3.135 \(\int (c+d x)^{-4-m} (e+f x)^m (g+h x) \, dx\)

Optimal. Leaf size=188 \[ -\frac{(d g-c h) (c+d x)^{-m-3} (e+f x)^{m+1}}{d (m+3) (d e-c f)}+\frac{(c+d x)^{-m-2} (e+f x)^{m+1} (c f h (m+1)+d (2 f g-e h (m+3)))}{d (m+2) (m+3) (d e-c f)^2}-\frac{f (c+d x)^{-m-1} (e+f x)^{m+1} (c f h (m+1)+d (2 f g-e h (m+3)))}{d (m+1) (m+2) (m+3) (d e-c f)^3} \]

[Out]

-(((d*g - c*h)*(c + d*x)^(-3 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)*(3 + m))) + ((c*f*h*(1 + m) + d*(2*f*g - e
*h*(3 + m)))*(c + d*x)^(-2 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)^2*(2 + m)*(3 + m)) - (f*(c*f*h*(1 + m) + d*(
2*f*g - e*h*(3 + m)))*(c + d*x)^(-1 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)^3*(1 + m)*(2 + m)*(3 + m))

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Rubi [A]  time = 0.0982305, antiderivative size = 186, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {79, 45, 37} \[ -\frac{(d g-c h) (c+d x)^{-m-3} (e+f x)^{m+1}}{d (m+3) (d e-c f)}+\frac{(c+d x)^{-m-2} (e+f x)^{m+1} (c f h (m+1)-d e h (m+3)+2 d f g)}{d (m+2) (m+3) (d e-c f)^2}-\frac{f (c+d x)^{-m-1} (e+f x)^{m+1} (c f h (m+1)-d e h (m+3)+2 d f g)}{d (m+1) (m+2) (m+3) (d e-c f)^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(-4 - m)*(e + f*x)^m*(g + h*x),x]

[Out]

-(((d*g - c*h)*(c + d*x)^(-3 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)*(3 + m))) + ((2*d*f*g + c*f*h*(1 + m) - d*
e*h*(3 + m))*(c + d*x)^(-2 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)^2*(2 + m)*(3 + m)) - (f*(2*d*f*g + c*f*h*(1
+ m) - d*e*h*(3 + m))*(c + d*x)^(-1 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)^3*(1 + m)*(2 + m)*(3 + m))

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 45

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

Rule 37

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

Rubi steps

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

Mathematica [A]  time = 0.119008, size = 182, normalized size = 0.97 \[ -\frac{(c h-d g) (c+d x)^{-m-3} (e+f x)^{m+1}}{d (-m-3) (d e-c f)}-\frac{\left (\frac{(c+d x)^{-m-2} (e+f x)^{m+1}}{(-m-2) (d e-c f)}+\frac{f (c+d x)^{-m-1} (e+f x)^{m+1}}{(-m-2) (-m-1) (d e-c f)^2}\right ) (-h (c f (m+1)+d e (-m-3))-2 d f g)}{d (-m-3) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(-4 - m)*(e + f*x)^m*(g + h*x),x]

[Out]

-(((-(d*g) + c*h)*(c + d*x)^(-3 - m)*(e + f*x)^(1 + m))/(d*(d*e - c*f)*(-3 - m))) - ((-2*d*f*g - h*(d*e*(-3 -
m) + c*f*(1 + m)))*(((c + d*x)^(-2 - m)*(e + f*x)^(1 + m))/((d*e - c*f)*(-2 - m)) + (f*(c + d*x)^(-1 - m)*(e +
f*x)^(1 + m))/((d*e - c*f)^2*(-2 - m)*(-1 - m))))/(d*(d*e - c*f)*(-3 - m))

