3.75 \(\int (3-x+2 x^2)^{5/2} (2+3 x+5 x^2) \, dx\)

Optimal. Leaf size=128 \[ \frac{5}{16} x \left (2 x^2-x+3\right )^{7/2}+\frac{141}{448} \left (2 x^2-x+3\right )^{7/2}-\frac{277 (1-4 x) \left (2 x^2-x+3\right )^{5/2}}{3072}-\frac{31855 (1-4 x) \left (2 x^2-x+3\right )^{3/2}}{98304}-\frac{732665 (1-4 x) \sqrt{2 x^2-x+3}}{524288}-\frac{16851295 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{1048576 \sqrt{2}} \]

[Out]

(-732665*(1 - 4*x)*Sqrt[3 - x + 2*x^2])/524288 - (31855*(1 - 4*x)*(3 - x + 2*x^2)^(3/2))/98304 - (277*(1 - 4*x
)*(3 - x + 2*x^2)^(5/2))/3072 + (141*(3 - x + 2*x^2)^(7/2))/448 + (5*x*(3 - x + 2*x^2)^(7/2))/16 - (16851295*A
rcSinh[(1 - 4*x)/Sqrt[23]])/(1048576*Sqrt[2])

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Rubi [A]  time = 0.0624409, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1661, 640, 612, 619, 215} \[ \frac{5}{16} x \left (2 x^2-x+3\right )^{7/2}+\frac{141}{448} \left (2 x^2-x+3\right )^{7/2}-\frac{277 (1-4 x) \left (2 x^2-x+3\right )^{5/2}}{3072}-\frac{31855 (1-4 x) \left (2 x^2-x+3\right )^{3/2}}{98304}-\frac{732665 (1-4 x) \sqrt{2 x^2-x+3}}{524288}-\frac{16851295 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{1048576 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-732665*(1 - 4*x)*Sqrt[3 - x + 2*x^2])/524288 - (31855*(1 - 4*x)*(3 - x + 2*x^2)^(3/2))/98304 - (277*(1 - 4*x
)*(3 - x + 2*x^2)^(5/2))/3072 + (141*(3 - x + 2*x^2)^(7/2))/448 + (5*x*(3 - x + 2*x^2)^(7/2))/16 - (16851295*A
rcSinh[(1 - 4*x)/Sqrt[23]])/(1048576*Sqrt[2])

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 (3-x+2 x^2\right )^{5/2} \left (2+3 x+5 x^2\right ) \, dx &=\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}+\frac{1}{16} \int \left (17+\frac{141 x}{2}\right ) \left (3-x+2 x^2\right )^{5/2} \, dx\\ &=\frac{141}{448} \left (3-x+2 x^2\right )^{7/2}+\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}+\frac{277}{128} \int \left (3-x+2 x^2\right )^{5/2} \, dx\\ &=-\frac{277 (1-4 x) \left (3-x+2 x^2\right )^{5/2}}{3072}+\frac{141}{448} \left (3-x+2 x^2\right )^{7/2}+\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}+\frac{31855 \int \left (3-x+2 x^2\right )^{3/2} \, dx}{6144}\\ &=-\frac{31855 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{98304}-\frac{277 (1-4 x) \left (3-x+2 x^2\right )^{5/2}}{3072}+\frac{141}{448} \left (3-x+2 x^2\right )^{7/2}+\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}+\frac{732665 \int \sqrt{3-x+2 x^2} \, dx}{65536}\\ &=-\frac{732665 (1-4 x) \sqrt{3-x+2 x^2}}{524288}-\frac{31855 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{98304}-\frac{277 (1-4 x) \left (3-x+2 x^2\right )^{5/2}}{3072}+\frac{141}{448} \left (3-x+2 x^2\right )^{7/2}+\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}+\frac{16851295 \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx}{1048576}\\ &=-\frac{732665 (1-4 x) \sqrt{3-x+2 x^2}}{524288}-\frac{31855 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{98304}-\frac{277 (1-4 x) \left (3-x+2 x^2\right )^{5/2}}{3072}+\frac{141}{448} \left (3-x+2 x^2\right )^{7/2}+\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}+\frac{\left (732665 \sqrt{\frac{23}{2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{1048576}\\ &=-\frac{732665 (1-4 x) \sqrt{3-x+2 x^2}}{524288}-\frac{31855 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{98304}-\frac{277 (1-4 x) \left (3-x+2 x^2\right )^{5/2}}{3072}+\frac{141}{448} \left (3-x+2 x^2\right )^{7/2}+\frac{5}{16} x \left (3-x+2 x^2\right )^{7/2}-\frac{16851295 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{1048576 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.107564, size = 75, normalized size = 0.59 \[ \frac{4 \sqrt{2 x^2-x+3} \left (27525120 x^7-13565952 x^6+118808576 x^5-1619968 x^4+172684416 x^3+67272352 x^2+148957444 x+58536675\right )-353877195 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{44040192} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(58536675 + 148957444*x + 67272352*x^2 + 172684416*x^3 - 1619968*x^4 + 118808576*x^5 -
13565952*x^6 + 27525120*x^7) - 353877195*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]])/44040192

