3.3098 \(\int (a+b x)^m (c+d x)^{-4-m} (e+f x)^2 \, dx\)

Optimal. Leaf size=353 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-2} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-2 a b d f (m+3) (c f (m+1)+d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )+2 c d e f (m+1)+2 d^2 e^2\right )\right )}{b d^2 (m+2) (m+3) (b c-a d)^2}+\frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-2 a b d f (m+3) (c f (m+1)+d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )+2 c d e f (m+1)+2 d^2 e^2\right )\right )}{d^2 (m+1) (m+2) (m+3) (b c-a d)^3}-\frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-3} (a d f (m+3)-b (c f (m+2)+d e))}{b d^2 (m+3) (b c-a d)}-\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-3}}{b d} \]

[Out]

-(((d*e - c*f)*(a*d*f*(3 + m) - b*(d*e + c*f*(2 + m)))*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m))/(b*d^2*(b*c - a*d
)*(3 + m))) + ((a^2*d^2*f^2*(6 + 5*m + m^2) - 2*a*b*d*f*(3 + m)*(d*e + c*f*(1 + m)) + b^2*(2*d^2*e^2 + 2*c*d*e
*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(b*d^2*(b*c - a*d)^2*(2 + m)*(3 +
m)) + ((a^2*d^2*f^2*(6 + 5*m + m^2) - 2*a*b*d*f*(3 + m)*(d*e + c*f*(1 + m)) + b^2*(2*d^2*e^2 + 2*c*d*e*f*(1 +
m) + c^2*f^2*(2 + 3*m + m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d^2*(b*c - a*d)^3*(1 + m)*(2 + m)*(3 +
m)) - (f*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m)*(e + f*x))/(b*d)

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Rubi [A]  time = 0.336113, antiderivative size = 351, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {90, 79, 45, 37} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-2} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-2 a b d f (m+3) (c f (m+1)+d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )+2 c d e f (m+1)+2 d^2 e^2\right )\right )}{b d^2 (m+2) (m+3) (b c-a d)^2}+\frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (a^2 d^2 f^2 \left (m^2+5 m+6\right )-2 a b d f (m+3) (c f (m+1)+d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )+2 c d e f (m+1)+2 d^2 e^2\right )\right )}{d^2 (m+1) (m+2) (m+3) (b c-a d)^3}+\frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-3} (-a d f (m+3)+b c f (m+2)+b d e)}{b d^2 (m+3) (b c-a d)}-\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-3}}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-4 - m)*(e + f*x)^2,x]

[Out]

((d*e - c*f)*(b*d*e + b*c*f*(2 + m) - a*d*f*(3 + m))*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m))/(b*d^2*(b*c - a*d)*
(3 + m)) + ((a^2*d^2*f^2*(6 + 5*m + m^2) - 2*a*b*d*f*(3 + m)*(d*e + c*f*(1 + m)) + b^2*(2*d^2*e^2 + 2*c*d*e*f*
(1 + m) + c^2*f^2*(2 + 3*m + m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(b*d^2*(b*c - a*d)^2*(2 + m)*(3 + m)
) + ((a^2*d^2*f^2*(6 + 5*m + m^2) - 2*a*b*d*f*(3 + m)*(d*e + c*f*(1 + m)) + b^2*(2*d^2*e^2 + 2*c*d*e*f*(1 + m)
+ c^2*f^2*(2 + 3*m + m^2)))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d^2*(b*c - a*d)^3*(1 + m)*(2 + m)*(3 + m))
- (f*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m)*(e + f*x))/(b*d)

