### 3.122 $$\int \frac{1+3 x+4 x^2}{(1+2 x)^2 \sqrt{2+3 x^2}} \, dx$$

Optimal. Leaf size=71 $-\frac{\sqrt{3 x^2+2}}{11 (2 x+1)}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{11 \sqrt{11}}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}}$

[Out]

-Sqrt[2 + 3*x^2]/(11*(1 + 2*x)) + ArcSinh[Sqrt[3/2]*x]/Sqrt[3] + (4*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2
])])/(11*Sqrt[11])

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Rubi [A]  time = 0.069962, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.172, Rules used = {1651, 844, 215, 725, 206} $-\frac{\sqrt{3 x^2+2}}{11 (2 x+1)}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{11 \sqrt{11}}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-Sqrt[2 + 3*x^2]/(11*(1 + 2*x)) + ArcSinh[Sqrt[3/2]*x]/Sqrt[3] + (4*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2
])])/(11*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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 215

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 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)^2 \sqrt{2+3 x^2}} \, dx &=-\frac{\sqrt{2+3 x^2}}{11 (1+2 x)}-\frac{1}{11} \int \frac{-7-22 x}{(1+2 x) \sqrt{2+3 x^2}} \, dx\\ &=-\frac{\sqrt{2+3 x^2}}{11 (1+2 x)}-\frac{4}{11} \int \frac{1}{(1+2 x) \sqrt{2+3 x^2}} \, dx+\int \frac{1}{\sqrt{2+3 x^2}} \, dx\\ &=-\frac{\sqrt{2+3 x^2}}{11 (1+2 x)}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}}+\frac{4}{11} \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}}{11 (1+2 x)}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{2+3 x^2}}\right )}{11 \sqrt{11}}\\ \end{align*}

Mathematica [A]  time = 0.106519, size = 64, normalized size = 0.9 $-\frac{\sqrt{3 x^2+2}}{22 x+11}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{33 x^2+22}}\right )}{11 \sqrt{11}}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{\sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sqrt[2 + 3*x^2]/(11 + 22*x)) + ArcSinh[Sqrt[3/2]*x]/Sqrt[3] + (4*ArcTanh[(4 - 3*x)/Sqrt[22 + 33*x^2]])/(11*S
qrt[11])

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Maple [A]  time = 0.058, size = 65, normalized size = 0.9 \begin{align*}{\frac{\sqrt{3}}{3}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) }-{\frac{1}{22}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}}+{\frac{4\,\sqrt{11}}{121}{\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 ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arcsinh(1/2*x*6^(1/2))*3^(1/2)-1/22/(x+1/2)*(3*(x+1/2)^2-3*x+5/4)^(1/2)+4/121*11^(1/2)*arctanh(2/11*(4-3*x
)*11^(1/2)/(12*(x+1/2)^2-12*x+5)^(1/2))

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Maxima [A]  time = 1.49284, size = 88, normalized size = 1.24 \begin{align*} \frac{1}{3} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) - \frac{4}{121} \, \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}}{11 \,{\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)^2/(3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*arcsinh(1/2*sqrt(6)*x) - 4/121*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x +
1)) - 1/11*sqrt(3*x^2 + 2)/(2*x + 1)

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Fricas [A]  time = 1.65382, size = 285, normalized size = 4.01 \begin{align*} \frac{121 \, \sqrt{3}{\left (2 \, x + 1\right )} \log \left (-\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + 12 \, \sqrt{11}{\left (2 \, 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 ) - 66 \, \sqrt{3 \, x^{2} + 2}}{726 \,{\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)^2/(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

1/726*(121*sqrt(3)*(2*x + 1)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + 12*sqrt(11)*(2*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)) - 66*sqrt(3*x^2 + 2))/(2*x + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4 x^{2} + 3 x + 1}{\left (2 x + 1\right )^{2} \sqrt{3 x^{2} + 2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((4*x**2 + 3*x + 1)/((2*x + 1)**2*sqrt(3*x**2 + 2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4 \, x^{2} + 3 \, x + 1}{\sqrt{3 \, x^{2} + 2}{\left (2 \, x + 1\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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