3.394 \(\int \frac{(1+4 x-7 x^2) (2+5 x+x^2)}{(3+2 x+5 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=82 \[ -\frac{7}{50} \sqrt{5 x^2+2 x+3} x-\frac{261}{250} \sqrt{5 x^2+2 x+3}-\frac{2 (2449 x+2321)}{875 \sqrt{5 x^2+2 x+3}}+\frac{149 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{25 \sqrt{5}} \]

[Out]

(-2*(2321 + 2449*x))/(875*Sqrt[3 + 2*x + 5*x^2]) - (261*Sqrt[3 + 2*x + 5*x^2])/250 - (7*x*Sqrt[3 + 2*x + 5*x^2
])/50 + (149*ArcSinh[(1 + 5*x)/Sqrt[14]])/(25*Sqrt[5])

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Rubi [A]  time = 0.0865919, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {1660, 1661, 640, 619, 215} \[ -\frac{7}{50} \sqrt{5 x^2+2 x+3} x-\frac{261}{250} \sqrt{5 x^2+2 x+3}-\frac{2 (2449 x+2321)}{875 \sqrt{5 x^2+2 x+3}}+\frac{149 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{25 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

Int[((1 + 4*x - 7*x^2)*(2 + 5*x + x^2))/(3 + 2*x + 5*x^2)^(3/2),x]

[Out]

(-2*(2321 + 2449*x))/(875*Sqrt[3 + 2*x + 5*x^2]) - (261*Sqrt[3 + 2*x + 5*x^2])/250 - (7*x*Sqrt[3 + 2*x + 5*x^2
])/50 + (149*ArcSinh[(1 + 5*x)/Sqrt[14]])/(25*Sqrt[5])

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

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

Rubi steps

\begin{align*} \int \frac{\left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx &=-\frac{2 (2321+2449 x)}{875 \sqrt{3+2 x+5 x^2}}+\frac{1}{28} \int \frac{\frac{15736}{125}-\frac{3948 x}{25}-\frac{196 x^2}{5}}{\sqrt{3+2 x+5 x^2}} \, dx\\ &=-\frac{2 (2321+2449 x)}{875 \sqrt{3+2 x+5 x^2}}-\frac{7}{50} x \sqrt{3+2 x+5 x^2}+\frac{1}{280} \int \frac{\frac{34412}{25}-\frac{7308 x}{5}}{\sqrt{3+2 x+5 x^2}} \, dx\\ &=-\frac{2 (2321+2449 x)}{875 \sqrt{3+2 x+5 x^2}}-\frac{261}{250} \sqrt{3+2 x+5 x^2}-\frac{7}{50} x \sqrt{3+2 x+5 x^2}+\frac{149}{25} \int \frac{1}{\sqrt{3+2 x+5 x^2}} \, dx\\ &=-\frac{2 (2321+2449 x)}{875 \sqrt{3+2 x+5 x^2}}-\frac{261}{250} \sqrt{3+2 x+5 x^2}-\frac{7}{50} x \sqrt{3+2 x+5 x^2}+\frac{149 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{56}}} \, dx,x,2+10 x\right )}{50 \sqrt{70}}\\ &=-\frac{2 (2321+2449 x)}{875 \sqrt{3+2 x+5 x^2}}-\frac{261}{250} \sqrt{3+2 x+5 x^2}-\frac{7}{50} x \sqrt{3+2 x+5 x^2}+\frac{149 \sinh ^{-1}\left (\frac{1+5 x}{\sqrt{14}}\right )}{25 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.155307, size = 55, normalized size = 0.67 \[ \frac{149 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{25 \sqrt{5}}-\frac{245 x^3+1925 x^2+2837 x+2953}{350 \sqrt{5 x^2+2 x+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 + 4*x - 7*x^2)*(2 + 5*x + x^2))/(3 + 2*x + 5*x^2)^(3/2),x]

[Out]

-(2953 + 2837*x + 1925*x^2 + 245*x^3)/(350*Sqrt[3 + 2*x + 5*x^2]) + (149*ArcSinh[(1 + 5*x)/Sqrt[14]])/(25*Sqrt
[5])

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Maple [A]  time = 0.05, size = 98, normalized size = 1.2 \begin{align*} -{\frac{7\,{x}^{3}}{10}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{11\,{x}^{2}}{2}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{149\,x}{25}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{1001}{125}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{7510\,x+1502}{3500}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}+{\frac{149\,\sqrt{5}}{125}{\it Arcsinh} \left ({\frac{5\,\sqrt{14}}{14} \left ( x+{\frac{1}{5}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-7*x^2+4*x+1)*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x)

[Out]

