Integrand size = 40, antiderivative size = 108 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {9897+2203 x}{119232 \sqrt {3-x+2 x^2}}-\frac {3667 \sqrt {3-x+2 x^2}}{10368 (5+2 x)}-\frac {5 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{8 \sqrt {2}}+\frac {25951 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{41472 \sqrt {2}} \]
-5/16*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+25951/82944*arctanh(1/24*(17- 22*x)*2^(1/2)/(2*x^2-x+3)^(1/2))*2^(1/2)+1/119232*(9897+2203*x)/(2*x^2-x+3 )^(1/2)-3667/10368*(2*x^2-x+3)^(1/2)/(5+2*x)
Time = 0.53 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {3-x+2 x^2} \left (51351-48653 x+53290 x^2\right )}{79488 \left (15+x+8 x^2+4 x^3\right )}-\frac {25951 \text {arctanh}\left (\frac {1}{6} \left (5+2 x-\sqrt {6-2 x+4 x^2}\right )\right )}{20736 \sqrt {2}}-\frac {5 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{8 \sqrt {2}} \]
-1/79488*(Sqrt[3 - x + 2*x^2]*(51351 - 48653*x + 53290*x^2))/(15 + x + 8*x ^2 + 4*x^3) - (25951*ArcTanh[(5 + 2*x - Sqrt[6 - 2*x + 4*x^2])/6])/(20736* Sqrt[2]) - (5*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/(8*Sqrt[2])
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {2177, 27, 2181, 27, 1269, 1090, 222, 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.
| \(\displaystyle \int \frac {5 x^4-x^3+3 x^2+x+2}{(2 x+5)^2 \left (2 x^2-x+3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2177 |
| \(\displaystyle \frac {2}{23} \int -\frac {23 \left (-25920 x^2-11410 x+1463\right )}{20736 (2 x+5)^2 \sqrt {2 x^2-x+3}}dx+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}-\frac {\int \frac {-25920 x^2-11410 x+1463}{(2 x+5)^2 \sqrt {2 x^2-x+3}}dx}{10368}\) |
| \(\Big \downarrow \) 2181 |
| \(\displaystyle \frac {\frac {1}{72} \int -\frac {108 (4351-8640 x)}{(2 x+5) \sqrt {2 x^2-x+3}}dx-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {3}{2} \int \frac {4351-8640 x}{(2 x+5) \sqrt {2 x^2-x+3}}dx-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {-\frac {3}{2} \left (25951 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-4320 \int \frac {1}{\sqrt {2 x^2-x+3}}dx\right )-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {-\frac {3}{2} \left (25951 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-2160 \sqrt {\frac {2}{23}} \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)\right )-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {-\frac {3}{2} \left (25951 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-2160 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )\right )-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {-\frac {3}{2} \left (-51902 \int \frac {1}{288-\frac {(17-22 x)^2}{2 x^2-x+3}}d\frac {17-22 x}{\sqrt {2 x^2-x+3}}-2160 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )\right )-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {3}{2} \left (-2160 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )-\frac {25951 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{6 \sqrt {2}}\right )-\frac {3667 \sqrt {2 x^2-x+3}}{2 x+5}}{10368}+\frac {2203 x+9897}{119232 \sqrt {2 x^2-x+3}}\) |
(9897 + 2203*x)/(119232*Sqrt[3 - x + 2*x^2]) + ((-3667*Sqrt[3 - x + 2*x^2] )/(5 + 2*x) - (3*(-2160*Sqrt[2]*ArcSinh[(-1 + 4*x)/Sqrt[23]] - (25951*ArcT anh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(6*Sqrt[2])))/2)/10368
