3.8.56 \(\int \frac {1}{(a+b x)^4 (a^2-b^2 x^2)} \, dx\) [756]

Optimal. Leaf size=86 \[ -\frac {1}{8 a b (a+b x)^4}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{16 a^3 b (a+b x)^2}-\frac {1}{16 a^4 b (a+b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{16 a^5 b} \]

[Out]

-1/8/a/b/(b*x+a)^4-1/12/a^2/b/(b*x+a)^3-1/16/a^3/b/(b*x+a)^2-1/16/a^4/b/(b*x+a)+1/16*arctanh(b*x/a)/a^5/b

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Rubi [A]
time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {641, 46, 214} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{16 a^5 b}-\frac {1}{16 a^4 b (a+b x)}-\frac {1}{16 a^3 b (a+b x)^2}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{8 a b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^4*(a^2 - b^2*x^2)),x]

[Out]

-1/8*1/(a*b*(a + b*x)^4) - 1/(12*a^2*b*(a + b*x)^3) - 1/(16*a^3*b*(a + b*x)^2) - 1/(16*a^4*b*(a + b*x)) + ArcT
anh[(b*x)/a]/(16*a^5*b)

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

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 {1}{(a+b x)^4 \left (a^2-b^2 x^2\right )} \, dx &=\int \frac {1}{(a-b x) (a+b x)^5} \, dx\\ &=\int \left (\frac {1}{2 a (a+b x)^5}+\frac {1}{4 a^2 (a+b x)^4}+\frac {1}{8 a^3 (a+b x)^3}+\frac {1}{16 a^4 (a+b x)^2}+\frac {1}{16 a^4 \left (a^2-b^2 x^2\right )}\right ) \, dx\\ &=-\frac {1}{8 a b (a+b x)^4}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{16 a^3 b (a+b x)^2}-\frac {1}{16 a^4 b (a+b x)}+\frac {\int \frac {1}{a^2-b^2 x^2} \, dx}{16 a^4}\\ &=-\frac {1}{8 a b (a+b x)^4}-\frac {1}{12 a^2 b (a+b x)^3}-\frac {1}{16 a^3 b (a+b x)^2}-\frac {1}{16 a^4 b (a+b x)}+\frac {\tanh ^{-1}\left (\frac {b x}{a}\right )}{16 a^5 b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 82, normalized size = 0.95 \begin {gather*} \frac {-2 a \left (16 a^3+19 a^2 b x+12 a b^2 x^2+3 b^3 x^3\right )-3 (a+b x)^4 \log (a-b x)+3 (a+b x)^4 \log (a+b x)}{96 a^5 b (a+b x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^4*(a^2 - b^2*x^2)),x]

[Out]

(-2*a*(16*a^3 + 19*a^2*b*x + 12*a*b^2*x^2 + 3*b^3*x^3) - 3*(a + b*x)^4*Log[a - b*x] + 3*(a + b*x)^4*Log[a + b*
x])/(96*a^5*b*(a + b*x)^4)

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Maple [A]
time = 0.46, size = 91, normalized size = 1.06

method	result	size
norman	\(\frac {-\frac {1}{3 b a}-\frac {b \,x^{2}}{4 a^{3}}-\frac {b^{2} x^{3}}{16 a^{4}}-\frac {19 x}{48 a^{2}}}{\left (b x +a \right )^{4}}-\frac {\ln \left (-b x +a \right )}{32 a^{5} b}+\frac {\ln \left (b x +a \right )}{32 a^{5} b}\)	\(74\)
risch	\(\frac {-\frac {1}{3 b a}-\frac {b \,x^{2}}{4 a^{3}}-\frac {b^{2} x^{3}}{16 a^{4}}-\frac {19 x}{48 a^{2}}}{\left (b x +a \right )^{4}}-\frac {\ln \left (-b x +a \right )}{32 a^{5} b}+\frac {\ln \left (b x +a \right )}{32 a^{5} b}\)	\(74\)
default	\(\frac {\ln \left (b x +a \right )}{32 a^{5} b}-\frac {1}{16 a^{4} b \left (b x +a \right )}-\frac {1}{16 a^{3} b \left (b x +a \right )^{2}}-\frac {1}{12 a^{2} b \left (b x +a \right )^{3}}-\frac {1}{8 a b \left (b x +a \right )^{4}}-\frac {\ln \left (-b x +a \right )}{32 a^{5} b}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^4/(-b^2*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

1/32/a^5/b*ln(b*x+a)-1/16/a^4/b/(b*x+a)-1/16/a^3/b/(b*x+a)^2-1/12/a^2/b/(b*x+a)^3-1/8/a/b/(b*x+a)^4-1/32/a^5/b
*ln(-b*x+a)

