3.2873 \(\int \frac{1}{(c+d x) (a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=59 \[ -\frac{\log \left (a+b (c+d x)^3\right )}{3 a^2 d}+\frac{\log (c+d x)}{a^2 d}+\frac{1}{3 a d \left (a+b (c+d x)^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.0571008, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {372, 266, 44} \[ -\frac{\log \left (a+b (c+d x)^3\right )}{3 a^2 d}+\frac{\log (c+d x)}{a^2 d}+\frac{1}{3 a d \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{(c+d x) \left (a+b (c+d x)^3\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x \left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x (a+b x)^2} \, dx,x,(c+d x)^3\right )}{3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a^2 x}-\frac{b}{a (a+b x)^2}-\frac{b}{a^2 (a+b x)}\right ) \, dx,x,(c+d x)^3\right )}{3 d}\\ &=\frac{1}{3 a d \left (a+b (c+d x)^3\right )}+\frac{\log (c+d x)}{a^2 d}-\frac{\log \left (a+b (c+d x)^3\right )}{3 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0223228, size = 48, normalized size = 0.81 \[ \frac{\frac{a}{a+b (c+d x)^3}-\log \left (a+b (c+d x)^3\right )+3 \log (c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 100, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( dx+c \right ) }{{a}^{2}d}}+{\frac{1}{3\,ad \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) }}-{\frac{\ln \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) }{3\,{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(d*x+c)/a^2/d+1/3/a/d/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)-1/3/a^2/d*ln(b*d^3*x^3+3*b*c*d^2*x^2+3*b
*c^2*d*x+b*c^3+a)

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Maxima [A]  time = 1.02442, size = 140, normalized size = 2.37 \begin{align*} \frac{1}{3 \,{\left (a b d^{4} x^{3} + 3 \, a b c d^{3} x^{2} + 3 \, a b c^{2} d^{2} x +{\left (a b c^{3} + a^{2}\right )} d\right )}} - \frac{\log \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}{3 \, a^{2} d} + \frac{\log \left (d x + c\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.54478, size = 367, normalized size = 6.22 \begin{align*} -\frac{{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \log \left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right ) - 3 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \log \left (d x + c\right ) - a}{3 \,{\left (a^{2} b d^{4} x^{3} + 3 \, a^{2} b c d^{3} x^{2} + 3 \, a^{2} b c^{2} d^{2} x +{\left (a^{2} b c^{3} + a^{3}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.253, size = 110, normalized size = 1.86 \begin{align*} \frac{1}{3 a^{2} d + 3 a b c^{3} d + 9 a b c^{2} d^{2} x + 9 a b c d^{3} x^{2} + 3 a b d^{4} x^{3}} + \frac{\log{\left (\frac{c}{d} + x \right )}}{a^{2} d} - \frac{\log{\left (\frac{3 c^{2} x}{d^{2}} + \frac{3 c x^{2}}{d} + x^{3} + \frac{a + b c^{3}}{b d^{3}} \right )}}{3 a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(3*a**2*d + 3*a*b*c**3*d + 9*a*b*c**2*d**2*x + 9*a*b*c*d**3*x**2 + 3*a*b*d**4*x**3) + log(c/d + x)/(a**2*d)
- log(3*c**2*x/d**2 + 3*c*x**2/d + x**3 + (a + b*c**3)/(b*d**3))/(3*a**2*d)

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Giac [A]  time = 1.13455, size = 136, normalized size = 2.31 \begin{align*} -\frac{\log \left ({\left | b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a \right |}\right )}{3 \, a^{2} d} + \frac{\log \left ({\left | d x + c \right |}\right )}{a^{2} d} + \frac{1}{3 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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