3.8.59 \(\int \frac {(a+b x)^5}{(a^2-b^2 x^2)^2} \, dx\) [759]

Optimal. Leaf size=44 \[ 5 a x+\frac {b x^2}{2}+\frac {8 a^3}{b (a-b x)}+\frac {12 a^2 \log (a-b x)}{b} \]

[Out]

5*a*x+1/2*b*x^2+8*a^3/b/(-b*x+a)+12*a^2*ln(-b*x+a)/b

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Rubi [A]
time = 0.02, antiderivative size = 44, 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^3}{b (a-b x)}+\frac {12 a^2 \log (a-b x)}{b}+5 a x+\frac {b x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^5/(a^2 - b^2*x^2)^2,x]

[Out]

5*a*x + (b*x^2)/2 + (8*a^3)/(b*(a - b*x)) + (12*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)^5}{\left (a^2-b^2 x^2\right )^2} \, dx &=\int \frac {(a+b x)^3}{(a-b x)^2} \, dx\\ &=\int \left (5 a+b x+\frac {8 a^3}{(a-b x)^2}-\frac {12 a^2}{a-b x}\right ) \, dx\\ &=5 a x+\frac {b x^2}{2}+\frac {8 a^3}{b (a-b x)}+\frac {12 a^2 \log (a-b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 45, normalized size = 1.02 \begin {gather*} 5 a x+\frac {b x^2}{2}-\frac {8 a^3}{b (-a+b x)}+\frac {12 a^2 \log (a-b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^5/(a^2 - b^2*x^2)^2,x]

[Out]

5*a*x + (b*x^2)/2 - (8*a^3)/(b*(-a + b*x)) + (12*a^2*Log[a - b*x])/b

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Maple [A]
time = 0.44, size = 43, normalized size = 0.98

method	result	size
default	\(5 a x +\frac {b \,x^{2}}{2}+\frac {8 a^{3}}{b \left (-b x +a \right )}+\frac {12 a^{2} \ln \left (-b x +a \right )}{b}\)	\(43\)
risch	\(5 a x +\frac {b \,x^{2}}{2}+\frac {8 a^{3}}{b \left (-b x +a \right )}+\frac {12 a^{2} \ln \left (-b x +a \right )}{b}\)	\(43\)
norman	\(\frac {13 a^{3} x -\frac {b^{3} x^{4}}{2}-5 a \,b^{2} x^{3}+\frac {17 a^{4}}{2 b}}{-b^{2} x^{2}+a^{2}}+\frac {12 a^{2} \ln \left (-b x +a \right )}{b}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^5/(-b^2*x^2+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

5*a*x+1/2*b*x^2+8*a^3/b/(-b*x+a)+12*a^2*ln(-b*x+a)/b

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Maxima [A]
time = 0.32, size = 44, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, b x^{2} - \frac {8 \, a^{3}}{b^{2} x - a b} + 5 \, a x + \frac {12 \, a^{2} \log \left (b x - a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(-b^2*x^2+a^2)^2,x, algorithm="maxima")

[Out]

1/2*b*x^2 - 8*a^3/(b^2*x - a*b) + 5*a*x + 12*a^2*log(b*x - a)/b

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Fricas [A]
time = 2.92, size = 65, normalized size = 1.48 \begin {gather*} \frac {b^{3} x^{3} + 9 \, a b^{2} x^{2} - 10 \, a^{2} b x - 16 \, a^{3} + 24 \, {\left (a^{2} b x - a^{3}\right )} \log \left (b x - a\right )}{2 \, {\left (b^{2} x - a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(-b^2*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/2*(b^3*x^3 + 9*a*b^2*x^2 - 10*a^2*b*x - 16*a^3 + 24*(a^2*b*x - a^3)*log(b*x - a))/(b^2*x - a*b)

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Sympy [A]
time = 0.09, size = 37, normalized size = 0.84 \begin {gather*} - \frac {8 a^{3}}{- a b + b^{2} x} + \frac {12 a^{2} \log {\left (- a + b x \right )}}{b} + 5 a x + \frac {b x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**5/(-b**2*x**2+a**2)**2,x)

[Out]

-8*a**3/(-a*b + b**2*x) + 12*a**2*log(-a + b*x)/b + 5*a*x + b*x**2/2

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Giac [A]
time = 0.97, size = 55, normalized size = 1.25 \begin {gather*} \frac {12 \, a^{2} \log \left ({\left | b x - a \right |}\right )}{b} - \frac {8 \, a^{3}}{{\left (b x - a\right )} b} + \frac {b^{5} x^{2} + 10 \, a b^{4} x}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(-b^2*x^2+a^2)^2,x, algorithm="giac")

[Out]

12*a^2*log(abs(b*x - a))/b - 8*a^3/((b*x - a)*b) + 1/2*(b^5*x^2 + 10*a*b^4*x)/b^4

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Mupad [B]
time = 0.05, size = 42, normalized size = 0.95 \begin {gather*} 5\,a\,x+\frac {b\,x^2}{2}+\frac {8\,a^3}{b\,\left (a-b\,x\right )}+\frac {12\,a^2\,\ln \left (a-b\,x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^5/(a^2 - b^2*x^2)^2,x)

[Out]

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

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