3.95.32 \(\int \frac {25+50 x+(75 x^2+150 x^3) \log (4)+(-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5) \log ^2(4)+(25 x^6+18 e^4 x^6+32 x^7) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx\)

Optimal. Leaf size=33 \[ x+x^2-\frac {9 x^2 \left (-e^4+x\right )^2}{25 \left (x+\frac {1}{x \log (4)}\right )^2} \]

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Rubi [B]  time = 0.21, antiderivative size = 108, normalized size of antiderivative = 3.27, number of steps used = 8, number of rules used = 4, integrand size = 105, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2073, 203, 639, 199} \begin {gather*} \frac {16 x^2}{25}+\frac {27 e^4 x}{25 \left (x^2 \log (4)+1\right )}-\frac {9 \left (7 e^4 x \log (4)+3-e^8 \log (16)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )}+\frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}+\frac {1}{25} \left (25+18 e^4\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 50*x + (75*x^2 + 150*x^3)*Log[4] + (-36*E^8*x^3 + 75*x^4 + 90*E^4*x^4 + 96*x^5)*Log[4]^2 + (25*x^6 +
18*E^4*x^6 + 32*x^7)*Log[4]^3)/(25 + 75*x^2*Log[4] + 75*x^4*Log[4]^2 + 25*x^6*Log[4]^3),x]

[Out]

((25 + 18*E^4)*x)/25 + (16*x^2)/25 + (9*(1 - E^8*Log[4] + 2*E^4*x*Log[4]))/(25*Log[4]*(1 + x^2*Log[4])^2) + (2
7*E^4*x)/(25*(1 + x^2*Log[4])) - (9*(3 + 7*E^4*x*Log[4] - E^8*Log[16]))/(25*Log[4]*(1 + x^2*Log[4]))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 639

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

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{25} \left (25+18 e^4\right )+\frac {32 x}{25}+\frac {36 e^4}{25 \left (1+x^2 \log (4)\right )}+\frac {36 \left (2 e^4-x \left (1-e^8 \log (4)\right )\right )}{25 \left (1+x^2 \log (4)\right )^3}+\frac {18 \left (-7 e^4+x \left (3-e^8 \log (16)\right )\right )}{25 \left (1+x^2 \log (4)\right )^2}\right ) \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}+\frac {18}{25} \int \frac {-7 e^4+x \left (3-e^8 \log (16)\right )}{\left (1+x^2 \log (4)\right )^2} \, dx+\frac {36}{25} \int \frac {2 e^4-x \left (1-e^8 \log (4)\right )}{\left (1+x^2 \log (4)\right )^3} \, dx+\frac {1}{25} \left (36 e^4\right ) \int \frac {1}{1+x^2 \log (4)} \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}+\frac {36 e^4 \tan ^{-1}\left (x \sqrt {\log (4)}\right )}{25 \sqrt {\log (4)}}+\frac {9 \left (1-e^8 \log (4)+2 e^4 x \log (4)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )^2}-\frac {9 \left (3+7 e^4 x \log (4)-e^8 \log (16)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )}+\frac {1}{25} \left (54 e^4\right ) \int \frac {1}{\left (1+x^2 \log (4)\right )^2} \, dx-\frac {1}{25} \left (63 e^4\right ) \int \frac {1}{1+x^2 \log (4)} \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}-\frac {27 e^4 \tan ^{-1}\left (x \sqrt {\log (4)}\right )}{25 \sqrt {\log (4)}}+\frac {9 \left (1-e^8 \log (4)+2 e^4 x \log (4)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )^2}+\frac {27 e^4 x}{25 \left (1+x^2 \log (4)\right )}-\frac {9 \left (3+7 e^4 x \log (4)-e^8 \log (16)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )}+\frac {1}{25} \left (27 e^4\right ) \int \frac {1}{1+x^2 \log (4)} \, dx\\ &=\frac {1}{25} \left (25+18 e^4\right ) x+\frac {16 x^2}{25}+\frac {9 \left (1-e^8 \log (4)+2 e^4 x \log (4)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )^2}+\frac {27 e^4 x}{25 \left (1+x^2 \log (4)\right )}-\frac {9 \left (3+7 e^4 x \log (4)-e^8 \log (16)\right )}{25 \log (4) \left (1+x^2 \log (4)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 84, normalized size = 2.55 \begin {gather*} \frac {1}{25} \left (\left (25+18 e^4\right ) x+16 x^2-\frac {9 \left (-1+e^8 \log (4)-2 e^4 x \log (4)\right )}{\log (4) \left (1+x^2 \log (4)\right )^2}+\frac {9 \left (-3+2 e^8 \log (4)-4 e^4 x \log (4)\right )}{\log (4) \left (1+x^2 \log (4)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 50*x + (75*x^2 + 150*x^3)*Log[4] + (-36*E^8*x^3 + 75*x^4 + 90*E^4*x^4 + 96*x^5)*Log[4]^2 + (25
*x^6 + 18*E^4*x^6 + 32*x^7)*Log[4]^3)/(25 + 75*x^2*Log[4] + 75*x^4*Log[4]^2 + 25*x^6*Log[4]^3),x]