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Maple [B]  time = 0.008, size = 509, normalized size = 2.7 \begin{align*} -{\frac{ \left ( dx+c \right ) ^{-3-m} \left ( fx+e \right ) ^{1+m} \left ( -{c}^{2}{f}^{2}h{m}^{2}x+2\,cdefh{m}^{2}x-cd{f}^{2}hm{x}^{2}-{d}^{2}{e}^{2}h{m}^{2}x+{d}^{2}efhm{x}^{2}-{c}^{2}{f}^{2}g{m}^{2}-4\,{c}^{2}{f}^{2}hmx+2\,cdefg{m}^{2}+8\,cdefhmx-2\,cd{f}^{2}gmx-cd{f}^{2}h{x}^{2}-{d}^{2}{e}^{2}g{m}^{2}-4\,{d}^{2}{e}^{2}hmx+2\,{d}^{2}efgmx+3\,{d}^{2}efh{x}^{2}-2\,{d}^{2}{f}^{2}g{x}^{2}+{c}^{2}efhm-5\,{c}^{2}{f}^{2}gm-3\,{c}^{2}{f}^{2}hx-cd{e}^{2}hm+8\,cdefgm+10\,cdefhx-6\,cd{f}^{2}gx-3\,{d}^{2}{e}^{2}gm-3\,{d}^{2}{e}^{2}hx+2\,{d}^{2}efgx+3\,{c}^{2}efh-6\,{c}^{2}{f}^{2}g-cd{e}^{2}h+6\,cdefg-2\,{d}^{2}{e}^{2}g \right ) }{{c}^{3}{f}^{3}{m}^{3}-3\,{c}^{2}de{f}^{2}{m}^{3}+3\,c{d}^{2}{e}^{2}f{m}^{3}-{d}^{3}{e}^{3}{m}^{3}+6\,{c}^{3}{f}^{3}{m}^{2}-18\,{c}^{2}de{f}^{2}{m}^{2}+18\,c{d}^{2}{e}^{2}f{m}^{2}-6\,{d}^{3}{e}^{3}{m}^{2}+11\,{c}^{3}{f}^{3}m-33\,{c}^{2}de{f}^{2}m+33\,c{d}^{2}{e}^{2}fm-11\,{d}^{3}{e}^{3}m+6\,{c}^{3}{f}^{3}-18\,{c}^{2}de{f}^{2}+18\,c{d}^{2}{e}^{2}f-6\,{d}^{3}{e}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(-4-m)*(f*x+e)^m*(h*x+g),x)

[Out]

-(d*x+c)^(-3-m)*(f*x+e)^(1+m)*(-c^2*f^2*h*m^2*x+2*c*d*e*f*h*m^2*x-c*d*f^2*h*m*x^2-d^2*e^2*h*m^2*x+d^2*e*f*h*m*
x^2-c^2*f^2*g*m^2-4*c^2*f^2*h*m*x+2*c*d*e*f*g*m^2+8*c*d*e*f*h*m*x-2*c*d*f^2*g*m*x-c*d*f^2*h*x^2-d^2*e^2*g*m^2-
4*d^2*e^2*h*m*x+2*d^2*e*f*g*m*x+3*d^2*e*f*h*x^2-2*d^2*f^2*g*x^2+c^2*e*f*h*m-5*c^2*f^2*g*m-3*c^2*f^2*h*x-c*d*e^
2*h*m+8*c*d*e*f*g*m+10*c*d*e*f*h*x-6*c*d*f^2*g*x-3*d^2*e^2*g*m-3*d^2*e^2*h*x+2*d^2*e*f*g*x+3*c^2*e*f*h-6*c^2*f
^2*g-c*d*e^2*h+6*c*d*e*f*g-2*d^2*e^2*g)/(c^3*f^3*m^3-3*c^2*d*e*f^2*m^3+3*c*d^2*e^2*f*m^3-d^3*e^3*m^3+6*c^3*f^3
*m^2-18*c^2*d*e*f^2*m^2+18*c*d^2*e^2*f*m^2-6*d^3*e^3*m^2+11*c^3*f^3*m-33*c^2*d*e*f^2*m+33*c*d^2*e^2*f*m-11*d^3
*e^3*m+6*c^3*f^3-18*c^2*d*e*f^2+18*c*d^2*e^2*f-6*d^3*e^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (h x + g\right )}{\left (d x + c\right )}^{-m - 4}{\left (f x + e\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(-4-m)*(f*x+e)^m*(h*x+g),x, algorithm="maxima")

