3.3152 \(\int (a+b x)^m (c+d x)^n (\frac {b c f+a d f+a d f m+b c f n}{b d (2+m+n)}+f x)^{-3-m-n} \, dx\)

Optimal. Leaf size=88 \[ \frac {b d (m+n+2) (a+b x)^{m+1} (c+d x)^{n+1} \left (\frac {f (a d (m+1)+b c (n+1))}{b d (m+n+2)}+f x\right )^{-m-n-2}}{f (m+1) (n+1) (b c-a d)^2} \]

[Out]

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)

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {95} \[ \frac {b d (m+n+2) (a+b x)^{m+1} (c+d x)^{n+1} \left (\frac {f (a d (m+1)+b c (n+1))}{b d (m+n+2)}+f x\right )^{-m-n-2}}{f (m+1) (n+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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]

[Out]

(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))

Rule 95

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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]

Rubi steps

\begin {align*} \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)}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 120, normalized size = 1.36 \[ \frac {b^3 d^3 (m+n+2)^3 (a+b x)^{m+1} (c+d x)^{n+1} \left (\frac {f (a d (m+1)+b c (n+1)+b d x (m+n+2))}{b d (m+n+2)}\right )^{-m-n}}{f^3 (m+1) (n+1) (b c-a d)^2 (a d (m+1)+b c (n+1)+b d x (m+n+2))^2} \]

Antiderivative was successfully verified.

[In]

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
]

[Out]

(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)

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fricas [B]  time = 1.07, size = 332, normalized size = 3.77 \[ \frac {{\left (a^{2} c d m + a b c^{2} n + a b c^{2} + a^{2} c d + {\left (b^{2} d^{2} m + b^{2} d^{2} n + 2 \, b^{2} d^{2}\right )} x^{3} + {\left (3 \, b^{2} c d + 3 \, a b d^{2} + {\left (b^{2} c d + 2 \, a b d^{2}\right )} m + {\left (2 \, b^{2} c d + a b d^{2}\right )} n\right )} x^{2} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2} + {\left (2 \, a b c d + a^{2} d^{2}\right )} m + {\left (b^{2} c^{2} + 2 \, a b c d\right )} n\right )} x\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} \left (\frac {a d f m + b c f n + {\left (b c + a d\right )} f + {\left (b d f m + b d f n + 2 \, b d f\right )} x}{b d m + b d n + 2 \, b d}\right )^{-m - n - 3}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m\right )} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + \frac {a d f m + b c f n + b c f + a d f}{b d {\left (m + n + 2\right )}}\right )}^{-m - n - 3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

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maple [B]  time = 0.01, size = 198, normalized size = 2.25 \[ \frac {\left (b d x m +b d x n +a d m +b c n +2 b d x +a d +b c \right ) \left (\frac {\left (b d x m +b d x n +a d m +b c n +2 b d x +a d +b c \right ) f}{\left (m +n +2\right ) b d}\right )^{-m -n -3} \left (b x +a \right )^{m +1} \left (d x +c \right )^{n +1}}{a^{2} d^{2} m n -2 a b c d m n +b^{2} c^{2} m n +a^{2} d^{2} m +a^{2} d^{2} n -2 a b c d m -2 a b c d n +b^{2} c^{2} m +b^{2} c^{2} n +a^{2} d^{2}-2 a b c d +b^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

(b*x+a)^(m+1)*(d*x+c)^(n+1)*(b*d*m*x+b*d*n*x+a*d*m+b*c*n+2*b*d*x+a*d+b*c)/(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)*(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)

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maxima [B]  time = 8.39, size = 1022, normalized size = 11.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

((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 + 1)*m^2 + n^3 + (n^3 + 5*n^2 + 7*n + 3)*m + 4*n^2 + 5*n + 2)*b
^4*c^3*d*f^(m + n + 3) + (m^3*(n + 1) - (n^2 - n - 2)*m^2 - 2*n^3 - (2*n^3 + 8*n^2 + 7*n + 1)*m - 7*n^2 - 7*n
- 2)*a*b^3*c^2*d^2*f^(m + n + 3) - (2*m^3*(n + 1) + (n^2 + 8*n + 7)*m^2 - n^3 - (n^3 + n^2 - 7*n - 7)*m - 2*n^
2 + n + 2)*a^2*b^2*c*d^3*f^(m + n + 3) + (m^3*(n + 1) + (n^2 + 5*n + 4)*m^2 + (2*n^2 + 7*n + 5)*m + n^2 + 3*n
+ 2)*a^3*b*d^4*f^(m + n + 3))*x)

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mupad [B]  time = 9.26, size = 299, normalized size = 3.40 \[ \frac {\frac {x\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (a^2\,d^2+b^2\,c^2+a^2\,d^2\,m+b^2\,c^2\,n+4\,a\,b\,c\,d+2\,a\,b\,c\,d\,m+2\,a\,b\,c\,d\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}+\frac {a\,c\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (a\,d+b\,c+a\,d\,m+b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}+\frac {b\,d\,x^2\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (3\,a\,d+3\,b\,c+2\,a\,d\,m+b\,c\,m+a\,d\,n+2\,b\,c\,n\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}+\frac {b^2\,d^2\,x^3\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n\,\left (m+n+2\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m+1\right )\,\left (n+1\right )}}{{\left (f\,x+\frac {a\,d\,f+b\,c\,f+a\,d\,f\,m+b\,c\,f\,n}{b\,d\,\left (m+n+2\right )}\right )}^{m+n+3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

((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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

Timed out

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