Integrand size = 22, antiderivative size = 82 \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=-\frac {2 (b c-a d) x}{b^2 \sqrt {a x+b x^2}}+\frac {d \sqrt {a x+b x^2}}{b^2}+\frac {(2 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{5/2}} \] Output:
-2*(-a*d+b*c)*x/b^2/(b*x^2+a*x)^(1/2)+d*(b*x^2+a*x)^(1/2)/b^2+(-3*a*d+2*b* c)*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/b^(5/2)
Time = 0.06 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.26 \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\frac {x^{3/2} \left (\sqrt {b} \sqrt {x} (a+b x) (-2 b c+3 a d+b d x)+2 (2 b c-3 a d) (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{b^{5/2} (x (a+b x))^{3/2}} \] Input:
Integrate[(x^2*(c + d*x))/(a*x + b*x^2)^(3/2),x]
Output:
(x^(3/2)*(Sqrt[b]*Sqrt[x]*(a + b*x)*(-2*b*c + 3*a*d + b*d*x) + 2*(2*b*c - 3*a*d)*(a + b*x)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x] )]))/(b^(5/2)*(x*(a + b*x))^(3/2))
Time = 0.40 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1211, 1160, 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.
\(\displaystyle \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1211 |
\(\displaystyle \frac {\int \frac {b c-a d+b d x}{\sqrt {b x^2+a x}}dx}{b^2}-\frac {2 x (b c-a d)}{b^2 \sqrt {a x+b x^2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\frac {1}{2} (2 b c-3 a d) \int \frac {1}{\sqrt {b x^2+a x}}dx+d \sqrt {a x+b x^2}}{b^2}-\frac {2 x (b c-a d)}{b^2 \sqrt {a x+b x^2}}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {(2 b c-3 a d) \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}+d \sqrt {a x+b x^2}}{b^2}-\frac {2 x (b c-a d)}{b^2 \sqrt {a x+b x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right ) (2 b c-3 a d)}{\sqrt {b}}+d \sqrt {a x+b x^2}}{b^2}-\frac {2 x (b c-a d)}{b^2 \sqrt {a x+b x^2}}\) |
Input:
Int[(x^2*(c + d*x))/(a*x + b*x^2)^(3/2),x]
Output:
(-2*(b*c - a*d)*x)/(b^2*Sqrt[a*x + b*x^2]) + (d*Sqrt[a*x + b*x^2] + ((2*b* c - 3*a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/Sqrt[b])/b^2
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[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]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2)) Int[ExpandToSum[((2*c*d - b *e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) *(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 0.66 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (a d -b c \right ) x}{\sqrt {x \left (b x +a \right )}}+d \sqrt {x \left (b x +a \right )}-\frac {\left (3 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )}{\sqrt {b}}}{b^{2}}\) | \(68\) |
risch | \(\frac {d x \left (b x +a \right )}{b^{2} \sqrt {x \left (b x +a \right )}}-\frac {\frac {3 a d \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{\sqrt {b}}-2 \sqrt {b}\, c \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )-\frac {4 \left (a d -b c \right ) \sqrt {\left (x +\frac {a}{b}\right )^{2} b -a \left (x +\frac {a}{b}\right )}}{b \left (x +\frac {a}{b}\right )}}{2 b^{2}}\) | \(135\) |
default | \(c \left (-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}\right )+d \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a x}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )\) | \(217\) |
Input:
int(x^2*(d*x+c)/(b*x^2+a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/b^2*(2*(a*d-b*c)*x/(x*(b*x+a))^(1/2)+d*(x*(b*x+a))^(1/2)-(3*a*d-2*b*c)/b ^(1/2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2)))
