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

Optimal. Leaf size=55 \[ \frac {16 a^4}{b (a-b x)}+\frac {32 a^3 \log (a-b x)}{b}+17 a^2 x+3 a b x^2+\frac {b^2 x^3}{3} \]

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Rubi [A]  time = 0.04, antiderivative size = 55, 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 {16 a^4}{b (a-b x)}+\frac {32 a^3 \log (a-b x)}{b}+17 a^2 x+3 a b x^2+\frac {b^2 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^6/(a^2 - b^2*x^2)^2,x]

[Out]

17*a^2*x + 3*a*b*x^2 + (b^2*x^3)/3 + (16*a^4)/(b*(a - b*x)) + (32*a^3*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 )^2} \, dx &=\int \frac {(a+b x)^4}{(a-b x)^2} \, dx\\ &=\int \left (17 a^2+6 a b x+b^2 x^2+\frac {16 a^4}{(a-b x)^2}-\frac {32 a^3}{a-b x}\right ) \, dx\\ &=17 a^2 x+3 a b x^2+\frac {b^2 x^3}{3}+\frac {16 a^4}{b (a-b x)}+\frac {32 a^3 \log (a-b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 1.02 \begin {gather*} -\frac {16 a^4}{b (b x-a)}+\frac {32 a^3 \log (a-b x)}{b}+17 a^2 x+3 a b x^2+\frac {b^2 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

17*a^2*x + 3*a*b*x^2 + (b^2*x^3)/3 - (16*a^4)/(b*(-a + b*x)) + (32*a^3*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 )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 76, normalized size = 1.38 \begin {gather*} \frac {b^{4} x^{4} + 8 \, a b^{3} x^{3} + 42 \, a^{2} b^{2} x^{2} - 51 \, a^{3} b x - 48 \, a^{4} + 96 \, {\left (a^{3} b x - a^{4}\right )} \log \left (b x - a\right )}{3 \, {\left (b^{2} x - a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^4*x^4 + 8*a*b^3*x^3 + 42*a^2*b^2*x^2 - 51*a^3*b*x - 48*a^4 + 96*(a^3*b*x - a^4)*log(b*x - a))/(b^2*x -
a*b)

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giac [A]  time = 0.19, size = 66, normalized size = 1.20 \begin {gather*} \frac {32 \, a^{3} \log \left ({\left | b x - a \right |}\right )}{b} - \frac {16 \, a^{4}}{{\left (b x - a\right )} b} + \frac {b^{8} x^{3} + 9 \, a b^{7} x^{2} + 51 \, a^{2} b^{6} x}{3 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32*a^3*log(abs(b*x - a))/b - 16*a^4/((b*x - a)*b) + 1/3*(b^8*x^3 + 9*a*b^7*x^2 + 51*a^2*b^6*x)/b^6

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maple [A]  time = 0.05, size = 56, normalized size = 1.02 \begin {gather*} \frac {b^{2} x^{3}}{3}+3 a b \,x^{2}-\frac {16 a^{4}}{\left (b x -a \right ) b}+\frac {32 a^{3} \ln \left (b x -a \right )}{b}+17 a^{2} 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)^2,x)

[Out]

1/3*b^2*x^3+3*a*b*x^2+17*a^2*x+32*a^3/b*ln(b*x-a)-16*a^4/b/(b*x-a)

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maxima [A]  time = 1.39, size = 55, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} + 3 \, a b x^{2} - \frac {16 \, a^{4}}{b^{2} x - a b} + 17 \, a^{2} x + \frac {32 \, a^{3} \log \left (b x - a\right )}{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)^2,x, algorithm="maxima")

[Out]

1/3*b^2*x^3 + 3*a*b*x^2 - 16*a^4/(b^2*x - a*b) + 17*a^2*x + 32*a^3*log(b*x - a)/b

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mupad [B]  time = 0.39, size = 53, normalized size = 0.96 \begin {gather*} 17\,a^2\,x+\frac {b^2\,x^3}{3}+\frac {16\,a^4}{b\,\left (a-b\,x\right )}+\frac {32\,a^3\,\ln \left (a-b\,x\right )}{b}+3\,a\,b\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^6/(a^2 - b^2*x^2)^2,x)

[Out]

17*a^2*x + (b^2*x^3)/3 + (16*a^4)/(b*(a - b*x)) + (32*a^3*log(a - b*x))/b + 3*a*b*x^2

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sympy [A]  time = 0.28, size = 49, normalized size = 0.89 \begin {gather*} - \frac {16 a^{4}}{- a b + b^{2} x} + \frac {32 a^{3} \log {\left (- a + b x \right )}}{b} + 17 a^{2} x + 3 a b x^{2} + \frac {b^{2} x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*a**4/(-a*b + b**2*x) + 32*a**3*log(-a + b*x)/b + 17*a**2*x + 3*a*b*x**2 + b**2*x**3/3

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