Integrand size = 15, antiderivative size = 149 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\frac {3 a^4 \sqrt {a x+b x^2}}{128 b^3}-\frac {a^3 x \sqrt {a x+b x^2}}{64 b^2}+\frac {a^2 x^2 \sqrt {a x+b x^2}}{80 b}+\frac {11}{40} a x^3 \sqrt {a x+b x^2}+\frac {1}{5} b x^4 \sqrt {a x+b x^2}-\frac {3 a^5 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{128 b^{7/2}} \] Output:
3/128*a^4*(b*x^2+a*x)^(1/2)/b^3-1/64*a^3*x*(b*x^2+a*x)^(1/2)/b^2+1/80*a^2* x^2*(b*x^2+a*x)^(1/2)/b+11/40*a*x^3*(b*x^2+a*x)^(1/2)+1/5*b*x^4*(b*x^2+a*x )^(1/2)-3/128*a^5*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/b^(7/2)
Time = 0.50 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (15 a^4-10 a^3 b x+8 a^2 b^2 x^2+176 a b^3 x^3+128 b^4 x^4\right )+\frac {30 a^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{640 b^{7/2}} \] Input:
Integrate[x*(a*x + b*x^2)^(3/2),x]
Output:
(Sqrt[x*(a + b*x)]*(Sqrt[b]*(15*a^4 - 10*a^3*b*x + 8*a^2*b^2*x^2 + 176*a*b ^3*x^3 + 128*b^4*x^4) + (30*a^5*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[ a + b*x])])/(Sqrt[x]*Sqrt[a + b*x])))/(640*b^(7/2))
Time = 0.38 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1160, 1087, 1087, 1091, 219}
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.
\(\displaystyle \int x \left (a x+b x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2}}{5 b}-\frac {a \int \left (b x^2+a x\right )^{3/2}dx}{2 b}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2}}{5 b}-\frac {a \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \int \sqrt {b x^2+a x}dx}{16 b}\right )}{2 b}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2}}{5 b}-\frac {a \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{\sqrt {b x^2+a x}}dx}{8 b}\right )}{16 b}\right )}{2 b}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2}}{5 b}-\frac {a \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{4 b}\right )}{16 b}\right )}{2 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (a x+b x^2\right )^{5/2}}{5 b}-\frac {a \left (\frac {(a+2 b x) \left (a x+b x^2\right )^{3/2}}{8 b}-\frac {3 a^2 \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{3/2}}\right )}{16 b}\right )}{2 b}\) |
Input:
Int[x*(a*x + b*x^2)^(3/2),x]
Output:
(a*x + b*x^2)^(5/2)/(5*b) - (a*(((a + 2*b*x)*(a*x + b*x^2)^(3/2))/(8*b) - (3*a^2*(((a + 2*b*x)*Sqrt[a*x + b*x^2])/(4*b) - (a^2*ArcTanh[(Sqrt[b]*x)/S qrt[a*x + b*x^2]])/(4*b^(3/2))))/(16*b)))/(2*b)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.56
method | result | size |
pseudoelliptic | \(-\frac {3 \left (\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{5}-\left (\sqrt {b}\, a^{4}-\frac {2 b^{\frac {3}{2}} a^{3} x}{3}+\frac {8 b^{\frac {5}{2}} a^{2} x^{2}}{15}+\frac {176 b^{\frac {7}{2}} a \,x^{3}}{15}+\frac {128 b^{\frac {9}{2}} x^{4}}{15}\right ) \sqrt {x \left (b x +a \right )}\right )}{128 b^{\frac {7}{2}}}\) | \(84\) |
risch | \(\frac {\left (128 b^{4} x^{4}+176 a \,b^{3} x^{3}+8 a^{2} b^{2} x^{2}-10 a^{3} b x +15 a^{4}\right ) x \left (b x +a \right )}{640 b^{3} \sqrt {x \left (b x +a \right )}}-\frac {3 a^{5} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{256 b^{\frac {7}{2}}}\) | \(95\) |
default | \(\frac {\left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{5 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{8 b}-\frac {3 a^{2} \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{16 b}\right )}{2 b}\) | \(110\) |
Input:
int(x*(b*x^2+a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-3/128/b^(7/2)*(arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*a^5-(b^(1/2)*a^4-2/3* b^(3/2)*a^3*x+8/15*b^(5/2)*a^2*x^2+176/15*b^(7/2)*a*x^3+128/15*b^(9/2)*x^4 )*(x*(b*x+a))^(1/2))
Time = 0.08 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.29 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\left [\frac {15 \, a^{5} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (128 \, b^{5} x^{4} + 176 \, a b^{4} x^{3} + 8 \, a^{2} b^{3} x^{2} - 10 \, a^{3} b^{2} x + 15 \, a^{4} b\right )} \sqrt {b x^{2} + a x}}{1280 \, b^{4}}, \frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (128 \, b^{5} x^{4} + 176 \, a b^{4} x^{3} + 8 \, a^{2} b^{3} x^{2} - 10 \, a^{3} b^{2} x + 15 \, a^{4} b\right )} \sqrt {b x^{2} + a x}}{640 \, b^{4}}\right ] \] Input:
