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

Optimal. Leaf size=124 \[ \frac {1825}{64} \sqrt {2 x^2-x+3} x^2+\frac {15565}{512} \sqrt {2 x^2-x+3} x-\frac {181561 \sqrt {2 x^2-x+3}}{2048}-\frac {1331 (17-45 x)}{368 \sqrt {2 x^2-x+3}}+\frac {125}{16} \sqrt {2 x^2-x+3} x^3+\frac {1168881 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}} \]

[Out]

1168881/8192*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-1331/368*(17-45*x)/(2*x^2-x+3)^(1/2)-181561/2048*(2*x^2-x+
3)^(1/2)+15565/512*x*(2*x^2-x+3)^(1/2)+1825/64*x^2*(2*x^2-x+3)^(1/2)+125/16*x^3*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1660, 1661, 640, 619, 215} \[ \frac {125}{16} \sqrt {2 x^2-x+3} x^3+\frac {1825}{64} \sqrt {2 x^2-x+3} x^2+\frac {15565}{512} \sqrt {2 x^2-x+3} x-\frac {181561 \sqrt {2 x^2-x+3}}{2048}-\frac {1331 (17-45 x)}{368 \sqrt {2 x^2-x+3}}+\frac {1168881 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1331*(17 - 45*x))/(368*Sqrt[3 - x + 2*x^2]) - (181561*Sqrt[3 - x + 2*x^2])/2048 + (15565*x*Sqrt[3 - x + 2*x^
2])/512 + (1825*x^2*Sqrt[3 - x + 2*x^2])/64 + (125*x^3*Sqrt[3 - x + 2*x^2])/16 + (1168881*ArcSinh[(1 - 4*x)/Sq
rt[23]])/(4096*Sqrt[2])

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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (2+3 x+5 x^2\right )^3}{\left (3-x+2 x^2\right )^{3/2}} \, dx &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}+\frac {2}{23} \int \frac {-\frac {110285}{64}-\frac {19067 x}{32}+\frac {22195 x^2}{16}+\frac {13225 x^3}{8}+\frac {2875 x^4}{4}}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}+\frac {125}{16} x^3 \sqrt {3-x+2 x^2}+\frac {1}{92} \int \frac {-\frac {110285}{8}-\frac {19067 x}{4}+\frac {18515 x^2}{4}+\frac {125925 x^3}{8}}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}+\frac {1825}{64} x^2 \sqrt {3-x+2 x^2}+\frac {125}{16} x^3 \sqrt {3-x+2 x^2}+\frac {1}{552} \int \frac {-\frac {330855}{4}-\frac {492177 x}{4}+\frac {1073985 x^2}{16}}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}+\frac {15565}{512} x \sqrt {3-x+2 x^2}+\frac {1825}{64} x^2 \sqrt {3-x+2 x^2}+\frac {125}{16} x^3 \sqrt {3-x+2 x^2}+\frac {\int \frac {-\frac {8515635}{16}-\frac {12527709 x}{32}}{\sqrt {3-x+2 x^2}} \, dx}{2208}\\ &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}-\frac {181561 \sqrt {3-x+2 x^2}}{2048}+\frac {15565}{512} x \sqrt {3-x+2 x^2}+\frac {1825}{64} x^2 \sqrt {3-x+2 x^2}+\frac {125}{16} x^3 \sqrt {3-x+2 x^2}-\frac {1168881 \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx}{4096}\\ &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}-\frac {181561 \sqrt {3-x+2 x^2}}{2048}+\frac {15565}{512} x \sqrt {3-x+2 x^2}+\frac {1825}{64} x^2 \sqrt {3-x+2 x^2}+\frac {125}{16} x^3 \sqrt {3-x+2 x^2}-\frac {1168881 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{4096 \sqrt {46}}\\ &=-\frac {1331 (17-45 x)}{368 \sqrt {3-x+2 x^2}}-\frac {181561 \sqrt {3-x+2 x^2}}{2048}+\frac {15565}{512} x \sqrt {3-x+2 x^2}+\frac {1825}{64} x^2 \sqrt {3-x+2 x^2}+\frac {125}{16} x^3 \sqrt {3-x+2 x^2}+\frac {1168881 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 65, normalized size = 0.52 \[ \frac {\frac {4 \left (736000 x^5+2318400 x^4+2624760 x^3-5754186 x^2+16138403 x-15423965\right )}{\sqrt {2 x^2-x+3}}-26884263 \sqrt {2} \sinh ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{188416} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*(-15423965 + 16138403*x - 5754186*x^2 + 2624760*x^3 + 2318400*x^4 + 736000*x^5))/Sqrt[3 - x + 2*x^2] - 268
84263*Sqrt[2]*ArcSinh[(-1 + 4*x)/Sqrt[23]])/188416