[Out]

((25 + 18*E^4)*x + 16*x^2 - (9*(-1 + E^8*Log[4] - 2*E^4*x*Log[4]))/(Log[4]*(1 + x^2*Log[4])^2) + (9*(-3 + 2*E^
8*Log[4] - 4*E^4*x*Log[4]))/(Log[4]*(1 + x^2*Log[4])))/25

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fricas [B]  time = 0.91, size = 92, normalized size = 2.79 \begin {gather*} \frac {4 \, {\left (16 \, x^{6} + 18 \, x^{5} e^{4} + 25 \, x^{5}\right )} \log \relax (2)^{3} + 4 \, {\left (16 \, x^{4} + 25 \, x^{3} + 9 \, x^{2} e^{8}\right )} \log \relax (2)^{2} - {\left (11 \, x^{2} - 25 \, x - 9 \, e^{8}\right )} \log \relax (2) - 9}{25 \, {\left (4 \, x^{4} \log \relax (2)^{3} + 4 \, x^{2} \log \relax (2)^{2} + \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90*x^4*exp(4)+96*x^5+75*x^4)*log(2)^2+
2*(150*x^3+75*x^2)*log(2)+50*x+25)/(200*x^6*log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x, algorithm="fricas"
)

[Out]

1/25*(4*(16*x^6 + 18*x^5*e^4 + 25*x^5)*log(2)^3 + 4*(16*x^4 + 25*x^3 + 9*x^2*e^8)*log(2)^2 - (11*x^2 - 25*x -
9*e^8)*log(2) - 9)/(4*x^4*log(2)^3 + 4*x^2*log(2)^2 + log(2))

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giac [B]  time = 0.21, size = 94, normalized size = 2.85 \begin {gather*} -\frac {9 \, {\left (8 \, x^{3} e^{4} \log \relax (2)^{2} - 4 \, x^{2} e^{8} \log \relax (2)^{2} + 3 \, x^{2} \log \relax (2) + 2 \, x e^{4} \log \relax (2) - e^{8} \log \relax (2) + 1\right )}}{25 \, {\left (2 \, x^{2} \log \relax (2) + 1\right )}^{2} \log \relax (2)} + \frac {16 \, x^{2} \log \relax (2)^{6} + 18 \, x e^{4} \log \relax (2)^{6} + 25 \, x \log \relax (2)^{6}}{25 \, \log \relax (2)^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90*x^4*exp(4)+96*x^5+75*x^4)*log(2)^2+
2*(150*x^3+75*x^2)*log(2)+50*x+25)/(200*x^6*log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x, algorithm="giac")

[Out]

