3.7.10 \(\int \frac {(a+b x)^6}{(a^2-b^2 x^2)^3} \, dx\)

Optimal. Leaf size=49 \[ \frac {4 a^3}{b (a-b x)^2}-\frac {12 a^2}{b (a-b x)}-\frac {6 a \log (a-b x)}{b}-x \]

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {627, 43} \begin {gather*} \frac {4 a^3}{b (a-b x)^2}-\frac {12 a^2}{b (a-b x)}-\frac {6 a \log (a-b x)}{b}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^6/(a^2 - b^2*x^2)^3,x]

[Out]

-x + (4*a^3)/(b*(a - b*x)^2) - (12*a^2)/(b*(a - b*x)) - (6*a*Log[a - b*x])/b

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rubi steps

\begin {align*} \int \frac {(a+b x)^6}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {(a+b x)^3}{(a-b x)^3} \, dx\\ &=\int \left (-1+\frac {8 a^3}{(a-b x)^3}-\frac {12 a^2}{(a-b x)^2}+\frac {6 a}{a-b x}\right ) \, dx\\ &=-x+\frac {4 a^3}{b (a-b x)^2}-\frac {12 a^2}{b (a-b x)}-\frac {6 a \log (a-b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 41, normalized size = 0.84 \begin {gather*} \frac {4 a^2 (3 b x-2 a)}{b (a-b x)^2}-\frac {6 a \log (a-b x)}{b}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^6/(a^2 - b^2*x^2)^3,x]

[Out]

-x + (4*a^2*(-2*a + 3*b*x))/(b*(a - b*x)^2) - (6*a*Log[a - b*x])/b

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^6}{\left (a^2-b^2 x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^6/(a^2 - b^2*x^2)^3,x]

[Out]

IntegrateAlgebraic[(a + b*x)^6/(a^2 - b^2*x^2)^3, x]

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fricas [A]  time = 0.40, size = 82, normalized size = 1.67 \begin {gather*} -\frac {b^{3} x^{3} - 2 \, a b^{2} x^{2} - 11 \, a^{2} b x + 8 \, a^{3} + 6 \, {\left (a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \log \left (b x - a\right )}{b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^6/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

-(b^3*x^3 - 2*a*b^2*x^2 - 11*a^2*b*x + 8*a^3 + 6*(a*b^2*x^2 - 2*a^2*b*x + a^3)*log(b*x - a))/(b^3*x^2 - 2*a*b^
2*x + a^2*b)

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giac [A]  time = 0.17, size = 46, normalized size = 0.94 \begin {gather*} -x - \frac {6 \, a \log \left ({\left | b x - a \right |}\right )}{b} + \frac {4 \, {\left (3 \, a^{2} b x - 2 \, a^{3}\right )}}{{\left (b x - a\right )}^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^6/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

-x - 6*a*log(abs(b*x - a))/b + 4*(3*a^2*b*x - 2*a^3)/((b*x - a)^2*b)

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maple [A]  time = 0.04, size = 53, normalized size = 1.08 \begin {gather*} \frac {4 a^{3}}{\left (b x -a \right )^{2} b}+\frac {12 a^{2}}{\left (b x -a \right ) b}-\frac {6 a \ln \left (b x -a \right )}{b}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^6/(-b^2*x^2+a^2)^3,x)

[Out]

-x-6*a/b*ln(b*x-a)+4*a^3/b/(b*x-a)^2+12/(b*x-a)*a^2/b

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maxima [A]  time = 1.32, size = 55, normalized size = 1.12 \begin {gather*} -x - \frac {6 \, a \log \left (b x - a\right )}{b} + \frac {4 \, {\left (3 \, a^{2} b x - 2 \, a^{3}\right )}}{b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^6/(-b^2*x^2+a^2)^3,x, algorithm="maxima")

[Out]

-x - 6*a*log(b*x - a)/b + 4*(3*a^2*b*x - 2*a^3)/(b^3*x^2 - 2*a*b^2*x + a^2*b)

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mupad [B]  time = 0.06, size = 52, normalized size = 1.06 \begin {gather*} \frac {12\,a^2\,x-\frac {8\,a^3}{b}}{a^2-2\,a\,b\,x+b^2\,x^2}-x-\frac {6\,a\,\ln \left (b\,x-a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^6/(a^2 - b^2*x^2)^3,x)

[Out]

(12*a^2*x - (8*a^3)/b)/(a^2 + b^2*x^2 - 2*a*b*x) - x - (6*a*log(b*x - a))/b

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sympy [A]  time = 0.31, size = 48, normalized size = 0.98 \begin {gather*} - \frac {6 a \log {\left (- a + b x \right )}}{b} - x - \frac {8 a^{3} - 12 a^{2} b x}{a^{2} b - 2 a b^{2} x + b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**6/(-b**2*x**2+a**2)**3,x)

[Out]

-6*a*log(-a + b*x)/b - x - (8*a**3 - 12*a**2*b*x)/(a**2*b - 2*a*b**2*x + b**3*x**2)

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