\(\int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+(-30-27 x+3 x^3) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx\) [3675]

Optimal result

Integrand size = 56, antiderivative size = 34 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=-1+2 x+\frac {\left (\frac {3}{x}-x\right ) (-x+\log (4))}{2 x (5+3 x)} \]

[Out]

(2*ln(2)-x)/x*(3/x-x)/(6*x+10)+2*x-1

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1608, 27, 12, 1634} \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=\frac {3 \log (4)}{10 x^2}+2 x+\frac {5+\log (64)}{75 (3 x+5)}-\frac {3 (5+\log (64))}{50 x} \]

[In]

Int[(15*x + 18*x^2 + 105*x^3 + 120*x^4 + 36*x^5 + (-30 - 27*x + 3*x^3)*Log[4])/(50*x^3 + 60*x^4 + 18*x^5),x]

[Out]

2*x + (3*Log[4])/(10*x^2) - (3*(5 + Log[64]))/(50*x) + (5 + Log[64])/(75*(5 + 3*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

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

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{x^3 \left (50+60 x+18 x^2\right )} \, dx \\ & = \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{2 x^3 (5+3 x)^2} \, dx \\ & = \frac {1}{2} \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{x^3 (5+3 x)^2} \, dx \\ & = \frac {1}{2} \int \left (4-\frac {6 \log (4)}{5 x^3}+\frac {3 (5+\log (64))}{25 x^2}-\frac {2 (5+\log (64))}{25 (5+3 x)^2}\right ) \, dx \\ & = 2 x+\frac {3 \log (4)}{10 x^2}-\frac {3 (5+\log (64))}{50 x}+\frac {5+\log (64)}{75 (5+3 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=\frac {-9 x+60 x^3+36 x^4+9 \log (4)-x^2 (5+\log (64))}{6 x^2 (5+3 x)} \]

[In]

Integrate[(15*x + 18*x^2 + 105*x^3 + 120*x^4 + 36*x^5 + (-30 - 27*x + 3*x^3)*Log[4])/(50*x^3 + 60*x^4 + 18*x^5
),x]

[Out]

(-9*x + 60*x^3 + 36*x^4 + 9*Log[4] - x^2*(5 + Log[64]))/(6*x^2*(5 + 3*x))

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97

method	result	size
risch	\(2 x +\frac {3 \left (-\frac {\ln \left (2\right )}{3}-\frac {5}{18}\right ) x^{2}-\frac {3 x}{2}+3 \ln \left (2\right )}{x^{2} \left (3 x +5\right )}\)	\(33\)
norman	\(\frac {\left (-\frac {35}{2}-\ln \left (2\right )\right ) x^{2}-\frac {3 x}{2}+6 x^{4}+3 \ln \left (2\right )}{x^{2} \left (3 x +5\right )}\)	\(35\)
gosper	\(-\frac {-12 x^{4}+2 x^{2} \ln \left (2\right )+35 x^{2}-6 \ln \left (2\right )+3 x}{2 x^{2} \left (3 x +5\right )}\)	\(38\)
default	\(2 x +\frac {3 \ln \left (2\right )}{5 x^{2}}-\frac {3 \left (\frac {6 \ln \left (2\right )}{25}+\frac {1}{5}\right )}{2 x}-\frac {3 \left (-\frac {4 \ln \left (2\right )}{75}-\frac {2}{45}\right )}{2 \left (3 x +5\right )}\)	\(38\)
parallelrisch	\(-\frac {-36 x^{4}+6 x^{2} \ln \left (2\right )+105 x^{2}-18 \ln \left (2\right )+9 x}{6 x^{2} \left (3 x +5\right )}\)	\(38\)

[In]

int((2*(3*x^3-27*x-30)*ln(2)+36*x^5+120*x^4+105*x^3+18*x^2+15*x)/(18*x^5+60*x^4+50*x^3),x,method=_RETURNVERBOS
E)

[Out]

2*x+3*((-1/3*ln(2)-5/18)*x^2-1/2*x+ln(2))/x^2/(3*x+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=\frac {36 \, x^{4} + 60 \, x^{3} - 5 \, x^{2} - 6 \, {\left (x^{2} - 3\right )} \log \left (2\right ) - 9 \, x}{6 \, {\left (3 \, x^{3} + 5 \, x^{2}\right )}} \]

[In]

integrate((2*(3*x^3-27*x-30)*log(2)+36*x^5+120*x^4+105*x^3+18*x^2+15*x)/(18*x^5+60*x^4+50*x^3),x, algorithm="f
ricas")

[Out]

1/6*(36*x^4 + 60*x^3 - 5*x^2 - 6*(x^2 - 3)*log(2) - 9*x)/(3*x^3 + 5*x^2)

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2 x + \frac {x^{2} \left (-5 - 6 \log {\left (2 \right )}\right ) - 9 x + 18 \log {\left (2 \right )}}{18 x^{3} + 30 x^{2}} \]

[In]

integrate((2*(3*x**3-27*x-30)*ln(2)+36*x**5+120*x**4+105*x**3+18*x**2+15*x)/(18*x**5+60*x**4+50*x**3),x)

[Out]

2*x + (x**2*(-5 - 6*log(2)) - 9*x + 18*log(2))/(18*x**3 + 30*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2 \, x - \frac {x^{2} {\left (6 \, \log \left (2\right ) + 5\right )} + 9 \, x - 18 \, \log \left (2\right )}{6 \, {\left (3 \, x^{3} + 5 \, x^{2}\right )}} \]

[In]

integrate((2*(3*x^3-27*x-30)*log(2)+36*x^5+120*x^4+105*x^3+18*x^2+15*x)/(18*x^5+60*x^4+50*x^3),x, algorithm="m
axima")

[Out]

2*x - 1/6*(x^2*(6*log(2) + 5) + 9*x - 18*log(2))/(3*x^3 + 5*x^2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2 \, x + \frac {6 \, \log \left (2\right ) + 5}{75 \, {\left (3 \, x + 5\right )}} - \frac {3 \, {\left (6 \, x \log \left (2\right ) + 5 \, x - 10 \, \log \left (2\right )\right )}}{50 \, x^{2}} \]

[In]

integrate((2*(3*x^3-27*x-30)*log(2)+36*x^5+120*x^4+105*x^3+18*x^2+15*x)/(18*x^5+60*x^4+50*x^3),x, algorithm="g
iac")

[Out]

2*x + 1/75*(6*log(2) + 5)/(3*x + 5) - 3/50*(6*x*log(2) + 5*x - 10*log(2))/x^2

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.21 \[ \int \frac {15 x+18 x^2+105 x^3+120 x^4+36 x^5+\left (-30-27 x+3 x^3\right ) \log (4)}{50 x^3+60 x^4+18 x^5} \, dx=2\,x-\frac {3\,x}{6\,x^3+10\,x^2}-\frac {5\,x^2}{3\,\left (6\,x^3+10\,x^2\right )}+\frac {6\,\ln \left (2\right )}{6\,x^3+10\,x^2}-\frac {2\,x^2\,\ln \left (2\right )}{6\,x^3+10\,x^2} \]

[In]

int((15*x - 2*log(2)*(27*x - 3*x^3 + 30) + 18*x^2 + 105*x^3 + 120*x^4 + 36*x^5)/(50*x^3 + 60*x^4 + 18*x^5),x)

[Out]

2*x - (3*x)/(10*x^2 + 6*x^3) - (5*x^2)/(3*(10*x^2 + 6*x^3)) + (6*log(2))/(10*x^2 + 6*x^3) - (2*x^2*log(2))/(10
*x^2 + 6*x^3)