[Out]

integrate((h*x + g)*(d*x + c)^(-m - 4)*(f*x + e)^m, x)

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Fricas [B]  time = 1.50452, size = 1805, normalized size = 9.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(-4-m)*(f*x+e)^m*(h*x+g),x, algorithm="fricas")

[Out]

-((2*d^3*f^3*g - (d^3*e*f^2 - c*d^2*f^3)*h*m - (3*d^3*e*f^2 - c*d^2*f^3)*h)*x^4 + (c*d^2*e^3 - 2*c^2*d*e^2*f +
c^3*e*f^2)*g*m^2 + (8*c*d^2*f^3*g + (d^3*e^2*f - 2*c*d^2*e*f^2 + c^2*d*f^3)*h*m^2 - 4*(3*c*d^2*e*f^2 - c^2*d*
f^3)*h - (2*(d^3*e*f^2 - c*d^2*f^3)*g - (3*d^3*e^2*f - 8*c*d^2*e*f^2 + 5*c^2*d*f^3)*h)*m)*x^3 + (12*c^2*d*f^3*
g + ((d^3*e^2*f - 2*c*d^2*e*f^2 + c^2*d*f^3)*g + (d^3*e^3 - c*d^2*e^2*f - c^2*d*e*f^2 + c^3*f^3)*h)*m^2 + 3*(d
^3*e^3 - 3*c*d^2*e^2*f - 3*c^2*d*e*f^2 + c^3*f^3)*h + ((d^3*e^2*f - 8*c*d^2*e*f^2 + 7*c^2*d*f^3)*g + 4*(d^3*e^
3 - c*d^2*e^2*f - c^2*d*e*f^2 + c^3*f^3)*h)*m)*x^2 + 2*(c*d^2*e^3 - 3*c^2*d*e^2*f + 3*c^3*e*f^2)*g + (c^2*d*e^
3 - 3*c^3*e^2*f)*h + ((3*c*d^2*e^3 - 8*c^2*d*e^2*f + 5*c^3*e*f^2)*g + (c^2*d*e^3 - c^3*e^2*f)*h)*m + (((d^3*e^
3 - c*d^2*e^2*f - c^2*d*e*f^2 + c^3*f^3)*g + (c*d^2*e^3 - 2*c^2*d*e^2*f + c^3*e*f^2)*h)*m^2 + 2*(d^3*e^3 - 3*c
*d^2*e^2*f + 3*c^2*d*e*f^2 + 3*c^3*f^3)*g + 4*(c*d^2*e^3 - 3*c^2*d*e^2*f)*h + ((3*d^3*e^3 - 7*c*d^2*e^2*f - c^
2*d*e*f^2 + 5*c^3*f^3)*g + (5*c*d^2*e^3 - 8*c^2*d*e^2*f + 3*c^3*e*f^2)*h)*m)*x)*(d*x + c)^(-m - 4)*(f*x + e)^m
/(6*d^3*e^3 - 18*c*d^2*e^2*f + 18*c^2*d*e*f^2 - 6*c^3*f^3 + (d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3
)*m^3 + 6*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*m^2 + 11*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^
2 - c^3*f^3)*m)

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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((d*x+c)**(-4-m)*(f*x+e)**m*(h*x+g),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (h x + g\right )}{\left (d x + c\right )}^{-m - 4}{\left (f x + e\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(-4-m)*(f*x+e)^m*(h*x+g),x, algorithm="giac")

[Out]

integrate((h*x + g)*(d*x + c)^(-m - 4)*(f*x + e)^m, x)