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

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

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 )} \end {gather*}

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 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])

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 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]

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*}

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Mathematica [A]  time = 0.03, size = 48, normalized size = 0.81 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x) \left (a+b (c+d x)^3\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 1.04, size = 169, normalized size = 2.86 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.19, size = 101, normalized size = 1.71 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.02, size = 100, normalized size = 1.69 \begin {gather*} \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}+\frac {\ln \left (d x +c \right )}{a^{2} d}-\frac {\ln \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} \end {gather*}

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 = 0.56, size = 104, normalized size = 1.76 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.22, size = 105, normalized size = 1.78 \begin {gather*} \frac {1}{3\,\left (a^2\,d+b\,a\,c^3\,d+3\,b\,a\,c^2\,d^2\,x+3\,b\,a\,c\,d^3\,x^2+b\,a\,d^4\,x^3\right )}-\frac {\ln \left (b\,c^3+3\,b\,c^2\,d\,x+3\,b\,c\,d^2\,x^2+b\,d^3\,x^3+a\right )}{3\,a^2\,d}+\frac {\ln \left (c+d\,x\right )}{a^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.48, size = 110, normalized size = 1.86 \begin {gather*} \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 {gather*}

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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