3.11.62 \(\int \frac {1}{(a+b x)^2 (a c-b c x)^3} \, dx\) [1062]

Optimal. Leaf size=83 \[ \frac {1}{8 a^2 b c^3 (a-b x)^2}+\frac {1}{4 a^3 b c^3 (a-b x)}-\frac {1}{8 a^3 b c^3 (a+b x)}+\frac {3 \tanh ^{-1}\left (\frac {b x}{a}\right )}{8 a^4 b c^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {46, 214} \begin {gather*} \frac {3 \tanh ^{-1}\left (\frac {b x}{a}\right )}{8 a^4 b c^3}+\frac {1}{4 a^3 b c^3 (a-b x)}-\frac {1}{8 a^3 b c^3 (a+b x)}+\frac {1}{8 a^2 b c^3 (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^2*(a*c - b*c*x)^3),x]

[Out]

1/(8*a^2*b*c^3*(a - b*x)^2) + 1/(4*a^3*b*c^3*(a - b*x)) - 1/(8*a^3*b*c^3*(a + b*x)) + (3*ArcTanh[(b*x)/a])/(8*
a^4*b*c^3)

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 68, normalized size = 0.82 \begin {gather*} \frac {\frac {2 a \left (2 a^2+3 a b x-3 b^2 x^2\right )}{(a-b x)^2 (a+b x)}-3 \log (a-b x)+3 \log (a+b x)}{16 a^4 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^2*(a*c - b*c*x)^3),x]

[Out]

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

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Mathics [A]
time = 2.78, size = 117, normalized size = 1.41 \begin {gather*} \frac {2 a \left (2 a^2+3 a b x-3 b^2 x^2\right )+3 \left (a^3-a^2 b x-a b^2 x^2+b^3 x^3\right ) \left (\text {Log}\left [\frac {a+b x}{b}\right ]-\text {Log}\left [\frac {-a+b x}{b}\right ]\right )}{16 a^4 b c^3 \left (a^3-a^2 b x-a b^2 x^2+b^3 x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(a + b*x)^2/(a*c - b*c*x)^3,x]')

[Out]

(2 a (2 a ^ 2 + 3 a b x - 3 b ^ 2 x ^ 2) + 3 (a ^ 3 - a ^ 2 b x - a b ^ 2 x ^ 2 + b ^ 3 x ^ 3) (Log[(a + b x)
/ b] - Log[(-a + b x) / b])) / (16 a ^ 4 b c ^ 3 (a ^ 3 - a ^ 2 b x - a b ^ 2 x ^ 2 + b ^ 3 x ^ 3))

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Maple [A]
time = 0.18, size = 82, normalized size = 0.99

method	result	size
risch	\(\frac {-\frac {3 b \,x^{2}}{8 a^{3}}+\frac {3 x}{8 a^{2}}+\frac {1}{4 a b}}{\left (b x +a \right ) c^{3} \left (-b x +a \right )^{2}}-\frac {3 \ln \left (-b x +a \right )}{16 a^{4} c^{3} b}+\frac {3 \ln \left (b x +a \right )}{16 a^{4} c^{3} b}\)	\(80\)
default	\(\frac {-\frac {3 \ln \left (-b x +a \right )}{16 a^{4} b}+\frac {1}{4 a^{3} b \left (-b x +a \right )}+\frac {1}{8 a^{2} b \left (-b x +a \right )^{2}}+\frac {3 \ln \left (b x +a \right )}{16 a^{4} b}-\frac {1}{8 a^{3} b \left (b x +a \right )}}{c^{3}}\)	\(82\)
norman	\(\frac {\frac {1}{4 a b c}+\frac {3 x}{8 a^{2} c}-\frac {3 b \,x^{2}}{8 a^{3} c}}{\left (b x +a \right ) c^{2} \left (-b x +a \right )^{2}}-\frac {3 \ln \left (-b x +a \right )}{16 a^{4} c^{3} b}+\frac {3 \ln \left (b x +a \right )}{16 a^{4} c^{3} b}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^2/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 108, normalized size = 1.30 \begin {gather*} -\frac {3 \, b^{2} x^{2} - 3 \, a b x - 2 \, a^{2}}{8 \, {\left (a^{3} b^{4} c^{3} x^{3} - a^{4} b^{3} c^{3} x^{2} - a^{5} b^{2} c^{3} x + a^{6} b c^{3}\right )}} + \frac {3 \, \log \left (b x + a\right )}{16 \, a^{4} b c^{3}} - \frac {3 \, \log \left (b x - a\right )}{16 \, a^{4} b c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c)^3,x, algorithm="maxima")

