3.72.86 \(\int \frac {300-120 x^2+12 x^4+2^{-\frac {x}{-5+x^2}} (-75+150 x-45 x^2-60 x^3+27 x^4+6 x^5-3 x^6+(-15 x+30 x^2-18 x^3+6 x^4-3 x^5) \log (2))}{100-200 x+60 x^2+80 x^3-36 x^4-8 x^5+4 x^6} \, dx\)

Optimal. Leaf size=34 \[ 3 \left (6-2^{-2+\frac {1}{\frac {5}{x}-x}} x-\frac {x}{-x+x^2}\right ) \]

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Rubi [B]  time = 1.76, antiderivative size = 250, normalized size of antiderivative = 7.35, number of steps used = 24, number of rules used = 11, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6741, 12, 6688, 6742, 741, 801, 633, 31, 1647, 1629, 2288} \begin {gather*} -\frac {15 (x+5)}{8 (1-x) \left (5-x^2\right )}-\frac {15 (3 x+5)}{16 \left (5-x^2\right )}-\frac {3\ 2^{\frac {x}{5-x^2}-2} \left (x^3 \log (2)+x \log (32)\right )}{\left (5-x^2\right )^2 \left (\frac {2 x^2}{\left (5-x^2\right )^2}+\frac {1}{5-x^2}\right ) \log (2)}+\frac {93}{16 (1-x)}-\frac {3}{32} \left (25+13 \sqrt {5}\right ) \log \left (\sqrt {5}-x\right )+\frac {3}{16} \left (15+7 \sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {3}{32} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {3}{32} \left (5-\sqrt {5}\right ) \log \left (x+\sqrt {5}\right )+\frac {3}{16} \left (15-7 \sqrt {5}\right ) \log \left (x+\sqrt {5}\right )-\frac {3}{32} \left (25-13 \sqrt {5}\right ) \log \left (x+\sqrt {5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(300 - 120*x^2 + 12*x^4 + (-75 + 150*x - 45*x^2 - 60*x^3 + 27*x^4 + 6*x^5 - 3*x^6 + (-15*x + 30*x^2 - 18*x
^3 + 6*x^4 - 3*x^5)*Log[2])/2^(x/(-5 + x^2)))/(100 - 200*x + 60*x^2 + 80*x^3 - 36*x^4 - 8*x^5 + 4*x^6),x]

[Out]

93/(16*(1 - x)) - (15*(5 + x))/(8*(1 - x)*(5 - x^2)) - (15*(5 + 3*x))/(16*(5 - x^2)) - (3*2^(-2 + x/(5 - x^2))
*(x^3*Log[2] + x*Log[32]))/((5 - x^2)^2*((2*x^2)/(5 - x^2)^2 + (5 - x^2)^(-1))*Log[2]) - (3*(5 + Sqrt[5])*Log[
Sqrt[5] - x])/32 + (3*(15 + 7*Sqrt[5])*Log[Sqrt[5] - x])/16 - (3*(25 + 13*Sqrt[5])*Log[Sqrt[5] - x])/32 - (3*(
25 - 13*Sqrt[5])*Log[Sqrt[5] + x])/32 + (3*(15 - 7*Sqrt[5])*Log[Sqrt[5] + x])/16 - (3*(5 - Sqrt[5])*Log[Sqrt[5
] + x])/32