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fricas [A]  time = 0.94, size = 102, normalized size = 0.82 \[ \frac {26884263 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (736000 \, x^{5} + 2318400 \, x^{4} + 2624760 \, x^{3} - 5754186 \, x^{2} + 16138403 \, x - 15423965\right )} \sqrt {2 \, x^{2} - x + 3}}{376832 \, {\left (2 \, x^{2} - x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/376832*(26884263*sqrt(2)*(2*x^2 - x + 3)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) +
8*(736000*x^5 + 2318400*x^4 + 2624760*x^3 - 5754186*x^2 + 16138403*x - 15423965)*sqrt(2*x^2 - x + 3))/(2*x^2
- x + 3)

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giac [A]  time = 0.28, size = 72, normalized size = 0.58 \[ \frac {1168881}{8192} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left (46 \, {\left (20 \, {\left (40 \, {\left (20 \, x + 63\right )} x + 2853\right )} x - 125091\right )} x + 16138403\right )} x - 15423965}{47104 \, \sqrt {2 \, x^{2} - x + 3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1168881/8192*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/47104*((46*(20*(40*(20*x + 63)*
x + 2853)*x - 125091)*x + 16138403)*x - 15423965)/sqrt(2*x^2 - x + 3)

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maple [A]  time = 0.01, size = 132, normalized size = 1.06 \[ \frac {125 x^{5}}{8 \sqrt {2 x^{2}-x +3}}+\frac {1575 x^{4}}{32 \sqrt {2 x^{2}-x +3}}+\frac {14265 x^{3}}{256 \sqrt {2 x^{2}-x +3}}-\frac {125091 x^{2}}{1024 \sqrt {2 x^{2}-x +3}}+\frac {1168881 x}{4096 \sqrt {2 x^{2}-x +3}}-\frac {1168881 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{8192}-\frac {5130399}{16384 \sqrt {2 x^{2}-x +3}}+\frac {\frac {5392543 x}{94208}-\frac {5392543}{376832}}{\sqrt {2 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5130399/16384/(2*x^2-x+3)^(1/2)+5392543/376832*(4*x-1)/(2*x^2-x+3)^(1/2)-1168881/8192*2^(1/2)*arcsinh(4/23*23
^(1/2)*(x-1/4))+125/8/(2*x^2-x+3)^(1/2)*x^5+1575/32/(2*x^2-x+3)^(1/2)*x^4+14265/256/(2*x^2-x+3)^(1/2)*x^3-1250
91/1024/(2*x^2-x+3)^(1/2)*x^2+1168881/4096/(2*x^2-x+3)^(1/2)*x

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maxima [A]  time = 0.97, size = 114, normalized size = 0.92 \[ \frac {125 \, x^{5}}{8 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {1575 \, x^{4}}{32 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {14265 \, x^{3}}{256 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {125091 \, x^{2}}{1024 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {1168881}{8192} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {16138403 \, x}{47104 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {15423965}{47104 \, \sqrt {2 \, x^{2} - x + 3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/8*x^5/sqrt(2*x^2 - x + 3) + 1575/32*x^4/sqrt(2*x^2 - x + 3) + 14265/256*x^3/sqrt(2*x^2 - x + 3) - 125091/1
024*x^2/sqrt(2*x^2 - x + 3) - 1168881/8192*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) + 16138403/47104*x/sqrt(2*
x^2 - x + 3) - 15423965/47104/sqrt(2*x^2 - x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x^{2} + 3 x + 2\right )^{3}}{\left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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