[Out]

-1/8*(3*b^2*x^2 - 3*a*b*x - 2*a^2)/(a^3*b^4*c^3*x^3 - a^4*b^3*c^3*x^2 - a^5*b^2*c^3*x + a^6*b*c^3) + 3/16*log(
b*x + a)/(a^4*b*c^3) - 3/16*log(b*x - a)/(a^4*b*c^3)

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Fricas [A]
time = 0.30, size = 146, normalized size = 1.76 \begin {gather*} -\frac {6 \, a b^{2} x^{2} - 6 \, a^{2} b x - 4 \, a^{3} - 3 \, {\left (b^{3} x^{3} - a b^{2} x^{2} - a^{2} b x + a^{3}\right )} \log \left (b x + a\right ) + 3 \, {\left (b^{3} x^{3} - a b^{2} x^{2} - a^{2} b x + a^{3}\right )} \log \left (b x - a\right )}{16 \, {\left (a^{4} b^{4} c^{3} x^{3} - a^{5} b^{3} c^{3} x^{2} - a^{6} b^{2} c^{3} x + a^{7} b c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c)^3,x, algorithm="fricas")

[Out]

-1/16*(6*a*b^2*x^2 - 6*a^2*b*x - 4*a^3 - 3*(b^3*x^3 - a*b^2*x^2 - a^2*b*x + a^3)*log(b*x + a) + 3*(b^3*x^3 - a
*b^2*x^2 - a^2*b*x + a^3)*log(b*x - a))/(a^4*b^4*c^3*x^3 - a^5*b^3*c^3*x^2 - a^6*b^2*c^3*x + a^7*b*c^3)

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Sympy [A]
time = 0.25, size = 104, normalized size = 1.25 \begin {gather*} - \frac {- 2 a^{2} - 3 a b x + 3 b^{2} x^{2}}{8 a^{6} b c^{3} - 8 a^{5} b^{2} c^{3} x - 8 a^{4} b^{3} c^{3} x^{2} + 8 a^{3} b^{4} c^{3} x^{3}} - \frac {\frac {3 \log {\left (- \frac {a}{b} + x \right )}}{16} - \frac {3 \log {\left (\frac {a}{b} + x \right )}}{16}}{a^{4} b c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**2/(-b*c*x+a*c)**3,x)

[Out]

-(-2*a**2 - 3*a*b*x + 3*b**2*x**2)/(8*a**6*b*c**3 - 8*a**5*b**2*c**3*x - 8*a**4*b**3*c**3*x**2 + 8*a**3*b**4*c
**3*x**3) - (3*log(-a/b + x)/16 - 3*log(a/b + x)/16)/(a**4*b*c**3)

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Giac [A]
time = 0.00, size = 93, normalized size = 1.12 \begin {gather*} \frac {3 \ln \left |x b+a\right |}{16 b a^{4} c^{3}}-\frac {3 \ln \left |x b-a\right |}{16 b a^{4} c^{3}}+\frac {\frac {1}{32} \left (-12 b^{2} a x^{2}+12 b a^{2} x+8 a^{3}\right )}{a^{4} c^{3} b \left (-x b+a\right )^{2} \left (x b+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c)^3,x)

[Out]

3/16*log(abs(b*x + a))/(a^4*b*c^3) - 3/16*log(abs(b*x - a))/(a^4*b*c^3) - 1/8*(3*a*b^2*x^2 - 3*a^2*b*x - 2*a^3
)/((b*x + a)*(b*x - a)^2*a^4*b*c^3)

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Mupad [B]
time = 0.10, size = 86, normalized size = 1.04 \begin {gather*} \frac {\frac {3\,x}{8\,a^2}+\frac {1}{4\,a\,b}-\frac {3\,b\,x^2}{8\,a^3}}{a^3\,c^3-a^2\,b\,c^3\,x-a\,b^2\,c^3\,x^2+b^3\,c^3\,x^3}+\frac {3\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{8\,a^4\,b\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*c - b*c*x)^3*(a + b*x)^2),x)

[Out]

((3*x)/(8*a^2) + 1/(4*a*b) - (3*b*x^2)/(8*a^3))/(a^3*c^3 + b^3*c^3*x^3 - a*b^2*c^3*x^2 - a^2*b*c^3*x) + (3*ata
nh((b*x)/a))/(8*a^4*b*c^3)

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