3.10.57 \(\int \frac {(a+b x)^5}{(a c+b c x)^8} \, dx\)

Optimal. Leaf size=17 \[ -\frac {1}{2 b c^8 (a+b x)^2} \]

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Rubi [A]  time = 0.00, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {21, 32} \begin {gather*} -\frac {1}{2 b c^8 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^8,x]

[Out]

-1/(2*b*c^8*(a + b*x)^2)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {align*} \int \frac {(a+b x)^5}{(a c+b c x)^8} \, dx &=\frac {\int \frac {1}{(a+b x)^3} \, dx}{c^8}\\ &=-\frac {1}{2 b c^8 (a+b x)^2}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} -\frac {1}{2 b c^8 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^8,x]

[Out]

-1/2*1/(b*c^8*(a + b*x)^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^5}{(a c+b c x)^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^5/(a*c + b*c*x)^8,x]

[Out]

IntegrateAlgebraic[(a + b*x)^5/(a*c + b*c*x)^8, x]

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fricas [B]  time = 1.36, size = 33, normalized size = 1.94 \begin {gather*} -\frac {1}{2 \, {\left (b^{3} c^{8} x^{2} + 2 \, a b^{2} c^{8} x + a^{2} b c^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^8,x, algorithm="fricas")

[Out]

-1/2/(b^3*c^8*x^2 + 2*a*b^2*c^8*x + a^2*b*c^8)

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giac [A]  time = 1.05, size = 15, normalized size = 0.88 \begin {gather*} -\frac {1}{2 \, {\left (b x + a\right )}^{2} b c^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^8,x, algorithm="giac")

[Out]

-1/2/((b*x + a)^2*b*c^8)

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maple [A]  time = 0.00, size = 16, normalized size = 0.94 \begin {gather*} -\frac {1}{2 \left (b x +a \right )^{2} b \,c^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^5/(b*c*x+a*c)^8,x)

[Out]

-1/2/b/c^8/(b*x+a)^2

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maxima [B]  time = 1.33, size = 33, normalized size = 1.94 \begin {gather*} -\frac {1}{2 \, {\left (b^{3} c^{8} x^{2} + 2 \, a b^{2} c^{8} x + a^{2} b c^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^8,x, algorithm="maxima")

[Out]

-1/2/(b^3*c^8*x^2 + 2*a*b^2*c^8*x + a^2*b*c^8)

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mupad [B]  time = 0.15, size = 35, normalized size = 2.06 \begin {gather*} -\frac {1}{2\,a^2\,b\,c^8+4\,a\,b^2\,c^8\,x+2\,b^3\,c^8\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^5/(a*c + b*c*x)^8,x)

[Out]

-1/(2*a^2*b*c^8 + 2*b^3*c^8*x^2 + 4*a*b^2*c^8*x)

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sympy [B]  time = 0.26, size = 36, normalized size = 2.12 \begin {gather*} - \frac {1}{2 a^{2} b c^{8} + 4 a b^{2} c^{8} x + 2 b^{3} c^{8} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**5/(b*c*x+a*c)**8,x)

[Out]

-1/(2*a**2*b*c**8 + 4*a*b**2*c**8*x + 2*b**3*c**8*x**2)

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