3.813 \(\int \frac{1+x}{(4+x^2) \sqrt{9+x^2}} \, dx\)

Optimal. Leaf size=53 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{x^2+9}}\right )}{2 \sqrt{5}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+9}}{\sqrt{5}}\right )}{\sqrt{5}} \]

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[9 + x^2])]/(2*Sqrt[5]) - ArcTanh[Sqrt[9 + x^2]/Sqrt[5]]/Sqrt[5]

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Rubi [A]  time = 0.0307952, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1010, 377, 203, 444, 63, 207} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{x^2+9}}\right )}{2 \sqrt{5}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+9}}{\sqrt{5}}\right )}{\sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)/((4 + x^2)*Sqrt[9 + x^2]),x]

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[9 + x^2])]/(2*Sqrt[5]) - ArcTanh[Sqrt[9 + x^2]/Sqrt[5]]/Sqrt[5]

Rule 1010

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{1+x}{\left (4+x^2\right ) \sqrt{9+x^2}} \, dx &=\int \frac{1}{\left (4+x^2\right ) \sqrt{9+x^2}} \, dx+\int \frac{x}{\left (4+x^2\right ) \sqrt{9+x^2}} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(4+x) \sqrt{9+x}} \, dx,x,x^2\right )+\operatorname{Subst}\left (\int \frac{1}{4+5 x^2} \, dx,x,\frac{x}{\sqrt{9+x^2}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{9+x^2}}\right )}{2 \sqrt{5}}+\operatorname{Subst}\left (\int \frac{1}{-5+x^2} \, dx,x,\sqrt{9+x^2}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{5} x}{2 \sqrt{9+x^2}}\right )}{2 \sqrt{5}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{9+x^2}}{\sqrt{5}}\right )}{\sqrt{5}}\\ \end{align*}

Mathematica [C]  time = 0.0455921, size = 64, normalized size = 1.21 \[ -\frac{(2+i) \tanh ^{-1}\left (\frac{9-2 i x}{\sqrt{5} \sqrt{x^2+9}}\right )+(2-i) \tanh ^{-1}\left (\frac{9+2 i x}{\sqrt{5} \sqrt{x^2+9}}\right )}{4 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/((4 + x^2)*Sqrt[9 + x^2]),x]

[Out]

-((2 + I)*ArcTanh[(9 - (2*I)*x)/(Sqrt[5]*Sqrt[9 + x^2])] + (2 - I)*ArcTanh[(9 + (2*I)*x)/(Sqrt[5]*Sqrt[9 + x^2
])])/(4*Sqrt[5])

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Maple [A]  time = 0.013, size = 39, normalized size = 0.7 \begin{align*}{\frac{\sqrt{5}}{10}\arctan \left ({\frac{x\sqrt{5}}{2}{\frac{1}{\sqrt{{x}^{2}+9}}}} \right ) }-{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{\sqrt{5}}{5}\sqrt{{x}^{2}+9}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)/(x^2+4)/(x^2+9)^(1/2),x)

[Out]

1/10*arctan(1/2*x*5^(1/2)/(x^2+9)^(1/2))*5^(1/2)-1/5*arctanh(1/5*(x^2+9)^(1/2)*5^(1/2))*5^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x + 1}{\sqrt{x^{2} + 9}{\left (x^{2} + 4\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x^2+4)/(x^2+9)^(1/2),x, algorithm="maxima")

[Out]

integrate((x + 1)/(sqrt(x^2 + 9)*(x^2 + 4)), x)