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

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 (a+b x)^m (c+d x)^{-4-m} (e+f x)^2 \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{-3-m} (e+f x)}{b d}-\frac{\int (a+b x)^m (c+d x)^{-4-m} \left (-b e (d e+c f (1+m))-a f (c f-d e (3+m))-(b c-a d) f^2 (2+m) x\right ) \, dx}{b d}\\ &=\frac{(d e-c f) (b d e+b c f (2+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-3-m}}{b d^2 (b c-a d) (3+m)}-\frac{f (a+b x)^{1+m} (c+d x)^{-3-m} (e+f x)}{b d}+\frac{\left ((b c-a d) f^2 (2+m) (a d (-3-m)+b c (1+m))-2 b d (-b e (d e+c f (1+m))-a f (c f-d e (3+m)))\right ) \int (a+b x)^m (c+d x)^{-3-m} \, dx}{b d^2 (-b c+a d) (-3-m)}\\ &=\frac{(d e-c f) (b d e+b c f (2+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-3-m}}{b d^2 (b c-a d) (3+m)}+\frac{\left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-2 a b d f (3+m) (d e+c f (1+m))+b^2 \left (2 d^2 e^2+2 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-2-m}}{b d^2 (b c-a d)^2 (2+m) (3+m)}-\frac{f (a+b x)^{1+m} (c+d x)^{-3-m} (e+f x)}{b d}+\frac{\left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-2 a b d f (3+m) (d e+c f (1+m))+b^2 \left (2 d^2 e^2+2 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) \int (a+b x)^m (c+d x)^{-2-m} \, dx}{d^2 (b c-a d)^2 (2+m) (3+m)}\\ &=\frac{(d e-c f) (b d e+b c f (2+m)-a d f (3+m)) (a+b x)^{1+m} (c+d x)^{-3-m}}{b d^2 (b c-a d) (3+m)}+\frac{\left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-2 a b d f (3+m) (d e+c f (1+m))+b^2 \left (2 d^2 e^2+2 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-2-m}}{b d^2 (b c-a d)^2 (2+m) (3+m)}+\frac{\left (a^2 d^2 f^2 \left (6+5 m+m^2\right )-2 a b d f (3+m) (d e+c f (1+m))+b^2 \left (2 d^2 e^2+2 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^2 (b c-a d)^3 (1+m) (2+m) (3+m)}-\frac{f (a+b x)^{1+m} (c+d x)^{-3-m} (e+f x)}{b d}\\ \end{align*}

Mathematica [A]  time = 0.237863, size = 286, normalized size = 0.81 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m-3} \left (a^2 \left (2 c^2 f^2+2 c d f (e (m+1)+f (m+3) x)+d^2 \left (e^2 \left (m^2+3 m+2\right )+2 e f \left (m^2+4 m+3\right ) x+f^2 \left (m^2+5 m+6\right ) x^2\right )\right )-2 a b \left (c^2 f (e (m+3)+f (m+1) x)+c d \left (e^2 \left (m^2+4 m+3\right )+2 e f \left (m^2+4 m+5\right ) x+f^2 \left (m^2+4 m+3\right ) x^2\right )+d^2 e x (e (m+1)+f (m+3) x)\right )+b^2 \left (c^2 \left (e^2 \left (m^2+5 m+6\right )+2 e f \left (m^2+4 m+3\right ) x+f^2 \left (m^2+3 m+2\right ) x^2\right )+2 c d e x (e (m+3)+f (m+1) x)+2 d^2 e^2 x^2\right )\right )}{(m+1) (m+2) (m+3) (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^(-4 - m)*(e + f*x)^2,x]

[Out]

-(((a + b*x)^(1 + m)*(c + d*x)^(-3 - m)*(b^2*(2*d^2*e^2*x^2 + 2*c*d*e*x*(e*(3 + m) + f*(1 + m)*x) + c^2*(e^2*(
6 + 5*m + m^2) + 2*e*f*(3 + 4*m + m^2)*x + f^2*(2 + 3*m + m^2)*x^2)) - 2*a*b*(c^2*f*(e*(3 + m) + f*(1 + m)*x)
+ d^2*e*x*(e*(1 + m) + f*(3 + m)*x) + c*d*(e^2*(3 + 4*m + m^2) + 2*e*f*(5 + 4*m + m^2)*x + f^2*(3 + 4*m + m^2)
*x^2)) + a^2*(2*c^2*f^2 + 2*c*d*f*(e*(1 + m) + f*(3 + m)*x) + d^2*(e^2*(2 + 3*m + m^2) + 2*e*f*(3 + 4*m + m^2)
*x + f^2*(6 + 5*m + m^2)*x^2))))/((-(b*c) + a*d)^3*(1 + m)*(2 + m)*(3 + m)))

