\(\int \frac {1+3 x+4 x^2}{(1+2 x) (2+3 x^2)^{5/2}} \, dx\) [133]

Optimal result

Integrand size = 29, antiderivative size = 73 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {-38+21 x}{198 \left (2+3 x^2\right )^{3/2}}+\frac {24+95 x}{726 \sqrt {2+3 x^2}}-\frac {8 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {2+3 x^2}}\right )}{121 \sqrt {11}} \] Output:

1/198*(-38+21*x)/(3*x^2+2)^(3/2)+1/726*(24+95*x)/(3*x^2+2)^(1/2)-8/1331*ar 
ctanh(1/11*(4-3*x)*11^(1/2)/(3*x^2+2)^(1/2))*11^(1/2)
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {-274+801 x+216 x^2+855 x^3}{2178 \left (2+3 x^2\right )^{3/2}}-\frac {8 \text {arctanh}\left (\frac {4-3 x}{\sqrt {22+33 x^2}}\right )}{121 \sqrt {11}} \] Input:

Integrate[(1 + 3*x + 4*x^2)/((1 + 2*x)*(2 + 3*x^2)^(5/2)),x]
 

Output:

(-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])
 

Rubi [A] (verified)

Time = 0.26 (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

\(\displaystyle \frac {1}{33} \left (\frac {95 x+24}{22 \sqrt {3 x^2+2}}-\frac {24 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {3 x^2+2}}\right )}{11 \sqrt {11}}\right )-\frac {38-21 x}{198 \left (3 x^2+2\right )^{3/2}}\)

Input:

Int[(1 + 3*x + 4*x^2)/((1 + 2*x)*(2 + 3*x^2)^(5/2)),x]
 

Output:

-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
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 488

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]
 

rule 686

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])
 

rule 2178

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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.91 (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	\(\frac {x}{12 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {x}{12 \sqrt {3 x^{2}+2}}-\frac {2}{9 \left (3 x^{2}+2\right )^{\frac {3}{2}}}+\frac {1}{33 \left (3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x \right )^{\frac {3}{2}}}+\frac {x}{44 \left (3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x \right )^{\frac {3}{2}}}+\frac {23 x}{484 \sqrt {3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x}}+\frac {4}{121 \sqrt {3 \left (\frac {1}{2}+x \right )^{2}+\frac {5}{4}-3 x}}-\frac {8 \sqrt {11}\, \operatorname {arctanh}\left (\frac {2 \left (4-3 x \right ) \sqrt {11}}{11 \sqrt {12 \left (\frac {1}{2}+x \right )^{2}+5-12 x}}\right )}{1331}\)	\(133\)

Input:

int((4*x^2+3*x+1)/(1+2*x)/(3*x^2+2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.41 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {72 \, \sqrt {11} {\left (9 \, x^{4} + 12 \, x^{2} + 4\right )} \log \left (-\frac {\sqrt {11} \sqrt {3 \, x^{2} + 2} {\left (3 \, x - 4\right )} + 21 \, x^{2} - 12 \, x + 19}{4 \, x^{2} + 4 \, x + 1}\right ) + 11 \, {\left (855 \, x^{3} + 216 \, x^{2} + 801 \, x - 274\right )} \sqrt {3 \, x^{2} + 2}}{23958 \, {\left (9 \, x^{4} + 12 \, x^{2} + 4\right )}} \] Input:

integrate((4*x^2+3*x+1)/(1+2*x)/(3*x^2+2)^(5/2),x, algorithm="fricas")
 

Output:

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)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((4*x**2+3*x+1)/(1+2*x)/(3*x**2+2)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.11 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {8}{1331} \, \sqrt {11} \operatorname {arsinh}\left (\frac {\sqrt {6} x}{2 \, {\left | 2 \, x + 1 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 1 \right |}}\right ) + \frac {95 \, x}{726 \, \sqrt {3 \, x^{2} + 2}} + \frac {4}{121 \, \sqrt {3 \, x^{2} + 2}} + \frac {7 \, x}{66 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} - \frac {19}{99 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \] Input:

integrate((4*x^2+3*x+1)/(1+2*x)/(3*x^2+2)^(5/2),x, algorithm="maxima")
 

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {8}{1331} \, \sqrt {11} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {11} - \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {11} + \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) + \frac {9 \, {\left ({\left (95 \, x + 24\right )} x + 89\right )} x - 274}{2178 \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}}} \] Input:

integrate((4*x^2+3*x+1)/(1+2*x)/(3*x^2+2)^(5/2),x, algorithm="giac")
 

Output:

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)
 

Mupad [B] (verification not implemented)

Time = 16.91 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.99 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {\sqrt {11}\,\left (8\,\ln \left (x+\frac {1}{2}\right )-8\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {11}\,\sqrt {x^2+\frac {2}{3}}}{3}-\frac {4}{3}\right )\right )}{1331}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {-\frac {21}{176}+\frac {\sqrt {6}\,19{}\mathrm {i}}{176}}{x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}}+\frac {\sqrt {6}\,\left (-\frac {7}{88}+\frac {\sqrt {6}\,19{}\mathrm {i}}{264}\right )\,1{}\mathrm {i}}{2\,{\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}^2}\right )}{27}+\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {\frac {21}{176}+\frac {\sqrt {6}\,19{}\mathrm {i}}{176}}{x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}}-\frac {\sqrt {6}\,\left (\frac {7}{88}+\frac {\sqrt {6}\,19{}\mathrm {i}}{264}\right )\,1{}\mathrm {i}}{2\,{\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}^2}\right )}{27}-\frac {\sqrt {3}\,\sqrt {6}\,\left (-288+\sqrt {6}\,303{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{104544\,\left (x+\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )}-\frac {\sqrt {3}\,\sqrt {6}\,\left (288+\sqrt {6}\,303{}\mathrm {i}\right )\,\sqrt {x^2+\frac {2}{3}}\,1{}\mathrm {i}}{104544\,\left (x-\frac {\sqrt {6}\,1{}\mathrm {i}}{3}\right )} \] Input:

int((3*x + 4*x^2 + 1)/((2*x + 1)*(3*x^2 + 2)^(5/2)),x)
 

