3.121 \(\int \frac {1}{\sqrt {-7+2 x+5 x^2} (8+12 x+5 x^2)} \, dx\)

Optimal. Leaf size=51 \[ \frac {1}{10} \tan ^{-1}\left (\frac {5 (x+2)}{2 \sqrt {5 x^2+2 x-7}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 (x+1)}{\sqrt {5 x^2+2 x-7}}\right ) \]

[Out]

1/10*arctan(5/2*(2+x)/(5*x^2+2*x-7)^(1/2))+1/5*arctanh(5*(1+x)/(5*x^2+2*x-7)^(1/2))

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Rubi [A]  time = 0.07, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {986, 1029, 203, 207} \[ \frac {1}{10} \tan ^{-1}\left (\frac {5 (x+2)}{2 \sqrt {5 x^2+2 x-7}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 (x+1)}{\sqrt {5 x^2+2 x-7}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-7 + 2*x + 5*x^2]*(8 + 12*x + 5*x^2)),x]

[Out]

ArcTan[(5*(2 + x))/(2*Sqrt[-7 + 2*x + 5*x^2])]/10 + ArcTanh[(5*(1 + x))/Sqrt[-7 + 2*x + 5*x^2]]/5

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 986

Int[1/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt
[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[(c*d - a*f + q + (c*e - b*f)*x)/((a + b*x + c
*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[(c*d - a*f - q + (c*e - b*f)*x)/((a + b*x + c*x^2)*Sq
rt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] &&
NeQ[c*e - b*f, 0] && NegQ[b^2 - 4*a*c]

Rule 1029

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> Dist[-2*g*(g*b - 2*a*h), Subst[Int[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - (b*d - a*e)*x^2, x], x], x, S
imp[g*b - 2*a*h - (b*h - 2*g*c)*x, x]/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && NeQ[b*d - a*e, 0] && EqQ[h^2*(b*d - a*e) - 2*g*h*(c*d - a*f) + g^2*(
c*e - b*f), 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx &=-\left (\frac {1}{50} \int \frac {-100-50 x}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx\right )+\frac {1}{50} \int \frac {-50-50 x}{\sqrt {-7+2 x+5 x^2} \left (8+12 x+5 x^2\right )} \, dx\\ &=400 \operatorname {Subst}\left (\int \frac {1}{160000+100 x^2} \, dx,x,\frac {200+100 x}{\sqrt {-7+2 x+5 x^2}}\right )+1600 \operatorname {Subst}\left (\int \frac {1}{-640000+100 x^2} \, dx,x,\frac {-400-400 x}{\sqrt {-7+2 x+5 x^2}}\right )\\ &=\frac {1}{10} \tan ^{-1}\left (\frac {5 (2+x)}{2 \sqrt {-7+2 x+5 x^2}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5 (1+x)}{\sqrt {-7+2 x+5 x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 81, normalized size = 1.59 \[ \left (\frac {1}{10}-\frac {i}{20}\right ) \tanh ^{-1}\left (\frac {\left (\frac {1}{100}+\frac {i}{50}\right ) ((100-40 i) x+(164-8 i))}{\sqrt {5 x^2+2 x-7}}\right )-\left (\frac {1}{20}-\frac {i}{10}\right ) \tan ^{-1}\left (\frac {\left (\frac {1}{50}+\frac {i}{100}\right ) ((-100-40 i) x-(164+8 i))}{\sqrt {5 x^2+2 x-7}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[-7 + 2*x + 5*x^2]*(8 + 12*x + 5*x^2)),x]

[Out]

(-1/20 + I/10)*ArcTan[((1/50 + I/100)*((-164 - 8*I) - (100 + 40*I)*x))/Sqrt[-7 + 2*x + 5*x^2]] + (1/10 - I/20)
*ArcTanh[((1/100 + I/50)*((164 - 8*I) + (100 - 40*I)*x))/Sqrt[-7 + 2*x + 5*x^2]]

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fricas [B]  time = 0.76, size = 154, normalized size = 3.02 \[ \frac {1}{20} \, \arctan \left (\frac {27 \, x^{2} + 20 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 2\right )} + 36 \, x}{31 \, x^{2} + 16 \, x - 56}\right ) + \frac {1}{20} \, \arctan \left (-\frac {27 \, x^{2} - 20 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 2\right )} + 36 \, x}{31 \, x^{2} + 16 \, x - 56}\right ) + \frac {1}{20} \, \log \left (\frac {15 \, x^{2} + 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 1\right )} + 26 \, x + 9}{x^{2}}\right ) - \frac {1}{20} \, \log \left (\frac {15 \, x^{2} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} {\left (x + 1\right )} + 26 \, x + 9}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+12*x+8)/(5*x^2+2*x-7)^(1/2),x, algorithm="fricas")

[Out]