-9/25*(8*x^3*e^4*log(2)^2 - 4*x^2*e^8*log(2)^2 + 3*x^2*log(2) + 2*x*e^4*log(2) - e^8*log(2) + 1)/((2*x^2*log(2
) + 1)^2*log(2)) + 1/25*(16*x^2*log(2)^6 + 18*x*e^4*log(2)^6 + 25*x*log(2)^6)/log(2)^6

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maple [B]  time = 0.14, size = 66, normalized size = 2.00

method	result	size
default	\(\frac {16 x^{2}}{25}+\frac {18 x \,{\mathrm e}^{4}}{25}+x +\frac {-\frac {72 x^{3} {\mathrm e}^{4} \ln \relax (2)}{25}+\frac {72 \left (\frac {{\mathrm e}^{8} \ln \relax (2)}{2}-\frac {3}{8}\right ) x^{2}}{25}-\frac {18 x \,{\mathrm e}^{4}}{25}+\frac {9 \left ({\mathrm e}^{8} \ln \relax (2)-1\right )}{25 \ln \relax (2)}}{\left (2 x^{2} \ln \relax (2)+1\right )^{2}}\)	\(66\)
norman	\(\frac {x +x^{2}+\left (4 \ln \relax (2)^{2}+\frac {72 \,{\mathrm e}^{4} \ln \relax (2)^{2}}{25}\right ) x^{5}+\left (-\frac {36 \,{\mathrm e}^{8} \ln \relax (2)^{2}}{25}+4 \ln \relax (2)\right ) x^{4}+4 x^{3} \ln \relax (2)+\frac {64 x^{6} \ln \relax (2)^{2}}{25}}{\left (2 x^{2} \ln \relax (2)+1\right )^{2}}\)	\(72\)
risch	\(\frac {18 x \,{\mathrm e}^{4}}{25}+\frac {16 x^{2}}{25}+x +\frac {-\frac {18 x^{3} {\mathrm e}^{4} \ln \relax (2)}{25}+\frac {\left (9 \,{\mathrm e}^{8} \ln \relax (2)-\frac {27}{4}\right ) x^{2}}{25}-\frac {9 x \,{\mathrm e}^{4}}{50}+\frac {\frac {9 \,{\mathrm e}^{8} \ln \relax (2)}{100}-\frac {9}{100}}{\ln \relax (2)}}{x^{4} \ln \relax (2)^{2}+x^{2} \ln \relax (2)+\frac {1}{4}}\)	\(73\)
gosper	\(-\frac {x \left (36 \ln \relax (2)^{2} {\mathrm e}^{8} x^{3}-72 \ln \relax (2)^{2} {\mathrm e}^{4} x^{4}-64 x^{5} \ln \relax (2)^{2}-100 x^{4} \ln \relax (2)^{2}-100 x^{3} \ln \relax (2)-100 x^{2} \ln \relax (2)-25 x -25\right )}{25 \left (4 x^{4} \ln \relax (2)^{2}+4 x^{2} \ln \relax (2)+1\right )}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*(18*x^6*exp(4)+32*x^7+25*x^6)*ln(2)^3+4*(-36*x^3*exp(4)^2+90*x^4*exp(4)+96*x^5+75*x^4)*ln(2)^2+2*(150*x
^3+75*x^2)*ln(2)+50*x+25)/(200*x^6*ln(2)^3+300*x^4*ln(2)^2+150*x^2*ln(2)+25),x,method=_RETURNVERBOSE)

[Out]

16/25*x^2+18/25*x*exp(4)+x+72/25*(-x^3*exp(4)*ln(2)+(1/2*exp(8)*ln(2)-3/8)*x^2-1/4*x*exp(4)+1/8/ln(2)*(exp(8)*
ln(2)-1))/(2*x^2*ln(2)+1)^2