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Maple [B]  time = 0.007, size = 741, normalized size = 2.1 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+m} \left ( dx+c \right ) ^{-3-m} \left ({a}^{2}{d}^{2}{f}^{2}{m}^{2}{x}^{2}-2\,abcd{f}^{2}{m}^{2}{x}^{2}+{b}^{2}{c}^{2}{f}^{2}{m}^{2}{x}^{2}+2\,{a}^{2}{d}^{2}ef{m}^{2}x+5\,{a}^{2}{d}^{2}{f}^{2}m{x}^{2}-4\,abcdef{m}^{2}x-8\,abcd{f}^{2}m{x}^{2}-2\,ab{d}^{2}efm{x}^{2}+2\,{b}^{2}{c}^{2}ef{m}^{2}x+3\,{b}^{2}{c}^{2}{f}^{2}m{x}^{2}+2\,{b}^{2}cdefm{x}^{2}+2\,{a}^{2}cd{f}^{2}mx+{a}^{2}{d}^{2}{e}^{2}{m}^{2}+8\,{a}^{2}{d}^{2}efmx+6\,{a}^{2}{d}^{2}{f}^{2}{x}^{2}-2\,ab{c}^{2}{f}^{2}mx-2\,abcd{e}^{2}{m}^{2}-16\,abcdefmx-6\,abcd{f}^{2}{x}^{2}-2\,ab{d}^{2}{e}^{2}mx-6\,ab{d}^{2}ef{x}^{2}+{b}^{2}{c}^{2}{e}^{2}{m}^{2}+8\,{b}^{2}{c}^{2}efmx+2\,{b}^{2}{c}^{2}{f}^{2}{x}^{2}+2\,{b}^{2}cd{e}^{2}mx+2\,{b}^{2}cdef{x}^{2}+2\,{b}^{2}{d}^{2}{e}^{2}{x}^{2}+2\,{a}^{2}cdefm+6\,{a}^{2}cd{f}^{2}x+3\,{a}^{2}{d}^{2}{e}^{2}m+6\,{a}^{2}{d}^{2}efx-2\,ab{c}^{2}efm-2\,ab{c}^{2}{f}^{2}x-8\,abcd{e}^{2}m-20\,abcdefx-2\,ab{d}^{2}{e}^{2}x+5\,{b}^{2}{c}^{2}{e}^{2}m+6\,{b}^{2}{c}^{2}efx+6\,{b}^{2}cd{e}^{2}x+2\,{a}^{2}{c}^{2}{f}^{2}+2\,{a}^{2}cdef+2\,{a}^{2}{d}^{2}{e}^{2}-6\,ab{c}^{2}ef-6\,abcd{e}^{2}+6\,{b}^{2}{c}^{2}{e}^{2} \right ) }{{a}^{3}{d}^{3}{m}^{3}-3\,{a}^{2}bc{d}^{2}{m}^{3}+3\,a{b}^{2}{c}^{2}d{m}^{3}-{b}^{3}{c}^{3}{m}^{3}+6\,{a}^{3}{d}^{3}{m}^{2}-18\,{a}^{2}bc{d}^{2}{m}^{2}+18\,a{b}^{2}{c}^{2}d{m}^{2}-6\,{b}^{3}{c}^{3}{m}^{2}+11\,{a}^{3}{d}^{3}m-33\,{a}^{2}bc{d}^{2}m+33\,a{b}^{2}{c}^{2}dm-11\,{b}^{3}{c}^{3}m+6\,{a}^{3}{d}^{3}-18\,{a}^{2}cb{d}^{2}+18\,a{b}^{2}{c}^{2}d-6\,{b}^{3}{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^2,x)

[Out]

-(b*x+a)^(1+m)*(d*x+c)^(-3-m)*(a^2*d^2*f^2*m^2*x^2-2*a*b*c*d*f^2*m^2*x^2+b^2*c^2*f^2*m^2*x^2+2*a^2*d^2*e*f*m^2
*x+5*a^2*d^2*f^2*m*x^2-4*a*b*c*d*e*f*m^2*x-8*a*b*c*d*f^2*m*x^2-2*a*b*d^2*e*f*m*x^2+2*b^2*c^2*e*f*m^2*x+3*b^2*c
^2*f^2*m*x^2+2*b^2*c*d*e*f*m*x^2+2*a^2*c*d*f^2*m*x+a^2*d^2*e^2*m^2+8*a^2*d^2*e*f*m*x+6*a^2*d^2*f^2*x^2-2*a*b*c
^2*f^2*m*x-2*a*b*c*d*e^2*m^2-16*a*b*c*d*e*f*m*x-6*a*b*c*d*f^2*x^2-2*a*b*d^2*e^2*m*x-6*a*b*d^2*e*f*x^2+b^2*c^2*
e^2*m^2+8*b^2*c^2*e*f*m*x+2*b^2*c^2*f^2*x^2+2*b^2*c*d*e^2*m*x+2*b^2*c*d*e*f*x^2+2*b^2*d^2*e^2*x^2+2*a^2*c*d*e*
f*m+6*a^2*c*d*f^2*x+3*a^2*d^2*e^2*m+6*a^2*d^2*e*f*x-2*a*b*c^2*e*f*m-2*a*b*c^2*f^2*x-8*a*b*c*d*e^2*m-20*a*b*c*d
*e*f*x-2*a*b*d^2*e^2*x+5*b^2*c^2*e^2*m+6*b^2*c^2*e*f*x+6*b^2*c*d*e^2*x+2*a^2*c^2*f^2+2*a^2*c*d*e*f+2*a^2*d^2*e
^2-6*a*b*c^2*e*f-6*a*b*c*d*e^2+6*b^2*c^2*e^2)/(a^3*d^3*m^3-3*a^2*b*c*d^2*m^3+3*a*b^2*c^2*d*m^3-b^3*c^3*m^3+6*a
^3*d^3*m^2-18*a^2*b*c*d^2*m^2+18*a*b^2*c^2*d*m^2-6*b^3*c^3*m^2+11*a^3*d^3*m-33*a^2*b*c*d^2*m+33*a*b^2*c^2*d*m-
11*b^3*c^3*m+6*a^3*d^3-18*a^2*b*c*d^2+18*a*b^2*c^2*d-6*b^3*c^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}^{2}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((f*x + e)^2*(b*x + a)^m*(d*x + c)^(-m - 4), x)

