Integrand size = 40, antiderivative size = 110 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {9897+2203 x}{357696 \left (3-x+2 x^2\right )^{3/2}}-\frac {1255878-62021 x}{24681024 \sqrt {3-x+2 x^2}}-\frac {3667 \sqrt {3-x+2 x^2}}{186624 (5+2 x)}-\frac {2821 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{2239488 \sqrt {2}} \]
1/357696*(9897+2203*x)/(2*x^2-x+3)^(3/2)-2821/4478976*arctanh(1/24*(17-22* x)*2^(1/2)/(2*x^2-x+3)^(1/2))*2^(1/2)+1/24681024*(-1255878+62021*x)/(2*x^2 -x+3)^(1/2)-3667/186624*(2*x^2-x+3)^(1/2)/(5+2*x)
Time = 0.65 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.74 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {-\frac {12 \left (79153407-18840090 x+63941915 x^2+10350004 x^3+6767036 x^4\right )}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}}+1492309 \sqrt {2} \text {arctanh}\left (\frac {1}{6} \left (5+2 x-\sqrt {6-2 x+4 x^2}\right )\right )}{1184689152} \]
((-12*(79153407 - 18840090*x + 63941915*x^2 + 10350004*x^3 + 6767036*x^4)) /((5 + 2*x)*(3 - x + 2*x^2)^(3/2)) + 1492309*Sqrt[2]*ArcTanh[(5 + 2*x - Sq rt[6 - 2*x + 4*x^2])/6])/1184689152
Time = 0.39 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2177, 27, 2177, 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.
| \(\displaystyle \int \frac {5 x^4-x^3+3 x^2+x+2}{(2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2177 |
| \(\displaystyle \frac {2}{69} \int \frac {1823728 x^2+963530 x+119353}{20736 (2 x+5)^2 \left (2 x^2-x+3\right )^{3/2}}dx+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {1823728 x^2+963530 x+119353}{(2 x+5)^2 \left (2 x^2-x+3\right )^{3/2}}dx}{715392}+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
| \(\Big \downarrow \) 2177 |
| \(\displaystyle \frac {\frac {2}{23} \int \frac {529 (19111-18758 x)}{6 (2 x+5)^2 \sqrt {2 x^2-x+3}}dx-\frac {2 (1255878-62021 x)}{69 \sqrt {2 x^2-x+3}}}{715392}+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {23}{3} \int \frac {19111-18758 x}{(2 x+5)^2 \sqrt {2 x^2-x+3}}dx-\frac {2 (1255878-62021 x)}{69 \sqrt {2 x^2-x+3}}}{715392}+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {\frac {23}{3} \left (\frac {2821}{4} \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-\frac {3667 \sqrt {2 x^2-x+3}}{2 (2 x+5)}\right )-\frac {2 (1255878-62021 x)}{69 \sqrt {2 x^2-x+3}}}{715392}+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\frac {23}{3} \left (-\frac {2821}{2} \int \frac {1}{288-\frac {(17-22 x)^2}{2 x^2-x+3}}d\frac {17-22 x}{\sqrt {2 x^2-x+3}}-\frac {3667 \sqrt {2 x^2-x+3}}{2 (2 x+5)}\right )-\frac {2 (1255878-62021 x)}{69 \sqrt {2 x^2-x+3}}}{715392}+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {23}{3} \left (-\frac {2821 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{24 \sqrt {2}}-\frac {3667 \sqrt {2 x^2-x+3}}{2 (2 x+5)}\right )-\frac {2 (1255878-62021 x)}{69 \sqrt {2 x^2-x+3}}}{715392}+\frac {2203 x+9897}{357696 \left (2 x^2-x+3\right )^{3/2}}\) |
(9897 + 2203*x)/(357696*(3 - x + 2*x^2)^(3/2)) + ((-2*(1255878 - 62021*x)) /(69*Sqrt[3 - x + 2*x^2]) + (23*((-3667*Sqrt[3 - x + 2*x^2])/(2*(5 + 2*x)) - (2821*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(24*Sqrt[2 ])))/3)/715392
3.4.62.3.1 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]
Timed out.
