3.3099 \(\int (a+b x)^m (c+d x)^{-4-m} (e+f x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0875537, antiderivative size = 184, normalized size of antiderivative = 0.98, 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} \[ \frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-3}}{d (m+3) (b c-a d)}+\frac{(a+b x)^{m+1} (c+d x)^{-m-2} (-a d f (m+3)+b c f (m+1)+2 b d e)}{d (m+2) (m+3) (b c-a d)^2}+\frac{b (a+b x)^{m+1} (c+d x)^{-m-1} (-a d f (m+3)+b c f (m+1)+2 b d e)}{d (m+1) (m+2) (m+3) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.132642, size = 179, normalized size = 0.95 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-3} \left (a^2 d (m+1) (c f+d e (m+2)+d f (m+3) x)-a b \left (c^2 f (m+3)+2 c d \left (e \left (m^2+4 m+3\right )+f \left (m^2+4 m+5\right ) x\right )+d^2 x (2 e (m+1)+f (m+3) x)\right )+b^2 \left (c^2 (m+3) (e (m+2)+f (m+1) x)+c d x (2 e (m+3)+f (m+1) x)+2 d^2 e x^2\right )\right )}{(m+1) (m+2) (m+3) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 503, normalized size = 2.7 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+m} \left ( dx+c \right ) ^{-3-m} \left ({a}^{2}{d}^{2}f{m}^{2}x-2\,abcdf{m}^{2}x-ab{d}^{2}fm{x}^{2}+{b}^{2}{c}^{2}f{m}^{2}x+{b}^{2}cdfm{x}^{2}+{a}^{2}{d}^{2}e{m}^{2}+4\,{a}^{2}{d}^{2}fmx-2\,abcde{m}^{2}-8\,abcdfmx-2\,ab{d}^{2}emx-3\,ab{d}^{2}f{x}^{2}+{b}^{2}{c}^{2}e{m}^{2}+4\,{b}^{2}{c}^{2}fmx+2\,{b}^{2}cdemx+{b}^{2}cdf{x}^{2}+2\,{b}^{2}{d}^{2}e{x}^{2}+{a}^{2}cdfm+3\,{a}^{2}{d}^{2}em+3\,{a}^{2}{d}^{2}fx-ab{c}^{2}fm-8\,abcdem-10\,abcdfx-2\,ab{d}^{2}ex+5\,{b}^{2}{c}^{2}em+3\,{b}^{2}{c}^{2}fx+6\,{b}^{2}cdex+{a}^{2}cdf+2\,{a}^{2}{d}^{2}e-3\,ab{c}^{2}f-6\,abcde+6\,{b}^{2}{c}^{2}e \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),x)

[Out]

-(b*x+a)^(1+m)*(d*x+c)^(-3-m)*(a^2*d^2*f*m^2*x-2*a*b*c*d*f*m^2*x-a*b*d^2*f*m*x^2+b^2*c^2*f*m^2*x+b^2*c*d*f*m*x
^2+a^2*d^2*e*m^2+4*a^2*d^2*f*m*x-2*a*b*c*d*e*m^2-8*a*b*c*d*f*m*x-2*a*b*d^2*e*m*x-3*a*b*d^2*f*x^2+b^2*c^2*e*m^2
+4*b^2*c^2*f*m*x+2*b^2*c*d*e*m*x+b^2*c*d*f*x^2+2*b^2*d^2*e*x^2+a^2*c*d*f*m+3*a^2*d^2*e*m+3*a^2*d^2*f*x-a*b*c^2
*f*m-8*a*b*c*d*e*m-10*a*b*c*d*f*x-2*a*b*d^2*e*x+5*b^2*c^2*e*m+3*b^2*c^2*f*x+6*b^2*c*d*e*x+a^2*c*d*f+2*a^2*d^2*
e-3*a*b*c^2*f-6*a*b*c*d*e+6*b^2*c^2*e)/(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 )}{\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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.70251, size = 1804, 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((b*x+a)^m*(d*x+c)^(-4-m)*(f*x+e),x, algorithm="fricas")

[Out]

((2*b^3*d^3*e + (b^3*c*d^2 - a*b^2*d^3)*f*m + (b^3*c*d^2 - 3*a*b^2*d^3)*f)*x^4 + (a*b^2*c^3 - 2*a^2*b*c^2*d +
a^3*c*d^2)*e*m^2 + (8*b^3*c*d^2*e + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*f*m^2 + 4*(b^3*c^2*d - 3*a*b^2*c*d
^2)*f + (2*(b^3*c*d^2 - a*b^2*d^3)*e + (5*b^3*c^2*d - 8*a*b^2*c*d^2 + 3*a^2*b*d^3)*f)*m)*x^3 + (12*b^3*c^2*d*e
+ ((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*e + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*f)*m^2 + 3*(b^
3*c^3 - 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*f + ((7*b^3*c^2*d - 8*a*b^2*c*d^2 + a^2*b*d^3)*e + 4*(b^3*c^3
- a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*f)*m)*x^2 + 2*(3*a*b^2*c^3 - 3*a^2*b*c^2*d + a^3*c*d^2)*e - (3*a^2*b*c
^3 - a^3*c^2*d)*f + ((5*a*b^2*c^3 - 8*a^2*b*c^2*d + 3*a^3*c*d^2)*e - (a^2*b*c^3 - a^3*c^2*d)*f)*m + (((b^3*c^3
- a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*e + (a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*f)*m^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 - 4*(3*a^2*b*c^2*d - a^3*c*d^2)*f + ((5*b^3*c^3 - a*b^2*c^2*d - 7*a^2
*b*c*d^2 + 3*a^3*d^3)*e + (3*a*b^2*c^3 - 8*a^2*b*c^2*d + 5*a^3*c*d^2)*f)*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),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 (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),x, algorithm="giac")

[Out]

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