integrate(x*(b*x^2+a*x)^(3/2),x, algorithm="fricas")
Output:
[1/1280*(15*a^5*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*( 128*b^5*x^4 + 176*a*b^4*x^3 + 8*a^2*b^3*x^2 - 10*a^3*b^2*x + 15*a^4*b)*sqr t(b*x^2 + a*x))/b^4, 1/640*(15*a^5*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt( -b)/(b*x + a)) + (128*b^5*x^4 + 176*a*b^4*x^3 + 8*a^2*b^3*x^2 - 10*a^3*b^2 *x + 15*a^4*b)*sqrt(b*x^2 + a*x))/b^4]
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\begin {cases} - \frac {3 a^{5} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{256 b^{3}} + \sqrt {a x + b x^{2}} \cdot \left (\frac {3 a^{4}}{128 b^{3}} - \frac {a^{3} x}{64 b^{2}} + \frac {a^{2} x^{2}}{80 b} + \frac {11 a x^{3}}{40} + \frac {b x^{4}}{5}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {7}{2}}}{7 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x*(b*x**2+a*x)**(3/2),x)
Output:
Piecewise((-3*a**5*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x )/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(256*b**3) + sqrt(a*x + b*x**2)*(3*a**4/(128*b**3) - a** 3*x/(64*b**2) + a**2*x**2/(80*b) + 11*a*x**3/40 + b*x**4/5), Ne(b, 0)), (2 *(a*x)**(7/2)/(7*a**2), Ne(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.83 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\frac {3 \, \sqrt {b x^{2} + a x} a^{3} x}{64 \, b^{2}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x}{8 \, b} - \frac {3 \, a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, \sqrt {b x^{2} + a x} a^{4}}{128 \, b^{3}} - \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}}{16 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}{5 \, b} \] Input:
integrate(x*(b*x^2+a*x)^(3/2),x, algorithm="maxima")
Output:
3/64*sqrt(b*x^2 + a*x)*a^3*x/b^2 - 1/8*(b*x^2 + a*x)^(3/2)*a*x/b - 3/256*a ^5*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 3/128*sqrt(b*x^2 + a*x)*a^4/b^3 - 1/16*(b*x^2 + a*x)^(3/2)*a^2/b^2 + 1/5*(b*x^2 + a*x)^(5/ 2)/b
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.62 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\frac {3 \, a^{5} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{256 \, b^{\frac {7}{2}}} + \frac {1}{640} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, b x + 11 \, a\right )} x + \frac {a^{2}}{b}\right )} x - \frac {5 \, a^{3}}{b^{2}}\right )} x + \frac {15 \, a^{4}}{b^{3}}\right )} \] Input:
integrate(x*(b*x^2+a*x)^(3/2),x, algorithm="giac")
Output:
3/256*a^5*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(7/2) + 1/640*sqrt(b*x^2 + a*x)*(2*(4*(2*(8*b*x + 11*a)*x + a^2/b)*x - 5*a^3/b^2 )*x + 15*a^4/b^3)
Time = 9.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\frac {{\left (b\,x^2+a\,x\right )}^{5/2}}{5\,b}-\frac {a\,\left (\frac {x\,{\left (b\,x^2+a\,x\right )}^{3/2}}{4}+\frac {a\,{\left (b\,x^2+a\,x\right )}^{3/2}}{8\,b}-\frac {3\,a^2\,\left (\frac {\sqrt {b\,x^2+a\,x}\,\left (a+2\,b\,x\right )}{4\,b}-\frac {a^2\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{8\,b^{3/2}}\right )}{16\,b}\right )}{2\,b} \] Input:
int(x*(a*x + b*x^2)^(3/2),x)
Output:
(a*x + b*x^2)^(5/2)/(5*b) - (a*((x*(a*x + b*x^2)^(3/2))/4 + (a*(a*x + b*x^ 2)^(3/2))/(8*b) - (3*a^2*(((a*x + b*x^2)^(1/2)*(a + 2*b*x))/(4*b) - (a^2*l og((a/2 + b*x)/b^(1/2) + (a*x + b*x^2)^(1/2)))/(8*b^(3/2))))/(16*b)))/(2*b )
Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.77 \[ \int x \left (a x+b x^2\right )^{3/2} \, dx=\frac {15 \sqrt {x}\, \sqrt {b x +a}\, a^{4} b -10 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{3} x^{2}+176 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{4} x^{3}+128 \sqrt {x}\, \sqrt {b x +a}\, b^{5} x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{5}}{640 b^{4}} \] Input:
int(x*(b*x^2+a*x)^(3/2),x)
Output:
(15*sqrt(x)*sqrt(a + b*x)*a**4*b - 10*sqrt(x)*sqrt(a + b*x)*a**3*b**2*x + 8*sqrt(x)*sqrt(a + b*x)*a**2*b**3*x**2 + 176*sqrt(x)*sqrt(a + b*x)*a*b**4* x**3 + 128*sqrt(x)*sqrt(a + b*x)*b**5*x**4 - 15*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**5)/(640*b**4)