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Maxima [A]
time = 0.29, size = 112, normalized size = 1.30 \begin {gather*} -\frac {3 \, b^{3} x^{3} + 12 \, a b^{2} x^{2} + 19 \, a^{2} b x + 16 \, a^{3}}{48 \, {\left (a^{4} b^{5} x^{4} + 4 \, a^{5} b^{4} x^{3} + 6 \, a^{6} b^{3} x^{2} + 4 \, a^{7} b^{2} x + a^{8} b\right )}} + \frac {\log \left (b x + a\right )}{32 \, a^{5} b} - \frac {\log \left (b x - a\right )}{32 \, a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^4/(-b^2*x^2+a^2),x, algorithm="maxima")

[Out]

-1/48*(3*b^3*x^3 + 12*a*b^2*x^2 + 19*a^2*b*x + 16*a^3)/(a^4*b^5*x^4 + 4*a^5*b^4*x^3 + 6*a^6*b^3*x^2 + 4*a^7*b^
2*x + a^8*b) + 1/32*log(b*x + a)/(a^5*b) - 1/32*log(b*x - a)/(a^5*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (76) = 152\).
time = 2.72, size = 178, normalized size = 2.07 \begin {gather*} -\frac {6 \, a b^{3} x^{3} + 24 \, a^{2} b^{2} x^{2} + 38 \, a^{3} b x + 32 \, a^{4} - 3 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x - a\right )}{96 \, {\left (a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{3} + 6 \, a^{7} b^{3} x^{2} + 4 \, a^{8} b^{2} x + a^{9} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^4/(-b^2*x^2+a^2),x, algorithm="fricas")

[Out]

-1/96*(6*a*b^3*x^3 + 24*a^2*b^2*x^2 + 38*a^3*b*x + 32*a^4 - 3*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b
*x + a^4)*log(b*x + a) + 3*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*log(b*x - a))/(a^5*b^5*x^
4 + 4*a^6*b^4*x^3 + 6*a^7*b^3*x^2 + 4*a^8*b^2*x + a^9*b)

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Sympy [A]
time = 0.25, size = 107, normalized size = 1.24 \begin {gather*} - \frac {16 a^{3} + 19 a^{2} b x + 12 a b^{2} x^{2} + 3 b^{3} x^{3}}{48 a^{8} b + 192 a^{7} b^{2} x + 288 a^{6} b^{3} x^{2} + 192 a^{5} b^{4} x^{3} + 48 a^{4} b^{5} x^{4}} - \frac {\frac {\log {\left (- \frac {a}{b} + x \right )}}{32} - \frac {\log {\left (\frac {a}{b} + x \right )}}{32}}{a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**4/(-b**2*x**2+a**2),x)

[Out]

-(16*a**3 + 19*a**2*b*x + 12*a*b**2*x**2 + 3*b**3*x**3)/(48*a**8*b + 192*a**7*b**2*x + 288*a**6*b**3*x**2 + 19
2*a**5*b**4*x**3 + 48*a**4*b**5*x**4) - (log(-a/b + x)/32 - log(a/b + x)/32)/(a**5*b)

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Giac [A]
time = 0.82, size = 81, normalized size = 0.94 \begin {gather*} \frac {\log \left ({\left | b x + a \right |}\right )}{32 \, a^{5} b} - \frac {\log \left ({\left | b x - a \right |}\right )}{32 \, a^{5} b} - \frac {3 \, a b^{3} x^{3} + 12 \, a^{2} b^{2} x^{2} + 19 \, a^{3} b x + 16 \, a^{4}}{48 \, {\left (b x + a\right )}^{4} a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^4/(-b^2*x^2+a^2),x, algorithm="giac")

[Out]

1/32*log(abs(b*x + a))/(a^5*b) - 1/32*log(abs(b*x - a))/(a^5*b) - 1/48*(3*a*b^3*x^3 + 12*a^2*b^2*x^2 + 19*a^3*
b*x + 16*a^4)/((b*x + a)^4*a^5*b)

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Mupad [B]
time = 0.44, size = 93, normalized size = 1.08 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{16\,a^5\,b}-\frac {\frac {19\,x}{48\,a^2}+\frac {1}{3\,a\,b}+\frac {b\,x^2}{4\,a^3}+\frac {b^2\,x^3}{16\,a^4}}{a^4+4\,a^3\,b\,x+6\,a^2\,b^2\,x^2+4\,a\,b^3\,x^3+b^4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a^2 - b^2*x^2)*(a + b*x)^4),x)

[Out]

atanh((b*x)/a)/(16*a^5*b) - ((19*x)/(48*a^2) + 1/(3*a*b) + (b*x^2)/(4*a^3) + (b^2*x^3)/(16*a^4))/(a^4 + b^4*x^
4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x)

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