\(\int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx\) [121]

Optimal result

Integrand size = 27, antiderivative size = 140 \[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=-\frac {(d e-c f) (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f) (f g-e h) (1+m)}+\frac {(d g-c h) (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {h (a+b x)}{b g-a h}\right )}{(b g-a h) (f g-e h) (1+m)} \]

[Out]

-(-c*f+d*e)*(b*x+a)^(1+m)*hypergeom([1, 1+m],[2+m],-f*(b*x+a)/(-a*f+b*e))/(-a*f+b*e)/(-e*h+f*g)/(1+m)+(-c*h+d*
g)*(b*x+a)^(1+m)*hypergeom([1, 1+m],[2+m],-h*(b*x+a)/(-a*h+b*g))/(-a*h+b*g)/(-e*h+f*g)/(1+m)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {162, 70} \[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\frac {(a+b x)^{m+1} (d g-c h) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (f g-e h)}-\frac {(a+b x)^{m+1} (d e-c f) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (f g-e h)} \]

[In]

Int[((a + b*x)^m*(c + d*x))/((e + f*x)*(g + h*x)),x]

[Out]

-(((d*e - c*f)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((f*(a + b*x))/(b*e - a*f))])/((b*e - a*f
)*(f*g - e*h)*(1 + m))) + ((d*g - c*h)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((h*(a + b*x))/(b
*g - a*h))])/((b*g - a*h)*(f*g - e*h)*(1 + m))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(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]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\frac {(a+b x)^{1+m} \left (-\frac {(d e-c f) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {f (a+b x)}{-b e+a f}\right )}{b e-a f}+\frac {(d g-c h) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {h (a+b x)}{-b g+a h}\right )}{b g-a h}\right )}{(f g-e h) (1+m)} \]

[In]

Integrate[((a + b*x)^m*(c + d*x))/((e + f*x)*(g + h*x)),x]

[Out]

((a + b*x)^(1 + m)*(-(((d*e - c*f)*Hypergeometric2F1[1, 1 + m, 2 + m, (f*(a + b*x))/(-(b*e) + a*f)])/(b*e - a*
f)) + ((d*g - c*h)*Hypergeometric2F1[1, 1 + m, 2 + m, (h*(a + b*x))/(-(b*g) + a*h)])/(b*g - a*h)))/((f*g - e*h
)*(1 + m))

Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )}{\left (f x +e \right ) \left (h x +g \right )}d x\]

[In]

int((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x)

[Out]

int((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x)

Fricas [F]

\[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\int { \frac {{\left (d x + c\right )} {\left (b x + a\right )}^{m}}{{\left (f x + e\right )} {\left (h x + g\right )}} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="fricas")

[Out]

integral((d*x + c)*(b*x + a)^m/(f*h*x^2 + e*g + (f*g + e*h)*x), x)

Sympy [F]

\[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\int \frac {\left (a + b x\right )^{m} \left (c + d x\right )}{\left (e + f x\right ) \left (g + h x\right )}\, dx \]

[In]

integrate((b*x+a)**m*(d*x+c)/(f*x+e)/(h*x+g),x)

[Out]

Integral((a + b*x)**m*(c + d*x)/((e + f*x)*(g + h*x)), x)

Maxima [F]

\[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\int { \frac {{\left (d x + c\right )} {\left (b x + a\right )}^{m}}{{\left (f x + e\right )} {\left (h x + g\right )}} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="maxima")

[Out]

integrate((d*x + c)*(b*x + a)^m/((f*x + e)*(h*x + g)), x)

Giac [F]

\[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\int { \frac {{\left (d x + c\right )} {\left (b x + a\right )}^{m}}{{\left (f x + e\right )} {\left (h x + g\right )}} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="giac")

[Out]

integrate((d*x + c)*(b*x + a)^m/((f*x + e)*(h*x + g)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx=\int \frac {{\left (a+b\,x\right )}^m\,\left (c+d\,x\right )}{\left (e+f\,x\right )\,\left (g+h\,x\right )} \,d x \]

[In]

int(((a + b*x)^m*(c + d*x))/((e + f*x)*(g + h*x)),x)

[Out]

int(((a + b*x)^m*(c + d*x))/((e + f*x)*(g + h*x)), x)