\(\int \frac {(1+2 x)^2 (1+3 x+4 x^2)}{\sqrt {2+3 x^2}} \, dx\) [119]

Optimal result

Integrand size = 29, antiderivative size = 82 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {5}{18} (1+2 x)^2 \sqrt {2+3 x^2}+\frac {1}{6} (1+2 x)^3 \sqrt {2+3 x^2}-\frac {1}{27} (61+3 x) \sqrt {2+3 x^2}-\sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ) \]

[Out]

-arcsinh(1/2*x*6^(1/2))*3^(1/2)+5/18*(1+2*x)^2*(3*x^2+2)^(1/2)+1/6*(1+2*x)^3*(3*x^2+2)^(1/2)-1/27*(61+3*x)*(3*
x^2+2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1668, 847, 794, 221} \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=-\sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )+\frac {1}{6} \sqrt {3 x^2+2} (2 x+1)^3+\frac {5}{18} \sqrt {3 x^2+2} (2 x+1)^2-\frac {1}{27} (3 x+61) \sqrt {3 x^2+2} \]

[In]

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

[Out]

(5*(1 + 2*x)^2*Sqrt[2 + 3*x^2])/18 + ((1 + 2*x)^3*Sqrt[2 + 3*x^2])/6 - ((61 + 3*x)*Sqrt[2 + 3*x^2])/27 - Sqrt[
3]*ArcSinh[Sqrt[3/2]*x]

Rule 221

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]

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 1668

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + 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)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} (1+2 x)^3 \sqrt {2+3 x^2}+\frac {1}{48} \int \frac {(1+2 x)^2 (-48+120 x)}{\sqrt {2+3 x^2}} \, dx \\ & = \frac {5}{18} (1+2 x)^2 \sqrt {2+3 x^2}+\frac {1}{6} (1+2 x)^3 \sqrt {2+3 x^2}+\frac {1}{432} \int \frac {(-1392-144 x) (1+2 x)}{\sqrt {2+3 x^2}} \, dx \\ & = \frac {5}{18} (1+2 x)^2 \sqrt {2+3 x^2}+\frac {1}{6} (1+2 x)^3 \sqrt {2+3 x^2}-\frac {1}{27} (61+3 x) \sqrt {2+3 x^2}-3 \int \frac {1}{\sqrt {2+3 x^2}} \, dx \\ & = \frac {5}{18} (1+2 x)^2 \sqrt {2+3 x^2}+\frac {1}{6} (1+2 x)^3 \sqrt {2+3 x^2}-\frac {1}{27} (61+3 x) \sqrt {2+3 x^2}-\sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {1}{27} \sqrt {2+3 x^2} \left (-49+54 x+84 x^2+36 x^3\right )+\sqrt {3} \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right ) \]

[In]

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

[Out]

(Sqrt[2 + 3*x^2]*(-49 + 54*x + 84*x^2 + 36*x^3))/27 + Sqrt[3]*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^2]]

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.49

method	result	size
risch	\(\frac {\left (36 x^{3}+84 x^{2}+54 x -49\right ) \sqrt {3 x^{2}+2}}{27}-\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}\)	\(40\)
trager	\(\left (\frac {4}{3} x^{3}+\frac {28}{9} x^{2}+2 x -\frac {49}{27}\right ) \sqrt {3 x^{2}+2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )\)	\(56\)
default	\(-\operatorname {arcsinh}\left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}-\frac {49 \sqrt {3 x^{2}+2}}{27}+\frac {4 x^{3} \sqrt {3 x^{2}+2}}{3}+2 x \sqrt {3 x^{2}+2}+\frac {28 x^{2} \sqrt {3 x^{2}+2}}{9}\)	\(65\)
meijerg	\(\frac {\sqrt {3}\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )}{3}+\frac {20 \sqrt {3}\, \left (\frac {\sqrt {\pi }\, x \sqrt {3}\, \sqrt {2}\, \sqrt {\frac {3 x^{2}}{2}+1}}{2}-\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )\right )}{9 \sqrt {\pi }}+\frac {7 \sqrt {2}\, \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {\frac {3 x^{2}}{2}+1}\right )}{6 \sqrt {\pi }}+\frac {28 \sqrt {2}\, \left (\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-6 x^{2}+8\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{6}\right )}{9 \sqrt {\pi }}+\frac {32 \sqrt {3}\, \left (-\frac {\sqrt {\pi }\, x \sqrt {3}\, \sqrt {2}\, \left (-15 x^{2}+15\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{40}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {3}\, \sqrt {2}}{2}\right )}{4}\right )}{27 \sqrt {\pi }}\)	\(177\)

[In]

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

[Out]

1/27*(36*x^3+84*x^2+54*x-49)*(3*x^2+2)^(1/2)-arcsinh(1/2*x*6^(1/2))*3^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.66 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {1}{27} \, {\left (36 \, x^{3} + 84 \, x^{2} + 54 \, x - 49\right )} \sqrt {3 \, x^{2} + 2} + \frac {1}{2} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \]

[In]

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

[Out]

1/27*(36*x^3 + 84*x^2 + 54*x - 49)*sqrt(3*x^2 + 2) + 1/2*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.91 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {4 x^{3} \sqrt {3 x^{2} + 2}}{3} + \frac {28 x^{2} \sqrt {3 x^{2} + 2}}{9} + 2 x \sqrt {3 x^{2} + 2} - \frac {49 \sqrt {3 x^{2} + 2}}{27} - \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )} \]

[In]

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

[Out]

4*x**3*sqrt(3*x**2 + 2)/3 + 28*x**2*sqrt(3*x**2 + 2)/9 + 2*x*sqrt(3*x**2 + 2) - 49*sqrt(3*x**2 + 2)/27 - sqrt(
3)*asinh(sqrt(6)*x/2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {4}{3} \, \sqrt {3 \, x^{2} + 2} x^{3} + \frac {28}{9} \, \sqrt {3 \, x^{2} + 2} x^{2} + 2 \, \sqrt {3 \, x^{2} + 2} x - \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) - \frac {49}{27} \, \sqrt {3 \, x^{2} + 2} \]

[In]

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

[Out]

4/3*sqrt(3*x^2 + 2)*x^3 + 28/9*sqrt(3*x^2 + 2)*x^2 + 2*sqrt(3*x^2 + 2)*x - sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 49
/27*sqrt(3*x^2 + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.59 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {1}{27} \, {\left (6 \, {\left (2 \, {\left (3 \, x + 7\right )} x + 9\right )} x - 49\right )} \sqrt {3 \, x^{2} + 2} + \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \]

[In]

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

[Out]

1/27*(6*(2*(3*x + 7)*x + 9)*x - 49)*sqrt(3*x^2 + 2) + sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))

Mupad [B] (verification not implemented)

Time = 12.85 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.49 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2+3 x^2}} \, dx=\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (4\,x^3+\frac {28\,x^2}{3}+6\,x-\frac {49}{9}\right )}{3}-\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right ) \]

[In]

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

[Out]

(3^(1/2)*(x^2 + 2/3)^(1/2)*(6*x + (28*x^2)/3 + 4*x^3 - 49/9))/3 - 3^(1/2)*asinh((6^(1/2)*x)/2)