1/20*arctan((27*x^2 + 20*sqrt(5*x^2 + 2*x - 7)*(x + 2) + 36*x)/(31*x^2 + 16*x - 56)) + 1/20*arctan(-(27*x^2 -
20*sqrt(5*x^2 + 2*x - 7)*(x + 2) + 36*x)/(31*x^2 + 16*x - 56)) + 1/20*log((15*x^2 + 5*sqrt(5*x^2 + 2*x - 7)*(x
+ 1) + 26*x + 9)/x^2) - 1/20*log((15*x^2 - 5*sqrt(5*x^2 + 2*x - 7)*(x + 1) + 26*x + 9)/x^2)

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giac [B]  time = 0.28, size = 205, normalized size = 4.02 \[ -\frac {1}{10} \, \arctan \left (-\frac {5 \, \sqrt {5} x + 6 \, \sqrt {5} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} + 5}{2 \, {\left (\sqrt {5} + 5\right )}}\right ) - \frac {1}{10} \, \arctan \left (\frac {5 \, \sqrt {5} x + 6 \, \sqrt {5} - 5 \, \sqrt {5 \, x^{2} + 2 \, x - 7} - 5}{2 \, {\left (\sqrt {5} - 5\right )}}\right ) + \frac {1}{10} \, \log \left (5 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )}^{2} + 2 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )} {\left (6 \, \sqrt {5} + 5\right )} + 20 \, \sqrt {5} + 65\right ) - \frac {1}{10} \, \log \left (5 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )}^{2} + 2 \, {\left (\sqrt {5} x - \sqrt {5 \, x^{2} + 2 \, x - 7}\right )} {\left (6 \, \sqrt {5} - 5\right )} - 20 \, \sqrt {5} + 65\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+12*x+8)/(5*x^2+2*x-7)^(1/2),x, algorithm="giac")

[Out]

-1/10*arctan(-1/2*(5*sqrt(5)*x + 6*sqrt(5) - 5*sqrt(5*x^2 + 2*x - 7) + 5)/(sqrt(5) + 5)) - 1/10*arctan(1/2*(5*
sqrt(5)*x + 6*sqrt(5) - 5*sqrt(5*x^2 + 2*x - 7) - 5)/(sqrt(5) - 5)) + 1/10*log(5*(sqrt(5)*x - sqrt(5*x^2 + 2*x
- 7))^2 + 2*(sqrt(5)*x - sqrt(5*x^2 + 2*x - 7))*(6*sqrt(5) + 5) + 20*sqrt(5) + 65) - 1/10*log(5*(sqrt(5)*x -
sqrt(5*x^2 + 2*x - 7))^2 + 2*(sqrt(5)*x - sqrt(5*x^2 + 2*x - 7))*(6*sqrt(5) - 5) - 20*sqrt(5) + 65)

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maple [B]  time = 0.02, size = 144, normalized size = 2.82 \[ -\frac {\sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}+9}\, \left (2 \arctanh \left (\frac {\sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}+9}}{5}\right )+\arctan \left (\frac {5 \sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}+9}\, \left (x +2\right )}{2 \left (\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}-9\right ) \left (-x -1\right )}\right )\right )}{10 \sqrt {-\frac {\frac {4 \left (x +2\right )^{2}}{\left (-x -1\right )^{2}}-9}{\left (1+\frac {x +2}{-x -1}\right )^{2}}}\, \left (1+\frac {x +2}{-x -1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^2+12*x+8)/(5*x^2+2*x-7)^(1/2),x)

[Out]

-1/10*(-4*(x+2)^2/(-1-x)^2+9)^(1/2)*(2*arctanh(1/5*(-4*(x+2)^2/(-1-x)^2+9)^(1/2))+arctan(5/2*(-4*(x+2)^2/(-1-x
)^2+9)^(1/2)/(4*(x+2)^2/(-1-x)^2-9)*(x+2)/(-1-x)))/(-(4*(x+2)^2/(-1-x)^2-9)/(1+(x+2)/(-1-x))^2)^(1/2)/(1+(x+2)
/(-1-x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (5 \, x^{2} + 12 \, x + 8\right )} \sqrt {5 \, x^{2} + 2 \, x - 7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^2+12*x+8)/(5*x^2+2*x-7)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((5*x^2 + 12*x + 8)*sqrt(5*x^2 + 2*x - 7)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {5\,x^2+2\,x-7}\,\left (5\,x^2+12\,x+8\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x + 5*x^2 - 7)^(1/2)*(12*x + 5*x^2 + 8)),x)

[Out]

int(1/((2*x + 5*x^2 - 7)^(1/2)*(12*x + 5*x^2 + 8)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\left (x - 1\right ) \left (5 x + 7\right )} \left (5 x^{2} + 12 x + 8\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x**2+12*x+8)/(5*x**2+2*x-7)**(1/2),x)

[Out]

Integral(1/(sqrt((x - 1)*(5*x + 7))*(5*x**2 + 12*x + 8)), x)

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