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rule 6741

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

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 \frac {300-120 x^2+12 x^4+2^{-\frac {x}{-5+x^2}} \left (-75+150 x-45 x^2-60 x^3+27 x^4+6 x^5-3 x^6+\left (-15 x+30 x^2-18 x^3+6 x^4-3 x^5\right ) \log (2)\right )}{4 \left (5-5 x-x^2+x^3\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {300-120 x^2+12 x^4+2^{-\frac {x}{-5+x^2}} \left (-75+150 x-45 x^2-60 x^3+27 x^4+6 x^5-3 x^6+\left (-15 x+30 x^2-18 x^3+6 x^4-3 x^5\right ) \log (2)\right )}{\left (5-5 x-x^2+x^3\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {300-120 x^2+12 x^4-3\ 2^{-\frac {x}{-5+x^2}} (-1+x)^2 \left (25-10 x^2+x^4+x^3 \log (2)+x \log (32)\right )}{\left (5-5 x-x^2+x^3\right )^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {300}{(-1+x)^2 \left (-5+x^2\right )^2}-\frac {120 x^2}{(-1+x)^2 \left (-5+x^2\right )^2}+\frac {12 x^4}{(-1+x)^2 \left (-5+x^2\right )^2}-\frac {3\ 2^{-\frac {x}{-5+x^2}} \left (25-10 x^2+x^4+x^3 \log (2)+x \log (32)\right )}{\left (-5+x^2\right )^2}\right ) \, dx\\ &=-\left (\frac {3}{4} \int \frac {2^{-\frac {x}{-5+x^2}} \left (25-10 x^2+x^4+x^3 \log (2)+x \log (32)\right )}{\left (-5+x^2\right )^2} \, dx\right )+3 \int \frac {x^4}{(-1+x)^2 \left (-5+x^2\right )^2} \, dx-30 \int \frac {x^2}{(-1+x)^2 \left (-5+x^2\right )^2} \, dx+75 \int \frac {1}{(-1+x)^2 \left (-5+x^2\right )^2} \, dx\\ &=-\frac {15 (5+x)}{8 (1-x) \left (5-x^2\right )}-\frac {15 (5+3 x)}{16 \left (5-x^2\right )}-\frac {3\ 2^{-2+\frac {x}{5-x^2}} \left (x^3 \log (2)+x \log (32)\right )}{\left (5-x^2\right )^2 \left (\frac {2 x^2}{\left (5-x^2\right )^2}+\frac {1}{5-x^2}\right ) \log (2)}+\frac {3}{10} \int \frac {\frac {75}{8}-\frac {25 x}{2}+\frac {5 x^2}{8}}{(-1+x)^2 \left (-5+x^2\right )} \, dx-\frac {15}{8} \int \frac {14+2 x}{(-1+x)^2 \left (-5+x^2\right )} \, dx-3 \int \frac {\frac {15}{8}-\frac {5 x}{2}-\frac {15 x^2}{8}}{(-1+x)^2 \left (-5+x^2\right )} \, dx\\ &=-\frac {15 (5+x)}{8 (1-x) \left (5-x^2\right )}-\frac {15 (5+3 x)}{16 \left (5-x^2\right )}-\frac {3\ 2^{-2+\frac {x}{5-x^2}} \left (x^3 \log (2)+x \log (32)\right )}{\left (5-x^2\right )^2 \left (\frac {2 x^2}{\left (5-x^2\right )^2}+\frac {1}{5-x^2}\right ) \log (2)}+\frac {3}{10} \int \left (\frac {5}{8 (-1+x)^2}+\frac {25}{8 (-1+x)}-\frac {25 (1+x)}{8 \left (-5+x^2\right )}\right ) \, dx-\frac {15}{8} \int \left (-\frac {4}{(-1+x)^2}-\frac {5}{2 (-1+x)}+\frac {13+5 x}{2 \left (-5+x^2\right )}\right ) \, dx-3 \int \left (\frac {5}{8 (-1+x)^2}+\frac {15}{8 (-1+x)}-\frac {5 (7+3 x)}{8 \left (-5+x^2\right )}\right ) \, dx\\ &=\frac {93}{16 (1-x)}-\frac {15 (5+x)}{8 (1-x) \left (5-x^2\right )}-\frac {15 (5+3 x)}{16 \left (5-x^2\right )}-\frac {3\ 2^{-2+\frac {x}{5-x^2}} \left (x^3 \log (2)+x \log (32)\right )}{\left (5-x^2\right )^2 \left (\frac {2 x^2}{\left (5-x^2\right )^2}+\frac {1}{5-x^2}\right ) \log (2)}-\frac {15}{16} \int \frac {1+x}{-5+x^2} \, dx-\frac {15}{16} \int \frac {13+5 x}{-5+x^2} \, dx+\frac {15}{8} \int \frac {7+3 x}{-5+x^2} \, dx\\ &=\frac {93}{16 (1-x)}-\frac {15 (5+x)}{8 (1-x) \left (5-x^2\right )}-\frac {15 (5+3 x)}{16 \left (5-x^2\right )}-\frac {3\ 2^{-2+\frac {x}{5-x^2}} \left (x^3 \log (2)+x \log (32)\right )}{\left (5-x^2\right )^2 \left (\frac {2 x^2}{\left (5-x^2\right )^2}+\frac {1}{5-x^2}\right ) \log (2)}-\frac {1}{32} \left (3 \left (25-13 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt {5}+x} \, dx+\frac {1}{16} \left (3 \left (15-7 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt {5}+x} \, dx-\frac {1}{32} \left (3 \left (5-\sqrt {5}\right )\right ) \int \frac {1}{\sqrt {5}+x} \, dx-\frac {1}{32} \left (3 \left (5+\sqrt {5}\right )\right ) \int \frac {1}{-\sqrt {5}+x} \, dx+\frac {1}{16} \left (3 \left (15+7 \sqrt {5}\right )\right ) \int \frac {1}{-\sqrt {5}+x} \, dx-\frac {1}{32} \left (3 \left (25+13 \sqrt {5}\right )\right ) \int \frac {1}{-\sqrt {5}+x} \, dx\\ &=\frac {93}{16 (1-x)}-\frac {15 (5+x)}{8 (1-x) \left (5-x^2\right )}-\frac {15 (5+3 x)}{16 \left (5-x^2\right )}-\frac {3\ 2^{-2+\frac {x}{5-x^2}} \left (x^3 \log (2)+x \log (32)\right )}{\left (5-x^2\right )^2 \left (\frac {2 x^2}{\left (5-x^2\right )^2}+\frac {1}{5-x^2}\right ) \log (2)}-\frac {3}{32} \left (5+\sqrt {5}\right ) \log \left (\sqrt {5}-x\right )+\frac {3}{16} \left (15+7 \sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {3}{32} \left (25+13 \sqrt {5}\right ) \log \left (\sqrt {5}-x\right )-\frac {3}{32} \left (25-13 \sqrt {5}\right ) \log \left (\sqrt {5}+x\right )+\frac {3}{16} \left (15-7 \sqrt {5}\right ) \log \left (\sqrt {5}+x\right )-\frac {3}{32} \left (5-\sqrt {5}\right ) \log \left (\sqrt {5}+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.40, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {300-120 x^2+12 x^4+2^{-\frac {x}{-5+x^2}} \left (-75+150 x-45 x^2-60 x^3+27 x^4+6 x^5-3 x^6+\left (-15 x+30 x^2-18 x^3+6 x^4-3 x^5\right ) \log (2)\right )}{100-200 x+60 x^2+80 x^3-36 x^4-8 x^5+4 x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(300 - 120*x^2 + 12*x^4 + (-75 + 150*x - 45*x^2 - 60*x^3 + 27*x^4 + 6*x^5 - 3*x^6 + (-15*x + 30*x^2
- 18*x^3 + 6*x^4 - 3*x^5)*Log[2])/2^(x/(-5 + x^2)))/(100 - 200*x + 60*x^2 + 80*x^3 - 36*x^4 - 8*x^5 + 4*x^6),x
]

