3.376 \(\int (1+4 x-7 x^2) (2+5 x+x^2) \sqrt{3+2 x+5 x^2} \, dx\)

Optimal. Leaf size=124 \[ -\frac{7}{30} \left (5 x^2+2 x+3\right )^{3/2} x^3-\frac{289}{250} \left (5 x^2+2 x+3\right )^{3/2} x^2+\frac{2149 \left (5 x^2+2 x+3\right )^{3/2} x}{2500}+\frac{7819 \left (5 x^2+2 x+3\right )^{3/2}}{7500}-\frac{4633 (5 x+1) \sqrt{5 x^2+2 x+3}}{12500}-\frac{32431 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{6250 \sqrt{5}} \]

[Out]

(-4633*(1 + 5*x)*Sqrt[3 + 2*x + 5*x^2])/12500 + (7819*(3 + 2*x + 5*x^2)^(3/2))/7500 + (2149*x*(3 + 2*x + 5*x^2
)^(3/2))/2500 - (289*x^2*(3 + 2*x + 5*x^2)^(3/2))/250 - (7*x^3*(3 + 2*x + 5*x^2)^(3/2))/30 - (32431*ArcSinh[(1
+ 5*x)/Sqrt[14]])/(6250*Sqrt[5])

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Rubi [A]  time = 0.115772, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {1661, 640, 612, 619, 215} \[ -\frac{7}{30} \left (5 x^2+2 x+3\right )^{3/2} x^3-\frac{289}{250} \left (5 x^2+2 x+3\right )^{3/2} x^2+\frac{2149 \left (5 x^2+2 x+3\right )^{3/2} x}{2500}+\frac{7819 \left (5 x^2+2 x+3\right )^{3/2}}{7500}-\frac{4633 (5 x+1) \sqrt{5 x^2+2 x+3}}{12500}-\frac{32431 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{6250 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4633*(1 + 5*x)*Sqrt[3 + 2*x + 5*x^2])/12500 + (7819*(3 + 2*x + 5*x^2)^(3/2))/7500 + (2149*x*(3 + 2*x + 5*x^2
)^(3/2))/2500 - (289*x^2*(3 + 2*x + 5*x^2)^(3/2))/250 - (7*x^3*(3 + 2*x + 5*x^2)^(3/2))/30 - (32431*ArcSinh[(1
+ 5*x)/Sqrt[14]])/(6250*Sqrt[5])

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 612

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

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 \left (1+4 x-7 x^2\right ) \left (2+5 x+x^2\right ) \sqrt{3+2 x+5 x^2} \, dx &=-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}+\frac{1}{30} \int \sqrt{3+2 x+5 x^2} \left (60+390 x+273 x^2-867 x^3\right ) \, dx\\ &=-\frac{289}{250} x^2 \left (3+2 x+5 x^2\right )^{3/2}-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}+\frac{1}{750} \int \sqrt{3+2 x+5 x^2} \left (1500+14952 x+12894 x^2\right ) \, dx\\ &=\frac{2149 x \left (3+2 x+5 x^2\right )^{3/2}}{2500}-\frac{289}{250} x^2 \left (3+2 x+5 x^2\right )^{3/2}-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}+\frac{\int (-8682+234570 x) \sqrt{3+2 x+5 x^2} \, dx}{15000}\\ &=\frac{7819 \left (3+2 x+5 x^2\right )^{3/2}}{7500}+\frac{2149 x \left (3+2 x+5 x^2\right )^{3/2}}{2500}-\frac{289}{250} x^2 \left (3+2 x+5 x^2\right )^{3/2}-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}-\frac{4633 \int \sqrt{3+2 x+5 x^2} \, dx}{1250}\\ &=-\frac{4633 (1+5 x) \sqrt{3+2 x+5 x^2}}{12500}+\frac{7819 \left (3+2 x+5 x^2\right )^{3/2}}{7500}+\frac{2149 x \left (3+2 x+5 x^2\right )^{3/2}}{2500}-\frac{289}{250} x^2 \left (3+2 x+5 x^2\right )^{3/2}-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}-\frac{32431 \int \frac{1}{\sqrt{3+2 x+5 x^2}} \, dx}{6250}\\ &=-\frac{4633 (1+5 x) \sqrt{3+2 x+5 x^2}}{12500}+\frac{7819 \left (3+2 x+5 x^2\right )^{3/2}}{7500}+\frac{2149 x \left (3+2 x+5 x^2\right )^{3/2}}{2500}-\frac{289}{250} x^2 \left (3+2 x+5 x^2\right )^{3/2}-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}-\frac{\left (4633 \sqrt{\frac{7}{10}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{56}}} \, dx,x,2+10 x\right )}{12500}\\ &=-\frac{4633 (1+5 x) \sqrt{3+2 x+5 x^2}}{12500}+\frac{7819 \left (3+2 x+5 x^2\right )^{3/2}}{7500}+\frac{2149 x \left (3+2 x+5 x^2\right )^{3/2}}{2500}-\frac{289}{250} x^2 \left (3+2 x+5 x^2\right )^{3/2}-\frac{7}{30} x^3 \left (3+2 x+5 x^2\right )^{3/2}-\frac{32431 \sinh ^{-1}\left (\frac{1+5 x}{\sqrt{14}}\right )}{6250 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.107163, size = 65, normalized size = 0.52 \[ \frac{5 \sqrt{5 x^2+2 x+3} \left (-43750 x^5-234250 x^4+48225 x^3+129895 x^2+105400 x+103386\right )-194586 \sqrt{5} \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{187500} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[3 + 2*x + 5*x^2]*(103386 + 105400*x + 129895*x^2 + 48225*x^3 - 234250*x^4 - 43750*x^5) - 194586*Sqrt[5
]*ArcSinh[(1 + 5*x)/Sqrt[14]])/187500

