Integrand size = 17, antiderivative size = 67 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=\frac {15}{4} x \sqrt {2+3 x^2}+\frac {5}{4} x \left (2+3 x^2\right )^{3/2}-\frac {1}{15} \left (2+3 x^2\right )^{5/2}+\frac {5}{2} \sqrt {3} \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ) \] Output:
15/4*x*(3*x^2+2)^(1/2)+5/4*x*(3*x^2+2)^(3/2)-1/15*(3*x^2+2)^(5/2)+5/2*arcs inh(1/2*x*6^(1/2))*3^(1/2)
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=-\frac {1}{60} \sqrt {2+3 x^2} \left (16-375 x+48 x^2-225 x^3+36 x^4\right )-\frac {5}{2} \sqrt {3} \log \left (-\sqrt {3} x+\sqrt {2+3 x^2}\right ) \] Input:
Integrate[(5 - x)*(2 + 3*x^2)^(3/2),x]
Output:
-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
Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {455, 211, 211, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (5-x) \left (3 x^2+2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle 5 \int \left (3 x^2+2\right )^{3/2}dx-\frac {1}{15} \left (3 x^2+2\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle 5 \left (\frac {3}{2} \int \sqrt {3 x^2+2}dx+\frac {1}{4} x \left (3 x^2+2\right )^{3/2}\right )-\frac {1}{15} \left (3 x^2+2\right )^{5/2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle 5 \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {3 x^2+2}}dx+\frac {1}{2} \sqrt {3 x^2+2} x\right )+\frac {1}{4} x \left (3 x^2+2\right )^{3/2}\right )-\frac {1}{15} \left (3 x^2+2\right )^{5/2}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle 5 \left (\frac {3}{2} \left (\frac {\text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {3}}+\frac {1}{2} \sqrt {3 x^2+2} x\right )+\frac {1}{4} x \left (3 x^2+2\right )^{3/2}\right )-\frac {1}{15} \left (3 x^2+2\right )^{5/2}\) |
Input:
Int[(5 - x)*(2 + 3*x^2)^(3/2),x]
Output:
-1/15*(2 + 3*x^2)^(5/2) + 5*((x*(2 + 3*x^2)^(3/2))/4 + (3*((x*Sqrt[2 + 3*x ^2])/2 + ArcSinh[Sqrt[3/2]*x]/Sqrt[3]))/2)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Time = 0.62 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67
method | result | size |
risch | \(-\frac {\left (36 x^{4}-225 x^{3}+48 x^{2}-375 x +16\right ) \sqrt {3 x^{2}+2}}{60}+\frac {5 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{2}\) | \(45\) |
default | \(\frac {15 x \sqrt {3 x^{2}+2}}{4}+\frac {5 x \left (3 x^{2}+2\right )^{\frac {3}{2}}}{4}-\frac {\left (3 x^{2}+2\right )^{\frac {5}{2}}}{15}+\frac {5 \,\operatorname {arcsinh}\left (\frac {\sqrt {6}\, x}{2}\right ) \sqrt {3}}{2}\) | \(49\) |
trager | \(\left (-\frac {3}{5} x^{4}+\frac {15}{4} x^{3}-\frac {4}{5} x^{2}+\frac {25}{4} x -\frac {4}{15}\right ) \sqrt {3 x^{2}+2}+\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \sqrt {3 x^{2}+2}+3 x \right )}{2}\) | \(61\) |
meijerg | \(\frac {5 \sqrt {3}\, \left (\frac {4 \sqrt {\pi }\, x \sqrt {2}\, \sqrt {3}\, \left (\frac {3 x^{2}}{8}+\frac {5}{8}\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{3}+\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )\right )}{2 \sqrt {\pi }}-\frac {\sqrt {2}\, \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (\frac {9}{2} x^{4}+6 x^{2}+2\right ) \sqrt {\frac {3 x^{2}}{2}+1}}{15}\right )}{2 \sqrt {\pi }}\) | \(93\) |
Input:
int((5-x)*(3*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/60*(36*x^4-225*x^3+48*x^2-375*x+16)*(3*x^2+2)^(1/2)+5/2*arcsinh(1/2*6^( 1/2)*x)*3^(1/2)
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=-\frac {1}{60} \, {\left (36 \, x^{4} - 225 \, x^{3} + 48 \, x^{2} - 375 \, x + 16\right )} \sqrt {3 \, x^{2} + 2} + \frac {5}{4} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) \] Input:
integrate((5-x)*(3*x^2+2)^(3/2),x, algorithm="fricas")
Output:
-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)
Time = 0.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.45 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=- \frac {3 x^{4} \sqrt {3 x^{2} + 2}}{5} + \frac {15 x^{3} \sqrt {3 x^{2} + 2}}{4} - \frac {4 x^{2} \sqrt {3 x^{2} + 2}}{5} + \frac {25 x \sqrt {3 x^{2} + 2}}{4} - \frac {4 \sqrt {3 x^{2} + 2}}{15} + \frac {5 \sqrt {3} \operatorname {asinh}{\left (\frac {\sqrt {6} x}{2} \right )}}{2} \] Input:
integrate((5-x)*(3*x**2+2)**(3/2),x)
Output:
-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)*as inh(sqrt(6)*x/2)/2
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.72 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=-\frac {1}{15} \, {\left (3 \, x^{2} + 2\right )}^{\frac {5}{2}} + \frac {5}{4} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} x + \frac {15}{4} \, \sqrt {3 \, x^{2} + 2} x + \frac {5}{2} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) \] Input:
integrate((5-x)*(3*x^2+2)^(3/2),x, algorithm="maxima")
Output:
-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)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=-\frac {1}{60} \, {\left (3 \, {\left ({\left (3 \, {\left (4 \, x - 25\right )} x + 16\right )} x - 125\right )} x + 16\right )} \sqrt {3 \, x^{2} + 2} - \frac {5}{2} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) \] Input:
integrate((5-x)*(3*x^2+2)^(3/2),x, algorithm="giac")
Output:
-1/60*(3*((3*(4*x - 25)*x + 16)*x - 125)*x + 16)*sqrt(3*x^2 + 2) - 5/2*sqr t(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))
Time = 5.98 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=\frac {5\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {6}\,x}{2}\right )}{2}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}\,\left (\frac {9\,x^4}{5}-\frac {45\,x^3}{4}+\frac {12\,x^2}{5}-\frac {75\,x}{4}+\frac {4}{5}\right )}{3} \] Input:
int(-(3*x^2 + 2)^(3/2)*(x - 5),x)
Output:
(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
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.25 \[ \int (5-x) \left (2+3 x^2\right )^{3/2} \, dx=-\frac {3 \sqrt {3 x^{2}+2}\, x^{4}}{5}+\frac {15 \sqrt {3 x^{2}+2}\, x^{3}}{4}-\frac {4 \sqrt {3 x^{2}+2}\, x^{2}}{5}+\frac {25 \sqrt {3 x^{2}+2}\, x}{4}-\frac {4 \sqrt {3 x^{2}+2}}{15}+\frac {5 \sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {3 x^{2}+2}+\sqrt {3}\, x}{\sqrt {2}}\right )}{2} \] Input:
int((5-x)*(3*x^2+2)^(3/2),x)
Output:
( - 36*sqrt(3*x**2 + 2)*x**4 + 225*sqrt(3*x**2 + 2)*x**3 - 48*sqrt(3*x**2 + 2)*x**2 + 375*sqrt(3*x**2 + 2)*x - 16*sqrt(3*x**2 + 2) + 150*sqrt(3)*log ((sqrt(3*x**2 + 2) + sqrt(3)*x)/sqrt(2)))/60