Integrand size = 29, antiderivative size = 73 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {-38+21 x}{198 \left (2+3 x^2\right )^{3/2}}+\frac {24+95 x}{726 \sqrt {2+3 x^2}}-\frac {8 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {2+3 x^2}}\right )}{121 \sqrt {11}} \]
1/198*(-38+21*x)/(3*x^2+2)^(3/2)-8/1331*arctanh(1/11*(4-3*x)*11^(1/2)/(3*x ^2+2)^(1/2))*11^(1/2)+1/726*(24+95*x)/(3*x^2+2)^(1/2)
Time = 0.60 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {-274+801 x+216 x^2+855 x^3}{2178 \left (2+3 x^2\right )^{3/2}}-\frac {8 \text {arctanh}\left (\frac {4-3 x}{\sqrt {22+33 x^2}}\right )}{121 \sqrt {11}} \]
(-274 + 801*x + 216*x^2 + 855*x^3)/(2178*(2 + 3*x^2)^(3/2)) - (8*ArcTanh[( 4 - 3*x)/Sqrt[22 + 33*x^2]])/(121*Sqrt[11])
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2178, 27, 686, 27, 488, 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.
\(\displaystyle \int \frac {4 x^2+3 x+1}{(2 x+1) \left (3 x^2+2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2178 |
\(\displaystyle -\frac {1}{18} \int -\frac {6 (14 x+13)}{11 (2 x+1) \left (3 x^2+2\right )^{3/2}}dx-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{33} \int \frac {14 x+13}{(2 x+1) \left (3 x^2+2\right )^{3/2}}dx-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\) |
\(\Big \downarrow \) 686 |
\(\displaystyle \frac {1}{33} \left (\frac {95 x+24}{22 \sqrt {3 x^2+2}}-\frac {1}{66} \int -\frac {144}{(2 x+1) \sqrt {3 x^2+2}}dx\right )-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{33} \left (\frac {24}{11} \int \frac {1}{(2 x+1) \sqrt {3 x^2+2}}dx+\frac {95 x+24}{22 \sqrt {3 x^2+2}}\right )-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {1}{33} \left (\frac {95 x+24}{22 \sqrt {3 x^2+2}}-\frac {24}{11} \int \frac {1}{11-\frac {(4-3 x)^2}{3 x^2+2}}d\frac {4-3 x}{\sqrt {3 x^2+2}}\right )-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{33} \left (\frac {95 x+24}{22 \sqrt {3 x^2+2}}-\frac {24 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {3 x^2+2}}\right )}{11 \sqrt {11}}\right )-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\) |
-1/198*(38 - 21*x)/(2 + 3*x^2)^(3/2) + ((24 + 95*x)/(22*Sqrt[2 + 3*x^2]) - (24*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2])])/(11*Sqrt[11]))/33
3.2.33.3.1 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/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x )^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.56 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01
method | result | size |
trager | \(\frac {855 x^{3}+216 x^{2}+801 x -274}{2178 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right ) x +11 \sqrt {3 x^{2}+2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right )}{1+2 x}\right )}{1331}\) | \(74\) |
default | \(\frac {x}{12 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {x}{12 \sqrt {3 x^{2}+2}}-\frac {2}{9 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {1}{33 \left (3 \left (x +\frac {1}{2}\right )^{2}-3 x +\frac {5}{4}\right )^{\frac {3}{2}}}+\frac {x}{44 \left (3 \left (x +\frac {1}{2}\right )^{2}-3 x +\frac {5}{4}\right )^{\frac {3}{2}}}+\frac {23 x}{484 \sqrt {3 \left (x +\frac {1}{2}\right )^{2}-3 x +\frac {5}{4}}}+\frac {4}{121 \sqrt {3 \left (x +\frac {1}{2}\right )^{2}-3 x +\frac {5}{4}}}-\frac {8 \sqrt {11}\, \operatorname {arctanh}\left (\frac {2 \left (4-3 x \right ) \sqrt {11}}{11 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-12 x +5}}\right )}{1331}\) | \(133\) |
1/2178*(855*x^3+216*x^2+801*x-274)/(3*x^2+2)^(3/2)+8/1331*RootOf(_Z^2-11)* ln((3*RootOf(_Z^2-11)*x+11*(3*x^2+2)^(1/2)-4*RootOf(_Z^2-11))/(1+2*x))