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Fricas [B]  time = 1.76801, size = 2583, normalized size = 7.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^2,x, algorithm="fricas")

[Out]

(2*a^3*c^3*f^2 + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e^2*m^2 + (2*b^3*d^3*e^2 + (b^3*c^2*d - 2*a*b^2*c*d^2
+ a^2*b*d^3)*f^2*m^2 + 2*(b^3*c*d^2 - 3*a*b^2*d^3)*e*f + 2*(b^3*c^2*d - 3*a*b^2*c*d^2 + 3*a^2*b*d^3)*f^2 + (2
*(b^3*c*d^2 - a*b^2*d^3)*e*f + (3*b^3*c^2*d - 8*a*b^2*c*d^2 + 5*a^2*b*d^3)*f^2)*m)*x^4 + (8*b^3*c*d^2*e^2 + 8*
(b^3*c^2*d - 3*a*b^2*c*d^2)*e*f + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 3*a^3*d^3)*f^2 + (2*(b^3*c^2*d
- 2*a*b^2*c*d^2 + a^2*b*d^3)*e*f + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*f^2)*m^2 + (2*(b^3*c*d^2 -
a*b^2*d^3)*e^2 + 2*(5*b^3*c^2*d - 8*a*b^2*c*d^2 + 3*a^2*b*d^3)*e*f + (3*b^3*c^3 - 7*a*b^2*c^2*d - a^2*b*c*d^2
+ 5*a^3*d^3)*f^2)*m)*x^3 + 2*(3*a*b^2*c^3 - 3*a^2*b*c^2*d + a^3*c*d^2)*e^2 - 2*(3*a^2*b*c^3 - a^3*c^2*d)*e*f +
(12*b^3*c^2*d*e^2 + 12*a^3*c*d^2*f^2 + 6*(b^3*c^3 - 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*e*f + ((b^3*c^2*
d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e^2 + 2*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e*f + (a*b^2*c^3 - 2*a^
2*b*c^2*d + a^3*c*d^2)*f^2)*m^2 + ((7*b^3*c^2*d - 8*a*b^2*c*d^2 + a^2*b*d^3)*e^2 + 8*(b^3*c^3 - a*b^2*c^2*d -
a^2*b*c*d^2 + a^3*d^3)*e*f + (a*b^2*c^3 - 8*a^2*b*c^2*d + 7*a^3*c*d^2)*f^2)*m)*x^2 + ((5*a*b^2*c^3 - 8*a^2*b*c
^2*d + 3*a^3*c*d^2)*e^2 - 2*(a^2*b*c^3 - a^3*c^2*d)*e*f)*m + (8*a^3*c^2*d*f^2 + 2*(3*b^3*c^3 + 3*a*b^2*c^2*d -
3*a^2*b*c*d^2 + a^3*d^3)*e^2 - 8*(3*a^2*b*c^2*d - a^3*c*d^2)*e*f + ((b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^
3*d^3)*e^2 + 2*(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*e*f)*m^2 + ((5*b^3*c^3 - a*b^2*c^2*d - 7*a^2*b*c*d^2 +
3*a^3*d^3)*e^2 + 2*(3*a*b^2*c^3 - 8*a^2*b*c^2*d + 5*a^3*c*d^2)*e*f - 2*(a^2*b*c^3 - a^3*c^2*d)*f^2)*m)*x)*(b*x
+ a)^m*(d*x + c)^(-m - 4)/(6*b^3*c^3 - 18*a*b^2*c^2*d + 18*a^2*b*c*d^2 - 6*a^3*d^3 + (b^3*c^3 - 3*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 - a^3*d^3)*m^3 + 6*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*m^2 + 11*(b^3*c^3 - 3*
a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^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((b*x+a)**m*(d*x+c)**(-4-m)*(f*x+e)**2,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 )}^{2}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*(b*x + a)^m*(d*x + c)^(-m - 4), x)