3.1.68 \(\int (3-x+2 x^2)^{3/2} (2+3 x+5 x^2) \, dx\) [68]

Optimal. Leaf size=105 \[ -\frac {4117 (1-4 x) \sqrt {3-x+2 x^2}}{8192}-\frac {179 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{1536}+\frac {107}{240} \left (3-x+2 x^2\right )^{5/2}+\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}-\frac {94691 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{16384 \sqrt {2}} \]

[Out]

-179/1536*(1-4*x)*(2*x^2-x+3)^(3/2)+107/240*(2*x^2-x+3)^(5/2)+5/12*x*(2*x^2-x+3)^(5/2)-94691/32768*arcsinh(1/2
3*(1-4*x)*23^(1/2))*2^(1/2)-4117/8192*(1-4*x)*(2*x^2-x+3)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1675, 654, 626, 633, 221} \begin {gather*} \frac {5}{12} x \left (2 x^2-x+3\right )^{5/2}+\frac {107}{240} \left (2 x^2-x+3\right )^{5/2}-\frac {179 (1-4 x) \left (2 x^2-x+3\right )^{3/2}}{1536}-\frac {4117 (1-4 x) \sqrt {2 x^2-x+3}}{8192}-\frac {94691 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{16384 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4117*(1 - 4*x)*Sqrt[3 - x + 2*x^2])/8192 - (179*(1 - 4*x)*(3 - x + 2*x^2)^(3/2))/1536 + (107*(3 - x + 2*x^2)
^(5/2))/240 + (5*x*(3 - x + 2*x^2)^(5/2))/12 - (94691*ArcSinh[(1 - 4*x)/Sqrt[23]])/(16384*Sqrt[2])

Rule 221

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]

Rule 626

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 633

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 654

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 1675

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]

Rubi steps

\begin {align*} \int \left (3-x+2 x^2\right )^{3/2} \left (2+3 x+5 x^2\right ) \, dx &=\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}+\frac {1}{12} \int \left (9+\frac {107 x}{2}\right ) \left (3-x+2 x^2\right )^{3/2} \, dx\\ &=\frac {107}{240} \left (3-x+2 x^2\right )^{5/2}+\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}+\frac {179}{96} \int \left (3-x+2 x^2\right )^{3/2} \, dx\\ &=-\frac {179 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{1536}+\frac {107}{240} \left (3-x+2 x^2\right )^{5/2}+\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}+\frac {4117 \int \sqrt {3-x+2 x^2} \, dx}{1024}\\ &=-\frac {4117 (1-4 x) \sqrt {3-x+2 x^2}}{8192}-\frac {179 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{1536}+\frac {107}{240} \left (3-x+2 x^2\right )^{5/2}+\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}+\frac {94691 \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx}{16384}\\ &=-\frac {4117 (1-4 x) \sqrt {3-x+2 x^2}}{8192}-\frac {179 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{1536}+\frac {107}{240} \left (3-x+2 x^2\right )^{5/2}+\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}+\frac {\left (4117 \sqrt {\frac {23}{2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{16384}\\ &=-\frac {4117 (1-4 x) \sqrt {3-x+2 x^2}}{8192}-\frac {179 (1-4 x) \left (3-x+2 x^2\right )^{3/2}}{1536}+\frac {107}{240} \left (3-x+2 x^2\right )^{5/2}+\frac {5}{12} x \left (3-x+2 x^2\right )^{5/2}-\frac {94691 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{16384 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 75, normalized size = 0.71 \begin {gather*} \frac {4 \sqrt {3-x+2 x^2} \left (388341+565276 x+319072 x^2+561024 x^3+14336 x^4+204800 x^5\right )-1420365 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{491520} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(388341 + 565276*x + 319072*x^2 + 561024*x^3 + 14336*x^4 + 204800*x^5) - 1420365*Sqrt[2
]*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]])/491520

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Maple [A]
time = 0.10, size = 83, normalized size = 0.79

method	result	size
risch	\(\frac {\left (204800 x^{5}+14336 x^{4}+561024 x^{3}+319072 x^{2}+565276 x +388341\right ) \sqrt {2 x^{2}-x +3}}{122880}+\frac {94691 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{32768}\)	\(55\)
trager	\(\left (\frac {5}{3} x^{5}+\frac {7}{60} x^{4}+\frac {1461}{320} x^{3}+\frac {9971}{3840} x^{2}+\frac {141319}{30720} x +\frac {129447}{40960}\right ) \sqrt {2 x^{2}-x +3}-\frac {94691 \RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-4 \RootOf \left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {2 x^{2}-x +3}+\RootOf \left (\textit {\_Z}^{2}-2\right )\right )}{32768}\)	\(79\)
default	\(\frac {5 x \left (2 x^{2}-x +3\right )^{\frac {5}{2}}}{12}+\frac {107 \left (2 x^{2}-x +3\right )^{\frac {5}{2}}}{240}+\frac {179 \left (4 x -1\right ) \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{1536}+\frac {4117 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{8192}+\frac {94691 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{32768}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/12*x*(2*x^2-x+3)^(5/2)+107/240*(2*x^2-x+3)^(5/2)+179/1536*(4*x-1)*(2*x^2-x+3)^(3/2)+4117/8192*(4*x-1)*(2*x^2
-x+3)^(1/2)+94691/32768*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))

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Maxima [A]
time = 0.54, size = 104, normalized size = 0.99 \begin {gather*} \frac {5}{12} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} x + \frac {107}{240} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} + \frac {179}{384} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x - \frac {179}{1536} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {4117}{2048} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {94691}{32768} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {4117}{8192} \, \sqrt {2 \, x^{2} - x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/12*(2*x^2 - x + 3)^(5/2)*x + 107/240*(2*x^2 - x + 3)^(5/2) + 179/384*(2*x^2 - x + 3)^(3/2)*x - 179/1536*(2*x
^2 - x + 3)^(3/2) + 4117/2048*sqrt(2*x^2 - x + 3)*x + 94691/32768*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 4
117/8192*sqrt(2*x^2 - x + 3)

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Fricas [A]
time = 2.18, size = 78, normalized size = 0.74 \begin {gather*} \frac {1}{122880} \, {\left (204800 \, x^{5} + 14336 \, x^{4} + 561024 \, x^{3} + 319072 \, x^{2} + 565276 \, x + 388341\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {94691}{65536} \, \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/122880*(204800*x^5 + 14336*x^4 + 561024*x^3 + 319072*x^2 + 565276*x + 388341)*sqrt(2*x^2 - x + 3) + 94691/65
536*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (2 x^{2} - x + 3\right )^{\frac {3}{2}} \cdot \left (5 x^{2} + 3 x + 2\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.89, size = 73, normalized size = 0.70 \begin {gather*} \frac {1}{122880} \, {\left (4 \, {\left (8 \, {\left (4 \, {\left (16 \, {\left (100 \, x + 7\right )} x + 4383\right )} x + 9971\right )} x + 141319\right )} x + 388341\right )} \sqrt {2 \, x^{2} - x + 3} - \frac {94691}{32768} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/122880*(4*(8*(4*(16*(100*x + 7)*x + 4383)*x + 9971)*x + 141319)*x + 388341)*sqrt(2*x^2 - x + 3) - 94691/3276
8*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (2\,x^2-x+3\right )}^{3/2}\,\left (5\,x^2+3\,x+2\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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