3.16.74 \(\int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx\)

Optimal. Leaf size=95 \[ \frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-1}}{a^2 b c^2 (m+1) (m+2)}-\frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)} \]

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Rubi [A]  time = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {45, 37} \begin {gather*} \frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-1}}{a^2 b c^2 (m+1) (m+2)}-\frac {(a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m-2}}{a b c (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(a*c*(1 + m) + b*c*(2 + m)*x)^(-3 - m),x]

[Out]

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

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Mathematica [A]  time = 0.07, size = 54, normalized size = 0.57 \begin {gather*} \frac {x (a+b x)^{m+1} (a c (m+1)+b c (m+2) x)^{-m}}{a^2 c^3 (m+1) (a (m+1)+b (m+2) x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(a*c*(1 + m) + b*c*(2 + m)*x)^(-3 - m),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x)^m (a c (1+m)+b c (2+m) x)^{-3-m} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^m*(a*c*(1 + m) + b*c*(2 + m)*x)^(-3 - m),x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x)^m*(a*c*(1 + m) + b*c*(2 + m)*x)^(-3 - m), x]

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fricas [A]  time = 1.32, size = 85, normalized size = 0.89 \begin {gather*} \frac {{\left ({\left (b^{2} m + 2 \, b^{2}\right )} x^{3} + {\left (2 \, a b m + 3 \, a b\right )} x^{2} + {\left (a^{2} m + a^{2}\right )} x\right )} {\left (a c m + a c + {\left (b c m + 2 \, b c\right )} x\right )}^{-m - 3} {\left (b x + a\right )}^{m}}{a^{2} m + a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b c {\left (m + 2\right )} x + a c {\left (m + 1\right )}\right )}^{-m - 3} {\left (b x + a\right )}^{m}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x, algorithm="giac")

[Out]

integrate((b*c*(m + 2)*x + a*c*(m + 1))^(-m - 3)*(b*x + a)^m, x)

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maple [A]  time = 0.01, size = 57, normalized size = 0.60 \begin {gather*} \frac {\left (b x m +a m +2 b x +a \right ) x \left (b x +a \right )^{m +1} \left (b c x m +a c m +2 b c x +a c \right )^{-m -3}}{\left (m +1\right ) a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(a*c*(m+1)+b*c*(m+2)*x)^(-3-m),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b c {\left (m + 2\right )} x + a c {\left (m + 1\right )}\right )}^{-m - 3} {\left (b x + a\right )}^{m}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(a*c*(1+m)+b*c*(2+m)*x)^(-3-m),x, algorithm="maxima")

[Out]

integrate((b*c*(m + 2)*x + a*c*(m + 1))^(-m - 3)*(b*x + a)^m, x)

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mupad [B]  time = 1.04, size = 81, normalized size = 0.85 \begin {gather*} \frac {x\,{\left (a+b\,x\right )}^m+\frac {b\,x^2\,\left (2\,m+3\right )\,{\left (a+b\,x\right )}^m}{a\,\left (m+1\right )}+\frac {b^2\,x^3\,\left (m+2\right )\,{\left (a+b\,x\right )}^m}{a^2\,\left (m+1\right )}}{{\left (a\,c\,\left (m+1\right )+b\,c\,x\,\left (m+2\right )\right )}^{m+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m/(a*c*(m + 1) + b*c*x*(m + 2))^(m + 3),x)

[Out]

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

[Out]

Timed out

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