3.129 \(\int (a+b x)^m (c+d x)^{-3-m} (e+f x) (g+h x) \, dx\)

Optimal. Leaf size=246 \[ \frac{f h (a+b x)^{m+3} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m+3,m+3;m+4;-\frac{d (a+b x)}{b c-a d}\right )}{(m+3) (b c-a d)^3}-\frac{(a+b x)^{m+1} (c+d x)^{-m-2} \left (-b x \left (a^2 d f h (2 m+3)-a b (2 c f h (m+1)+d (m+2) (e h+f g))+b^2 (c (m+1) (e h+f g)+d e g)\right )+a^2 b c f h m+a^3 (-d) f h (m+1)+a b^2 (c (e h+f g)+d e g (m+1))-b^3 c e g (m+2)\right )}{b^2 (m+1) (m+2) (b c-a d)^2} \]

[Out]

-(((a + b*x)^(1 + m)*(c + d*x)^(-2 - m)*(a^2*b*c*f*h*m - a^3*d*f*h*(1 + m) - b^3*c*e*g*(2 + m) + a*b^2*(c*(f*g
+ e*h) + d*e*g*(1 + m)) - b*(a^2*d*f*h*(3 + 2*m) + b^2*(d*e*g + c*(f*g + e*h)*(1 + m)) - a*b*(2*c*f*h*(1 + m)
+ d*(f*g + e*h)*(2 + m)))*x))/(b^2*(b*c - a*d)^2*(1 + m)*(2 + m))) + (f*h*(a + b*x)^(3 + m)*((b*(c + d*x))/(b
*c - a*d))^m*Hypergeometric2F1[3 + m, 3 + m, 4 + m, -((d*(a + b*x))/(b*c - a*d))])/((b*c - a*d)^3*(3 + m)*(c +
d*x)^m)

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Rubi [A]  time = 0.159763, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {145, 70, 69} \[ \frac{f h (a+b x)^{m+3} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m+3,m+3;m+4;-\frac{d (a+b x)}{b c-a d}\right )}{(m+3) (b c-a d)^3}-\frac{(a+b x)^{m+1} (c+d x)^{-m-2} \left (-b x \left (a^2 d f h (2 m+3)-a b (2 c f h (m+1)+d (m+2) (e h+f g))+b^2 (c (m+1) (e h+f g)+d e g)\right )+a^2 b c f h m+a^3 (-d) f h (m+1)+a b^2 (c (e h+f g)+d e g (m+1))-b^3 c e g (m+2)\right )}{b^2 (m+1) (m+2) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-3 - m)*(e + f*x)*(g + h*x),x]

[Out]

-(((a + b*x)^(1 + m)*(c + d*x)^(-2 - m)*(a^2*b*c*f*h*m - a^3*d*f*h*(1 + m) - b^3*c*e*g*(2 + m) + a*b^2*(c*(f*g
+ e*h) + d*e*g*(1 + m)) - b*(a^2*d*f*h*(3 + 2*m) + b^2*(d*e*g + c*(f*g + e*h)*(1 + m)) - a*b*(2*c*f*h*(1 + m)
+ d*(f*g + e*h)*(2 + m)))*x))/(b^2*(b*c - a*d)^2*(1 + m)*(2 + m))) + (f*h*(a + b*x)^(3 + m)*((b*(c + d*x))/(b
*c - a*d))^m*Hypergeometric2F1[3 + m, 3 + m, 4 + m, -((d*(a + b*x))/(b*c - a*d))])/((b*c - a*d)^3*(3 + m)*(c +
d*x)^m)

