3.2.21 \(\int \frac {1}{\sqrt {-7+2 x+5 x^2} (8+12 x+5 x^2)} \, dx\) [121]

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

[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 = 0.04, 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 = {1000, 1043, 209, 213} \begin {gather*} \frac {1}{10} \text {ArcTan}\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 ) \end {gather*}

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 209

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

Rule 213

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

Rule 1000

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 1043

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 \text {Subst}\left (\int \frac {1}{160000+100 x^2} \, dx,x,\frac {200+100 x}{\sqrt {-7+2 x+5 x^2}}\right )+1600 \text {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 [A]
time = 0.16, size = 53, normalized size = 1.04 \begin {gather*} \frac {1}{10} \tan ^{-1}\left (\frac {5+\frac {5 x}{2}}{\sqrt {-7+2 x+5 x^2}}\right )+\frac {1}{5} \tanh ^{-1}\left (\frac {5+5 x}{\sqrt {-7+2 x+5 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(143\) vs. \(2(41)=82\).
time = 0.39, size = 144, normalized size = 2.82

method	result	size
default	\(-\frac {\sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-1-x \right )^{2}}+9}\, \left (2 \arctanh \left (\frac {\sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-1-x \right )^{2}}+9}}{5}\right )+\arctan \left (\frac {5 \sqrt {-\frac {4 \left (x +2\right )^{2}}{\left (-1-x \right )^{2}}+9}\, \left (x +2\right )}{2 \left (\frac {4 \left (x +2\right )^{2}}{\left (-1-x \right )^{2}}-9\right ) \left (-1-x \right )}\right )\right )}{10 \sqrt {-\frac {\frac {4 \left (x +2\right )^{2}}{\left (-1-x \right )^{2}}-9}{\left (1+\frac {x +2}{-1-x}\right )^{2}}}\, \left (1+\frac {x +2}{-1-x}\right )}\)	\(144\)
trager	\(\RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) \ln \left (\frac {129600 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x -8750 \sqrt {5 x^{2}+2 x -7}\, \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-18630 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x +30330 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )+155 \sqrt {5 x^{2}+2 x -7}-1105 x -5729}{20 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -5 x -4}\right )+\frac {\ln \left (\frac {129600 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x +8750 \sqrt {5 x^{2}+2 x -7}\, \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-33210 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -30330 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-1595 \sqrt {5 x^{2}+2 x -7}+353 x +337}{20 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x +x +4}\right )}{5}-\ln \left (\frac {129600 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )^{2} x +8750 \sqrt {5 x^{2}+2 x -7}\, \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-33210 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x -30330 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )-1595 \sqrt {5 x^{2}+2 x -7}+353 x +337}{20 \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right ) x +x +4}\right ) \RootOf \left (80 \textit {\_Z}^{2}-16 \textit {\_Z} +1\right )\)	\(355\)

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,method=_RETURNVERBOSE)

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (41) = 82\).
time = 3.34, size = 154, normalized size = 3.02 \begin {gather*} \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 ) \end {gather*}

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

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (41) = 82\).
time = 4.96, size = 205, normalized size = 4.02 \begin {gather*} -\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 ) \end {gather*}

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

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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