Integrand size = 60, antiderivative size = 88 \[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\frac {b d (2+m+n) (a+b x)^{1+m} (c+d x)^{1+n} \left (\frac {f (a d (1+m)+b c (1+n))}{b d (2+m+n)}+f x\right )^{-2-m-n}}{(b c-a d)^2 f (1+m) (1+n)} \]
b*d*(2+m+n)*(b*x+a)^(1+m)*(d*x+c)^(1+n)*(f*(a*d*(1+m)+b*c*(1+n))/b/d/(2+m+ n)+f*x)^(-2-m-n)/(-a*d+b*c)^2/f/(1+m)/(1+n)
Time = 0.54 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\frac {b^3 d^3 (2+m+n)^3 (a+b x)^{1+m} (c+d x)^{1+n} \left (\frac {f (a d (1+m)+b c (1+n)+b d (2+m+n) x)}{b d (2+m+n)}\right )^{-m-n}}{(b c-a d)^2 f^3 (1+m) (1+n) (a d (1+m)+b c (1+n)+b d (2+m+n) x)^2} \]
Integrate[(a + b*x)^m*(c + d*x)^n*((b*c*f + a*d*f + a*d*f*m + b*c*f*n)/(b* d*(2 + m + n)) + f*x)^(-3 - m - n),x]
(b^3*d^3*(2 + m + n)^3*(a + b*x)^(1 + m)*(c + d*x)^(1 + n)*((f*(a*d*(1 + m ) + b*c*(1 + n) + b*d*(2 + m + n)*x))/(b*d*(2 + m + n)))^(-m - n))/((b*c - a*d)^2*f^3*(1 + m)*(1 + n)*(a*d*(1 + m) + b*c*(1 + n) + b*d*(2 + m + n)*x )^2)
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {106}
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.
\(\displaystyle \int (a+b x)^m (c+d x)^n \left (\frac {a d f m+a d f+b c f n+b c f}{b d (m+n+2)}+f x\right )^{-m-n-3} \, dx\) |
\(\Big \downarrow \) 106 |
\(\displaystyle \frac {b d (m+n+2) (a+b x)^{m+1} (c+d x)^{n+1} \left (\frac {f (a d (m+1)+b c (n+1))}{b d (m+n+2)}+f x\right )^{-m-n-2}}{f (m+1) (n+1) (b c-a d)^2}\) |
Int[(a + b*x)^m*(c + d*x)^n*((b*c*f + a*d*f + a*d*f*m + b*c*f*n)/(b*d*(2 + m + n)) + f*x)^(-3 - m - n),x]
(b*d*(2 + m + n)*(a + b*x)^(1 + m)*(c + d*x)^(1 + n)*((f*(a*d*(1 + m) + b* c*(1 + n)))/(b*d*(2 + m + n)) + f*x)^(-2 - m - n))/((b*c - a*d)^2*f*(1 + m )*(1 + n))
3.32.52.3.1 Defintions of rubi rules used
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] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(197\) vs. \(2(88)=176\).
Time = 195.68 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.25
method | result | size |
gosper | \(\frac {\left (b x +a \right )^{1+m} \left (d x +c \right )^{1+n} \left (b d x m +b d x n +a d m +b c n +2 b d x +a d +b c \right ) \left (\frac {f \left (b d x m +b d x n +a d m +b c n +2 b d x +a d +b c \right )}{b d \left (2+m +n \right )}\right )^{-3-m -n}}{a^{2} d^{2} m n -2 a b c d m n +b^{2} c^{2} m n +a^{2} d^{2} m +a^{2} d^{2} n -2 a b c d m -2 a b c d n +b^{2} c^{2} m +b^{2} c^{2} n +a^{2} d^{2}-2 a b c d +b^{2} c^{2}}\) | \(198\) |
parallelrisch | \(\text {Expression too large to display}\) | \(3494\) |
int((b*x+a)^m*(d*x+c)^n*((a*d*f*m+b*c*f*n+a*d*f+b*c*f)/b/d/(2+m+n)+f*x)^(- 3-m-n),x,method=_RETURNVERBOSE)
(b*x+a)^(1+m)*(d*x+c)^(1+n)*(b*d*m*x+b*d*n*x+a*d*m+b*c*n+2*b*d*x+a*d+b*c)* (f*(b*d*m*x+b*d*n*x+a*d*m+b*c*n+2*b*d*x+a*d+b*c)/b/d/(2+m+n))^(-3-m-n)/(a^ 2*d^2*m*n-2*a*b*c*d*m*n+b^2*c^2*m*n+a^2*d^2*m+a^2*d^2*n-2*a*b*c*d*m-2*a*b* c*d*n+b^2*c^2*m+b^2*c^2*n+a^2*d^2-2*a*b*c*d+b^2*c^2)
Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (88) = 176\).
