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

Optimal. Leaf size=124 \[ \frac{49}{100} \sqrt{5 x^2+2 x+3} x^3+\frac{203}{100} \sqrt{5 x^2+2 x+3} x^2-\frac{8749 \sqrt{5 x^2+2 x+3} x}{1250}-\frac{5086 \sqrt{5 x^2+2 x+3}}{3125}-\frac{8 (136602 x+12983)}{21875 \sqrt{5 x^2+2 x+3}}+\frac{89583 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{1250 \sqrt{5}} \]

[Out]

(-8*(12983 + 136602*x))/(21875*Sqrt[3 + 2*x + 5*x^2]) - (5086*Sqrt[3 + 2*x + 5*x^2])/3125 - (8749*x*Sqrt[3 + 2
*x + 5*x^2])/1250 + (203*x^2*Sqrt[3 + 2*x + 5*x^2])/100 + (49*x^3*Sqrt[3 + 2*x + 5*x^2])/100 + (89583*ArcSinh[
(1 + 5*x)/Sqrt[14]])/(1250*Sqrt[5])

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Rubi [A]  time = 0.160835, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1660, 1661, 640, 619, 215} \[ \frac{49}{100} \sqrt{5 x^2+2 x+3} x^3+\frac{203}{100} \sqrt{5 x^2+2 x+3} x^2-\frac{8749 \sqrt{5 x^2+2 x+3} x}{1250}-\frac{5086 \sqrt{5 x^2+2 x+3}}{3125}-\frac{8 (136602 x+12983)}{21875 \sqrt{5 x^2+2 x+3}}+\frac{89583 \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{1250 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(12983 + 136602*x))/(21875*Sqrt[3 + 2*x + 5*x^2]) - (5086*Sqrt[3 + 2*x + 5*x^2])/3125 - (8749*x*Sqrt[3 + 2
*x + 5*x^2])/1250 + (203*x^2*Sqrt[3 + 2*x + 5*x^2])/100 + (49*x^3*Sqrt[3 + 2*x + 5*x^2])/100 + (89583*ArcSinh[
(1 + 5*x)/Sqrt[14]])/(1250*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 )^2 \left (2+5 x+x^2\right )}{\left (3+2 x+5 x^2\right )^{3/2}} \, dx &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}+\frac{1}{28} \int \frac{\frac{4291112}{3125}-\frac{296716 x}{625}-\frac{194012 x^2}{125}+\frac{23716 x^3}{25}+\frac{1372 x^4}{5}}{\sqrt{3+2 x+5 x^2}} \, dx\\ &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}+\frac{49}{100} x^3 \sqrt{3+2 x+5 x^2}+\frac{1}{560} \int \frac{\frac{17164448}{625}-\frac{1186864 x}{125}-\frac{837788 x^2}{25}+17052 x^3}{\sqrt{3+2 x+5 x^2}} \, dx\\ &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}+\frac{203}{100} x^2 \sqrt{3+2 x+5 x^2}+\frac{49}{100} x^3 \sqrt{3+2 x+5 x^2}+\frac{\int \frac{\frac{51493344}{125}-\frac{6118392 x}{25}-\frac{2939664 x^2}{5}}{\sqrt{3+2 x+5 x^2}} \, dx}{8400}\\ &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}-\frac{8749 x \sqrt{3+2 x+5 x^2}}{1250}+\frac{203}{100} x^2 \sqrt{3+2 x+5 x^2}+\frac{49}{100} x^3 \sqrt{3+2 x+5 x^2}+\frac{\int \frac{\frac{147081648}{25}-\frac{3417792 x}{5}}{\sqrt{3+2 x+5 x^2}} \, dx}{84000}\\ &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}-\frac{5086 \sqrt{3+2 x+5 x^2}}{3125}-\frac{8749 x \sqrt{3+2 x+5 x^2}}{1250}+\frac{203}{100} x^2 \sqrt{3+2 x+5 x^2}+\frac{49}{100} x^3 \sqrt{3+2 x+5 x^2}+\frac{89583 \int \frac{1}{\sqrt{3+2 x+5 x^2}} \, dx}{1250}\\ &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}-\frac{5086 \sqrt{3+2 x+5 x^2}}{3125}-\frac{8749 x \sqrt{3+2 x+5 x^2}}{1250}+\frac{203}{100} x^2 \sqrt{3+2 x+5 x^2}+\frac{49}{100} x^3 \sqrt{3+2 x+5 x^2}+\frac{89583 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{56}}} \, dx,x,2+10 x\right )}{2500 \sqrt{70}}\\ &=-\frac{8 (12983+136602 x)}{21875 \sqrt{3+2 x+5 x^2}}-\frac{5086 \sqrt{3+2 x+5 x^2}}{3125}-\frac{8749 x \sqrt{3+2 x+5 x^2}}{1250}+\frac{203}{100} x^2 \sqrt{3+2 x+5 x^2}+\frac{49}{100} x^3 \sqrt{3+2 x+5 x^2}+\frac{89583 \sinh ^{-1}\left (\frac{1+5 x}{\sqrt{14}}\right )}{1250 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.274322, size = 65, normalized size = 0.52 \[ \frac{\frac{5 \left (42875 x^5+194775 x^4-515655 x^3-280805 x^2-1298674 x-168536\right )}{\sqrt{5 x^2+2 x+3}}+1254162 \sqrt{5} \sinh ^{-1}\left (\frac{5 x+1}{\sqrt{14}}\right )}{87500} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*(-168536 - 1298674*x - 280805*x^2 - 515655*x^3 + 194775*x^4 + 42875*x^5))/Sqrt[3 + 2*x + 5*x^2] + 1254162*
Sqrt[5]*ArcSinh[(1 + 5*x)/Sqrt[14]])/87500

