a*d+b*c)^2/(d*x+c)+b^3*x^(1+m)*hypergeom([1, 1+m],[2+m],-b*x/a)/a/(- a*d+b*c)^3/(1+m)+1/2*d*(a^2*d^2*(1-m)*m-2*a*b*c*d*(2-m)*m-b^2*c^2*(m^2-3*m +2))*x^(1+m)*hypergeom([1, 1+m],[2+m],-d*x/c)/c^3/(-a*d+b*c)^3/(1+m)
Time = 0.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.82 \[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\frac {x^{1+m} \left (\frac {d}{(c+d x)^2}+\frac {d (-b c (-3+m)+a d (-1+m))}{c (b c-a d) (c+d x)}+\frac {-2 b^3 c^3 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )+a d \left (-2 a b c d (-2+m) m+a^2 d^2 (-1+m) m+b^2 c^2 \left (2-3 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d x}{c}\right )}{a c^2 (b c-a d)^2 (1+m)}\right )}{2 c (-b c+a d)} \] Input:
Integrate[x^m/((a + b*x)*(c + d*x)^3),x]
Output:
(x^(1 + m)*(d/(c + d*x)^2 + (d*(-(b*c*(-3 + m)) + a*d*(-1 + m)))/(c*(b*c - a*d)*(c + d*x)) + (-2*b^3*c^3*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/ a)] + a*d*(-2*a*b*c*d*(-2 + m)*m + a^2*d^2*(-1 + m)*m + b^2*c^2*(2 - 3*m + m^2))*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*x)/c)])/(a*c^2*(b*c - a*d)^ 2*(1 + m))))/(2*c*(-(b*c) + a*d))
Time = 0.40 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {114, 25, 168, 25, 174, 74}
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.
\(\displaystyle \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx\) |
\(\Big \downarrow \) 114 |
\(\displaystyle -\frac {\int -\frac {x^m (2 b c-a d (1-m)-b d (1-m) x)}{(a+b x) (c+d x)^2}dx}{2 c (b c-a d)}-\frac {d x^{m+1}}{2 c (c+d x)^2 (b c-a d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {x^m (2 b c-a d (1-m)-b d (1-m) x)}{(a+b x) (c+d x)^2}dx}{2 c (b c-a d)}-\frac {d x^{m+1}}{2 c (c+d x)^2 (b c-a d)}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {\frac {d x^{m+1} (a d (1-m)-b c (3-m))}{c (c+d x) (b c-a d)}-\frac {\int -\frac {x^m \left (2 b^2 c^2+a b d (3-m) m c-a^2 d^2 (1-m) m-b d (a d (1-m)-b c (3-m)) m x\right )}{(a+b x) (c+d x)}dx}{c (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1}}{2 c (c+d x)^2 (b c-a d)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {x^m \left (2 b^2 c^2+a b d (3-m) m c-a^2 d^2 (1-m) m-b d (a d (1-m)-b c (3-m)) m x\right )}{(a+b x) (c+d x)}dx}{c (b c-a d)}+\frac {d x^{m+1} (a d (1-m)-b c (3-m))}{c (c+d x) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1}}{2 c (c+d x)^2 (b c-a d)}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {\frac {\frac {d \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (m^2-3 m+2\right )\right ) \int \frac {x^m}{c+d x}dx}{b c-a d}+\frac {2 b^3 c^2 \int \frac {x^m}{a+b x}dx}{b c-a d}}{c (b c-a d)}+\frac {d x^{m+1} (a d (1-m)-b c (3-m))}{c (c+d x) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1}}{2 c (c+d x)^2 (b c-a d)}\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {\frac {\frac {d x^{m+1} \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (m^2-3 m+2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d x}{c}\right )}{c (m+1) (b c-a d)}+\frac {2 b^3 c^2 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b x}{a}\right )}{a (m+1) (b c-a d)}}{c (b c-a d)}+\frac {d x^{m+1} (a d (1-m)-b c (3-m))}{c (c+d x) (b c-a d)}}{2 c (b c-a d)}-\frac {d x^{m+1}}{2 c (c+d x)^2 (b c-a d)}\) |
Input:
Int[x^m/((a + b*x)*(c + d*x)^3),x]
Output:
-1/2*(d*x^(1 + m))/(c*(b*c - a*d)*(c + d*x)^2) + ((d*(a*d*(1 - m) - b*c*(3 - m))*x^(1 + m))/(c*(b*c - a*d)*(c + d*x)) + ((2*b^3*c^2*x^(1 + m)*Hyperg eometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a*(b*c - a*d)*(1 + m)) + (d*(a^ 2*d^2*(1 - m)*m - 2*a*b*c*d*(2 - m)*m - b^2*c^2*(2 - 3*m + m^2))*x^(1 + m) *Hypergeometric2F1[1, 1 + m, 2 + m, -((d*x)/c)])/(c*(b*c - a*d)*(1 + m)))/ (c*(b*c - a*d)))/(2*c*(b*c - a*d))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
\[\int \frac {x^{m}}{\left (b x +a \right ) \left (x d +c \right )^{3}}d x\]
Input:
int(x^m/(b*x+a)/(d*x+c)^3,x)
Output:
int(x^m/(b*x+a)/(d*x+c)^3,x)
\[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\int { \frac {x^{m}}{{\left (b x + a\right )} {\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate(x^m/(b*x+a)/(d*x+c)^3,x, algorithm="fricas")
Output:
integral(x^m/(b*d^3*x^4 + a*c^3 + (3*b*c*d^2 + a*d^3)*x^3 + 3*(b*c^2*d + a *c*d^2)*x^2 + (b*c^3 + 3*a*c^2*d)*x), x)
Result contains complex when optimal does not.