Time = 0.32 (sec) , antiderivative size = 332, normalized size of antiderivative = 3.77 \[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\frac {{\left (a^{2} c d m + a b c^{2} n + a b c^{2} + a^{2} c d + {\left (b^{2} d^{2} m + b^{2} d^{2} n + 2 \, b^{2} d^{2}\right )} x^{3} + {\left (3 \, b^{2} c d + 3 \, a b d^{2} + {\left (b^{2} c d + 2 \, a b d^{2}\right )} m + {\left (2 \, b^{2} c d + a b d^{2}\right )} n\right )} x^{2} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2} + {\left (2 \, a b c d + a^{2} d^{2}\right )} m + {\left (b^{2} c^{2} + 2 \, a b c d\right )} n\right )} x\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} \left (\frac {a d f m + b c f n + {\left (b c + a d\right )} f + {\left (b d f m + b d f n + 2 \, b d f\right )} x}{b d m + b d n + 2 \, b d}\right )^{-m - n - 3}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m\right )} n} \]
integrate((b*x+a)^m*(d*x+c)^n*((a*d*f*m+b*c*f*n+a*d*f+b*c*f)/b/d/(2+m+n)+f *x)^(-3-m-n),x, algorithm="fricas")
(a^2*c*d*m + a*b*c^2*n + a*b*c^2 + a^2*c*d + (b^2*d^2*m + b^2*d^2*n + 2*b^ 2*d^2)*x^3 + (3*b^2*c*d + 3*a*b*d^2 + (b^2*c*d + 2*a*b*d^2)*m + (2*b^2*c*d + a*b*d^2)*n)*x^2 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2 + (2*a*b*c*d + a^2*d^2 )*m + (b^2*c^2 + 2*a*b*c*d)*n)*x)*(b*x + a)^m*(d*x + c)^n*((a*d*f*m + b*c* f*n + (b*c + a*d)*f + (b*d*f*m + b*d*f*n + 2*b*d*f)*x)/(b*d*m + b*d*n + 2* b*d))^(-m - n - 3)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*m + (b^2*c^2 - 2*a*b*c*d + a^2*d^2 + (b^2*c^2 - 2*a*b*c*d + a^2* d^2)*m)*n)
\[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\int \left (\frac {f \left (a d m + a d + b c n + b c + b d m x + b d n x + 2 b d x\right )}{b d m + b d n + 2 b d}\right )^{- m - n - 3} \left (a + b x\right )^{m} \left (c + d x\right )^{n}\, dx \]
Integral((f*(a*d*m + a*d + b*c*n + b*c + b*d*m*x + b*d*n*x + 2*b*d*x)/(b*d *m + b*d*n + 2*b*d))**(-m - n - 3)*(a + b*x)**m*(c + d*x)**n, x)
Leaf count of result is larger than twice the leaf count of optimal. 1022 vs. \(2 (88) = 176\).
Time = 1.41 (sec) , antiderivative size = 1022, normalized size of antiderivative = 11.61 \[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\text {Too large to display} \]
integrate((b*x+a)^m*(d*x+c)^n*((a*d*f*m+b*c*f*n+a*d*f+b*c*f)/b/d/(2+m+n)+f *x)^(-3-m-n),x, algorithm="maxima")
((m^3 + 3*m^2*(n + 2) + n^3 + 3*(n^2 + 4*n + 4)*m + 6*n^2 + 12*n + 8)*a*b^ (m + n + 3)*c*d^(m + n + 3)*(m + n + 2)^(m + n) + (m^3 + 3*m^2*(n + 2) + n ^3 + 3*(n^2 + 4*n + 4)*m + 6*n^2 + 12*n + 8)*b^(m + n + 4)*d^(m + n + 4)*( m + n + 2)^(m + n)*x^2 + ((m^3 + 3*m^2*(n + 2) + n^3 + 3*(n^2 + 4*n + 4)*m + 6*n^2 + 12*n + 8)*a*b^(m + n + 3)*d^(m + n + 4) + (m^3 + 3*m^2*(n + 2) + n^3 + 3*(n^2 + 4*n + 4)*m + 6*n^2 + 12*n + 8)*b^(m + n + 4)*c*d^(m + n + 3))*(m + n + 2)^(m + n)*x)*e^(-m*log(a*d*m + b*c*n + b*c + a*d + (b*d*m + b*d*n + 2*b*d)*x) - n*log(a*d*m + b*c*n + b*c + a*d + (b*d*m + b*d*n + 2* b*d)*x) + m*log(b*x + a) + n*log(d*x + c))/((n^3 + (n^3 + 3*n^2 + 3*n + 1) *m + 3*n^2 + 3*n + 1)*b^4*c^4*f^(m + n + 3) + 2*((n^2 + 2*n + 1)*m^2 - n^3 - (n^3 + n^2 - n - 1)*m - 2*n^2 - n)*a*b^3*c^3*d*f^(m + n + 3) + (m^3*(n + 1) - (4*n^2 + 5*n + 1)*m^2 + n^3 + (n^3 - 5*n^2 - 10*n - 4)*m - n^2 - 4* n - 2)*a^2*b^2*c^2*d^2*f^(m + n + 3) - 2*(m^3*(n + 1) - (n^2 - n - 2)*m^2 - (2*n^2 + n - 1)*m - n^2 - n)*a^3*b*c*d^3*f^(m + n + 3) + (m^3*(n + 1) + 3*m^2*(n + 1) + 3*m*(n + 1) + n + 1)*a^4*d^4*f^(m + n + 3) + ((m^3*(n + 1) + (2*n^2 + 7*n + 5)*m^2 + n^3 + (n^3 + 7*n^2 + 14*n + 8)*m + 5*n^2 + 8*n + 4)*b^4*c^2*d^2*f^(m + n + 3) - 2*(m^3*(n + 1) + (2*n^2 + 7*n + 5)*m^2 + n^3 + (n^3 + 7*n^2 + 14*n + 8)*m + 5*n^2 + 8*n + 4)*a*b^3*c*d^3*f^(m + n + 3) + (m^3*(n + 1) + (2*n^2 + 7*n + 5)*m^2 + n^3 + (n^3 + 7*n^2 + 14*n + 8 )*m + 5*n^2 + 8*n + 4)*a^2*b^2*d^4*f^(m + n + 3))*x^2 + 2*(((n^2 + 2*n ...
\[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + \frac {a d f m + b c f n + b c f + a d f}{b d {\left (m + n + 2\right )}}\right )}^{-m - n - 3} \,d x } \]
integrate((b*x+a)^m*(d*x+c)^n*((a*d*f*m+b*c*f*n+a*d*f+b*c*f)/b/d/(2+m+n)+f *x)^(-3-m-n),x, algorithm="giac")
integrate((b*x + a)^m*(d*x + c)^n*(f*x + (a*d*f*m + b*c*f*n + b*c*f + a*d* f)/(b*d*(m + n + 2)))^(-m - n - 3), x)
Time = 10.06 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.40 \[ \int (a+b x)^m (c+d x)^n \left (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x\right )^{-3-m-n} \, dx=\frac {\frac {x\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (a^2\,d^2+b^2\,c^2+a^2\,d^2\,m+b^2\,c^2\,n+4\,a\,b\,c\,d+2\,a\,b\,c\,d\,m+2\,a\,b\,c\,d\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}+\frac {a\,c\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (a\,d+b\,c+a\,d\,m+b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}+\frac {b\,d\,x^2\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (3\,a\,d+3\,b\,c+2\,a\,d\,m+b\,c\,m+a\,d\,n+2\,b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}+\frac {b^2\,d^2\,x^3\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (m+n+2\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}}{{\left (f\,x+\frac {a\,d\,f+b\,c\,f+a\,d\,f\,m+b\,c\,f\,n}{b\,d\,\left (m+n+2\right )}\right )}^{m+n+3}} \]
int(((a + b*x)^m*(c + d*x)^n)/(f*x + (a*d*f + b*c*f + a*d*f*m + b*c*f*n)/( b*d*(m + n + 2)))^(m + n + 3),x)
((x*(a + b*x)^m*(c + d*x)^n*(a^2*d^2 + b^2*c^2 + a^2*d^2*m + b^2*c^2*n + 4 *a*b*c*d + 2*a*b*c*d*m + 2*a*b*c*d*n))/((a*d - b*c)^2*(m + 1)*(n + 1)) + ( a*c*(a + b*x)^m*(c + d*x)^n*(a*d + b*c + a*d*m + b*c*n))/((a*d - b*c)^2*(m + 1)*(n + 1)) + (b*d*x^2*(a + b*x)^m*(c + d*x)^n*(3*a*d + 3*b*c + 2*a*d*m + b*c*m + a*d*n + 2*b*c*n))/((a*d - b*c)^2*(m + 1)*(n + 1)) + (b^2*d^2*x^ 3*(a + b*x)^m*(c + d*x)^n*(m + n + 2))/((a*d - b*c)^2*(m + 1)*(n + 1)))/(f *x + (a*d*f + b*c*f + a*d*f*m + b*c*f*n)/(b*d*(m + n + 2)))^(m + n + 3)