3.14.72 \(\int (5-x) (2+3 x^2)^{3/2} \, dx\) [1372]

Optimal. Leaf size=67 \[ \frac {15}{4} x \sqrt {2+3 x^2}+\frac {5}{4} x \left (2+3 x^2\right )^{3/2}-\frac {1}{15} \left (2+3 x^2\right )^{5/2}+\frac {5}{2} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right ) \]

[Out]

5/4*x*(3*x^2+2)^(3/2)-1/15*(3*x^2+2)^(5/2)+5/2*arcsinh(1/2*x*6^(1/2))*3^(1/2)+15/4*x*(3*x^2+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {655, 201, 221} \begin {gather*} -\frac {1}{15} \left (3 x^2+2\right )^{5/2}+\frac {5}{4} x \left (3 x^2+2\right )^{3/2}+\frac {15}{4} x \sqrt {3 x^2+2}+\frac {5}{2} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - x)*(2 + 3*x^2)^(3/2),x]

[Out]

(15*x*Sqrt[2 + 3*x^2])/4 + (5*x*(2 + 3*x^2)^(3/2))/4 - (2 + 3*x^2)^(5/2)/15 + (5*Sqrt[3]*ArcSinh[Sqrt[3/2]*x])
/2

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

\begin {align*} \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx &=-\frac {1}{15} \left (2+3 x^2\right )^{5/2}+5 \int \left (2+3 x^2\right )^{3/2} \, dx\\ &=\frac {5}{4} x \left (2+3 x^2\right )^{3/2}-\frac {1}{15} \left (2+3 x^2\right )^{5/2}+\frac {15}{2} \int \sqrt {2+3 x^2} \, dx\\ &=\frac {15}{4} x \sqrt {2+3 x^2}+\frac {5}{4} x \left (2+3 x^2\right )^{3/2}-\frac {1}{15} \left (2+3 x^2\right )^{5/2}+\frac {15}{2} \int \frac {1}{\sqrt {2+3 x^2}} \, dx\\ &=\frac {15}{4} x \sqrt {2+3 x^2}+\frac {5}{4} x \left (2+3 x^2\right )^{3/2}-\frac {1}{15} \left (2+3 x^2\right )^{5/2}+\frac {5}{2} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 66, normalized size = 0.99 \begin {gather*} -\frac {1}{60} \sqrt {2+3 x^2} \left (16-375 x+48 x^2-225 x^3+36 x^4\right )-\frac {5}{2} \sqrt {3} \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Sqrt[2 + 3*x^2]*(16 - 375*x + 48*x^2 - 225*x^3 + 36*x^4)) - (5*Sqrt[3]*Log[-(Sqrt[3]*x) + Sqrt[2 + 3*x^
2]])/2

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Maple [A]
time = 0.61, size = 49, normalized size = 0.73

method	result	size
risch	\(-\frac {\left (36 x^{4}-225 x^{3}+48 x^{2}-375 x +16\right ) \sqrt {3 x^{2}+2}}{60}+\frac {5 \arcsinh \left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{2}\)	\(45\)
default	\(\frac {5 x \left (3 x^{2}+2\right )^{\frac {3}{2}}}{4}-\frac {\left (3 x^{2}+2\right )^{\frac {5}{2}}}{15}+\frac {5 \arcsinh \left (\frac {x \sqrt {6}}{2}\right ) \sqrt {3}}{2}+\frac {15 x \sqrt {3 x^{2}+2}}{4}\)	\(49\)
trager	\(\left (-\frac {3}{5} x^{4}+\frac {15}{4} x^{3}-\frac {4}{5} x^{2}+\frac {25}{4} x -\frac {4}{15}\right ) \sqrt {3 x^{2}+2}+\frac {5 \RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{2}\)	\(61\)
meijerg	\(\frac {5 \sqrt {3}\, \left (\frac {4 \sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}\, \left (\frac {3 x^{2}}{8}+\frac {5}{8}\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{3}+\sqrt {\pi }\, \arcsinh \left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )\right )}{2 \sqrt {\pi }}-\frac {\sqrt {2}\, \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (\frac {9}{2} x^{4}+6 x^{2}+2\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{15}\right )}{2 \sqrt {\pi }}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*x*(3*x^2+2)^(3/2)-1/15*(3*x^2+2)^(5/2)+5/2*arcsinh(1/2*x*6^(1/2))*3^(1/2)+15/4*x*(3*x^2+2)^(1/2)

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Maxima [A]
time = 0.48, size = 48, normalized size = 0.72 \begin {gather*} -\frac {1}{15} \, {\left (3 \, x^{2} + 2\right )}^{\frac {5}{2}} + \frac {5}{4} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} x + \frac {15}{4} \, \sqrt {3 \, x^{2} + 2} x + \frac {5}{2} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*x^2 + 2)^(5/2) + 5/4*(3*x^2 + 2)^(3/2)*x + 15/4*sqrt(3*x^2 + 2)*x + 5/2*sqrt(3)*arcsinh(1/2*sqrt(6)*x
)

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Fricas [A]
time = 2.94, size = 60, normalized size = 0.90 \begin {gather*} -\frac {1}{60} \, {\left (36 \, x^{4} - 225 \, x^{3} + 48 \, x^{2} - 375 \, x + 16\right )} \sqrt {3 \, x^{2} + 2} + \frac {5}{4} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(36*x^4 - 225*x^3 + 48*x^2 - 375*x + 16)*sqrt(3*x^2 + 2) + 5/4*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x -
3*x^2 - 1)

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Sympy [A]
time = 0.46, size = 97, normalized size = 1.45 \begin {gather*} - \frac {3 x^{4} \sqrt {3 x^{2} + 2}}{5} + \frac {15 x^{3} \sqrt {3 x^{2} + 2}}{4} - \frac {4 x^{2} \sqrt {3 x^{2} + 2}}{5} + \frac {25 x \sqrt {3 x^{2} + 2}}{4} - \frac {4 \sqrt {3 x^{2} + 2}}{15} + \frac {5 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x**4*sqrt(3*x**2 + 2)/5 + 15*x**3*sqrt(3*x**2 + 2)/4 - 4*x**2*sqrt(3*x**2 + 2)/5 + 25*x*sqrt(3*x**2 + 2)/4
- 4*sqrt(3*x**2 + 2)/15 + 5*sqrt(3)*asinh(sqrt(6)*x/2)/2

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Giac [A]
time = 0.98, size = 53, normalized size = 0.79 \begin {gather*} -\frac {1}{60} \, {\left (3 \, {\left ({\left (3 \, {\left (4 \, x - 25\right )} x + 16\right )} x - 125\right )} x + 16\right )} \sqrt {3 \, x^{2} + 2} - \frac {5}{2} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(3*((3*(4*x - 25)*x + 16)*x - 125)*x + 16)*sqrt(3*x^2 + 2) - 5/2*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2
))

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Mupad [B]
time = 0.04, size = 45, normalized size = 0.67 \begin {gather*} \frac {5\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right )}{2}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {9\,x^4}{5}-\frac {45\,x^3}{4}+\frac {12\,x^2}{5}-\frac {75\,x}{4}+\frac {4}{5}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^2 + 2)^(3/2)*(x - 5),x)

[Out]

(5*3^(1/2)*asinh((6^(1/2)*x)/2))/2 - (3^(1/2)*(x^2 + 2/3)^(1/2)*((12*x^2)/5 - (75*x)/4 - (45*x^3)/4 + (9*x^4)/
5 + 4/5))/3

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