-7/10*x^3/(5*x^2+2*x+3)^(1/2)-11/2*x^2/(5*x^2+2*x+3)^(1/2)-149/25*x/(5*x^2+2*x+3)^(1/2)-1001/125/(5*x^2+2*x+3)
^(1/2)-751/3500*(10*x+2)/(5*x^2+2*x+3)^(1/2)+149/125*5^(1/2)*arcsinh(5/14*14^(1/2)*(x+1/5))

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Maxima [A]  time = 1.5036, size = 108, normalized size = 1.32 \begin{align*} -\frac{7 \, x^{3}}{10 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} - \frac{11 \, x^{2}}{2 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} + \frac{149}{125} \, \sqrt{5} \operatorname{arsinh}\left (\frac{1}{14} \, \sqrt{14}{\left (5 \, x + 1\right )}\right ) - \frac{2837 \, x}{350 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} - \frac{2953}{350 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7*x^2+4*x+1)*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x, algorithm="maxima")

[Out]

-7/10*x^3/sqrt(5*x^2 + 2*x + 3) - 11/2*x^2/sqrt(5*x^2 + 2*x + 3) + 149/125*sqrt(5)*arcsinh(1/14*sqrt(14)*(5*x
+ 1)) - 2837/350*x/sqrt(5*x^2 + 2*x + 3) - 2953/350/sqrt(5*x^2 + 2*x + 3)

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Fricas [A]  time = 1.38676, size = 254, normalized size = 3.1 \begin{align*} \frac{1043 \, \sqrt{5}{\left (5 \, x^{2} + 2 \, x + 3\right )} \log \left (-\sqrt{5} \sqrt{5 \, x^{2} + 2 \, x + 3}{\left (5 \, x + 1\right )} - 25 \, x^{2} - 10 \, x - 8\right ) - 5 \,{\left (245 \, x^{3} + 1925 \, x^{2} + 2837 \, x + 2953\right )} \sqrt{5 \, x^{2} + 2 \, x + 3}}{1750 \,{\left (5 \, x^{2} + 2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7*x^2+4*x+1)*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x, algorithm="fricas")

[Out]

1/1750*(1043*sqrt(5)*(5*x^2 + 2*x + 3)*log(-sqrt(5)*sqrt(5*x^2 + 2*x + 3)*(5*x + 1) - 25*x^2 - 10*x - 8) - 5*(
245*x^3 + 1925*x^2 + 2837*x + 2953)*sqrt(5*x^2 + 2*x + 3))/(5*x^2 + 2*x + 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{13 x}{5 x^{2} \sqrt{5 x^{2} + 2 x + 3} + 2 x \sqrt{5 x^{2} + 2 x + 3} + 3 \sqrt{5 x^{2} + 2 x + 3}}\, dx - \int - \frac{7 x^{2}}{5 x^{2} \sqrt{5 x^{2} + 2 x + 3} + 2 x \sqrt{5 x^{2} + 2 x + 3} + 3 \sqrt{5 x^{2} + 2 x + 3}}\, dx - \int \frac{31 x^{3}}{5 x^{2} \sqrt{5 x^{2} + 2 x + 3} + 2 x \sqrt{5 x^{2} + 2 x + 3} + 3 \sqrt{5 x^{2} + 2 x + 3}}\, dx - \int \frac{7 x^{4}}{5 x^{2} \sqrt{5 x^{2} + 2 x + 3} + 2 x \sqrt{5 x^{2} + 2 x + 3} + 3 \sqrt{5 x^{2} + 2 x + 3}}\, dx - \int - \frac{2}{5 x^{2} \sqrt{5 x^{2} + 2 x + 3} + 2 x \sqrt{5 x^{2} + 2 x + 3} + 3 \sqrt{5 x^{2} + 2 x + 3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7*x**2+4*x+1)*(x**2+5*x+2)/(5*x**2+2*x+3)**(3/2),x)

[Out]

-Integral(-13*x/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x) -
Integral(-7*x**2/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x) -
Integral(31*x**3/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x)
- Integral(7*x**4/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x)
- Integral(-2/(5*x**2*sqrt(5*x**2 + 2*x + 3) + 2*x*sqrt(5*x**2 + 2*x + 3) + 3*sqrt(5*x**2 + 2*x + 3)), x)

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Giac [A]  time = 1.18432, size = 84, normalized size = 1.02 \begin{align*} -\frac{149}{125} \, \sqrt{5} \log \left (-\sqrt{5}{\left (\sqrt{5} x - \sqrt{5 \, x^{2} + 2 \, x + 3}\right )} - 1\right ) - \frac{{\left (35 \,{\left (7 \, x + 55\right )} x + 2837\right )} x + 2953}{350 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7*x^2+4*x+1)*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x, algorithm="giac")

[Out]

-149/125*sqrt(5)*log(-sqrt(5)*(sqrt(5)*x - sqrt(5*x^2 + 2*x + 3)) - 1) - 1/350*((35*(7*x + 55)*x + 2837)*x + 2
953)/sqrt(5*x^2 + 2*x + 3)