Time = 0.10 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.50 \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (2 \, a b c - 3 \, a^{2} d + {\left (2 \, b^{2} c - 3 \, a b d\right )} x\right )} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (b^{2} d x - 2 \, b^{2} c + 3 \, a b d\right )} \sqrt {b x^{2} + a x}}{2 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {{\left (2 \, a b c - 3 \, a^{2} d + {\left (2 \, b^{2} c - 3 \, a b d\right )} x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (b^{2} d x - 2 \, b^{2} c + 3 \, a b d\right )} \sqrt {b x^{2} + a x}}{b^{4} x + a b^{3}}\right ] \] Input:
integrate(x^2*(d*x+c)/(b*x^2+a*x)^(3/2),x, algorithm="fricas")
Output:
[-1/2*((2*a*b*c - 3*a^2*d + (2*b^2*c - 3*a*b*d)*x)*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(b^2*d*x - 2*b^2*c + 3*a*b*d)*sqrt(b*x^2 + a*x))/(b^4*x + a*b^3), -((2*a*b*c - 3*a^2*d + (2*b^2*c - 3*a*b*d)*x)*sq rt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) - (b^2*d*x - 2*b^2*c + 3*a*b*d)*sqrt(b*x^2 + a*x))/(b^4*x + a*b^3)]
\[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\int \frac {x^{2} \left (c + d x\right )}{\left (x \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**2*(d*x+c)/(b*x**2+a*x)**(3/2),x)
Output:
Integral(x**2*(c + d*x)/(x*(a + b*x))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.40 \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\frac {d x^{2}}{\sqrt {b x^{2} + a x} b} - \frac {2 \, c x}{\sqrt {b x^{2} + a x} b} + \frac {3 \, a d x}{\sqrt {b x^{2} + a x} b^{2}} + \frac {c \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {3}{2}}} - \frac {3 \, a d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {5}{2}}} \] Input:
integrate(x^2*(d*x+c)/(b*x^2+a*x)^(3/2),x, algorithm="maxima")
Output:
d*x^2/(sqrt(b*x^2 + a*x)*b) - 2*c*x/(sqrt(b*x^2 + a*x)*b) + 3*a*d*x/(sqrt( b*x^2 + a*x)*b^2) + c*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) - 3/2*a*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2)
Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\frac {\sqrt {b x^{2} + a x} d}{b^{2}} - \frac {{\left (2 \, b c - 3 \, a d\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{2 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (a b^{\frac {3}{2}} c - a^{2} \sqrt {b} d\right )}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a\right )} b^{3}} \] Input:
integrate(x^2*(d*x+c)/(b*x^2+a*x)^(3/2),x, algorithm="giac")
Output:
sqrt(b*x^2 + a*x)*d/b^2 - 1/2*(2*b*c - 3*a*d)*log(abs(2*(sqrt(b)*x - sqrt( b*x^2 + a*x))*sqrt(b) + a))/b^(5/2) - 2*(a*b^(3/2)*c - a^2*sqrt(b)*d)/(((s qrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a)*b^3)
Timed out. \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\left (c+d\,x\right )}{{\left (b\,x^2+a\,x\right )}^{3/2}} \,d x \] Input:
int((x^2*(c + d*x))/(a*x + b*x^2)^(3/2),x)
Output:
int((x^2*(c + d*x))/(a*x + b*x^2)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.49 \[ \int \frac {x^2 (c+d x)}{\left (a x+b x^2\right )^{3/2}} \, dx=\frac {-12 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a d +8 \sqrt {b}\, \sqrt {b x +a}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) b c +9 \sqrt {b}\, \sqrt {b x +a}\, a d -8 \sqrt {b}\, \sqrt {b x +a}\, b c +12 \sqrt {x}\, a b d -8 \sqrt {x}\, b^{2} c +4 \sqrt {x}\, b^{2} d x}{4 \sqrt {b x +a}\, b^{3}} \] Input:
int(x^2*(d*x+c)/(b*x^2+a*x)^(3/2),x)
Output:
( - 12*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a) )*a*d + 8*sqrt(b)*sqrt(a + b*x)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt (a))*b*c + 9*sqrt(b)*sqrt(a + b*x)*a*d - 8*sqrt(b)*sqrt(a + b*x)*b*c + 12* sqrt(x)*a*b*d - 8*sqrt(x)*b**2*c + 4*sqrt(x)*b**2*d*x)/(4*sqrt(a + b*x)*b* *3)