Integrand size = 40, antiderivative size = 112 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {65991-8779 x}{4292352 \sqrt {3-x+2 x^2}}-\frac {3667 \sqrt {3-x+2 x^2}}{20736 (5+2 x)^2}+\frac {115369 \sqrt {3-x+2 x^2}}{1492992 (5+2 x)}-\frac {52631 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{5971968 \sqrt {2}} \]
-52631/11943936*arctanh(1/24*(17-22*x)*2^(1/2)/(2*x^2-x+3)^(1/2))*2^(1/2)+ 1/4292352*(65991-8779*x)/(2*x^2-x+3)^(1/2)-3667/20736*(2*x^2-x+3)^(1/2)/(5 +2*x)^2+115369/1492992*(2*x^2-x+3)^(1/2)/(5+2*x)
Time = 0.49 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.68 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {\frac {12 \left (11594283+5842933 x+3263288 x^2+3444340 x^3\right )}{(5+2 x)^2 \sqrt {3-x+2 x^2}}+1210513 \sqrt {2} \text {arctanh}\left (\frac {1}{6} \left (5+2 x-\sqrt {6-2 x+4 x^2}\right )\right )}{137355264} \]
((12*(11594283 + 5842933*x + 3263288*x^2 + 3444340*x^3))/((5 + 2*x)^2*Sqrt [3 - x + 2*x^2]) + 1210513*Sqrt[2]*ArcTanh[(5 + 2*x - Sqrt[6 - 2*x + 4*x^2 ])/6])/137355264
Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2177, 27, 2181, 27, 1228, 1154, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {5 x^4-x^3+3 x^2+x+2}{(2 x+5)^3 \left (2 x^2-x+3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2177 |
| \(\displaystyle \frac {2}{23} \int \frac {23 \left (977500 x^2+632660 x+224707\right )}{746496 (2 x+5)^3 \sqrt {2 x^2-x+3}}dx+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {977500 x^2+632660 x+224707}{(2 x+5)^3 \sqrt {2 x^2-x+3}}dx}{373248}+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 2181 |
| \(\displaystyle \frac {-\frac {1}{144} \int \frac {288 (73238-178369 x)}{(2 x+5)^2 \sqrt {2 x^2-x+3}}dx-\frac {66006 \sqrt {2 x^2-x+3}}{(2 x+5)^2}}{373248}+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-2 \int \frac {73238-178369 x}{(2 x+5)^2 \sqrt {2 x^2-x+3}}dx-\frac {66006 \sqrt {2 x^2-x+3}}{(2 x+5)^2}}{373248}+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {-2 \left (-\frac {157893}{16} \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-\frac {115369 \sqrt {2 x^2-x+3}}{8 (2 x+5)}\right )-\frac {66006 \sqrt {2 x^2-x+3}}{(2 x+5)^2}}{373248}+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {-2 \left (\frac {157893}{8} \int \frac {1}{288-\frac {(17-22 x)^2}{2 x^2-x+3}}d\frac {17-22 x}{\sqrt {2 x^2-x+3}}-\frac {115369 \sqrt {2 x^2-x+3}}{8 (2 x+5)}\right )-\frac {66006 \sqrt {2 x^2-x+3}}{(2 x+5)^2}}{373248}+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-2 \left (\frac {52631 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{32 \sqrt {2}}-\frac {115369 \sqrt {2 x^2-x+3}}{8 (2 x+5)}\right )-\frac {66006 \sqrt {2 x^2-x+3}}{(2 x+5)^2}}{373248}+\frac {65991-8779 x}{4292352 \sqrt {2 x^2-x+3}}\) |
(65991 - 8779*x)/(4292352*Sqrt[3 - x + 2*x^2]) + ((-66006*Sqrt[3 - x + 2*x ^2])/(5 + 2*x)^2 - 2*((-115369*Sqrt[3 - x + 2*x^2])/(8*(5 + 2*x)) + (52631 *ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(32*Sqrt[2])))/373 248
3.4.56.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x + c* x^2, x], R = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + (2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^ m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Qx)/(d + e*x )^m - ((2*p + 3)*(2*c*R - b*S))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a* e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = Polynomi alRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*x^2) ^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Qx + c*d*R*(m + 1) - b*e*R*(m + p + 2) - c*e*R *(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]
Timed out.
hanged
Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {1210513 \, \sqrt {2} {\left (8 \, x^{4} + 36 \, x^{3} + 42 \, x^{2} + 35 \, x + 75\right )} \log \left (-\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) + 48 \, {\left (3444340 \, x^{3} + 3263288 \, x^{2} + 5842933 \, x + 11594283\right )} \sqrt {2 \, x^{2} - x + 3}}{549421056 \, {\left (8 \, x^{4} + 36 \, x^{3} + 42 \, x^{2} + 35 \, x + 75\right )}} \]
1/549421056*(1210513*sqrt(2)*(8*x^4 + 36*x^3 + 42*x^2 + 35*x + 75)*log(-(2 4*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) + 1060*x^2 - 1036*x + 1153)/(4*x ^2 + 20*x + 25)) + 48*(3444340*x^3 + 3263288*x^2 + 5842933*x + 11594283)*s qrt(2*x^2 - x + 3))/(8*x^4 + 36*x^3 + 42*x^2 + 35*x + 75)
\[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x + 5\right )^{3} \left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.33 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {52631}{11943936} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) + \frac {861085 \, x}{11446272 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {1163201}{3815424 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {3667}{1152 \, {\left (4 \, \sqrt {2 \, x^{2} - x + 3} x^{2} + 20 \, \sqrt {2 \, x^{2} - x + 3} x + 25 \, \sqrt {2 \, x^{2} - x + 3}\right )}} + \frac {196043}{82944 \, {\left (2 \, \sqrt {2 \, x^{2} - x + 3} x + 5 \, \sqrt {2 \, x^{2} - x + 3}\right )}} \]
52631/11943936*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt( 23)/abs(2*x + 5)) + 861085/11446272*x/sqrt(2*x^2 - x + 3) - 1163201/381542 4/sqrt(2*x^2 - x + 3) - 3667/1152/(4*sqrt(2*x^2 - x + 3)*x^2 + 20*sqrt(2*x ^2 - x + 3)*x + 25*sqrt(2*x^2 - x + 3)) + 196043/82944/(2*sqrt(2*x^2 - x + 3)*x + 5*sqrt(2*x^2 - x + 3))
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (90) = 180\).
Time = 0.30 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.96 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=-\frac {52631}{11943936} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {52631}{11943936} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) - \frac {8779 \, x - 65991}{4292352 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {\sqrt {2} {\left (3594214 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{3} + 19874490 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} - 30140067 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 19989859\right )}}{2985984 \, {\left (2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} + 10 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} - 11\right )}^{2}} \]
-52631/11943936*sqrt(2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 52631/11943936*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt (2*x^2 - x + 3))) - 1/4292352*(8779*x - 65991)/sqrt(2*x^2 - x + 3) + 1/298 5984*sqrt(2)*(3594214*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^3 + 198744 90*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 - 30140067*sqrt(2)*(sqrt(2)*x - sqr t(2*x^2 - x + 3)) + 19989859)/(2*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 10* sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 11)^2
Timed out. \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5\,x^4-x^3+3\,x^2+x+2}{{\left (2\,x+5\right )}^3\,{\left (2\,x^2-x+3\right )}^{3/2}} \,d x \]