3.8.68 \(\int \frac {(a+b x)^7}{(a^2-b^2 x^2)^3} \, dx\) [768]

Optimal. Leaf size=60 \[ -7 a x-\frac {b x^2}{2}+\frac {8 a^4}{b (a-b x)^2}-\frac {32 a^3}{b (a-b x)}-\frac {24 a^2 \log (a-b x)}{b} \]

[Out]

-7*a*x-1/2*b*x^2+8*a^4/b/(-b*x+a)^2-32*a^3/b/(-b*x+a)-24*a^2*ln(-b*x+a)/b

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Rubi [A]
time = 0.03, antiderivative size = 60, 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 = {641, 45} \begin {gather*} \frac {8 a^4}{b (a-b x)^2}-\frac {32 a^3}{b (a-b x)}-\frac {24 a^2 \log (a-b x)}{b}-7 a x-\frac {b x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^7/(a^2 - b^2*x^2)^3,x]

[Out]

-7*a*x - (b*x^2)/2 + (8*a^4)/(b*(a - b*x)^2) - (32*a^3)/(b*(a - b*x)) - (24*a^2*Log[a - b*x])/b

Rule 45

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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)^7}{\left (a^2-b^2 x^2\right )^3} \, dx &=\int \frac {(a+b x)^4}{(a-b x)^3} \, dx\\ &=\int \left (-7 a-b x+\frac {16 a^4}{(a-b x)^3}-\frac {32 a^3}{(a-b x)^2}+\frac {24 a^2}{a-b x}\right ) \, dx\\ &=-7 a x-\frac {b x^2}{2}+\frac {8 a^4}{b (a-b x)^2}-\frac {32 a^3}{b (a-b x)}-\frac {24 a^2 \log (a-b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 62, normalized size = 1.03 \begin {gather*} -7 a x-\frac {b x^2}{2}+\frac {8 a^4}{b (-a+b x)^2}+\frac {32 a^3}{b (-a+b x)}-\frac {24 a^2 \log (a-b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^7/(a^2 - b^2*x^2)^3,x]

[Out]

-7*a*x - (b*x^2)/2 + (8*a^4)/(b*(-a + b*x)^2) + (32*a^3)/(b*(-a + b*x)) - (24*a^2*Log[a - b*x])/b

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Maple [A]
time = 0.45, size = 59, normalized size = 0.98

method	result	size
risch	\(-\frac {b \,x^{2}}{2}-7 a x +\frac {32 a^{3} x -\frac {24 a^{4}}{b}}{\left (-b x +a \right )^{2}}-\frac {24 a^{2} \ln \left (-b x +a \right )}{b}\)	\(51\)
default	\(-7 a x -\frac {b \,x^{2}}{2}+\frac {8 a^{4}}{b \left (-b x +a \right )^{2}}-\frac {32 a^{3}}{b \left (-b x +a \right )}-\frac {24 a^{2} \ln \left (-b x +a \right )}{b}\)	\(59\)
norman	\(\frac {-23 a^{5} x -\frac {b^{5} x^{6}}{2}-7 a \,b^{4} x^{5}+46 a^{3} b^{2} x^{3}-\frac {25 a^{6}}{b}+\frac {83 a^{4} b \,x^{2}}{2}}{\left (-b^{2} x^{2}+a^{2}\right )^{2}}-\frac {24 a^{2} \ln \left (-b x +a \right )}{b}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^7/(-b^2*x^2+a^2)^3,x,method=_RETURNVERBOSE)

[Out]

-7*a*x-1/2*b*x^2+8*a^4/b/(-b*x+a)^2-32*a^3/b/(-b*x+a)-24*a^2*ln(-b*x+a)/b

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Maxima [A]
time = 0.30, size = 64, normalized size = 1.07 \begin {gather*} -\frac {1}{2} \, b x^{2} - 7 \, a x - \frac {24 \, a^{2} \log \left (b x - a\right )}{b} + \frac {8 \, {\left (4 \, a^{3} b x - 3 \, a^{4}\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)^7/(-b^2*x^2+a^2)^3,x, algorithm="maxima")

[Out]

-1/2*b*x^2 - 7*a*x - 24*a^2*log(b*x - a)/b + 8*(4*a^3*b*x - 3*a^4)/(b^3*x^2 - 2*a*b^2*x + a^2*b)

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Fricas [A]
time = 2.42, size = 95, normalized size = 1.58 \begin {gather*} -\frac {b^{4} x^{4} + 12 \, a b^{3} x^{3} - 27 \, a^{2} b^{2} x^{2} - 50 \, a^{3} b x + 48 \, a^{4} + 48 \, {\left (a^{2} b^{2} x^{2} - 2 \, a^{3} b x + a^{4}\right )} \log \left (b x - a\right )}{2 \, {\left (b^{3} x^{2} - 2 \, a b^{2} x + a^{2} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^7/(-b^2*x^2+a^2)^3,x, algorithm="fricas")

[Out]

-1/2*(b^4*x^4 + 12*a*b^3*x^3 - 27*a^2*b^2*x^2 - 50*a^3*b*x + 48*a^4 + 48*(a^2*b^2*x^2 - 2*a^3*b*x + a^4)*log(b
*x - a))/(b^3*x^2 - 2*a*b^2*x + a^2*b)

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Sympy [A]
time = 0.18, size = 60, normalized size = 1.00 \begin {gather*} - \frac {24 a^{2} \log {\left (- a + b x \right )}}{b} - 7 a x - \frac {b x^{2}}{2} - \frac {24 a^{4} - 32 a^{3} 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)**7/(-b**2*x**2+a**2)**3,x)

[Out]

-24*a**2*log(-a + b*x)/b - 7*a*x - b*x**2/2 - (24*a**4 - 32*a**3*b*x)/(a**2*b - 2*a*b**2*x + b**3*x**2)

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Giac [A]
time = 0.89, size = 65, normalized size = 1.08 \begin {gather*} -\frac {24 \, a^{2} \log \left ({\left | b x - a \right |}\right )}{b} + \frac {8 \, {\left (4 \, a^{3} b x - 3 \, a^{4}\right )}}{{\left (b x - a\right )}^{2} b} - \frac {b^{7} x^{2} + 14 \, a b^{6} x}{2 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^7/(-b^2*x^2+a^2)^3,x, algorithm="giac")

[Out]

-24*a^2*log(abs(b*x - a))/b + 8*(4*a^3*b*x - 3*a^4)/((b*x - a)^2*b) - 1/2*(b^7*x^2 + 14*a*b^6*x)/b^6

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Mupad [B]
time = 0.05, size = 61, normalized size = 1.02 \begin {gather*} \frac {32\,a^3\,x-\frac {24\,a^4}{b}}{a^2-2\,a\,b\,x+b^2\,x^2}-7\,a\,x-\frac {b\,x^2}{2}-\frac {24\,a^2\,\ln \left (b\,x-a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^7/(a^2 - b^2*x^2)^3,x)

[Out]

(32*a^3*x - (24*a^4)/b)/(a^2 + b^2*x^2 - 2*a*b*x) - 7*a*x - (b*x^2)/2 - (24*a^2*log(b*x - a))/b

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