3.4.55.3.1 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/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 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[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 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 = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.45 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {596160 \, \sqrt {2} {\left (4 \, x^{3} + 8 \, x^{2} + x + 15\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 596873 \, \sqrt {2} {\left (4 \, x^{3} + 8 \, x^{2} + x + 15\right )} \log \left (\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} - 1060 \, x^{2} + 1036 \, x - 1153}{4 \, x^{2} + 20 \, x + 25}\right ) - 48 \, {\left (53290 \, x^{2} - 48653 \, x + 51351\right )} \sqrt {2 \, x^{2} - x + 3}}{3815424 \, {\left (4 \, x^{3} + 8 \, x^{2} + x + 15\right )}} \]
1/3815424*(596160*sqrt(2)*(4*x^3 + 8*x^2 + x + 15)*log(-4*sqrt(2)*sqrt(2*x ^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 596873*sqrt(2)*(4*x^3 + 8*x^ 2 + x + 15)*log((24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) - 1060*x^2 + 1 036*x - 1153)/(4*x^2 + 20*x + 25)) - 48*(53290*x^2 - 48653*x + 51351)*sqrt (2*x^2 - x + 3))/(4*x^3 + 8*x^2 + x + 15)
\[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x + 5\right )^{2} \left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {5}{16} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) - \frac {25951}{82944} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) - \frac {26645 \, x}{79488 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {30313}{26496 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {3667}{576 \, {\left (2 \, \sqrt {2 \, x^{2} - x + 3} x + 5 \, \sqrt {2 \, x^{2} - x + 3}\right )}} \]
5/16*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) - 25951/82944*sqrt(2 )*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) - 2 6645/79488*x/sqrt(2*x^2 - x + 3) + 30313/26496/sqrt(2*x^2 - x + 3) - 3667/ 576/(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. 225 vs. \(2 (85) = 170\).
Time = 0.35 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.08 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {1}{1907712} \, \sqrt {2} {\left (\frac {12 \, {\left (\frac {\frac {315103}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )} - \frac {1012092}{{\left (2 \, x + 5\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}}{2 \, x + 5} - \frac {26645}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}\right )}}{\sqrt {-\frac {11}{2 \, x + 5} + \frac {36}{{\left (2 \, x + 5\right )}^{2}} + 1}} + \frac {596873 \, \log \left (12 \, \sqrt {-\frac {11}{2 \, x + 5} + \frac {36}{{\left (2 \, x + 5\right )}^{2}} + 1} + \frac {72}{2 \, x + 5} - 11\right )}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )} + \frac {596160 \, \log \left ({\left | \sqrt {-\frac {11}{2 \, x + 5} + \frac {36}{{\left (2 \, x + 5\right )}^{2}} + 1} + \frac {6}{2 \, x + 5} + 1 \right |}\right )}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )} - \frac {596160 \, \log \left ({\left | \sqrt {-\frac {11}{2 \, x + 5} + \frac {36}{{\left (2 \, x + 5\right )}^{2}} + 1} + \frac {6}{2 \, x + 5} - 1 \right |}\right )}{\mathrm {sgn}\left (\frac {1}{2 \, x + 5}\right )}\right )} \]
1/1907712*sqrt(2)*(12*((315103/sgn(1/(2*x + 5)) - 1012092/((2*x + 5)*sgn(1 /(2*x + 5))))/(2*x + 5) - 26645/sgn(1/(2*x + 5)))/sqrt(-11/(2*x + 5) + 36/ (2*x + 5)^2 + 1) + 596873*log(12*sqrt(-11/(2*x + 5) + 36/(2*x + 5)^2 + 1) + 72/(2*x + 5) - 11)/sgn(1/(2*x + 5)) + 596160*log(abs(sqrt(-11/(2*x + 5) + 36/(2*x + 5)^2 + 1) + 6/(2*x + 5) + 1))/sgn(1/(2*x + 5)) - 596160*log(ab s(sqrt(-11/(2*x + 5) + 36/(2*x + 5)^2 + 1) + 6/(2*x + 5) - 1))/sgn(1/(2*x + 5)))
Timed out. \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5\,x^4-x^3+3\,x^2+x+2}{{\left (2\,x+5\right )}^2\,{\left (2\,x^2-x+3\right )}^{3/2}} \,d x \]