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

Optimal. Leaf size=363 \[ \frac {(c+d x)^{-m-2} (e+f x)^{m+1} \left (a d f (c f h (m+1)+d (2 f g-e h (m+3)))+b \left (c^2 f^2 h \left (m^2+3 m+2\right )+c d f (m+1) (f g-2 e h (m+3))+d^2 (-e) (m+3) (f g-e h (m+2))\right )\right )}{d^2 f (m+2) (m+3) (d e-c f)^2}-\frac {(c+d x)^{-m-1} (e+f x)^{m+1} \left (a d f (c f h (m+1)+d (2 f g-e h (m+3)))+b \left (c^2 f^2 h \left (m^2+3 m+2\right )+c d f (m+1) (f g-2 e h (m+3))+d^2 (-e) (m+3) (f g-e h (m+2))\right )\right )}{d^2 (m+1) (m+2) (m+3) (d e-c f)^3}-\frac {(c+d x)^{-m-3} (e+f x)^{m+1} (a d f (d g-c h)-b c (c f h (m+2)+d (f g-e h (m+3)))+b d h (m+3) x (d e-c f))}{d^2 f (m+3) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 360, normalized size of antiderivative = 0.99, 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 = {146, 45, 37} \[ \frac {(c+d x)^{-m-2} (e+f x)^{m+1} \left (a d f (c f h (m+1)-d e h (m+3)+2 d f g)+b \left (c^2 f^2 h \left (m^2+3 m+2\right )+c d f (m+1) (f g-2 e h (m+3))+d^2 (-e) (m+3) (f g-e h (m+2))\right )\right )}{d^2 f (m+2) (m+3) (d e-c f)^2}-\frac {(c+d x)^{-m-1} (e+f x)^{m+1} \left (a d f (c f h (m+1)-d e h (m+3)+2 d f g)+b \left (c^2 f^2 h \left (m^2+3 m+2\right )+c d f (m+1) (f g-2 e h (m+3))+d^2 (-e) (m+3) (f g-e h (m+2))\right )\right )}{d^2 (m+1) (m+2) (m+3) (d e-c f)^3}-\frac {(c+d x)^{-m-3} (e+f x)^{m+1} (a d f (d g-c h)-b c (c f h (m+2)-d e h (m+3)+d f g)+b d h (m+3) x (d e-c f))}{d^2 f (m+3) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 146

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

Rubi steps

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

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Mathematica [A]  time = 0.55, size = 227, normalized size = 0.63 \[ \frac {(c+d x)^{-m-3} (e+f x)^{m+1} \left (\frac {(c+d x) (c f (m+2)-d (e m+e-f x)) \left (a d f (c f h (m+1)-d e h (m+3)+2 d f g)+b \left (c^2 f^2 h \left (m^2+3 m+2\right )+c d f (m+1) (f g-2 e h (m+3))+d^2 e (m+3) (e h (m+2)-f g)\right )\right )}{(m+1) (m+2) (d e-c f)^2}+a d f (d g-c h)-b \left (c^2 f h (m+2)+c d (f (g+h (m+3) x)-e h (m+3))-d^2 e h (m+3) x\right )\right )}{d^2 f (m+3) (c f-d e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.05, size = 1608, normalized size = 4.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((b*d^3*e^2*f - 2*b*c*d^2*e*f^2 + b*c^2*d*f^3)*h*m^2 - (3*b*d^3*e*f^2 - (b*c*d^2 + 2*a*d^3)*f^3)*g + (6*b*d^
