3.2580 \(\int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx\)

Optimal. Leaf size=141 \[ -\frac {\sqrt {3 x^2+5 x+2} (x+21)}{3 \sqrt {2 x+3}}-\frac {161 \sqrt {-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {3 x^2+5 x+2}}+\frac {121 \sqrt {-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {3 x^2+5 x+2}} \]

[Out]

121/18*EllipticE(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)-161/18*El
lipticF(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)-1/3*(21+x)*(3*x^2+
5*x+2)^(1/2)/(3+2*x)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {812, 843, 718, 424, 419} \[ -\frac {\sqrt {3 x^2+5 x+2} (x+21)}{3 \sqrt {2 x+3}}-\frac {161 \sqrt {-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {3 x^2+5 x+2}}+\frac {121 \sqrt {-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((21 + x)*Sqrt[2 + 5*x + 3*x^2])/(3*Sqrt[3 + 2*x]) + (121*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqr
t[1 + x]], -2/3])/(6*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (161*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqr
t[1 + x]], -2/3])/(6*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 812

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

Rule 843

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

Rubi steps

\begin {align*} \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx &=-\frac {(21+x) \sqrt {2+5 x+3 x^2}}{3 \sqrt {3+2 x}}-\frac {1}{6} \int \frac {-101-121 x}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {(21+x) \sqrt {2+5 x+3 x^2}}{3 \sqrt {3+2 x}}+\frac {121}{12} \int \frac {\sqrt {3+2 x}}{\sqrt {2+5 x+3 x^2}} \, dx-\frac {161}{12} \int \frac {1}{\sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {(21+x) \sqrt {2+5 x+3 x^2}}{3 \sqrt {3+2 x}}+\frac {\left (121 \sqrt {-2-5 x-3 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 x^2}{3}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {6+6 x}}{\sqrt {2}}\right )}{6 \sqrt {3} \sqrt {2+5 x+3 x^2}}-\frac {\left (161 \sqrt {-2-5 x-3 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {2 x^2}{3}}} \, dx,x,\frac {\sqrt {6+6 x}}{\sqrt {2}}\right )}{6 \sqrt {3} \sqrt {2+5 x+3 x^2}}\\ &=-\frac {(21+x) \sqrt {2+5 x+3 x^2}}{3 \sqrt {3+2 x}}+\frac {121 \sqrt {-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {2+5 x+3 x^2}}-\frac {161 \sqrt {-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 185, normalized size = 1.31 \[ \frac {10 \left (-9 x^3+159 x^2+284 x+116\right )-122 \sqrt {5} \sqrt {\frac {x+1}{2 x+3}} \sqrt {\frac {3 x+2}{2 x+3}} (2 x+3)^{3/2} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {2 x+3}}\right )|\frac {3}{5}\right )+605 \sqrt {5} \sqrt {\frac {x+1}{2 x+3}} \sqrt {\frac {3 x+2}{2 x+3}} (2 x+3)^{3/2} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {2 x+3}}\right )|\frac {3}{5}\right )}{90 \sqrt {2 x+3} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(116 + 284*x + 159*x^2 - 9*x^3) + 605*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(3/2)*Sqrt[(2 + 3*x)/(3 +
2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] - 122*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(3/2)*Sq
rt[(2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(90*Sqrt[3 + 2*x]*Sqrt[2 + 5*x + 3*x^
2])

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {2 \, x + 3} {\left (x - 5\right )}}{4 \, x^{2} + 12 \, x + 9}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3)*(x - 5)/(4*x^2 + 12*x + 9), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 136, normalized size = 0.96 \[ -\frac {\sqrt {3 x^{2}+5 x +2}\, \sqrt {2 x +3}\, \left (180 x^{3}+4080 x^{2}+6420 x +121 \sqrt {2 x +3}\, \sqrt {15}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, \EllipticE \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+40 \sqrt {2 x +3}\, \sqrt {15}\, \sqrt {-2 x -2}\, \sqrt {-30 x -20}\, \EllipticF \left (\frac {\sqrt {30 x +45}}{5}, \frac {\sqrt {15}}{3}\right )+2520\right )}{180 \left (6 x^{3}+19 x^{2}+19 x +6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*(3*x^2+5*x+2)^(1/2)*(2*x+3)^(1/2)*(40*(2*x+3)^(1/2)*15^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)*EllipticF(
1/5*(30*x+45)^(1/2),1/3*15^(1/2))+121*(2*x+3)^(1/2)*15^(1/2)*(-2*x-2)^(1/2)*(-30*x-20)^(1/2)*EllipticE(1/5*(30
*x+45)^(1/2),1/3*15^(1/2))+180*x^3+4080*x^2+6420*x+2520)/(6*x^3+19*x^2+19*x+6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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