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.41 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {72 \, \sqrt {11} {\left (9 \, x^{4} + 12 \, x^{2} + 4\right )} \log \left (-\frac {\sqrt {11} \sqrt {3 \, x^{2} + 2} {\left (3 \, x - 4\right )} + 21 \, x^{2} - 12 \, x + 19}{4 \, x^{2} + 4 \, x + 1}\right ) + 11 \, {\left (855 \, x^{3} + 216 \, x^{2} + 801 \, x - 274\right )} \sqrt {3 \, x^{2} + 2}}{23958 \, {\left (9 \, x^{4} + 12 \, x^{2} + 4\right )}} \]
1/23958*(72*sqrt(11)*(9*x^4 + 12*x^2 + 4)*log(-(sqrt(11)*sqrt(3*x^2 + 2)*( 3*x - 4) + 21*x^2 - 12*x + 19)/(4*x^2 + 4*x + 1)) + 11*(855*x^3 + 216*x^2 + 801*x - 274)*sqrt(3*x^2 + 2))/(9*x^4 + 12*x^2 + 4)
Timed out. \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {8}{1331} \, \sqrt {11} \operatorname {arsinh}\left (\frac {\sqrt {6} x}{2 \, {\left | 2 \, x + 1 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 1 \right |}}\right ) + \frac {95 \, x}{726 \, \sqrt {3 \, x^{2} + 2}} + \frac {4}{121 \, \sqrt {3 \, x^{2} + 2}} + \frac {7 \, x}{66 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} - \frac {19}{99 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \]
8/1331*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x + 1)) + 95/726*x/sqrt(3*x^2 + 2) + 4/121/sqrt(3*x^2 + 2) + 7/66*x/(3*x^2 + 2)^(3/2) - 19/99/(3*x^2 + 2)^(3/2)
Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {8}{1331} \, \sqrt {11} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {11} - \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {11} + \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) + \frac {9 \, {\left ({\left (95 \, x + 24\right )} x + 89\right )} x - 274}{2178 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \]
8/1331*sqrt(11)*log(-abs(-2*sqrt(3)*x - sqrt(11) - sqrt(3) + 2*sqrt(3*x^2 + 2))/(2*sqrt(3)*x - sqrt(11) + sqrt(3) - 2*sqrt(3*x^2 + 2))) + 1/2178*(9* ((95*x + 24)*x + 89)*x - 274)/(3*x^2 + 2)^(3/2)
Time = 0.13 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.99 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {\sqrt {11}\,\left (8\,\ln \left (x+\frac {1}{2}\right )-8\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {11}\,\sqrt {x^2+\frac {2}{3}}}{3}-\frac {4}{3}\right )\right )}{1331}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {-\frac {21}{176}+\frac {\sqrt {6}\,19{}\mathrm {i}}{176}}{x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}}+\frac {\sqrt {6}\,\left (-\frac {7}{88}+\frac {\sqrt {6}\,19{}\mathrm {i}}{264}\right )\,1{}\mathrm {i}}{2\,{\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}^2}\right )}{27}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {\frac {21}{176}+\frac {\sqrt {6}\,19{}\mathrm {i}}{176}}{x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}}-\frac {\sqrt {6}\,\left (\frac {7}{88}+\frac {\sqrt {6}\,19{}\mathrm {i}}{264}\right )\,1{}\mathrm {i}}{2\,{\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}^2}\right )}{27}-\frac {\sqrt {3}\,\sqrt {6}\,\left (-288+\sqrt {6}\,303{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{104544\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}-\frac {\sqrt {3}\,\sqrt {6}\,\left (288+\sqrt {6}\,303{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{104544\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \]
(11^(1/2)*(8*log(x + 1/2) - 8*log(x - (3^(1/2)*11^(1/2)*(x^2 + 2/3)^(1/2)) /3 - 4/3)))/1331 - (3^(1/2)*(x^2 + 2/3)^(1/2)*(((6^(1/2)*19i)/176 - 21/176 )/(x + (6^(1/2)*1i)/3) + (6^(1/2)*((6^(1/2)*19i)/264 - 7/88)*1i)/(2*(x + ( 6^(1/2)*1i)/3)^2)))/27 + (3^(1/2)*(x^2 + 2/3)^(1/2)*(((6^(1/2)*19i)/176 + 21/176)/(x - (6^(1/2)*1i)/3) - (6^(1/2)*((6^(1/2)*19i)/264 + 7/88)*1i)/(2* (x - (6^(1/2)*1i)/3)^2)))/27 - (3^(1/2)*6^(1/2)*(6^(1/2)*303i - 288)*(x^2 + 2/3)^(1/2)*1i)/(104544*(x + (6^(1/2)*1i)/3)) - (3^(1/2)*6^(1/2)*(6^(1/2) *303i + 288)*(x^2 + 2/3)^(1/2)*1i)/(104544*(x - (6^(1/2)*1i)/3))