3*e^2*f - 3*(2*b*c*d^2 + a*d^3)*e*f^2 + (2*b*c^2*d + a*c*d^2)*f^3)*h - ((b*d^3*e*f^2 - b*c*d^2*f^3)*g - (5*b*d
^3*e^2*f - (8*b*c*d^2 + a*d^3)*e*f^2 + (3*b*c^2*d + a*c*d^2)*f^3)*h)*m)*x^4 + (a*c*d^2*e^3 - 2*a*c^2*d*e^2*f +
 a*c^3*e*f^2)*g*m^2 + (((b*d^3*e^2*f - 2*b*c*d^2*e*f^2 + b*c^2*d*f^3)*g + (b*d^3*e^3 - (b*c*d^2 - a*d^3)*e^2*f
 - (b*c^2*d + 2*a*c*d^2)*e*f^2 + (b*c^3 + a*c^2*d)*f^3)*h)*m^2 - 4*(3*b*c*d^2*e*f^2 - (b*c^2*d + 2*a*c*d^2)*f^
3)*g + 2*(3*b*d^3*e^3 + 3*b*c*d^2*e^2*f - 3*(b*c^2*d + 2*a*c*d^2)*e*f^2 + (b*c^3 + 2*a*c^2*d)*f^3)*h + ((3*b*d
^3*e^2*f - 2*(4*b*c*d^2 + a*d^3)*e*f^2 + (5*b*c^2*d + 2*a*c*d^2)*f^3)*g + (5*b*d^3*e^3 - (b*c*d^2 - 3*a*d^3)*e
^2*f - (7*b*c^2*d + 8*a*c*d^2)*e*f^2 + (3*b*c^3 + 5*a*c^2*d)*f^3)*h)*m)*x^3 + (((b*d^3*e^3 - (b*c*d^2 - a*d^3)
*e^2*f - (b*c^2*d + 2*a*c*d^2)*e*f^2 + (b*c^3 + a*c^2*d)*f^3)*g + (a*c^3*f^3 + (b*c*d^2 + a*d^3)*e^3 - (2*b*c^
2*d + a*c*d^2)*e^2*f + (b*c^3 - a*c^2*d)*e*f^2)*h)*m^2 + 3*(b*d^3*e^3 - 3*b*c*d^2*e^2*f - 3*b*c^2*d*e*f^2 + (b
*c^3 + 4*a*c^2*d)*f^3)*g - 3*(3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3 - (4*b*c*d^2 + a*d^3)*e^3)*h + ((4
*b*d^3*e^3 - (4*b*c*d^2 - a*d^3)*e^2*f - 4*(b*c^2*d + 2*a*c*d^2)*e*f^2 + (4*b*c^3 + 7*a*c^2*d)*f^3)*g + (4*a*c
^3*f^3 + (7*b*c*d^2 + 4*a*d^3)*e^3 - 4*(2*b*c^2*d + a*c*d^2)*e^2*f + (b*c^3 - 4*a*c^2*d)*e*f^2)*h)*m)*x^2 + (6
*a*c^3*e*f^2 + (b*c^2*d + 2*a*c*d^2)*e^3 - 3*(b*c^3 + 2*a*c^2*d)*e^2*f)*g - (3*a*c^3*e^2*f - (2*b*c^3 + a*c^2*
d)*e^3)*h + ((5*a*c^3*e*f^2 + (b*c^2*d + 3*a*c*d^2)*e^3 - (b*c^3 + 8*a*c^2*d)*e^2*f)*g + (a*c^2*d*e^3 - a*c^3*
e^2*f)*h)*m + (((a*c^3*f^3 + (b*c*d^2 + a*d^3)*e^3 - (2*b*c^2*d + a*c*d^2)*e^2*f + (b*c^3 - a*c^2*d)*e*f^2)*g
+ (a*c*d^2*e^3 - 2*a*c^2*d*e^2*f + a*c^3*e*f^2)*h)*m^2 + 2*(3*a*c^2*d*e*f^2 + 3*a*c^3*f^3 + (2*b*c*d^2 + a*d^3
)*e^3 - 3*(2*b*c^2*d + a*c*d^2)*e^2*f)*g - 4*(3*a*c^2*d*e^2*f - (2*b*c^2*d + a*c*d^2)*e^3)*h + ((5*a*c^3*f^3 +
 (5*b*c*d^2 + 3*a*d^3)*e^3 - (8*b*c^2*d + 7*a*c*d^2)*e^2*f + (3*b*c^3 - a*c^2*d)*e*f^2)*g + (3*a*c^3*e*f^2 + (