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Maple [A]  time = 0.049, size = 102, normalized size = 0.8 \begin{align*}{\frac{5\,x}{16} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{7}{2}}}}+{\frac{141}{448} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{7}{2}}}}+{\frac{-277+1108\,x}{3072} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{5}{2}}}}+{\frac{-31855+127420\,x}{98304} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{3}{2}}}}+{\frac{-732665+2930660\,x}{524288}\sqrt{2\,{x}^{2}-x+3}}+{\frac{16851295\,\sqrt{2}}{2097152}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*x*(2*x^2-x+3)^(7/2)+141/448*(2*x^2-x+3)^(7/2)+277/3072*(-1+4*x)*(2*x^2-x+3)^(5/2)+31855/98304*(-1+4*x)*(2
*x^2-x+3)^(3/2)+732665/524288*(-1+4*x)*(2*x^2-x+3)^(1/2)[phone]/2097152*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4
))

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Maxima [A]  time = 1.43838, size = 180, normalized size = 1.41 \begin{align*} \frac{5}{16} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{7}{2}} x + \frac{141}{448} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{7}{2}} + \frac{277}{768} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{5}{2}} x - \frac{277}{3072} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{5}{2}} + \frac{31855}{24576} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}} x - \frac{31855}{98304} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}} + \frac{732665}{131072} \, \sqrt{2 \, x^{2} - x + 3} x + \frac{16851295}{2097152} \, \sqrt{2} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) - \frac{732665}{524288} \, \sqrt{2 \, x^{2} - x + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*(2*x^2 - x + 3)^(7/2)*x + 141/448*(2*x^2 - x + 3)^(7/2) + 277/768*(2*x^2 - x + 3)^(5/2)*x - 277/3072*(2*x
^2 - x + 3)^(5/2) + 31855/24576*(2*x^2 - x + 3)^(3/2)*x - 31855/98304*(2*x^2 - x + 3)^(3/2) + 732665/131072*sq
rt(2*x^2 - x + 3)*x + 16851295/2097152*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 732665/524288*sqrt(2*x^2 - x
+ 3)

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Fricas [A]  time = 1.38597, size = 327, normalized size = 2.55 \begin{align*} \frac{1}{11010048} \,{\left (27525120 \, x^{7} - 13565952 \, x^{6} + 118808576 \, x^{5} - 1619968 \, x^{4} + 172684416 \, x^{3} + 67272352 \, x^{2} + 148957444 \, x + 58536675\right )} \sqrt{2 \, x^{2} - x + 3} + \frac{16851295}{4194304} \, \sqrt{2} \log \left (-4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11010048*(27525120*x^7 - 13565952*x^6 + 118808576*x^5 - 1619968*x^4 + 172684416*x^3 + 67272352*x^2 + 1489574
44*x + 58536675)*sqrt(2*x^2 - x + 3) + 16851295/4194304*sqrt(2)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) -
32*x^2 + 16*x - 25)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (2 x^{2} - x + 3\right )^{\frac{5}{2}} \left (5 x^{2} + 3 x + 2\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14499, size = 112, normalized size = 0.88 \begin{align*} \frac{1}{11010048} \,{\left (4 \,{\left (8 \,{\left (4 \,{\left (16 \,{\left (4 \,{\left (24 \,{\left (140 \, x - 69\right )} x + 14503\right )} x - 791\right )} x + 1349097\right )} x + 2102261\right )} x + 37239361\right )} x + 58536675\right )} \sqrt{2 \, x^{2} - x + 3} - \frac{16851295}{2097152} \, \sqrt{2} \log \left (-2 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11010048*(4*(8*(4*(16*(4*(24*(140*x - 69)*x + 14503)*x - 791)*x + 1349097)*x + 2102261)*x + 37239361)*x + 58
536675)*sqrt(2*x^2 - x + 3) - 16851295/2097152*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)