Rule 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

Rule 70

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

Rule 69

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

Rubi steps

\begin{align*} \int (a+b x)^m (c+d x)^{-3-m} (e+f x) (g+h x) \, dx &=-\frac{(a+b x)^{1+m} (c+d x)^{-2-m} \left (a^2 b c f h m-a^3 d f h (1+m)-b^3 c e g (2+m)+a b^2 (c (f g+e h)+d e g (1+m))-b \left (a^2 d f h (3+2 m)+b^2 (d e g+c (f g+e h) (1+m))-a b (2 c f h (1+m)+d (f g+e h) (2+m))\right ) x\right )}{b^2 (b c-a d)^2 (1+m) (2+m)}+\frac{(f h) \int (a+b x)^{2+m} (c+d x)^{-3-m} \, dx}{b^2}\\ &=-\frac{(a+b x)^{1+m} (c+d x)^{-2-m} \left (a^2 b c f h m-a^3 d f h (1+m)-b^3 c e g (2+m)+a b^2 (c (f g+e h)+d e g (1+m))-b \left (a^2 d f h (3+2 m)+b^2 (d e g+c (f g+e h) (1+m))-a b (2 c f h (1+m)+d (f g+e h) (2+m))\right ) x\right )}{b^2 (b c-a d)^2 (1+m) (2+m)}+\frac{\left (b f h (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^{2+m} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^{-3-m} \, dx}{(b c-a d)^3}\\ &=-\frac{(a+b x)^{1+m} (c+d x)^{-2-m} \left (a^2 b c f h m-a^3 d f h (1+m)-b^3 c e g (2+m)+a b^2 (c (f g+e h)+d e g (1+m))-b \left (a^2 d f h (3+2 m)+b^2 (d e g+c (f g+e h) (1+m))-a b (2 c f h (1+m)+d (f g+e h) (2+m))\right ) x\right )}{b^2 (b c-a d)^2 (1+m) (2+m)}+\frac{f h (a+b x)^{3+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (3+m,3+m;4+m;-\frac{d (a+b x)}{b c-a d}\right )}{(b c-a d)^3 (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.331029, size = 237, normalized size = 0.96 \[ -\frac{(a+b x)^m (c+d x)^{-m-2} \left (d^3 (a+b x) \left (a^2 b f h (c m-d (2 m+3) x)+a^3 (-d) f h (m+1)+a b^2 (c e h+c f (g+2 h (m+1) x)+d e g (m+1)+d e h (m+2) x+d f g (m+2) x)-b^3 (c (e g (m+2)+e h (m+1) x+f g (m+1) x)+d e g x)\right )+f h (m+1) (b c-a d)^4 \left (\frac{d (a+b x)}{a d-b c}\right )^{-m} \, _2F_1\left (-m-2,-m-2;-m-1;\frac{b (c+d x)}{b c-a d}\right )\right )}{b^2 d^3 (m+1) (m+2) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^(-3 - m)*(e + f*x)*(g + h*x),x]

[Out]

-(((a + b*x)^m*(c + d*x)^(-2 - m)*(d^3*(a + b*x)*(-(a^3*d*f*h*(1 + m)) + a^2*b*f*h*(c*m - d*(3 + 2*m)*x) + a*b
^2*(c*e*h + d*e*g*(1 + m) + d*f*g*(2 + m)*x + d*e*h*(2 + m)*x + c*f*(g + 2*h*(1 + m)*x)) - b^3*(d*e*g*x + c*(e
*g*(2 + m) + f*g*(1 + m)*x + e*h*(1 + m)*x))) + ((b*c - a*d)^4*f*h*(1 + m)*Hypergeometric2F1[-2 - m, -2 - m, -
1 - m, (b*(c + d*x))/(b*c - a*d)])/((d*(a + b*x))/(-(b*c) + a*d))^m))/(b^2*d^3*(b*c - a*d)^2*(1 + m)*(2 + m)))

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-3-m} \left ( fx+e \right ) \left ( hx+g \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)*(h*x+g),x)

[Out]

int((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)*(h*x+g),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}{\left (h x + g\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)*(h*x+g),x, algorithm="maxima")

[Out]

integrate((f*x + e)*(h*x + g)*(b*x + a)^m*(d*x + c)^(-m - 3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (f h x^{2} + e g +{\left (f g + e h\right )} x\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)*(h*x+g),x, algorithm="fricas")

[Out]

integral((f*h*x^2 + e*g + (f*g + e*h)*x)*(b*x + a)^m*(d*x + c)^(-m - 3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)**(-3-m)*(f*x+e)*(h*x+g),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}{\left (h x + g\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)*(h*x+g),x, algorithm="giac")

[Out]

integrate((f*x + e)*(h*x + g)*(b*x + a)^m*(d*x + c)^(-m - 3), x)