2*b*c^2*d + 5*a*c*d^2)*e^3 - 2*(b*c^3 + 4*a*c^2*d)*e^2*f)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )} {\left (h x + g\right )} {\left (d x + c\right )}^{-m - 4} {\left (f x + e\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 906, normalized size = 2.50 \[ -\frac {\left (-b \,c^{2} f^{2} h \,m^{2} x^{2}+2 b c d e f h \,m^{2} x^{2}-b \,d^{2} e^{2} h \,m^{2} x^{2}-a \,c^{2} f^{2} h \,m^{2} x +2 a c d e f h \,m^{2} x -a c d \,f^{2} h m \,x^{2}-a \,d^{2} e^{2} h \,m^{2} x +a \,d^{2} e f h m \,x^{2}-b \,c^{2} f^{2} g \,m^{2} x -3 b \,c^{2} f^{2} h m \,x^{2}+2 b c d e f g \,m^{2} x +8 b c d e f h m \,x^{2}-b c d \,f^{2} g m \,x^{2}-b \,d^{2} e^{2} g \,m^{2} x -5 b \,d^{2} e^{2} h m \,x^{2}+b \,d^{2} e f g m \,x^{2}-a \,c^{2} f^{2} g \,m^{2}-4 a \,c^{2} f^{2} h m x +2 a c d e f g \,m^{2}+8 a c d e f h m x -2 a c d \,f^{2} g m x -a c d \,f^{2} h \,x^{2}-a \,d^{2} e^{2} g \,m^{2}-4 a \,d^{2} e^{2} h m x +2 a \,d^{2} e f g m x +3 a \,d^{2} e f h \,x^{2}-2 a \,d^{2} f^{2} g \,x^{2}+2 b \,c^{2} e f h m x -4 b \,c^{2} f^{2} g m x -2 b \,c^{2} f^{2} h \,x^{2}-2 b c d \,e^{2} h m x +8 b c d e f g m x +6 b c d e f h \,x^{2}-b c d \,f^{2} g \,x^{2}-4 b \,d^{2} e^{2} g m x -6 b \,d^{2} e^{2} h \,x^{2}+3 b \,d^{2} e f g \,x^{2}+a \,c^{2} e f h m -5 a \,c^{2} f^{2} g m -3 a \,c^{2} f^{2} h x -a c d \,e^{2} h m +8 a c d e f g m +10 a c d e f h x -6 a c d \,f^{2} g x -3 a \,d^{2} e^{2} g m -3 a \,d^{2} e^{2} h x +2 a \,d^{2} e f g x +b \,c^{2} e f g m +2 b \,c^{2} e f h x -3 b \,c^{2} f^{2} g x -b c d \,e^{2} g m -6 b c d \,e^{2} h x +10 b c d e f g x -3 b \,d^{2} e^{2} g x +3 a \,c^{2} e f h -6 a \,c^{2} f^{2} g -a c d \,e^{2} h +6 a c d e f g -2 a \,d^{2} e^{2} g -2 b \,c^{2} e^{2} h +3 b \,c^{2} e f g -b c d \,e^{2} g \right ) \left (d x +c \right )^{-m -3} \left (f x +e \right )^{m +1}}{c^{3} f^{3} m^{3}-3 c^{2} d e \,f^{2} m^{3}+3 c \,d^{2} e^{2} f \,m^{3}-d^{3} e^{3} m^{3}+6 c^{3} f^{3} m^{2}-18 c^{2} d e \,f^{2} m^{2}+18 c \,d^{2} e^{2} f \,m^{2}-6 d^{3} e^{3} m^{2}+11 c^{3} f^{3} m -33 c^{2} d e \,f^{2} m +33 c \,d^{2} e^{2} f m -11 d^{3} e^{3} m +6 c^{3} f^{3}-18 c^{2} d e \,f^{2}+18 c \,d^{2} e^{2} f -6 d^{3} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x+c)^(-m-3)*(f*x+e)^(m+1)*(-b*c^2*f^2*h*m^2*x^2+2*b*c*d*e*f*h*m^2*x^2-b*d^2*e^2*h*m^2*x^2-a*c^2*f^2*h*m^2*
