\(\int \frac {1}{c x^2+d x^3} \, dx\) [24]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {1}{c x^2+d x^3} \, dx=-\frac {1}{c x}-\frac {d \log (x)}{c^2}+\frac {d \log (c+d x)}{c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1607, 46} \[ \int \frac {1}{c x^2+d x^3} \, dx=-\frac {d \log (x)}{c^2}+\frac {d \log (c+d x)}{c^2}-\frac {1}{c x} \]

[In]

Int[(c*x^2 + d*x^3)^(-1),x]

[Out]

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

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^2 (c+d x)} \, dx \\ & = \int \left (\frac {1}{c x^2}-\frac {d}{c^2 x}+\frac {d^2}{c^2 (c+d x)}\right ) \, dx \\ & = -\frac {1}{c x}-\frac {d \log (x)}{c^2}+\frac {d \log (c+d x)}{c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c x^2+d x^3} \, dx=-\frac {1}{c x}-\frac {d \log (x)}{c^2}+\frac {d \log (c+d x)}{c^2} \]

[In]

Integrate[(c*x^2 + d*x^3)^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(-\frac {d \ln \left (x \right ) x -d \ln \left (d x +c \right ) x +c}{c^{2} x}\)	\(26\)
default	\(-\frac {1}{c x}-\frac {d \ln \left (x \right )}{c^{2}}+\frac {d \ln \left (d x +c \right )}{c^{2}}\)	\(29\)
norman	\(-\frac {1}{c x}-\frac {d \ln \left (x \right )}{c^{2}}+\frac {d \ln \left (d x +c \right )}{c^{2}}\)	\(29\)
risch	\(-\frac {1}{c x}+\frac {d \ln \left (-d x -c \right )}{c^{2}}-\frac {d \ln \left (x \right )}{c^{2}}\)	\(32\)

[In]

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

[Out]

-(d*ln(x)*x-d*ln(d*x+c)*x+c)/c^2/x

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{c x^2+d x^3} \, dx=\frac {d x \log \left (d x + c\right ) - d x \log \left (x\right ) - c}{c^{2} x} \]

[In]

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

[Out]

(d*x*log(d*x + c) - d*x*log(x) - c)/(c^2*x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {1}{c x^2+d x^3} \, dx=- \frac {1}{c x} + \frac {d \left (- \log {\left (x \right )} + \log {\left (\frac {c}{d} + x \right )}\right )}{c^{2}} \]

[In]

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

[Out]

-1/(c*x) + d*(-log(x) + log(c/d + x))/c**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1}{c x^2+d x^3} \, dx=\frac {d \log \left (d x + c\right )}{c^{2}} - \frac {d \log \left (x\right )}{c^{2}} - \frac {1}{c x} \]

[In]

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

[Out]

d*log(d*x + c)/c^2 - d*log(x)/c^2 - 1/(c*x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1}{c x^2+d x^3} \, dx=\frac {d \log \left ({\left | d x + c \right |}\right )}{c^{2}} - \frac {d \log \left ({\left | x \right |}\right )}{c^{2}} - \frac {1}{c x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {1}{c x^2+d x^3} \, dx=\frac {2\,d\,\mathrm {atanh}\left (\frac {2\,d\,x}{c}+1\right )}{c^2}-\frac {1}{c\,x} \]

[In]

int(1/(c*x^2 + d*x^3),x)

[Out]

(2*d*atanh((2*d*x)/c + 1))/c^2 - 1/(c*x)