[Out]

Integrate[(300 - 120*x^2 + 12*x^4 + (-75 + 150*x - 45*x^2 - 60*x^3 + 27*x^4 + 6*x^5 - 3*x^6 + (-15*x + 30*x^2
- 18*x^3 + 6*x^4 - 3*x^5)*Log[2])/2^(x/(-5 + x^2)))/(100 - 200*x + 60*x^2 + 80*x^3 - 36*x^4 - 8*x^5 + 4*x^6),
x]

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fricas [A]  time = 0.52, size = 40, normalized size = 1.18 \begin {gather*} -\frac {3 \, {\left (x^{2} + 4 \cdot 2^{\frac {x}{x^{2} - 5}} - x\right )}}{4 \cdot 2^{\frac {x}{x^{2} - 5}} {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^5+6*x^4-18*x^3+30*x^2-15*x)*log(2)-3*x^6+6*x^5+27*x^4-60*x^3-45*x^2+150*x-75)*exp(-x*log(2)/
(x^2-5))+12*x^4-120*x^2+300)/(4*x^6-8*x^5-36*x^4+80*x^3+60*x^2-200*x+100),x, algorithm="fricas")

[Out]

-3/4*(x^2 + 4*2^(x/(x^2 - 5)) - x)/(2^(x/(x^2 - 5))*(x - 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, {\left (4 \, x^{4} - 40 \, x^{2} - \frac {x^{6} - 2 \, x^{5} - 9 \, x^{4} + 20 \, x^{3} + 15 \, x^{2} + {\left (x^{5} - 2 \, x^{4} + 6 \, x^{3} - 10 \, x^{2} + 5 \, x\right )} \log \relax (2) - 50 \, x + 25}{2^{\frac {x}{x^{2} - 5}}} + 100\right )}}{4 \, {\left (x^{6} - 2 \, x^{5} - 9 \, x^{4} + 20 \, x^{3} + 15 \, x^{2} - 50 \, x + 25\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^5+6*x^4-18*x^3+30*x^2-15*x)*log(2)-3*x^6+6*x^5+27*x^4-60*x^3-45*x^2+150*x-75)*exp(-x*log(2)/
(x^2-5))+12*x^4-120*x^2+300)/(4*x^6-8*x^5-36*x^4+80*x^3+60*x^2-200*x+100),x, algorithm="giac")

