3.767 \(\int \frac {(a+b x)^8}{(a^2-b^2 x^2)^3} \, dx\)

Optimal. Leaf size=71 \[ \frac {16 a^5}{b (a-b x)^2}-\frac {80 a^4}{b (a-b x)}-\frac {80 a^3 \log (a-b x)}{b}-31 a^2 x-4 a b x^2-\frac {b^2 x^3}{3} \]

[Out]

-31*a^2*x-4*a*b*x^2-1/3*b^2*x^3+16*a^5/b/(-b*x+a)^2-80*a^4/b/(-b*x+a)-80*a^3*ln(-b*x+a)/b

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Rubi [A]  time = 0.05, antiderivative size = 71, 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} \[ \frac {16 a^5}{b (a-b x)^2}-\frac {80 a^4}{b (a-b x)}-\frac {80 a^3 \log (a-b x)}{b}-31 a^2 x-4 a b x^2-\frac {b^2 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^8/(a^2 - b^2*x^2)^3,x]

[Out]

-31*a^2*x - 4*a*b*x^2 - (b^2*x^3)/3 + (16*a^5)/(b*(a - b*x)^2) - (80*a^4)/(b*(a - b*x)) - (80*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)^8}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {(a+b x)^5}{(a-b x)^3} \, dx\\ &=\int \left (-31 a^2-8 a b x-b^2 x^2+\frac {32 a^5}{(a-b x)^3}-\frac {80 a^4}{(a-b x)^2}+\frac {80 a^3}{a-b x}\right ) \, dx\\ &=-31 a^2 x-4 a b x^2-\frac {b^2 x^3}{3}+\frac {16 a^5}{b (a-b x)^2}-\frac {80 a^4}{b (a-b x)}-\frac {80 a^3 \log (a-b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 73, normalized size = 1.03 \[ \frac {16 a^5}{b (b x-a)^2}+\frac {80 a^4}{b (b x-a)}-\frac {80 a^3 \log (a-b x)}{b}-31 a^2 x-4 a b x^2-\frac {b^2 x^3}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^8/(a^2 - b^2*x^2)^3,x]

[Out]

-31*a^2*x - 4*a*b*x^2 - (b^2*x^3)/3 + (16*a^5)/(b*(-a + b*x)^2) + (80*a^4)/(b*(-a + b*x)) - (80*a^3*Log[a - b*
x])/b

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fricas [A]  time = 0.87, size = 106, normalized size = 1.49 \[ -\frac {b^{5} x^{5} + 10 \, a b^{4} x^{4} + 70 \, a^{2} b^{3} x^{3} - 174 \, a^{3} b^{2} x^{2} - 147 \, a^{4} b x + 192 \, a^{5} + 240 \, {\left (a^{3} b^{2} x^{2} - 2 \, a^{4} b x + a^{5}\right )} \log \left (b x - a\right )}{3 \, {\left (b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^8/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

-1/3*(b^5*x^5 + 10*a*b^4*x^4 + 70*a^2*b^3*x^3 - 174*a^3*b^2*x^2 - 147*a^4*b*x + 192*a^5 + 240*(a^3*b^2*x^2 - 2
*a^4*b*x + a^5)*log(b*x - a))/(b^3*x^2 - 2*a*b^2*x + a^2*b)

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giac [A]  time = 0.16, size = 76, normalized size = 1.07 \[ -\frac {80 \, a^{3} \log \left ({\left | b x - a \right |}\right )}{b} + \frac {16 \, {\left (5 \, a^{4} b x - 4 \, a^{5}\right )}}{{\left (b x - a\right )}^{2} b} - \frac {b^{11} x^{3} + 12 \, a b^{10} x^{2} + 93 \, a^{2} b^{9} x}{3 \, b^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^8/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

-80*a^3*log(abs(b*x - a))/b + 16*(5*a^4*b*x - 4*a^5)/((b*x - a)^2*b) - 1/3*(b^11*x^3 + 12*a*b^10*x^2 + 93*a^2*
b^9*x)/b^9

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maple [A]  time = 0.04, size = 73, normalized size = 1.03 \[ -\frac {b^{2} x^{3}}{3}+\frac {16 a^{5}}{\left (b x -a \right )^{2} b}-4 a b \,x^{2}+\frac {80 a^{4}}{\left (b x -a \right ) b}-\frac {80 a^{3} \ln \left (b x -a \right )}{b}-31 a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^8/(-b^2*x^2+a^2)^3,x)

[Out]

-1/3*b^2*x^3-4*a*b*x^2-31*a^2*x-80*a^3/b*ln(b*x-a)+16*a^5/b/(b*x-a)^2+80/(b*x-a)*a^4/b

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maxima [A]  time = 1.40, size = 75, normalized size = 1.06 \[ -\frac {1}{3} \, b^{2} x^{3} - 4 \, a b x^{2} - 31 \, a^{2} x - \frac {80 \, a^{3} \log \left (b x - a\right )}{b} + \frac {16 \, {\left (5 \, a^{4} b x - 4 \, a^{5}\right )}}{b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^8/(-b^2*x^2+a^2)^3,x, algorithm="maxima")

[Out]

-1/3*b^2*x^3 - 4*a*b*x^2 - 31*a^2*x - 80*a^3*log(b*x - a)/b + 16*(5*a^4*b*x - 4*a^5)/(b^3*x^2 - 2*a*b^2*x + a^
2*b)

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mupad [B]  time = 0.05, size = 72, normalized size = 1.01 \[ \frac {80\,a^4\,x-\frac {64\,a^5}{b}}{a^2-2\,a\,b\,x+b^2\,x^2}-31\,a^2\,x-\frac {b^2\,x^3}{3}-\frac {80\,a^3\,\ln \left (b\,x-a\right )}{b}-4\,a\,b\,x^2 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^8/(a^2 - b^2*x^2)^3,x)

[Out]

(80*a^4*x - (64*a^5)/b)/(a^2 + b^2*x^2 - 2*a*b*x) - 31*a^2*x - (b^2*x^3)/3 - (80*a^3*log(b*x - a))/b - 4*a*b*x
^2

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sympy [A]  time = 0.41, size = 71, normalized size = 1.00 \[ - \frac {80 a^{3} \log {\left (- a + b x \right )}}{b} - 31 a^{2} x - 4 a b x^{2} - \frac {b^{2} x^{3}}{3} - \frac {64 a^{5} - 80 a^{4} b x}{a^{2} b - 2 a b^{2} x + b^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**8/(-b**2*x**2+a**2)**3,x)

[Out]

-80*a**3*log(-a + b*x)/b - 31*a**2*x - 4*a*b*x**2 - b**2*x**3/3 - (64*a**5 - 80*a**4*b*x)/(a**2*b - 2*a*b**2*x
+ b**3*x**2)

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