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Fricas [B]  time = 1.7695, size = 574, normalized size = 10.83 \begin{align*} \frac{1}{5} \, \sqrt{5} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{x^{2} - \sqrt{x^{2} + 9}{\left (x + \sqrt{5}\right )} + \sqrt{5} x + 9} + \frac{1}{2} \, x + \frac{1}{2} \, \sqrt{5} - \frac{1}{2} \, \sqrt{x^{2} + 9}\right ) - \frac{1}{5} \, \sqrt{5} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{x^{2} - \sqrt{x^{2} + 9}{\left (x - \sqrt{5}\right )} - \sqrt{5} x + 9} + \frac{1}{2} \, x - \frac{1}{2} \, \sqrt{5} - \frac{1}{2} \, \sqrt{x^{2} + 9}\right ) + \frac{1}{10} \, \sqrt{5} \log \left (50 \, x^{2} - 50 \, \sqrt{x^{2} + 9}{\left (x + \sqrt{5}\right )} + 50 \, \sqrt{5} x + 450\right ) - \frac{1}{10} \, \sqrt{5} \log \left (50 \, x^{2} - 50 \, \sqrt{x^{2} + 9}{\left (x - \sqrt{5}\right )} - 50 \, \sqrt{5} x + 450\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x^2+4)/(x^2+9)^(1/2),x, algorithm="fricas")

[Out]

1/5*sqrt(5)*arctan(1/2*sqrt(2)*sqrt(x^2 - sqrt(x^2 + 9)*(x + sqrt(5)) + sqrt(5)*x + 9) + 1/2*x + 1/2*sqrt(5) -
1/2*sqrt(x^2 + 9)) - 1/5*sqrt(5)*arctan(1/2*sqrt(2)*sqrt(x^2 - sqrt(x^2 + 9)*(x - sqrt(5)) - sqrt(5)*x + 9) +
1/2*x - 1/2*sqrt(5) - 1/2*sqrt(x^2 + 9)) + 1/10*sqrt(5)*log(50*x^2 - 50*sqrt(x^2 + 9)*(x + sqrt(5)) + 50*sqrt
(5)*x + 450) - 1/10*sqrt(5)*log(50*x^2 - 50*sqrt(x^2 + 9)*(x - sqrt(5)) - 50*sqrt(5)*x + 450)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x + 1}{\left (x^{2} + 4\right ) \sqrt{x^{2} + 9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x**2+4)/(x**2+9)**(1/2),x)

[Out]

Integral((x + 1)/((x**2 + 4)*sqrt(x**2 + 9)), x)

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Giac [B]  time = 1.25997, size = 528, normalized size = 9.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x^2+4)/(x^2+9)^(1/2),x, algorithm="giac")

[Out]

1/40*(9*sqrt(5)*arctan(2/(sqrt(5) + 3)) + 9*sqrt(5)*arctan(2/(sqrt(5) - 3)) + 49*sqrt(5)*log(3/2*sqrt(5) + 9/2
) - 49*sqrt(5)*log(-3/2*sqrt(5) + 9/2) - 15*arctan(2/(sqrt(5) + 3)) + 15*arctan(2/(sqrt(5) - 3)) - 105*log(3/2
*sqrt(5) + 9/2) - 105*log(-3/2*sqrt(5) + 9/2))*sgn(x) - 1/10*(7*sqrt(5) + 15)*log((sqrt(9/x^2 + 1) - 3/x)^2 +
1/2*(3*sqrt(5)*sgn(x) + 7*sgn(x))/sgn(x))*sgn(x)/(7*abs(sgn(x))*sgn(x) + 3*sqrt(5)) + 1/10*(7*sqrt(5) - 15)*lo
g((sqrt(9/x^2 + 1) - 3/x)^2 - 1/2*(3*sqrt(5)*sgn(x) - 7*sgn(x))/sgn(x))*sgn(x)/(7*abs(sgn(x))*sgn(x) - 3*sqrt(
5)) - 1/20*(5*(sqrt(5) + 3)*abs(sgn(x)) + 3*(3*sqrt(5) + 5)*sgn(x))*arctan(2*sqrt(1/2)*(sqrt(9/x^2 + 1) - 3/x)
/sqrt((3*sqrt(5)*sgn(x) + 7*sgn(x))/sgn(x)))/(7*abs(sgn(x))*sgn(x) + 3*sqrt(5)) + 1/20*(5*(sqrt(5) - 3)*abs(sg
n(x)) + 3*(3*sqrt(5) - 5)*sgn(x))*arctan(2*sqrt(1/2)*(sqrt(9/x^2 + 1) - 3/x)/sqrt(-(3*sqrt(5)*sgn(x) - 7*sgn(x
))/sgn(x)))/(7*abs(sgn(x))*sgn(x) - 3*sqrt(5))