hanged
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.28 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {1492309 \, \sqrt {2} {\left (8 \, x^{5} + 12 \, x^{4} + 6 \, x^{3} + 53 \, x^{2} - 12 \, x + 45\right )} \log \left (-\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) - 48 \, {\left (6767036 \, x^{4} + 10350004 \, x^{3} + 63941915 \, x^{2} - 18840090 \, x + 79153407\right )} \sqrt {2 \, x^{2} - x + 3}}{4738756608 \, {\left (8 \, x^{5} + 12 \, x^{4} + 6 \, x^{3} + 53 \, x^{2} - 12 \, x + 45\right )}} \]
1/4738756608*(1492309*sqrt(2)*(8*x^5 + 12*x^4 + 6*x^3 + 53*x^2 - 12*x + 45 )*log(-(24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) + 1060*x^2 - 1036*x + 1 153)/(4*x^2 + 20*x + 25)) - 48*(6767036*x^4 + 10350004*x^3 + 63941915*x^2 - 18840090*x + 79153407)*sqrt(2*x^2 - x + 3))/(8*x^5 + 12*x^4 + 6*x^3 + 53 *x^2 - 12*x + 45)
\[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=\int \frac {5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x + 5\right )^{2} \left (2 x^{2} - x + 3\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=\frac {2821}{4478976} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) - \frac {1691759 \, x}{98724096 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {265339}{32908032 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {248617 \, x}{715392 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {3667}{576 \, {\left (2 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 5 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} + \frac {259621}{238464 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]
2821/4478976*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23 )/abs(2*x + 5)) - 1691759/98724096*x/sqrt(2*x^2 - x + 3) + 265339/32908032 /sqrt(2*x^2 - x + 3) - 248617/715392*x/(2*x^2 - x + 3)^(3/2) - 3667/576/(2 *(2*x^2 - x + 3)^(3/2)*x + 5*(2*x^2 - x + 3)^(3/2)) + 259621/238464/(2*x^2 - x + 3)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (88) = 176\).
Time = 0.35 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.87 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=-\frac {1}{2369378304} \, \sqrt {2} {\left (\frac {1492309 \, \log \left (12 \, \sqrt {-\frac {11}{2 \, x + 5} + \frac {36}{{\left (2 \, x + 5\right )}^{2}} + 1} + \frac {72}{2 \, x + 5} - 11\right )}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )} + \frac {12 \, {\left (\frac {\frac {\frac {48 \, {\left (\frac {23642785}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )} - \frac {52375761}{{\left (2 \, x + 5\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}\right )}}{2 \, x + 5} - \frac {240080735}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}}{2 \, x + 5} + \frac {28660178}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}}{2 \, x + 5} - \frac {1691759}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}\right )}}{{\left (\frac {11}{2 \, x + 5} - \frac {36}{{\left (2 \, x + 5\right )}^{2}} - 1\right )} \sqrt {-\frac {11}{2 \, x + 5} + \frac {36}{{\left (2 \, x + 5\right )}^{2}} + 1}} - 20301108 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )\right )} \]
-1/2369378304*sqrt(2)*(1492309*log(12*sqrt(-11/(2*x + 5) + 36/(2*x + 5)^2 + 1) + 72/(2*x + 5) - 11)/sgn(1/(2*x + 5)) + 12*(((48*(23642785/sgn(1/(2*x + 5)) - 52375761/((2*x + 5)*sgn(1/(2*x + 5))))/(2*x + 5) - 240080735/sgn( 1/(2*x + 5)))/(2*x + 5) + 28660178/sgn(1/(2*x + 5)))/(2*x + 5) - 1691759/s gn(1/(2*x + 5)))/((11/(2*x + 5) - 36/(2*x + 5)^2 - 1)*sqrt(-11/(2*x + 5) + 36/(2*x + 5)^2 + 1)) - 20301108*sgn(1/(2*x + 5)))
Timed out. \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}} \, dx=\int \frac {5\,x^4-x^3+3\,x^2+x+2}{{\left (2\,x+5\right )}^2\,{\left (2\,x^2-x+3\right )}^{5/2}} \,d x \]