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Maple [A]  time = 0.054, size = 132, normalized size = 1.1 \begin{align*} -{\frac{28506}{3125}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}+{\frac{49\,{x}^{5}}{20}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}+{\frac{1113\,{x}^{4}}{100}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{8023\,{x}^{2}}{500}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}+{\frac{89583\,\sqrt{5}}{6250}{\it Arcsinh} \left ({\frac{5\,\sqrt{14}}{14} \left ( x+{\frac{1}{5}} \right ) } \right ) }-{\frac{55640\,x+11128}{21875}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{89583\,x}{1250}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}}-{\frac{14733\,{x}^{3}}{500}{\frac{1}{\sqrt{5\,{x}^{2}+2\,x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-28506/3125/(5*x^2+2*x+3)^(1/2)+49/20*x^5/(5*x^2+2*x+3)^(1/2)+1113/100*x^4/(5*x^2+2*x+3)^(1/2)-8023/500*x^2/(5
*x^2+2*x+3)^(1/2)+89583/6250*5^(1/2)*arcsinh(5/14*14^(1/2)*(x+1/5))-5564/21875*(10*x+2)/(5*x^2+2*x+3)^(1/2)-89
583/1250*x/(5*x^2+2*x+3)^(1/2)-14733/500*x^3/(5*x^2+2*x+3)^(1/2)

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Maxima [A]  time = 1.57881, size = 154, normalized size = 1.24 \begin{align*} \frac{49 \, x^{5}}{20 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} + \frac{1113 \, x^{4}}{100 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} - \frac{14733 \, x^{3}}{500 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} - \frac{8023 \, x^{2}}{500 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} + \frac{89583}{6250} \, \sqrt{5} \operatorname{arsinh}\left (\frac{1}{14} \, \sqrt{14}{\left (5 \, x + 1\right )}\right ) - \frac{649337 \, x}{8750 \, \sqrt{5 \, x^{2} + 2 \, x + 3}} - \frac{42134}{4375 \, \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)^2*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x, algorithm="maxima")

[Out]

49/20*x^5/sqrt(5*x^2 + 2*x + 3) + 1113/100*x^4/sqrt(5*x^2 + 2*x + 3) - 14733/500*x^3/sqrt(5*x^2 + 2*x + 3) - 8
023/500*x^2/sqrt(5*x^2 + 2*x + 3) + 89583/6250*sqrt(5)*arcsinh(1/14*sqrt(14)*(5*x + 1)) - 649337/8750*x/sqrt(5
*x^2 + 2*x + 3) - 42134/4375/sqrt(5*x^2 + 2*x + 3)

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Fricas [A]  time = 1.30554, size = 305, normalized size = 2.46 \begin{align*} \frac{627081 \, \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 (42875 \, x^{5} + 194775 \, x^{4} - 515655 \, x^{3} - 280805 \, x^{2} - 1298674 \, x - 168536\right )} \sqrt{5 \, x^{2} + 2 \, x + 3}}{87500 \,{\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)^2*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x, algorithm="fricas")

[Out]

1/87500*(627081*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*(42875*x^5 + 194775*x^4 - 515655*x^3 - 280805*x^2 - 1298674*x - 168536)*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{\left (x^{2} + 5 x + 2\right ) \left (7 x^{2} - 4 x - 1\right )^{2}}{\left (5 x^{2} + 2 x + 3\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16135, size = 96, normalized size = 0.77 \begin{align*} -\frac{89583}{6250} \, \sqrt{5} \log \left (-\sqrt{5}{\left (\sqrt{5} x - \sqrt{5 \, x^{2} + 2 \, x + 3}\right )} - 1\right ) + \frac{{\left (35 \,{\left ({\left (35 \,{\left (35 \, x + 159\right )} x - 14733\right )} x - 8023\right )} x - 1298674\right )} x - 168536}{17500 \, \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)^2*(x^2+5*x+2)/(5*x^2+2*x+3)^(3/2),x, algorithm="giac")

[Out]

-89583/6250*sqrt(5)*log(-sqrt(5)*(sqrt(5)*x - sqrt(5*x^2 + 2*x + 3)) - 1) + 1/17500*((35*((35*(35*x + 159)*x -
14733)*x - 8023)*x - 1298674)*x - 168536)/sqrt(5*x^2 + 2*x + 3)