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

Optimal. Leaf size=53 \[ \frac {\log \left (x+\sqrt {2} \sqrt {3 x+4}+3\right )}{\sqrt {2}}-\frac {\log \left (x-\sqrt {2} \sqrt {3 x+4}+3\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {827, 1164, 628} \begin {gather*} \frac {\log \left (x+\sqrt {2} \sqrt {3 x+4}+3\right )}{\sqrt {2}}-\frac {\log \left (x-\sqrt {2} \sqrt {3 x+4}+3\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[3 + x - Sqrt[2]*Sqrt[4 + 3*x]]/Sqrt[2]) + Log[3 + x + Sqrt[2]*Sqrt[4 + 3*x]]/Sqrt[2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {1-3 x}{\sqrt {4+3 x} \left (1+x^2\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {15-3 x^2}{25-8 x^2+x^4} \, dx,x,\sqrt {4+3 x}\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {3 \sqrt {2}+2 x}{-5-3 \sqrt {2} x-x^2} \, dx,x,\sqrt {4+3 x}\right )}{\sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {3 \sqrt {2}-2 x}{-5+3 \sqrt {2} x-x^2} \, dx,x,\sqrt {4+3 x}\right )}{\sqrt {2}}\\ &=-\frac {\log \left (3+x-\sqrt {2} \sqrt {4+3 x}\right )}{\sqrt {2}}+\frac {\log \left (3+x+\sqrt {2} \sqrt {4+3 x}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 63, normalized size = 1.19 \begin {gather*} \frac {1}{5} \left ((3+i) \sqrt {4-3 i} \tanh ^{-1}\left (\frac {\sqrt {3 x+4}}{\sqrt {4-3 i}}\right )+(3-i) \sqrt {4+3 i} \tanh ^{-1}\left (\frac {\sqrt {3 x+4}}{\sqrt {4+3 i}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 + I)*Sqrt[4 - 3*I]*ArcTanh[Sqrt[4 + 3*x]/Sqrt[4 - 3*I]] + (3 - I)*Sqrt[4 + 3*I]*ArcTanh[Sqrt[4 + 3*x]/Sqrt
[4 + 3*I]])/5

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IntegrateAlgebraic [A]  time = 0.05, size = 30, normalized size = 0.57 \begin {gather*} \sqrt {2} \tanh ^{-1}\left (\frac {3 \sqrt {2} \sqrt {3 x+4}}{3 x+9}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[2]*ArcTanh[(3*Sqrt[2]*Sqrt[4 + 3*x])/(9 + 3*x)]

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fricas [A]  time = 0.42, size = 37, normalized size = 0.70 \begin {gather*} \frac {1}{2} \, \sqrt {2} \log \left (\frac {2 \, \sqrt {2} \sqrt {3 \, x + 4} {\left (x + 3\right )} + x^{2} + 12 \, x + 17}{x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log((2*sqrt(2)*sqrt(3*x + 4)*(x + 3) + x^2 + 12*x + 17)/(x^2 + 1))

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giac [A]  time = 0.22, size = 53, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, \sqrt {2} \log \left (\frac {3}{5} \cdot 25^{\frac {1}{4}} \sqrt {10} \sqrt {3 \, x + 4} + 3 \, x + 9\right ) - \frac {1}{2} \, \sqrt {2} \log \left (-\frac {3}{5} \cdot 25^{\frac {1}{4}} \sqrt {10} \sqrt {3 \, x + 4} + 3 \, x + 9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(3/5*25^(1/4)*sqrt(10)*sqrt(3*x + 4) + 3*x + 9) - 1/2*sqrt(2)*log(-3/5*25^(1/4)*sqrt(10)*sqrt(3
*x + 4) + 3*x + 9)

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maple [A]  time = 0.07, size = 48, normalized size = 0.91 \begin {gather*} -\frac {\sqrt {2}\, \ln \left (3 x +9-3 \sqrt {2}\, \sqrt {3 x +4}\right )}{2}+\frac {\sqrt {2}\, \ln \left (3 x +9+3 \sqrt {2}\, \sqrt {3 x +4}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*2^(1/2)*ln(3*x+9+3*2^(1/2)*(3*x+4)^(1/2))-1/2*2^(1/2)*ln(3*x+9-3*2^(1/2)*(3*x+4)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {3 \, x - 1}{{\left (x^{2} + 1\right )} \sqrt {3 \, x + 4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.84, size = 21, normalized size = 0.40 \begin {gather*} \sqrt {2}\,\mathrm {atanh}\left (\frac {24\,\sqrt {6\,x+8}}{24\,x+72}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^(1/2)*atanh((24*(6*x + 8)^(1/2))/(24*x + 72))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {3 x}{x^{2} \sqrt {3 x + 4} + \sqrt {3 x + 4}}\, dx - \int \left (- \frac {1}{x^{2} \sqrt {3 x + 4} + \sqrt {3 x + 4}}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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