\(\int ((45-15 x^2) \log (2)-15 x^2 \log (2) \log (x^3)) \, dx\) [10048]

Optimal result

Integrand size = 22, antiderivative size = 18 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=\frac {5}{2} x \log (2) \left (18-2 x^2 \log \left (x^3\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2341} \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=45 x \log (2)-5 x^3 \log (2) \log \left (x^3\right ) \]

[In]

Int[(45 - 15*x^2)*Log[2] - 15*x^2*Log[2]*Log[x^3],x]

[Out]

45*x*Log[2] - 5*x^3*Log[2]*Log[x^3]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps \begin{align*} \text {integral}& = \log (2) \int \left (45-15 x^2\right ) \, dx-(15 \log (2)) \int x^2 \log \left (x^3\right ) \, dx \\ & = 45 x \log (2)-5 x^3 \log (2) \log \left (x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=45 x \log (2)-5 x^3 \log (2) \log \left (x^3\right ) \]

[In]

Integrate[(45 - 15*x^2)*Log[2] - 15*x^2*Log[2]*Log[x^3],x]

[Out]

45*x*Log[2] - 5*x^3*Log[2]*Log[x^3]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
norman	\(45 x \ln \left (2\right )-5 \ln \left (2\right ) x^{3} \ln \left (x^{3}\right )\)	\(18\)
risch	\(45 x \ln \left (2\right )-5 \ln \left (2\right ) x^{3} \ln \left (x^{3}\right )\)	\(18\)
parallelrisch	\(45 x \ln \left (2\right )-5 \ln \left (2\right ) x^{3} \ln \left (x^{3}\right )\)	\(18\)
default	\(15 \ln \left (2\right ) \left (-\frac {1}{3} x^{3}+3 x \right )-5 \ln \left (2\right ) x^{3} \ln \left (x^{3}\right )+5 x^{3} \ln \left (2\right )\)	\(33\)
parts	\(-15 \ln \left (2\right ) \left (\frac {1}{3} x^{3}-3 x \right )-5 \ln \left (2\right ) x^{3} \ln \left (x^{3}\right )+5 x^{3} \ln \left (2\right )\)	\(33\)

[In]

int(-15*x^2*ln(2)*ln(x^3)+(-15*x^2+45)*ln(2),x,method=_RETURNVERBOSE)

[Out]

45*x*ln(2)-5*ln(2)*x^3*ln(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=-5 \, x^{3} \log \left (2\right ) \log \left (x^{3}\right ) + 45 \, x \log \left (2\right ) \]

[In]

integrate(-15*x^2*log(2)*log(x^3)+(-15*x^2+45)*log(2),x, algorithm="fricas")

[Out]

-5*x^3*log(2)*log(x^3) + 45*x*log(2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=- 5 x^{3} \log {\left (2 \right )} \log {\left (x^{3} \right )} + 45 x \log {\left (2 \right )} \]

[In]

integrate(-15*x**2*ln(2)*ln(x**3)+(-15*x**2+45)*ln(2),x)

[Out]

-5*x**3*log(2)*log(x**3) + 45*x*log(2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.67 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=-5 \, {\left (x^{3} \log \left (x^{3}\right ) - x^{3}\right )} \log \left (2\right ) - 5 \, {\left (x^{3} - 9 \, x\right )} \log \left (2\right ) \]

[In]

integrate(-15*x^2*log(2)*log(x^3)+(-15*x^2+45)*log(2),x, algorithm="maxima")

[Out]

-5*(x^3*log(x^3) - x^3)*log(2) - 5*(x^3 - 9*x)*log(2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.67 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=-5 \, {\left (x^{3} \log \left (x^{3}\right ) - x^{3}\right )} \log \left (2\right ) - 5 \, {\left (x^{3} - 9 \, x\right )} \log \left (2\right ) \]

[In]

integrate(-15*x^2*log(2)*log(x^3)+(-15*x^2+45)*log(2),x, algorithm="giac")

[Out]

-5*(x^3*log(x^3) - x^3)*log(2) - 5*(x^3 - 9*x)*log(2)

Mupad [B] (verification not implemented)

Time = 15.43 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \left (\left (45-15 x^2\right ) \log (2)-15 x^2 \log (2) \log \left (x^3\right )\right ) \, dx=-5\,x\,\ln \left (2\right )\,\left (x^2\,\ln \left (x^3\right )-9\right ) \]

[In]

int(- log(2)*(15*x^2 - 45) - 15*x^2*log(x^3)*log(2),x)

[Out]

-5*x*log(2)*(x^2*log(x^3) - 9)