\(\int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx\) [3144]

Optimal result

Integrand size = 33, antiderivative size = 65 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (b c-a d) (1+m)} \]

[Out]

(b*x+a)^(1+m)*(d*x+c)^(-1-m)*hypergeom([4, 1+m],[2+m],-d*(b*x+a)/b/(d*x+c))/b^4/(-a*d+b*c)/(1+m)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {133} \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\frac {(a+b x)^{m+1} (c+d x)^{-m-1} \operatorname {Hypergeometric2F1}\left (4,m+1,m+2,-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (m+1) (b c-a d)} \]

[In]

Int[((a + b*x)^m*(c + d*x)^(2 - m))/(b*c + a*d + 2*b*d*x)^4,x]

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[4, 1 + m, 2 + m, -((d*(a + b*x))/(b*(c + d*x)))])/(b^4
*(b*c - a*d)*(1 + m))

Rule 133

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> 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] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (4,1+m;2+m;-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (b c-a d) (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-1-m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,-\frac {d (a+b x)}{b (c+d x)}\right )}{b^4 (b c-a d) (1+m)} \]

[In]

Integrate[((a + b*x)^m*(c + d*x)^(2 - m))/(b*c + a*d + 2*b*d*x)^4,x]

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-1 - m)*Hypergeometric2F1[4, 1 + m, 2 + m, -((d*(a + b*x))/(b*(c + d*x)))])/(b^4
*(b*c - a*d)*(1 + m))

Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (2 b d x +a d +b c \right )^{4}}d x\]

[In]

int((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^4,x)

[Out]

int((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^4,x)

Fricas [F]

\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{4}} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^4,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 2)/(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)

Sympy [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((b*x+a)**m*(d*x+c)**(2-m)/(2*b*d*x+a*d+b*c)**4,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{4}} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^4,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2)/(2*b*d*x + b*c + a*d)^4, x)

Giac [F]

\[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{{\left (2 \, b d x + b c + a d\right )}^{4}} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^4,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2)/(2*b*d*x + b*c + a*d)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^4} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^4} \,d x \]

[In]

int(((a + b*x)^m*(c + d*x)^(2 - m))/(a*d + b*c + 2*b*d*x)^4,x)

[Out]

int(((a + b*x)^m*(c + d*x)^(2 - m))/(a*d + b*c + 2*b*d*x)^4, x)