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Maple [A]  time = 0.055, size = 98, normalized size = 0.8 \begin{align*} -{\frac{7\,{x}^{3}}{30} \left ( 5\,{x}^{2}+2\,x+3 \right ) ^{{\frac{3}{2}}}}-{\frac{289\,{x}^{2}}{250} \left ( 5\,{x}^{2}+2\,x+3 \right ) ^{{\frac{3}{2}}}}+{\frac{2149\,x}{2500} \left ( 5\,{x}^{2}+2\,x+3 \right ) ^{{\frac{3}{2}}}}+{\frac{7819}{7500} \left ( 5\,{x}^{2}+2\,x+3 \right ) ^{{\frac{3}{2}}}}-{\frac{46330\,x+9266}{25000}\sqrt{5\,{x}^{2}+2\,x+3}}-{\frac{32431\,\sqrt{5}}{31250}{\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)^(1/2),x)

[Out]

-7/30*x^3*(5*x^2+2*x+3)^(3/2)-289/250*x^2*(5*x^2+2*x+3)^(3/2)+2149/2500*x*(5*x^2+2*x+3)^(3/2)+7819/7500*(5*x^2
+2*x+3)^(3/2)-4633/25000*(10*x+2)*(5*x^2+2*x+3)^(1/2)-32431/31250*5^(1/2)*arcsinh(5/14*14^(1/2)*(x+1/5))

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Maxima [A]  time = 1.59115, size = 147, normalized size = 1.19 \begin{align*} -\frac{7}{30} \,{\left (5 \, x^{2} + 2 \, x + 3\right )}^{\frac{3}{2}} x^{3} - \frac{289}{250} \,{\left (5 \, x^{2} + 2 \, x + 3\right )}^{\frac{3}{2}} x^{2} + \frac{2149}{2500} \,{\left (5 \, x^{2} + 2 \, x + 3\right )}^{\frac{3}{2}} x + \frac{7819}{7500} \,{\left (5 \, x^{2} + 2 \, x + 3\right )}^{\frac{3}{2}} - \frac{4633}{2500} \, \sqrt{5 \, x^{2} + 2 \, x + 3} x - \frac{32431}{31250} \, \sqrt{5} \operatorname{arsinh}\left (\frac{1}{14} \, \sqrt{14}{\left (5 \, x + 1\right )}\right ) - \frac{4633}{12500} \, \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)^(1/2),x, algorithm="maxima")

[Out]

-7/30*(5*x^2 + 2*x + 3)^(3/2)*x^3 - 289/250*(5*x^2 + 2*x + 3)^(3/2)*x^2 + 2149/2500*(5*x^2 + 2*x + 3)^(3/2)*x
+ 7819/7500*(5*x^2 + 2*x + 3)^(3/2) - 4633/2500*sqrt(5*x^2 + 2*x + 3)*x - 32431/31250*sqrt(5)*arcsinh(1/14*sqr
t(14)*(5*x + 1)) - 4633/12500*sqrt(5*x^2 + 2*x + 3)

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Fricas [A]  time = 1.03031, size = 255, normalized size = 2.06 \begin{align*} -\frac{1}{37500} \,{\left (43750 \, x^{5} + 234250 \, x^{4} - 48225 \, x^{3} - 129895 \, x^{2} - 105400 \, x - 103386\right )} \sqrt{5 \, x^{2} + 2 \, x + 3} + \frac{32431}{62500} \, \sqrt{5} \log \left (\sqrt{5} \sqrt{5 \, x^{2} + 2 \, x + 3}{\left (5 \, x + 1\right )} - 25 \, x^{2} - 10 \, x - 8\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)^(1/2),x, algorithm="fricas")

[Out]

-1/37500*(43750*x^5 + 234250*x^4 - 48225*x^3 - 129895*x^2 - 105400*x - 103386)*sqrt(5*x^2 + 2*x + 3) + 32431/6
2500*sqrt(5)*log(sqrt(5)*sqrt(5*x^2 + 2*x + 3)*(5*x + 1) - 25*x^2 - 10*x - 8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - 13 x \sqrt{5 x^{2} + 2 x + 3}\, dx - \int - 7 x^{2} \sqrt{5 x^{2} + 2 x + 3}\, dx - \int 31 x^{3} \sqrt{5 x^{2} + 2 x + 3}\, dx - \int 7 x^{4} \sqrt{5 x^{2} + 2 x + 3}\, dx - \int - 2 \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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.24684, size = 97, normalized size = 0.78 \begin{align*} -\frac{1}{37500} \,{\left (5 \,{\left ({\left (5 \,{\left (10 \,{\left (175 \, x + 937\right )} x - 1929\right )} x - 25979\right )} x - 21080\right )} x - 103386\right )} \sqrt{5 \, x^{2} + 2 \, x + 3} + \frac{32431}{31250} \, \sqrt{5} \log \left (-\sqrt{5}{\left (\sqrt{5} x - \sqrt{5 \, x^{2} + 2 \, x + 3}\right )} - 1\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)^(1/2),x, algorithm="giac")

[Out]

-1/37500*(5*((5*(10*(175*x + 937)*x - 1929)*x - 25979)*x - 21080)*x - 103386)*sqrt(5*x^2 + 2*x + 3) + 32431/31
250*sqrt(5)*log(-sqrt(5)*(sqrt(5)*x - sqrt(5*x^2 + 2*x + 3)) - 1)