x+2*a*c*d*e*f*h*m^2*x-a*c*d*f^2*h*m*x^2-a*d^2*e^2*h*m^2*x+a*d^2*e*f*h*m*x^2-b*c^2*f^2*g*m^2*x-3*b*c^2*f^2*h*m*
x^2+2*b*c*d*e*f*g*m^2*x+8*b*c*d*e*f*h*m*x^2-b*c*d*f^2*g*m*x^2-b*d^2*e^2*g*m^2*x-5*b*d^2*e^2*h*m*x^2+b*d^2*e*f*
g*m*x^2-a*c^2*f^2*g*m^2-4*a*c^2*f^2*h*m*x+2*a*c*d*e*f*g*m^2+8*a*c*d*e*f*h*m*x-2*a*c*d*f^2*g*m*x-a*c*d*f^2*h*x^
2-a*d^2*e^2*g*m^2-4*a*d^2*e^2*h*m*x+2*a*d^2*e*f*g*m*x+3*a*d^2*e*f*h*x^2-2*a*d^2*f^2*g*x^2+2*b*c^2*e*f*h*m*x-4*
b*c^2*f^2*g*m*x-2*b*c^2*f^2*h*x^2-2*b*c*d*e^2*h*m*x+8*b*c*d*e*f*g*m*x+6*b*c*d*e*f*h*x^2-b*c*d*f^2*g*x^2-4*b*d^
2*e^2*g*m*x-6*b*d^2*e^2*h*x^2+3*b*d^2*e*f*g*x^2+a*c^2*e*f*h*m-5*a*c^2*f^2*g*m-3*a*c^2*f^2*h*x-a*c*d*e^2*h*m+8*
a*c*d*e*f*g*m+10*a*c*d*e*f*h*x-6*a*c*d*f^2*g*x-3*a*d^2*e^2*g*m-3*a*d^2*e^2*h*x+2*a*d^2*e*f*g*x+b*c^2*e*f*g*m+2
*b*c^2*e*f*h*x-3*b*c^2*f^2*g*x-b*c*d*e^2*g*m-6*b*c*d*e^2*h*x+10*b*c*d*e*f*g*x-3*b*d^2*e^2*g*x+3*a*c^2*e*f*h-6*
a*c^2*f^2*g-a*c*d*e^2*h+6*a*c*d*e*f*g-2*a*d^2*e^2*g-2*b*c^2*e^2*h+3*b*c^2*e*f*g-b*c*d*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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )} {\left (h x + g\right )} {\left (d x + c\right )}^{-m - 4} {\left (f x + e\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.28, size = 1890, normalized size = 5.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((e + f*x)^m*(2*b*c^3*e^3*h + 2*a*c*d^2*e^3*g + a*c^2*d*e^3*h + b*c^2*d*e^3*g + 6*a*c^3*e*f^2*g - 3*a*c^3*e^2*
f*h - 3*b*c^3*e^2*f*g - 6*a*c^2*d*e^2*f*g + 3*a*c*d^2*e^3*g*m + a*c^2*d*e^3*h*m + b*c^2*d*e^3*g*m + 5*a*c^3*e*
f^2*g*m - a*c^3*e^2*f*h*m - b*c^3*e^2*f*g*m + a*c*d^2*e^3*g*m^2 + a*c^3*e*f^2*g*m^2 - 2*a*c^2*d*e^2*f*g*m^2 -
8*a*c^2*d*e^2*f*g*m))/((c*f - d*e)^3*(c + d*x)^(m + 4)*(11*m + 6*m^2 + m^3 + 6)) + (x*(e + f*x)^m*(6*a*c^3*f^3
*g + 2*a*d^3*e^3*g + 4*a*c*d^2*e^3*h + 4*b*c*d^2*e^3*g + 8*b*c^2*d*e^3*h + 5*a*c^3*f^3*g*m + 3*a*d^3*e^3*g*m +
 a*c^3*f^3*g*m^2 + a*d^3*e^3*g*m^2 - 6*a*c*d^2*e^2*f*g + 6*a*c^2*d*e*f^2*g - 12*a*c^2*d*e^2*f*h - 12*b*c^2*d*e
^2*f*g + 5*a*c*d^2*e^3*h*m + 5*b*c*d^2*e^3*g*m + 2*b*c^2*d*e^3*h*m + 3*a*c^3*e*f^2*h*m + 3*b*c^3*e*f^2*g*m - 2
*b*c^3*e^2*f*h*m + a*c*d^2*e^3*h*m^2 + b*c*d^2*e^3*g*m^2 + a*c^3*e*f^2*h*m^2 + b*c^3*e*f^2*g*m^2 - a*c*d^2*e^2