[Out]

integrate(3/4*(4*x^4 - 40*x^2 - (x^6 - 2*x^5 - 9*x^4 + 20*x^3 + 15*x^2 + (x^5 - 2*x^4 + 6*x^3 - 10*x^2 + 5*x)*
log(2) - 50*x + 25)/2^(x/(x^2 - 5)) + 100)/(x^6 - 2*x^5 - 9*x^4 + 20*x^3 + 15*x^2 - 50*x + 25), x)

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maple [A]  time = 0.20, size = 23, normalized size = 0.68

method	result	size
risch	\(-\frac {3}{x -1}-\frac {3 \left (\frac {1}{2}\right )^{\frac {x}{x^{2}-5}} x}{4}\)	\(23\)
norman	\(\frac {-3 x^{2}+15-\frac {15 \,{\mathrm e}^{-\frac {x \ln \relax (2)}{x^{2}-5}} x}{4}+\frac {15 \,{\mathrm e}^{-\frac {x \ln \relax (2)}{x^{2}-5}} x^{2}}{4}+\frac {3 \,{\mathrm e}^{-\frac {x \ln \relax (2)}{x^{2}-5}} x^{3}}{4}-\frac {3 \,{\mathrm e}^{-\frac {x \ln \relax (2)}{x^{2}-5}} x^{4}}{4}}{x^{3}-x^{2}-5 x +5}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x^5+6*x^4-18*x^3+30*x^2-15*x)*ln(2)-3*x^6+6*x^5+27*x^4-60*x^3-45*x^2+150*x-75)*exp(-x*ln(2)/(x^2-5))
+12*x^4-120*x^2+300)/(4*x^6-8*x^5-36*x^4+80*x^3+60*x^2-200*x+100),x,method=_RETURNVERBOSE)

[Out]

-3/(x-1)-3/4*(1/2)^(x/(x^2-5))*x

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maxima [B]  time = 0.54, size = 94, normalized size = 2.76 \begin {gather*} -\frac {3 \, {\left (8 \, x^{2} + 5 \, x - 15\right )}}{8 \, {\left (x^{3} - x^{2} - 5 \, x + 5\right )}} - \frac {15 \, {\left (4 \, x^{2} + x - 15\right )}}{8 \, {\left (x^{3} - x^{2} - 5 \, x + 5\right )}} + \frac {15 \, {\left (2 \, x^{2} + x - 5\right )}}{4 \, {\left (x^{3} - x^{2} - 5 \, x + 5\right )}} - \frac {3 \, x}{4 \cdot 2^{\frac {x}{x^{2} - 5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^5+6*x^4-18*x^3+30*x^2-15*x)*log(2)-3*x^6+6*x^5+27*x^4-60*x^3-45*x^2+150*x-75)*exp(-x*log(2)/
(x^2-5))+12*x^4-120*x^2+300)/(4*x^6-8*x^5-36*x^4+80*x^3+60*x^2-200*x+100),x, algorithm="maxima")

[Out]

-3/8*(8*x^2 + 5*x - 15)/(x^3 - x^2 - 5*x + 5) - 15/8*(4*x^2 + x - 15)/(x^3 - x^2 - 5*x + 5) + 15/4*(2*x^2 + x
- 5)/(x^3 - x^2 - 5*x + 5) - 3/4*x/2^(x/(x^2 - 5))

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mupad [B]  time = 4.77, size = 24, normalized size = 0.71 \begin {gather*} -\frac {3}{x-1}-\frac {3\,x}{4\,2^{\frac {x}{x^2-5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x*log(2))/(x^2 - 5))*(45*x^2 - 150*x + 60*x^3 - 27*x^4 - 6*x^5 + 3*x^6 + log(2)*(15*x - 30*x^2 + 1
8*x^3 - 6*x^4 + 3*x^5) + 75) + 120*x^2 - 12*x^4 - 300)/(60*x^2 - 200*x + 80*x^3 - 36*x^4 - 8*x^5 + 4*x^6 + 100
),x)

[Out]

- 3/(x - 1) - (3*x)/(4*2^(x/(x^2 - 5)))

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sympy [A]  time = 0.48, size = 22, normalized size = 0.65 \begin {gather*} - \frac {3 x e^{- \frac {x \log {\relax (2 )}}{x^{2} - 5}}}{4} - \frac {3}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x**5+6*x**4-18*x**3+30*x**2-15*x)*ln(2)-3*x**6+6*x**5+27*x**4-60*x**3-45*x**2+150*x-75)*exp(-x
*ln(2)/(x**2-5))+12*x**4-120*x**2+300)/(4*x**6-8*x**5-36*x**4+80*x**3+60*x**2-200*x+100),x)

[Out]

-3*x*exp(-x*log(2)/(x**2 - 5))/4 - 3/(x - 1)

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