3.3100 \(\int (a+b x)^m (c+d x)^{-4-m} \, dx\)

Optimal. Leaf size=130 \[ \frac{2 b^2 (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (m+3) (b c-a d)^3}+\frac{(a+b x)^{m+1} (c+d x)^{-m-3}}{(m+3) (b c-a d)}+\frac{2 b (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (m+3) (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0335958, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{2 b^2 (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (m+3) (b c-a d)^3}+\frac{(a+b x)^{m+1} (c+d x)^{-m-3}}{(m+3) (b c-a d)}+\frac{2 b (a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (m+3) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-4 - m),x]

[Out]

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

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

Mathematica [A]  time = 0.0506881, size = 112, normalized size = 0.86 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-3} \left (a^2 d^2 \left (m^2+3 m+2\right )-2 a b d (m+1) (c (m+3)+d x)+b^2 \left (c^2 \left (m^2+5 m+6\right )+2 c d (m+3) x+2 d^2 x^2\right )\right )}{(m+1) (m+2) (m+3) (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 319, normalized size = 2.5 \begin{align*} -{\frac{ \left ( bx+a \right ) ^{1+m} \left ( dx+c \right ) ^{-3-m} \left ({a}^{2}{d}^{2}{m}^{2}-2\,abcd{m}^{2}-2\,ab{d}^{2}mx+{b}^{2}{c}^{2}{m}^{2}+2\,{b}^{2}cdmx+2\,{b}^{2}{d}^{2}{x}^{2}+3\,{a}^{2}{d}^{2}m-8\,abcdm-2\,ab{d}^{2}x+5\,{b}^{2}{c}^{2}m+6\,{b}^{2}cdx+2\,{a}^{2}{d}^{2}-6\,abcd+6\,{b}^{2}{c}^{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),x)

[Out]

-(b*x+a)^(1+m)*(d*x+c)^(-3-m)*(a^2*d^2*m^2-2*a*b*c*d*m^2-2*a*b*d^2*m*x+b^2*c^2*m^2+2*b^2*c*d*m*x+2*b^2*d^2*x^2
+3*a^2*d^2*m-8*a*b*c*d*m-2*a*b*d^2*x+5*b^2*c^2*m+6*b^2*c*d*x+2*a^2*d^2-6*a*b*c*d+6*b^2*c^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+1
1*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 (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),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.60948, size = 1021, normalized size = 7.85 \begin{align*} \frac{{\left (2 \, b^{3} d^{3} x^{4} + 6 \, a b^{2} c^{3} - 6 \, a^{2} b c^{2} d + 2 \, a^{3} c d^{2} + 2 \,{\left (4 \, b^{3} c d^{2} +{\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} m\right )} x^{3} +{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} m^{2} +{\left (12 \, b^{3} c^{2} d +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} m^{2} +{\left (7 \, b^{3} c^{2} d - 8 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} m\right )} x^{2} +{\left (5 \, a b^{2} c^{3} - 8 \, a^{2} b c^{2} d + 3 \, a^{3} c d^{2}\right )} m +{\left (6 \, b^{3} c^{3} + 6 \, a b^{2} c^{2} d - 6 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} m^{2} +{\left (5 \, b^{3} c^{3} - a b^{2} c^{2} d - 7 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} m\right )} x\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-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} +{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} m^{3} + 6 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} m^{2} + 11 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*d^3*x^4 + 6*a*b^2*c^3 - 6*a^2*b*c^2*d + 2*a^3*c*d^2 + 2*(4*b^3*c*d^2 + (b^3*c*d^2 - a*b^2*d^3)*m)*x^3 +
(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*m^2 + (12*b^3*c^2*d + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*m^2 + (
7*b^3*c^2*d - 8*a*b^2*c*d^2 + a^2*b*d^3)*m)*x^2 + (5*a*b^2*c^3 - 8*a^2*b*c^2*d + 3*a^3*c*d^2)*m + (6*b^3*c^3 +
6*a*b^2*c^2*d - 6*a^2*b*c*d^2 + 2*a^3*d^3 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*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)*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),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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),x, algorithm="giac")

[Out]

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