3.1046 \(\int (2-5 x) \sqrt {x} (2+5 x+3 x^2)^{3/2} \, dx\)

Optimal. Leaf size=210 \[ -\frac {10}{33} \sqrt {x} \left (3 x^2+5 x+2\right )^{5/2}+\frac {4}{231} \sqrt {x} (84 x+65) \left (3 x^2+5 x+2\right )^{3/2}-\frac {4}{385} \sqrt {x} (39 x+55) \sqrt {3 x^2+5 x+2}-\frac {424 \sqrt {x} (3 x+2)}{1155 \sqrt {3 x^2+5 x+2}}-\frac {36 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{77 \sqrt {3 x^2+5 x+2}}+\frac {424 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{1155 \sqrt {3 x^2+5 x+2}} \]

[Out]

4/231*(65+84*x)*(3*x^2+5*x+2)^(3/2)*x^(1/2)-10/33*(3*x^2+5*x+2)^(5/2)*x^(1/2)-424/1155*(2+3*x)*x^(1/2)/(3*x^2+
5*x+2)^(1/2)+424/1155*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*((2+3*x
)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)-36/77*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/
2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)-4/385*(55+39*x)*x^(1/2)*(3*x^2+5*x+2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {832, 814, 839, 1189, 1100, 1136} \[ -\frac {10}{33} \sqrt {x} \left (3 x^2+5 x+2\right )^{5/2}+\frac {4}{231} \sqrt {x} (84 x+65) \left (3 x^2+5 x+2\right )^{3/2}-\frac {4}{385} \sqrt {x} (39 x+55) \sqrt {3 x^2+5 x+2}-\frac {424 \sqrt {x} (3 x+2)}{1155 \sqrt {3 x^2+5 x+2}}-\frac {36 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{77 \sqrt {3 x^2+5 x+2}}+\frac {424 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{1155 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-424*Sqrt[x]*(2 + 3*x))/(1155*Sqrt[2 + 5*x + 3*x^2]) - (4*Sqrt[x]*(55 + 39*x)*Sqrt[2 + 5*x + 3*x^2])/385 + (4
*Sqrt[x]*(65 + 84*x)*(2 + 5*x + 3*x^2)^(3/2))/231 - (10*Sqrt[x]*(2 + 5*x + 3*x^2)^(5/2))/33 + (424*Sqrt[2]*(1
+ x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(1155*Sqrt[2 + 5*x + 3*x^2]) - (36*Sqrt[2]*(1 +
x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(77*Sqrt[2 + 5*x + 3*x^2])

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 839

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f +
g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1100

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b -
q)*x^2)*Sqrt[(2*a + (b + q)*x^2)/(2*a + (b - q)*x^2)]*EllipticF[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)
])/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^
2 - 4*a*c, 0]

Rule 1136

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b -
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*Sqrt[(2*a + (
b + q)*x^2)/(2*a + (b - q)*x^2)]*EllipticE[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)])/(2*c*Sqrt[a + b*x^
2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int (2-5 x) \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2} \, dx &=-\frac {10}{33} \sqrt {x} \left (2+5 x+3 x^2\right )^{5/2}+\frac {2}{33} \int \frac {(5+108 x) \left (2+5 x+3 x^2\right )^{3/2}}{\sqrt {x}} \, dx\\ &=\frac {4}{231} \sqrt {x} (65+84 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {10}{33} \sqrt {x} \left (2+5 x+3 x^2\right )^{5/2}-\frac {4 \int \frac {\left (270+\frac {1053 x}{2}\right ) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx}{2079}\\ &=-\frac {4}{385} \sqrt {x} (55+39 x) \sqrt {2+5 x+3 x^2}+\frac {4}{231} \sqrt {x} (65+84 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {10}{33} \sqrt {x} \left (2+5 x+3 x^2\right )^{5/2}+\frac {8 \int \frac {-\frac {10935}{2}-\frac {12879 x}{2}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx}{93555}\\ &=-\frac {4}{385} \sqrt {x} (55+39 x) \sqrt {2+5 x+3 x^2}+\frac {4}{231} \sqrt {x} (65+84 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {10}{33} \sqrt {x} \left (2+5 x+3 x^2\right )^{5/2}+\frac {16 \operatorname {Subst}\left (\int \frac {-\frac {10935}{2}-\frac {12879 x^2}{2}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )}{93555}\\ &=-\frac {4}{385} \sqrt {x} (55+39 x) \sqrt {2+5 x+3 x^2}+\frac {4}{231} \sqrt {x} (65+84 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {10}{33} \sqrt {x} \left (2+5 x+3 x^2\right )^{5/2}-\frac {72}{77} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-\frac {424}{385} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {424 \sqrt {x} (2+3 x)}{1155 \sqrt {2+5 x+3 x^2}}-\frac {4}{385} \sqrt {x} (55+39 x) \sqrt {2+5 x+3 x^2}+\frac {4}{231} \sqrt {x} (65+84 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {10}{33} \sqrt {x} \left (2+5 x+3 x^2\right )^{5/2}+\frac {424 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{1155 \sqrt {2+5 x+3 x^2}}-\frac {36 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{77 \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 173, normalized size = 0.82 \[ \frac {-424 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-2 \left (58 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+4725 x^7+16065 x^6+17775 x^5+3497 x^4-6140 x^3-3106 x^2+520 x+424\right )}{1155 \sqrt {x} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-424*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - 2*(424
+ 520*x - 3106*x^2 - 6140*x^3 + 3497*x^4 + 17775*x^5 + 16065*x^6 + 4725*x^7 + (58*I)*Sqrt[2]*Sqrt[1 + x^(-1)]
*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2]))/(1155*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (15 \, x^{3} + 19 \, x^{2} - 4\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(3/2)*x^(1/2),x, algorithm="fricas")

[Out]

integral(-(15*x^3 + 19*x^2 - 4)*sqrt(3*x^2 + 5*x + 2)*sqrt(x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(3/2)*x^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 132, normalized size = 0.63 \[ \frac {-\frac {90 x^{7}}{11}-\frac {306 x^{6}}{11}-\frac {2370 x^{5}}{77}-\frac {6994 x^{4}}{1155}+\frac {2456 x^{3}}{231}+\frac {7484 x^{2}}{1155}+\frac {72 x}{77}-\frac {212 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3465}+\frac {32 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{1155}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465/(3*x^2+5*x+2)^(1/2)/x^(1/2)*(-14175*x^7-48195*x^6-53325*x^5+48*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)
^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))-106*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2
*(6*x+4)^(1/2),I*2^(1/2))-10491*x^4+18420*x^3+11226*x^2+1620*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(3/2)*x^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- 4 \sqrt {x} \sqrt {3 x^{2} + 5 x + 2}\right )\, dx - \int 19 x^{\frac {5}{2}} \sqrt {3 x^{2} + 5 x + 2}\, dx - \int 15 x^{\frac {7}{2}} \sqrt {3 x^{2} + 5 x + 2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-4*sqrt(x)*sqrt(3*x**2 + 5*x + 2), x) - Integral(19*x**(5/2)*sqrt(3*x**2 + 5*x + 2), x) - Integral(1
5*x**(7/2)*sqrt(3*x**2 + 5*x + 2), x)

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