Output:

(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))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 372, normalized size of antiderivative = 5.10 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2+3 x^2\right )^{5/2}} \, dx=\frac {1296 \sqrt {11}\, \mathit {atan} \left (\frac {2 \sqrt {3 x^{2}+2}\, i +2 \sqrt {3}\, i x}{\sqrt {11}-\sqrt {3}}\right ) i \,x^{4}+1728 \sqrt {11}\, \mathit {atan} \left (\frac {2 \sqrt {3 x^{2}+2}\, i +2 \sqrt {3}\, i x}{\sqrt {11}-\sqrt {3}}\right ) i \,x^{2}+576 \sqrt {11}\, \mathit {atan} \left (\frac {2 \sqrt {3 x^{2}+2}\, i +2 \sqrt {3}\, i x}{\sqrt {11}-\sqrt {3}}\right ) i +9405 \sqrt {3 x^{2}+2}\, x^{3}+2376 \sqrt {3 x^{2}+2}\, x^{2}+8811 \sqrt {3 x^{2}+2}\, x -3014 \sqrt {3 x^{2}+2}+648 \sqrt {11}\, \mathrm {log}\left (4 \sqrt {3 x^{2}+2}\, \sqrt {3}\, x +\sqrt {33}+12 x^{2}-3\right ) x^{4}+864 \sqrt {11}\, \mathrm {log}\left (4 \sqrt {3 x^{2}+2}\, \sqrt {3}\, x +\sqrt {33}+12 x^{2}-3\right ) x^{2}+288 \sqrt {11}\, \mathrm {log}\left (4 \sqrt {3 x^{2}+2}\, \sqrt {3}\, x +\sqrt {33}+12 x^{2}-3\right )-1296 \sqrt {11}\, \mathrm {log}\left (\frac {2 \sqrt {3 x^{2}+2}+\sqrt {11}+2 \sqrt {3}\, x +\sqrt {3}}{\sqrt {2}}\right ) x^{4}-1728 \sqrt {11}\, \mathrm {log}\left (\frac {2 \sqrt {3 x^{2}+2}+\sqrt {11}+2 \sqrt {3}\, x +\sqrt {3}}{\sqrt {2}}\right ) x^{2}-576 \sqrt {11}\, \mathrm {log}\left (\frac {2 \sqrt {3 x^{2}+2}+\sqrt {11}+2 \sqrt {3}\, x +\sqrt {3}}{\sqrt {2}}\right )-8217 \sqrt {3}\, x^{4}-10956 \sqrt {3}\, x^{2}-3652 \sqrt {3}}{215622 x^{4}+287496 x^{2}+95832} \] Input:

int((4*x^2+3*x+1)/(1+2*x)/(3*x^2+2)^(5/2),x)
 

Output:

(1296*sqrt(11)*atan((2*sqrt(3*x**2 + 2)*i + 2*sqrt(3)*i*x)/(sqrt(11) - sqr 
t(3)))*i*x**4 + 1728*sqrt(11)*atan((2*sqrt(3*x**2 + 2)*i + 2*sqrt(3)*i*x)/ 
(sqrt(11) - sqrt(3)))*i*x**2 + 576*sqrt(11)*atan((2*sqrt(3*x**2 + 2)*i + 2 
*sqrt(3)*i*x)/(sqrt(11) - sqrt(3)))*i + 9405*sqrt(3*x**2 + 2)*x**3 + 2376* 
sqrt(3*x**2 + 2)*x**2 + 8811*sqrt(3*x**2 + 2)*x - 3014*sqrt(3*x**2 + 2) + 
648*sqrt(11)*log(4*sqrt(3*x**2 + 2)*sqrt(3)*x + sqrt(33) + 12*x**2 - 3)*x* 
*4 + 864*sqrt(11)*log(4*sqrt(3*x**2 + 2)*sqrt(3)*x + sqrt(33) + 12*x**2 - 
3)*x**2 + 288*sqrt(11)*log(4*sqrt(3*x**2 + 2)*sqrt(3)*x + sqrt(33) + 12*x* 
*2 - 3) - 1296*sqrt(11)*log((2*sqrt(3*x**2 + 2) + sqrt(11) + 2*sqrt(3)*x + 
 sqrt(3))/sqrt(2))*x**4 - 1728*sqrt(11)*log((2*sqrt(3*x**2 + 2) + sqrt(11) 
 + 2*sqrt(3)*x + sqrt(3))/sqrt(2))*x**2 - 576*sqrt(11)*log((2*sqrt(3*x**2 
+ 2) + sqrt(11) + 2*sqrt(3)*x + sqrt(3))/sqrt(2)) - 8217*sqrt(3)*x**4 - 10 
956*sqrt(3)*x**2 - 3652*sqrt(3))/(23958*(9*x**4 + 12*x**2 + 4))