Time = 9.91 (sec) , antiderivative size = 9442, normalized size of antiderivative = 45.83 \[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\text {Too large to display} \] Input:
integrate(x**m/(b*x+a)/(d*x+c)**3,x)
Output:
a**2*c**2*d**2*m**3*x**m*lerchphi(c*exp_polar(I*pi)/(d*x), 1, m*exp_polar( I*pi))*gamma(-m)/(2*a**3*c**4*d**3*gamma(1 - m) + 4*a**3*c**3*d**4*x*gamma (1 - m) + 2*a**3*c**2*d**5*x**2*gamma(1 - m) - 6*a**2*b*c**5*d**2*gamma(1 - m) - 12*a**2*b*c**4*d**3*x*gamma(1 - m) - 6*a**2*b*c**3*d**4*x**2*gamma( 1 - m) + 6*a*b**2*c**6*d*gamma(1 - m) + 12*a*b**2*c**5*d**2*x*gamma(1 - m) + 6*a*b**2*c**4*d**3*x**2*gamma(1 - m) - 2*b**3*c**7*gamma(1 - m) - 4*b** 3*c**6*d*x*gamma(1 - m) - 2*b**3*c**5*d**2*x**2*gamma(1 - m)) - a**2*c**2* d**2*m**2*x**m*lerchphi(c*exp_polar(I*pi)/(d*x), 1, m*exp_polar(I*pi))*gam ma(-m)/(2*a**3*c**4*d**3*gamma(1 - m) + 4*a**3*c**3*d**4*x*gamma(1 - m) + 2*a**3*c**2*d**5*x**2*gamma(1 - m) - 6*a**2*b*c**5*d**2*gamma(1 - m) - 12* a**2*b*c**4*d**3*x*gamma(1 - m) - 6*a**2*b*c**3*d**4*x**2*gamma(1 - m) + 6 *a*b**2*c**6*d*gamma(1 - m) + 12*a*b**2*c**5*d**2*x*gamma(1 - m) + 6*a*b** 2*c**4*d**3*x**2*gamma(1 - m) - 2*b**3*c**7*gamma(1 - m) - 4*b**3*c**6*d*x *gamma(1 - m) - 2*b**3*c**5*d**2*x**2*gamma(1 - m)) + 2*a**2*c*d**3*m**3*x *x**m*lerchphi(c*exp_polar(I*pi)/(d*x), 1, m*exp_polar(I*pi))*gamma(-m)/(2 *a**3*c**4*d**3*gamma(1 - m) + 4*a**3*c**3*d**4*x*gamma(1 - m) + 2*a**3*c* *2*d**5*x**2*gamma(1 - m) - 6*a**2*b*c**5*d**2*gamma(1 - m) - 12*a**2*b*c* *4*d**3*x*gamma(1 - m) - 6*a**2*b*c**3*d**4*x**2*gamma(1 - m) + 6*a*b**2*c **6*d*gamma(1 - m) + 12*a*b**2*c**5*d**2*x*gamma(1 - m) + 6*a*b**2*c**4*d* *3*x**2*gamma(1 - m) - 2*b**3*c**7*gamma(1 - m) - 4*b**3*c**6*d*x*gamma...
\[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\int { \frac {x^{m}}{{\left (b x + a\right )} {\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate(x^m/(b*x+a)/(d*x+c)^3,x, algorithm="maxima")
Output:
integrate(x^m/((b*x + a)*(d*x + c)^3), x)
\[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\int { \frac {x^{m}}{{\left (b x + a\right )} {\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate(x^m/(b*x+a)/(d*x+c)^3,x, algorithm="giac")
Output:
integrate(x^m/((b*x + a)*(d*x + c)^3), x)
Timed out. \[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\int \frac {x^m}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^3} \,d x \] Input:
int(x^m/((a + b*x)*(c + d*x)^3),x)
Output:
int(x^m/((a + b*x)*(c + d*x)^3), x)
\[ \int \frac {x^m}{(a+b x) (c+d x)^3} \, dx=\int \frac {x^{m}}{b \,d^{3} x^{4}+a \,d^{3} x^{3}+3 b c \,d^{2} x^{3}+3 a c \,d^{2} x^{2}+3 b \,c^{2} d \,x^{2}+3 a \,c^{2} d x +b \,c^{3} x +a \,c^{3}}d x \] Input:
int(x^m/(b*x+a)/(d*x+c)^3,x)
Output:
int(x**m/(a*c**3 + 3*a*c**2*d*x + 3*a*c*d**2*x**2 + a*d**3*x**3 + b*c**3*x + 3*b*c**2*d*x**2 + 3*b*c*d**2*x**3 + b*d**3*x**4),x)