3.26.37 \(\int (a+b x)^m (c+d x)^{-3-m} \, dx\)

Optimal. Leaf size=79 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d)}+\frac {b (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (b c-a d)^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {45, 37} \begin {gather*} \frac {(a+b x)^{m+1} (c+d x)^{-m-2}}{(m+2) (b c-a d)}+\frac {b (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (m+2) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-3 - m),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 59, normalized size = 0.75 \begin {gather*} \frac {(a+b x)^{m+1} (c+d x)^{-m-2} (-a d (m+1)+b c (m+2)+b d x)}{(m+1) (m+2) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^(-3 - m),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x)^m (c+d x)^{-3-m} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^m*(c + d*x)^(-3 - m),x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x)^m*(c + d*x)^(-3 - m), x]

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fricas [B]  time = 1.68, size = 205, normalized size = 2.59 \begin {gather*} \frac {{\left (b^{2} d^{2} x^{3} + 2 \, a b c^{2} - a^{2} c d + {\left (3 \, b^{2} c d + {\left (b^{2} c d - a b d^{2}\right )} m\right )} x^{2} + {\left (a b c^{2} - a^{2} c d\right )} m + {\left (2 \, b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} + {\left (b^{2} c^{2} - a^{2} d^{2}\right )} m\right )} x\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}}{2 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m),x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 3), x)

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maple [A]  time = 0.00, size = 124, normalized size = 1.57 \begin {gather*} -\frac {\left (a d m -b c m -b d x +a d -2 b c \right ) \left (b x +a \right )^{m +1} \left (d x +c \right )^{-m -2}}{a^{2} d^{2} m^{2}-2 a b c d \,m^{2}+b^{2} c^{2} m^{2}+3 a^{2} d^{2} m -6 a b c d m +3 b^{2} c^{2} m +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-m-3),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m),x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - 3), x)

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mupad [B]  time = 2.79, size = 214, normalized size = 2.71 \begin {gather*} \frac {\frac {x\,{\left (a+b\,x\right )}^m\,\left (2\,b^2\,c^2-a^2\,d^2-a^2\,d^2\,m+b^2\,c^2\,m+2\,a\,b\,c\,d\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m^2+3\,m+2\right )}-\frac {a\,c\,{\left (a+b\,x\right )}^m\,\left (a\,d-2\,b\,c+a\,d\,m-b\,c\,m\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m^2+3\,m+2\right )}+\frac {b^2\,d^2\,x^3\,{\left (a+b\,x\right )}^m}{{\left (a\,d-b\,c\right )}^2\,\left (m^2+3\,m+2\right )}+\frac {b\,d\,x^2\,{\left (a+b\,x\right )}^m\,\left (3\,b\,c-a\,d\,m+b\,c\,m\right )}{{\left (a\,d-b\,c\right )}^2\,\left (m^2+3\,m+2\right )}}{{\left (c+d\,x\right )}^{m+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m/(c + d*x)^(m + 3),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)**(-3-m),x)

[Out]

Timed out

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