Integrand size = 35, antiderivative size = 71 \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=\frac {(a+b x)^{2-n} (c+d x)^{-2+n} \operatorname {Hypergeometric2F1}\left (4,2-n,3-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (b c-a d) (2-n)} \] Output:
(b*x+a)^(2-n)*(d*x+c)^(-2+n)*hypergeom([4, 2-n],[3-n],-d*(b*x+a)/b/(d*x+c) )/b^4/(-a*d+b*c)/(2-n)
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=-\frac {(a+b x)^{2-n} (c+d x)^{-2+n} \operatorname {Hypergeometric2F1}\left (4,2-n,3-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (b c-a d) (-2+n)} \] Input:
Integrate[((a + b*x)^(1 - n)*(c + d*x)^(1 + n))/(b*c + a*d + 2*b*d*x)^4,x]
Output:
-(((a + b*x)^(2 - n)*(c + d*x)^(-2 + n)*Hypergeometric2F1[4, 2 - n, 3 - n, -((d*(a + b*x))/(b*(c + d*x)))])/(b^4*(b*c - a*d)*(-2 + n)))
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {141}
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.
| \(\displaystyle \int \frac {(a+b x)^{1-n} (c+d x)^{n+1}}{(a d+b c+2 b d x)^4} \, dx\) |
| \(\Big \downarrow \) 141 |
| \(\displaystyle \frac {(a+b x)^{2-n} (c+d x)^{n-2} \operatorname {Hypergeometric2F1}\left (4,2-n,3-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (2-n) (b c-a d)}\) |
Input:
Int[((a + b*x)^(1 - n)*(c + d*x)^(1 + n))/(b*c + a*d + 2*b*d*x)^4,x]
Output:
((a + b*x)^(2 - n)*(c + d*x)^(-2 + n)*Hypergeometric2F1[4, 2 - n, 3 - n, - ((d*(a + b*x))/(b*(c + d*x)))])/(b^4*(b*c - a*d)*(2 - n))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f ))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] || !Su mSimplerQ[p, 1]) && !ILtQ[m, 0]
\[\int \frac {\left (b x +a \right )^{-n +1} \left (x d +c \right )^{1+n}}{\left (2 b d x +a d +b c \right )^{4}}d x\]
Input:
int((b*x+a)^(-n+1)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^4,x)
Output:
int((b*x+a)^(-n+1)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^4,x)
\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{4}} \,d x } \] Input:
integrate((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^4,x, algorithm="fr icas")
Output:
integral((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(16*b^4*d^4*x^4 + b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 + a^4*d^4 + 32*(b^4*c*d^ 3 + a*b^3*d^4)*x^3 + 24*(b^4*c^2*d^2 + 2*a*b^3*c*d^3 + a^2*b^2*d^4)*x^2 + 8*(b^4*c^3*d + 3*a*b^3*c^2*d^2 + 3*a^2*b^2*c*d^3 + a^3*b*d^4)*x), x)
Timed out. \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=\text {Timed out} \] Input:
integrate((b*x+a)**(1-n)*(d*x+c)**(1+n)/(2*b*d*x+a*d+b*c)**4,x)
Output:
Timed out
\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{4}} \,d x } \] Input:
integrate((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^4,x, algorithm="ma xima")
Output:
integrate((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(2*b*d*x + b*c + a*d)^4, x)
\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{4}} \,d x } \] Input:
integrate((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^4,x, algorithm="gi ac")
Output:
integrate((b*x + a)^(-n + 1)*(d*x + c)^(n + 1)/(2*b*d*x + b*c + a*d)^4, x)
Timed out. \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx=\int \frac {{\left (a+b\,x\right )}^{1-n}\,{\left (c+d\,x\right )}^{n+1}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^4} \,d x \] Input:
int(((a + b*x)^(1 - n)*(c + d*x)^(n + 1))/(a*d + b*c + 2*b*d*x)^4,x)
Output:
int(((a + b*x)^(1 - n)*(c + d*x)^(n + 1))/(a*d + b*c + 2*b*d*x)^4, x)
\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^4} \, dx =\text {Too large to display} \] Input:
int((b*x+a)^(1-n)*(d*x+c)^(1+n)/(2*b*d*x+a*d+b*c)^4,x)
Output:
int((c + d*x)**n/((a + b*x)**n*a**4*d**4 + 4*(a + b*x)**n*a**3*b*c*d**3 + 8*(a + b*x)**n*a**3*b*d**4*x + 6*(a + b*x)**n*a**2*b**2*c**2*d**2 + 24*(a + b*x)**n*a**2*b**2*c*d**3*x + 24*(a + b*x)**n*a**2*b**2*d**4*x**2 + 4*(a + b*x)**n*a*b**3*c**3*d + 24*(a + b*x)**n*a*b**3*c**2*d**2*x + 48*(a + b*x )**n*a*b**3*c*d**3*x**2 + 32*(a + b*x)**n*a*b**3*d**4*x**3 + (a + b*x)**n* b**4*c**4 + 8*(a + b*x)**n*b**4*c**3*d*x + 24*(a + b*x)**n*b**4*c**2*d**2* x**2 + 32*(a + b*x)**n*b**4*c*d**3*x**3 + 16*(a + b*x)**n*b**4*d**4*x**4), x)*a*c + int(((c + d*x)**n*x**2)/((a + b*x)**n*a**4*d**4 + 4*(a + b*x)**n* a**3*b*c*d**3 + 8*(a + b*x)**n*a**3*b*d**4*x + 6*(a + b*x)**n*a**2*b**2*c* *2*d**2 + 24*(a + b*x)**n*a**2*b**2*c*d**3*x + 24*(a + b*x)**n*a**2*b**2*d **4*x**2 + 4*(a + b*x)**n*a*b**3*c**3*d + 24*(a + b*x)**n*a*b**3*c**2*d**2 *x + 48*(a + b*x)**n*a*b**3*c*d**3*x**2 + 32*(a + b*x)**n*a*b**3*d**4*x**3 + (a + b*x)**n*b**4*c**4 + 8*(a + b*x)**n*b**4*c**3*d*x + 24*(a + b*x)**n *b**4*c**2*d**2*x**2 + 32*(a + b*x)**n*b**4*c*d**3*x**3 + 16*(a + b*x)**n* b**4*d**4*x**4),x)*b*d + int(((c + d*x)**n*x)/((a + b*x)**n*a**4*d**4 + 4* (a + b*x)**n*a**3*b*c*d**3 + 8*(a + b*x)**n*a**3*b*d**4*x + 6*(a + b*x)**n *a**2*b**2*c**2*d**2 + 24*(a + b*x)**n*a**2*b**2*c*d**3*x + 24*(a + b*x)** n*a**2*b**2*d**4*x**2 + 4*(a + b*x)**n*a*b**3*c**3*d + 24*(a + b*x)**n*a*b **3*c**2*d**2*x + 48*(a + b*x)**n*a*b**3*c*d**3*x**2 + 32*(a + b*x)**n*a*b **3*d**4*x**3 + (a + b*x)**n*b**4*c**4 + 8*(a + b*x)**n*b**4*c**3*d*x +...