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

Optimal. Leaf size=134 \[ \frac{(a+b x)^{m+1} (d g-c h) (f g-e h) \, _2F_1\left (1,m+1;m+2;-\frac{h (a+b x)}{b g-a h}\right )}{h^2 (m+1) (b g-a h)}-\frac{(a+b x)^{m+1} (a d f h+b (m+2) (-c f h-d e h+d f g)-b d f h (m+1) x)}{b^2 h^2 (m+1) (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.0860064, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {147, 68} \[ \frac{(a+b x)^{m+1} (d g-c h) (f g-e h) \, _2F_1\left (1,m+1;m+2;-\frac{h (a+b x)}{b g-a h}\right )}{h^2 (m+1) (b g-a h)}-\frac{(a+b x)^{m+1} (a d f h+b (m+2) (-c f h-d e h+d f g)-b d f h (m+1) x)}{b^2 h^2 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.190052, size = 120, normalized size = 0.9 \[ \frac{(a+b x)^{m+1} \left (\frac{b (c f h+d e h-d f g)-a d f h}{b^2 (m+1)}+\frac{d f h (a+b x)}{b^2 (m+2)}+\frac{(d g-c h) (f g-e h) \, _2F_1\left (1,m+1;m+2;\frac{h (a+b x)}{a h-b g}\right )}{(m+1) (b g-a h)}\right )}{h^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) \left ( dx+c \right ) \left ( bx+a \right ) ^{m}}{hx+g}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}{\left (f x + e\right )}{\left (b x + a\right )}^{m}}{h x + g}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d f x^{2} + c e +{\left (d e + c f\right )} x\right )}{\left (b x + a\right )}^{m}}{h x + g}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{m} \left (c + d x\right ) \left (e + f x\right )}{g + h x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}{\left (f x + e\right )}{\left (b x + a\right )}^{m}}{h x + g}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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