3.86.73 \(\int \frac {45-3 x+x^2+(-30 x+2 x^2) \log (60-4 x)}{-15+x} \, dx\)

Optimal. Leaf size=16 \[ 5+x (-3+x \log (4 (15-x))) \]

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Rubi [A]  time = 0.08, antiderivative size = 14, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6742, 698, 2395, 43} \begin {gather*} x^2 \log (60-4 x)-3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(45 - 3*x + x^2 + (-30*x + 2*x^2)*Log[60 - 4*x])/(-15 + x),x]

[Out]

-3*x + x^2*Log[60 - 4*x]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {45-3 x+x^2}{-15+x}+2 x \log (60-4 x)\right ) \, dx\\ &=2 \int x \log (60-4 x) \, dx+\int \frac {45-3 x+x^2}{-15+x} \, dx\\ &=x^2 \log (60-4 x)+4 \int \frac {x^2}{60-4 x} \, dx+\int \left (12+\frac {225}{-15+x}+x\right ) \, dx\\ &=12 x+\frac {x^2}{2}+x^2 \log (60-4 x)+225 \log (15-x)+4 \int \left (-\frac {15}{4}-\frac {225}{4 (-15+x)}-\frac {x}{4}\right ) \, dx\\ &=-3 x+x^2 \log (60-4 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 14, normalized size = 0.88 \begin {gather*} -3 x+x^2 \log (60-4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45 - 3*x + x^2 + (-30*x + 2*x^2)*Log[60 - 4*x])/(-15 + x),x]

[Out]

-3*x + x^2*Log[60 - 4*x]

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fricas [A]  time = 0.55, size = 14, normalized size = 0.88 \begin {gather*} x^{2} \log \left (-4 \, x + 60\right ) - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-30*x)*log(-4*x+60)+x^2-3*x+45)/(x-15),x, algorithm="fricas")

[Out]

x^2*log(-4*x + 60) - 3*x

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giac [A]  time = 0.17, size = 14, normalized size = 0.88 \begin {gather*} x^{2} \log \left (-4 \, x + 60\right ) - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-30*x)*log(-4*x+60)+x^2-3*x+45)/(x-15),x, algorithm="giac")

[Out]

x^2*log(-4*x + 60) - 3*x

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maple [A]  time = 0.38, size = 15, normalized size = 0.94

method	result	size
norman	\(\ln \left (-4 x +60\right ) x^{2}-3 x\)	\(15\)
risch	\(\ln \left (-4 x +60\right ) x^{2}-3 x\)	\(15\)
derivativedivides	\(\frac {\ln \left (-4 x +60\right ) \left (-4 x +60\right )^{2}}{16}-\frac {15 \ln \left (-4 x +60\right ) \left (-4 x +60\right )}{2}-3 x +45+225 \ln \left (-4 x +60\right )\)	\(42\)
default	\(\frac {\ln \left (-4 x +60\right ) \left (-4 x +60\right )^{2}}{16}-\frac {15 \ln \left (-4 x +60\right ) \left (-4 x +60\right )}{2}-3 x +45+225 \ln \left (-4 x +60\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-30*x)*ln(-4*x+60)+x^2-3*x+45)/(x-15),x,method=_RETURNVERBOSE)

[Out]

ln(-4*x+60)*x^2-3*x

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maxima [B]  time = 0.34, size = 40, normalized size = 2.50 \begin {gather*} {\left (x^{2} + 30 \, x + 450 \, \log \left (x - 15\right )\right )} \log \left (-4 \, x + 60\right ) - 30 \, {\left (x + 15 \, \log \left (x - 15\right )\right )} \log \left (-4 \, x + 60\right ) - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-30*x)*log(-4*x+60)+x^2-3*x+45)/(x-15),x, algorithm="maxima")

[Out]

(x^2 + 30*x + 450*log(x - 15))*log(-4*x + 60) - 30*(x + 15*log(x - 15))*log(-4*x + 60) - 3*x

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mupad [B]  time = 5.22, size = 14, normalized size = 0.88 \begin {gather*} x^2\,\ln \left (60-4\,x\right )-3\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + log(60 - 4*x)*(30*x - 2*x^2) - x^2 - 45)/(x - 15),x)

[Out]

x^2*log(60 - 4*x) - 3*x

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sympy [A]  time = 0.11, size = 12, normalized size = 0.75 \begin {gather*} x^{2} \log {\left (60 - 4 x \right )} - 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-30*x)*ln(-4*x+60)+x**2-3*x+45)/(x-15),x)

[Out]

x**2*log(60 - 4*x) - 3*x

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