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

Optimal. Leaf size=100 \[ \frac {1}{24} \sqrt {3 x^2+5 x+2} (73-6 x)-\frac {311 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{48 \sqrt {3}}+\frac {13}{8} \sqrt {5} \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {814, 843, 621, 206, 724} \begin {gather*} \frac {1}{24} \sqrt {3 x^2+5 x+2} (73-6 x)-\frac {311 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{48 \sqrt {3}}+\frac {13}{8} \sqrt {5} \tanh ^{-1}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((73 - 6*x)*Sqrt[2 + 5*x + 3*x^2])/24 - (311*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(48*Sqrt[3]
) + (13*Sqrt[5]*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/8

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

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

Rule 724

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

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 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} \, dx &=\frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {1}{48} \int \frac {543+622 x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311}{48} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx+\frac {65}{8} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311}{24} \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )-\frac {65}{4} \operatorname {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=\frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311 \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{48 \sqrt {3}}+\frac {13}{8} \sqrt {5} \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 93, normalized size = 0.93 \begin {gather*} \frac {1}{144} \left (-6 \sqrt {3 x^2+5 x+2} (6 x-73)-234 \sqrt {5} \tanh ^{-1}\left (\frac {-8 x-7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )-311 \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {9 x^2+15 x+6}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(-73 + 6*x)*Sqrt[2 + 5*x + 3*x^2] - 234*Sqrt[5]*ArcTanh[(-7 - 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])] - 31
1*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 15*x + 9*x^2])])/144

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IntegrateAlgebraic [A]  time = 0.48, size = 94, normalized size = 0.94 \begin {gather*} \frac {1}{24} \sqrt {3 x^2+5 x+2} (73-6 x)-\frac {311 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {3} (x+1)}\right )}{24 \sqrt {3}}+\frac {13}{4} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {5} (x+1)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((73 - 6*x)*Sqrt[2 + 5*x + 3*x^2])/24 - (311*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[3]*(1 + x))])/(24*Sqrt[3]) +
(13*Sqrt[5]*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[5]*(1 + x))])/4

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fricas [A]  time = 0.41, size = 109, normalized size = 1.09 \begin {gather*} -\frac {1}{24} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 73\right )} + \frac {311}{288} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + \frac {13}{16} \, \sqrt {5} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) \end {gather*}

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),x, algorithm="fricas")

[Out]

-1/24*sqrt(3*x^2 + 5*x + 2)*(6*x - 73) + 311/288*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x
^2 + 120*x + 49) + 13/16*sqrt(5)*log((4*sqrt(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2
+ 12*x + 9))

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giac [A]  time = 0.29, size = 126, normalized size = 1.26 \begin {gather*} -\frac {1}{24} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 73\right )} + \frac {13}{8} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {311}{144} \, \sqrt {3} \log \left ({\left | -6 \, \sqrt {3} x - 5 \, \sqrt {3} + 6 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}\right ) \end {gather*}

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),x, algorithm="giac")

[Out]

-1/24*sqrt(3*x^2 + 5*x + 2)*(6*x - 73) + 13/8*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*
x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) + 311/144*sqrt(3)*log(abs
(-6*sqrt(3)*x - 5*sqrt(3) + 6*sqrt(3*x^2 + 5*x + 2)))

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maple [A]  time = 0.05, size = 127, normalized size = 1.27 \begin {gather*} -\frac {13 \sqrt {5}\, \arctanh \left (\frac {2 \left (-4 x -\frac {7}{2}\right ) \sqrt {5}}{5 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}\right )}{8}-\frac {13 \sqrt {3}\, \ln \left (\frac {\left (3 x +\frac {5}{2}\right ) \sqrt {3}}{3}+\sqrt {-4 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}\right )}{6}+\frac {\sqrt {3}\, \ln \left (\frac {\left (3 x +\frac {5}{2}\right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right )}{144}-\frac {\left (6 x +5\right ) \sqrt {3 x^{2}+5 x +2}}{24}+\frac {13 \sqrt {-16 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}{8} \end {gather*}

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),x)

[Out]

-1/24*(6*x+5)*(3*x^2+5*x+2)^(1/2)+1/144*3^(1/2)*ln(1/3*(3*x+5/2)*3^(1/2)+(3*x^2+5*x+2)^(1/2))+13/8*(12*(x+3/2)
^2-16*x-19)^(1/2)-13/6*ln(1/3*(3*x+5/2)*3^(1/2)+(3*(x+3/2)^2-4*x-19/4)^(1/2))*3^(1/2)-13/8*5^(1/2)*arctanh(2/5
*(-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))

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maxima [A]  time = 1.11, size = 99, normalized size = 0.99 \begin {gather*} -\frac {1}{4} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x - \frac {311}{144} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {13}{8} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {73}{24} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \end {gather*}

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),x, algorithm="maxima")

[Out]

-1/4*sqrt(3*x^2 + 5*x + 2)*x - 311/144*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 13/8*sqrt(5)*l
og(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) + 73/24*sqrt(3*x^2 + 5*x + 2)

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

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),x)

[Out]

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

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

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),x)

[Out]

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

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