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maxima [B]  time = 0.40, size = 84, normalized size = 2.55 \begin {gather*} \frac {16}{25} \, x^{2} + \frac {1}{25} \, x {\left (18 \, e^{4} + 25\right )} - \frac {9 \, {\left (8 \, x^{3} e^{4} \log \relax (2)^{2} - {\left (4 \, e^{8} \log \relax (2)^{2} - 3 \, \log \relax (2)\right )} x^{2} + 2 \, x e^{4} \log \relax (2) - e^{8} \log \relax (2) + 1\right )}}{25 \, {\left (4 \, x^{4} \log \relax (2)^{3} + 4 \, x^{2} \log \relax (2)^{2} + \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90*x^4*exp(4)+96*x^5+75*x^4)*log(2)^2+
2*(150*x^3+75*x^2)*log(2)+50*x+25)/(200*x^6*log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x, algorithm="maxima"
)

[Out]

16/25*x^2 + 1/25*x*(18*e^4 + 25) - 9/25*(8*x^3*e^4*log(2)^2 - (4*e^8*log(2)^2 - 3*log(2))*x^2 + 2*x*e^4*log(2)
- e^8*log(2) + 1)/(4*x^4*log(2)^3 + 4*x^2*log(2)^2 + log(2))

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mupad [B]  time = 9.01, size = 77, normalized size = 2.33 \begin {gather*} \frac {16\,x^2}{25}-\frac {72\,{\mathrm {e}}^4\,\ln \relax (2)\,x^3+\left (27-36\,{\mathrm {e}}^8\,\ln \relax (2)\right )\,x^2+18\,{\mathrm {e}}^4\,x-\frac {9\,\left ({\mathrm {e}}^8\,\ln \relax (2)-1\right )}{\ln \relax (2)}}{100\,{\ln \relax (2)}^2\,x^4+100\,\ln \relax (2)\,x^2+25}+x\,\left (\frac {18\,{\mathrm {e}}^4}{25}+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x + 4*log(2)^2*(90*x^4*exp(4) - 36*x^3*exp(8) + 75*x^4 + 96*x^5) + 8*log(2)^3*(18*x^6*exp(4) + 25*x^6
+ 32*x^7) + 2*log(2)*(75*x^2 + 150*x^3) + 25)/(300*x^4*log(2)^2 + 200*x^6*log(2)^3 + 150*x^2*log(2) + 25),x)

[Out]

(16*x^2)/25 - (18*x*exp(4) - (9*(exp(8)*log(2) - 1))/log(2) - x^2*(36*exp(8)*log(2) - 27) + 72*x^3*exp(4)*log(
2))/(100*x^4*log(2)^2 + 100*x^2*log(2) + 25) + x*((18*exp(4))/25 + 1)

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sympy [B]  time = 1.55, size = 94, normalized size = 2.85 \begin {gather*} \frac {16 x^{2}}{25} + x \left (1 + \frac {18 e^{4}}{25}\right ) + \frac {- 72 x^{3} e^{4} \log {\relax (2 )}^{2} + x^{2} \left (- 27 \log {\relax (2 )} + 36 e^{8} \log {\relax (2 )}^{2}\right ) - 18 x e^{4} \log {\relax (2 )} - 9 + 9 e^{8} \log {\relax (2 )}}{100 x^{4} \log {\relax (2 )}^{3} + 100 x^{2} \log {\relax (2 )}^{2} + 25 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*(18*x**6*exp(4)+32*x**7+25*x**6)*ln(2)**3+4*(-36*x**3*exp(4)**2+90*x**4*exp(4)+96*x**5+75*x**4)*l
n(2)**2+2*(150*x**3+75*x**2)*ln(2)+50*x+25)/(200*x**6*ln(2)**3+300*x**4*ln(2)**2+150*x**2*ln(2)+25),x)

[Out]

16*x**2/25 + x*(1 + 18*exp(4)/25) + (-72*x**3*exp(4)*log(2)**2 + x**2*(-27*log(2) + 36*exp(8)*log(2)**2) - 18*
x*exp(4)*log(2) - 9 + 9*exp(8)*log(2))/(100*x**4*log(2)**3 + 100*x**2*log(2)**2 + 25*log(2))

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