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]

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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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 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 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 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])

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 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 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]

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*}

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Mathematica [C]  time = 0.04, 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]

-1/4*((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])])/Sqrt[5]

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fricas [B]  time = 0.44, size = 182, normalized size = 3.43 \[ \frac {1}{5} \, \sqrt {5} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x^{2} - \sqrt {x^{2} + 9} {\left (x + \sqrt {5}\right )} + \sqrt {5} x + 9} + \frac {1}{2} \, x + \frac {1}{2} \, \sqrt {5} - \frac {1}{2} \, \sqrt {x^{2} + 9}\right ) - \frac {1}{5} \, \sqrt {5} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x^{2} - \sqrt {x^{2} + 9} {\left (x - \sqrt {5}\right )} - \sqrt {5} x + 9} + \frac {1}{2} \, x - \frac {1}{2} \, \sqrt {5} - \frac {1}{2} \, \sqrt {x^{2} + 9}\right ) + \frac {1}{10} \, \sqrt {5} \log \left (50 \, x^{2} - 50 \, \sqrt {x^{2} + 9} {\left (x + \sqrt {5}\right )} + 50 \, \sqrt {5} x + 450\right ) - \frac {1}{10} \, \sqrt {5} \log \left (50 \, x^{2} - 50 \, \sqrt {x^{2} + 9} {\left (x - \sqrt {5}\right )} - 50 \, \sqrt {5} x + 450\right ) \]

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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giac [B]  time = 0.44, size = 123, normalized size = 2.32 \[ -\frac {1}{10} \, \sqrt {5} \arctan \left (\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {5} - \frac {1}{2} \, \sqrt {x^{2} + 9}\right ) - \frac {1}{10} \, \sqrt {5} \arctan \left (-\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {5} + \frac {1}{2} \, \sqrt {x^{2} + 9}\right ) + \frac {1}{10} \, \sqrt {5} \log \left ({\left (x - \sqrt {x^{2} + 9}\right )}^{2} + 2 \, \sqrt {5} {\left (x - \sqrt {x^{2} + 9}\right )} + 9\right ) - \frac {1}{10} \, \sqrt {5} \log \left ({\left (x - \sqrt {x^{2} + 9}\right )}^{2} - 2 \, \sqrt {5} {\left (x - \sqrt {x^{2} + 9}\right )} + 9\right ) \]

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]

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

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maple [A]  time = 0.02, size = 39, normalized size = 0.74 \[ -\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {x^{2}+9}\, \sqrt {5}}{5}\right )}{5}+\frac {\sqrt {5}\, \arctan \left (\frac {\sqrt {5}\, x}{2 \sqrt {x^{2}+9}}\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+1)/(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x + 1}{\sqrt {x^{2} + 9} {\left (x^{2} + 4\right )}}\,{d x} \]

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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mupad [B]  time = 3.59, size = 67, normalized size = 1.26 \[ \sqrt {5}\,\left (\ln \left (x-2{}\mathrm {i}\right )-\ln \left (\sqrt {5}\,\sqrt {x^2+9}+9+x\,2{}\mathrm {i}\right )\right )\,\left (\frac {1}{10}-\frac {1}{20}{}\mathrm {i}\right )+\sqrt {5}\,\left (\ln \left (x+2{}\mathrm {i}\right )-\ln \left (\sqrt {5}\,\sqrt {x^2+9}+9-x\,2{}\mathrm {i}\right )\right )\,\left (\frac {1}{10}+\frac {1}{20}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5^(1/2)*(log(x - 2i) - log(x*2i + 5^(1/2)*(x^2 + 9)^(1/2) + 9))*(1/10 - 1i/20) + 5^(1/2)*(log(x + 2i) - log(5^
(1/2)*(x^2 + 9)^(1/2) - x*2i + 9))*(1/10 + 1i/20)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x + 1}{\left (x^{2} + 4\right ) \sqrt {x^{2} + 9}}\, dx \]

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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