*f*g*m^2 - a*c^2*d*e*f^2*g*m^2 - 2*a*c^2*d*e^2*f*h*m^2 - 2*b*c^2*d*e^2*f*g*m^2 - 7*a*c*d^2*e^2*f*g*m - a*c^2*d
*e*f^2*g*m - 8*a*c^2*d*e^2*f*h*m - 8*b*c^2*d*e^2*f*g*m))/((c*f - d*e)^3*(c + d*x)^(m + 4)*(11*m + 6*m^2 + m^3
+ 6)) + (x^4*(e + f*x)^m*(2*a*d^3*f^3*g + a*c*d^2*f^3*h + b*c*d^2*f^3*g + 2*b*c^2*d*f^3*h - 3*a*d^3*e*f^2*h -
3*b*d^3*e*f^2*g + 6*b*d^3*e^2*f*h - 6*b*c*d^2*e*f^2*h + a*c*d^2*f^3*h*m + b*c*d^2*f^3*g*m + 3*b*c^2*d*f^3*h*m
- a*d^3*e*f^2*h*m - b*d^3*e*f^2*g*m + 5*b*d^3*e^2*f*h*m + b*c^2*d*f^3*h*m^2 + b*d^3*e^2*f*h*m^2 - 2*b*c*d^2*e*
f^2*h*m^2 - 8*b*c*d^2*e*f^2*h*m))/((c*f - d*e)^3*(c + d*x)^(m + 4)*(11*m + 6*m^2 + m^3 + 6)) + (x^2*(e + f*x)^
m*(3*a*c^3*f^3*h + 3*a*d^3*e^3*h + 3*b*c^3*f^3*g + 3*b*d^3*e^3*g + 12*a*c^2*d*f^3*g + 12*b*c*d^2*e^3*h + 4*a*c
^3*f^3*h*m + 4*a*d^3*e^3*h*m + 4*b*c^3*f^3*g*m + 4*b*d^3*e^3*g*m + a*c^3*f^3*h*m^2 + a*d^3*e^3*h*m^2 + b*c^3*f
^3*g*m^2 + b*d^3*e^3*g*m^2 - 9*a*c*d^2*e^2*f*h - 9*a*c^2*d*e*f^2*h - 9*b*c*d^2*e^2*f*g - 9*b*c^2*d*e*f^2*g + 7
*a*c^2*d*f^3*g*m + 7*b*c*d^2*e^3*h*m + a*d^3*e^2*f*g*m + b*c^3*e*f^2*h*m + a*c^2*d*f^3*g*m^2 + b*c*d^2*e^3*h*m
^2 + a*d^3*e^2*f*g*m^2 + b*c^3*e*f^2*h*m^2 - 2*a*c*d^2*e*f^2*g*m^2 - a*c*d^2*e^2*f*h*m^2 - a*c^2*d*e*f^2*h*m^2
 - b*c*d^2*e^2*f*g*m^2 - b*c^2*d*e*f^2*g*m^2 - 2*b*c^2*d*e^2*f*h*m^2 - 8*a*c*d^2*e*f^2*g*m - 4*a*c*d^2*e^2*f*h
*m - 4*a*c^2*d*e*f^2*h*m - 4*b*c*d^2*e^2*f*g*m - 4*b*c^2*d*e*f^2*g*m - 8*b*c^2*d*e^2*f*h*m))/((c*f - d*e)^3*(c
 + d*x)^(m + 4)*(11*m + 6*m^2 + m^3 + 6)) + (x^3*(e + f*x)^m*(2*b*c^3*f^3*h + 6*b*d^3*e^3*h + 8*a*c*d^2*f^3*g
+ 4*a*c^2*d*f^3*h + 4*b*c^2*d*f^3*g + 3*b*c^3*f^3*h*m + 5*b*d^3*e^3*h*m + b*c^3*f^3*h*m^2 + b*d^3*e^3*h*m^2 -
12*a*c*d^2*e*f^2*h - 12*b*c*d^2*e*f^2*g + 6*b*c*d^2*e^2*f*h - 6*b*c^2*d*e*f^2*h + 2*a*c*d^2*f^3*g*m + 5*a*c^2*
d*f^3*h*m + 5*b*c^2*d*f^3*g*m - 2*a*d^3*e*f^2*g*m + 3*a*d^3*e^2*f*h*m + 3*b*d^3*e^2*f*g*m + a*c^2*d*f^3*h*m^2
+ b*c^2*d*f^3*g*m^2 + a*d^3*e^2*f*h*m^2 + b*d^3*e^2*f*g*m^2 - 2*a*c*d^2*e*f^2*h*m^2 - 2*b*c*d^2*e*f^2*g*m^2 -
b*c*d^2*e^2*f*h*m^2 - b*c^2*d*e*f^2*h*m^2 - 8*a*c*d^2*e*f^2*h*m - 8*b*c*d^2*e*f^2*g*m - b*c*d^2*e^2*f*h*m - 7*
b*c^2*d*e*f^2*h*m))/((c*f - d*e)^3*(c + d*x)^(m + 4)*(11*m + 6*m^2 + m^3 + 6))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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