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

Optimal. Leaf size=77 \[ \frac{13 \sqrt{3 x^2+2}}{242 (2 x+1)}-\frac{\sqrt{3 x^2+2}}{22 (2 x+1)^2}-\frac{103 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{121 \sqrt{11}} \]

[Out]

-Sqrt[2 + 3*x^2]/(22*(1 + 2*x)^2) + (13*Sqrt[2 + 3*x^2])/(242*(1 + 2*x)) - (103*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sq
rt[2 + 3*x^2])])/(121*Sqrt[11])

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Rubi [A]  time = 0.066702, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1651, 807, 725, 206} \[ \frac{13 \sqrt{3 x^2+2}}{242 (2 x+1)}-\frac{\sqrt{3 x^2+2}}{22 (2 x+1)^2}-\frac{103 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{121 \sqrt{11}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[2 + 3*x^2]/(22*(1 + 2*x)^2) + (13*Sqrt[2 + 3*x^2])/(242*(1 + 2*x)) - (103*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sq
rt[2 + 3*x^2])])/(121*Sqrt[11])

Rule 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1+3 x+4 x^2}{(1+2 x)^3 \sqrt{2+3 x^2}} \, dx &=-\frac{\sqrt{2+3 x^2}}{22 (1+2 x)^2}-\frac{1}{22} \int \frac{-14-41 x}{(1+2 x)^2 \sqrt{2+3 x^2}} \, dx\\ &=-\frac{\sqrt{2+3 x^2}}{22 (1+2 x)^2}+\frac{13 \sqrt{2+3 x^2}}{242 (1+2 x)}+\frac{103}{121} \int \frac{1}{(1+2 x) \sqrt{2+3 x^2}} \, dx\\ &=-\frac{\sqrt{2+3 x^2}}{22 (1+2 x)^2}+\frac{13 \sqrt{2+3 x^2}}{242 (1+2 x)}-\frac{103}{121} \operatorname{Subst}\left (\int \frac{1}{11-x^2} \, dx,x,\frac{4-3 x}{\sqrt{2+3 x^2}}\right )\\ &=-\frac{\sqrt{2+3 x^2}}{22 (1+2 x)^2}+\frac{13 \sqrt{2+3 x^2}}{242 (1+2 x)}-\frac{103 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{2+3 x^2}}\right )}{121 \sqrt{11}}\\ \end{align*}

Mathematica [A]  time = 0.0677386, size = 55, normalized size = 0.71 \[ \frac{\frac{11 (13 x+1) \sqrt{3 x^2+2}}{(2 x+1)^2}-103 \sqrt{11} \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{33 x^2+22}}\right )}{1331} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*(1 + 13*x)*Sqrt[2 + 3*x^2])/(1 + 2*x)^2 - 103*Sqrt[11]*ArcTanh[(4 - 3*x)/Sqrt[22 + 33*x^2]])/1331

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Maple [A]  time = 0.058, size = 74, normalized size = 1. \begin{align*}{\frac{13}{484}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}}-{\frac{103\,\sqrt{11}}{1331}{\it Artanh} \left ({\frac{ \left ( 8-6\,x \right ) \sqrt{11}}{11}{\frac{1}{\sqrt{12\, \left ( x+1/2 \right ) ^{2}-12\,x+5}}}} \right ) }-{\frac{1}{88}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}} \left ( x+{\frac{1}{2}} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/484/(x+1/2)*(3*(x+1/2)^2-3*x+5/4)^(1/2)-103/1331*11^(1/2)*arctanh(2/11*(4-3*x)*11^(1/2)/(12*(x+1/2)^2-12*x+
5)^(1/2))-1/88/(x+1/2)^2*(3*(x+1/2)^2-3*x+5/4)^(1/2)

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Maxima [A]  time = 1.47774, size = 103, normalized size = 1.34 \begin{align*} \frac{103}{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{\sqrt{3 \, x^{2} + 2}}{22 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} + \frac{13 \, \sqrt{3 \, x^{2} + 2}}{242 \,{\left (2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

103/1331*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x + 1)) - 1/22*sqrt(3*x^2 + 2)/(4*x^2
+ 4*x + 1) + 13/242*sqrt(3*x^2 + 2)/(2*x + 1)

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Fricas [A]  time = 1.57685, size = 234, normalized size = 3.04 \begin{align*} \frac{103 \, \sqrt{11}{\left (4 \, x^{2} + 4 \, x + 1\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 ) + 22 \, \sqrt{3 \, x^{2} + 2}{\left (13 \, x + 1\right )}}{2662 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2662*(103*sqrt(11)*(4*x^2 + 4*x + 1)*log(-(sqrt(11)*sqrt(3*x^2 + 2)*(3*x - 4) + 21*x^2 - 12*x + 19)/(4*x^2 +
4*x + 1)) + 22*sqrt(3*x^2 + 2)*(13*x + 1))/(4*x^2 + 4*x + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.26592, size = 243, normalized size = 3.16 \begin{align*} \frac{103}{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{72 \,{\left (\sqrt{3} x - \sqrt{3 \, x^{2} + 2}\right )}^{3} - 13 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} + 2}\right )}^{2} - 168 \, \sqrt{3} x + 104 \, \sqrt{3} + 168 \, \sqrt{3 \, x^{2} + 2}}{484 \,{\left ({\left (\sqrt{3} x - \sqrt{3 \, x^{2} + 2}\right )}^{2} + \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} + 2}\right )} - 2\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

103/1331*sqrt(11)*log(-abs(-2*sqrt(3)*x - sqrt(11) - sqrt(3) + 2*sqrt(3*x^2 + 2))/(2*sqrt(3)*x - sqrt(11) + sq
rt(3) - 2*sqrt(3*x^2 + 2))) + 1/484*(72*(sqrt(3)*x - sqrt(3*x^2 + 2))^3 - 13*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 +
2))^2 - 168*sqrt(3)*x + 104*sqrt(3) + 168*sqrt(3*x^2 + 2))/((sqrt(3)*x - sqrt(3*x^2 + 2))^2 + sqrt(3)*(sqrt(3
)*x - sqrt(3*x^2 + 2)) - 2)^2