Integrand size = 40, antiderivative size = 101 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {1191+917 x}{3312 \sqrt {3-x+2 x^2}}+\frac {5}{8} \sqrt {3-x+2 x^2}+\frac {39 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{16 \sqrt {2}}-\frac {3667 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{1728 \sqrt {2}} \]
39/32*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-3667/3456*arctanh(1/24*(17-22 *x)*2^(1/2)/(2*x^2-x+3)^(1/2))*2^(1/2)+1/3312*(1191+917*x)/(2*x^2-x+3)^(1/ 2)+5/8*(2*x^2-x+3)^(1/2)
Time = 0.47 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {\frac {12 \left (7401-1153 x+4140 x^2\right )}{\sqrt {3-x+2 x^2}}+84341 \sqrt {2} \text {arctanh}\left (\frac {1}{6} \left (5+2 x-\sqrt {6-2 x+4 x^2}\right )\right )+48438 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{39744} \]
((12*(7401 - 1153*x + 4140*x^2))/Sqrt[3 - x + 2*x^2] + 84341*Sqrt[2]*ArcTa nh[(5 + 2*x - Sqrt[6 - 2*x + 4*x^2])/6] + 48438*Sqrt[2]*Log[1 - 4*x + 2*Sq rt[6 - 2*x + 4*x^2]])/39744
Time = 0.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {2177, 27, 2184, 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.
| \(\displaystyle \int \frac {5 x^4-x^3+3 x^2+x+2}{(2 x+5) \left (2 x^2-x+3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2177 |
| \(\displaystyle \frac {2}{23} \int -\frac {23 \left (-720 x^2-216 x+293\right )}{576 (2 x+5) \sqrt {2 x^2-x+3}}dx+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}-\frac {1}{288} \int \frac {-720 x^2-216 x+293}{(2 x+5) \sqrt {2 x^2-x+3}}dx\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {1}{288} \left (180 \sqrt {2 x^2-x+3}-\frac {1}{8} \int -\frac {8 (157-1404 x)}{(2 x+5) \sqrt {2 x^2-x+3}}dx\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{288} \left (\int \frac {157-1404 x}{(2 x+5) \sqrt {2 x^2-x+3}}dx+180 \sqrt {2 x^2-x+3}\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {1}{288} \left (-702 \int \frac {1}{\sqrt {2 x^2-x+3}}dx+3667 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx+180 \sqrt {2 x^2-x+3}\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {1}{288} \left (3667 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-351 \sqrt {\frac {2}{23}} \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)+180 \sqrt {2 x^2-x+3}\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{288} \left (3667 \int \frac {1}{(2 x+5) \sqrt {2 x^2-x+3}}dx-351 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )+180 \sqrt {2 x^2-x+3}\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {1}{288} \left (-7334 \int \frac {1}{288-\frac {(17-22 x)^2}{2 x^2-x+3}}d\frac {17-22 x}{\sqrt {2 x^2-x+3}}-351 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )+180 \sqrt {2 x^2-x+3}\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{288} \left (-351 \sqrt {2} \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )-\frac {3667 \text {arctanh}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{6 \sqrt {2}}+180 \sqrt {2 x^2-x+3}\right )+\frac {917 x+1191}{3312 \sqrt {2 x^2-x+3}}\) |
(1191 + 917*x)/(3312*Sqrt[3 - x + 2*x^2]) + (180*Sqrt[3 - x + 2*x^2] - 351 *Sqrt[2]*ArcSinh[(-1 + 4*x)/Sqrt[23]] - (3667*ArcTanh[(17 - 22*x)/(12*Sqrt [2]*Sqrt[3 - x + 2*x^2])])/(6*Sqrt[2]))/288
3.4.54.3.1 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[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c *d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !(IGt Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Timed out.
hanged
Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.48 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {96876 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 84341 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (-\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) + 48 \, {\left (4140 \, x^{2} - 1153 \, x + 7401\right )} \sqrt {2 \, x^{2} - x + 3}}{158976 \, {\left (2 \, x^{2} - x + 3\right )}} \]
1/158976*(96876*sqrt(2)*(2*x^2 - x + 3)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)* (4*x - 1) - 32*x^2 + 16*x - 25) + 84341*sqrt(2)*(2*x^2 - x + 3)*log(-(24*s qrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) + 1060*x^2 - 1036*x + 1153)/(4*x^2 + 20*x + 25)) + 48*(4140*x^2 - 1153*x + 7401)*sqrt(2*x^2 - x + 3))/(2*x^2 - x + 3)
\[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x + 5\right ) \left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {5 \, x^{2}}{4 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {39}{32} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) + \frac {3667}{3456} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) - \frac {1153 \, x}{3312 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {2467}{1104 \, \sqrt {2 \, x^{2} - x + 3}} \]
5/4*x^2/sqrt(2*x^2 - x + 3) - 39/32*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23 *sqrt(23)) + 3667/3456*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/ 23*sqrt(23)/abs(2*x + 5)) - 1153/3312*x/sqrt(2*x^2 - x + 3) + 2467/1104/sq rt(2*x^2 - x + 3)
Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\frac {39}{32} \, \sqrt {2} \log \left (-4 \, \sqrt {2} x + \sqrt {2} + 4 \, \sqrt {2 \, x^{2} - x + 3}\right ) - \frac {3667}{3456} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {3667}{3456} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {{\left (4140 \, x - 1153\right )} x + 7401}{3312 \, \sqrt {2 \, x^{2} - x + 3}} \]
39/32*sqrt(2)*log(-4*sqrt(2)*x + sqrt(2) + 4*sqrt(2*x^2 - x + 3)) - 3667/3 456*sqrt(2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 366 7/3456*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 1/3312*((4140*x - 1153)*x + 7401)/sqrt(2*x^2 - x + 3)
Timed out. \[ \int \frac {2+x+3 x^2-x^3+5 x^4}{(5+2 x) \left (3-x+2 x^2\right )^{3/2}} \, dx=\int \frac {5\,x^4-x^3+3\,x^2+x+2}{\left (2\,x+5\right )\,{\left (2\,x^2-x+3\right )}^{3/2}} \,d x \]