Integrand size = 24, antiderivative size = 66 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 (b c-a d) x}{a b \sqrt {a x^2+b x^3}}-\frac {2 c \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{a^{3/2}} \] Output:
2*(-a*d+b*c)*x/a/b/(b*x^3+a*x^2)^(1/2)-2*c*arctanh((b*x^3+a*x^2)^(1/2)/a^( 1/2)/x)/a^(3/2)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 x \left (\sqrt {a} (-b c+a d)+b c \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{a^{3/2} b \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(x^2*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(-2*x*(Sqrt[a]*(-(b*c) + a*d) + b*c*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sq rt[a]]))/(a^(3/2)*b*Sqrt[x^2*(a + b*x)])
Time = 0.36 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1943, 1914, 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.
\(\displaystyle \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1943 |
\(\displaystyle \frac {c \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{a}+\frac {2 x (b c-a d)}{a b \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {2 x (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {2 c \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 x (b c-a d)}{a b \sqrt {a x^2+b x^3}}-\frac {2 c \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}\) |
Input:
Int[(x^2*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*(b*c - a*d)*x)/(a*b*Sqrt[a*x^2 + b*x^3]) - (2*c*ArcTanh[(Sqrt[a]*x)/Sqr t[a*x^2 + b*x^3]])/a^(3/2)
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[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1))) Int[(e*x)^(m - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, n}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 ] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
Time = 0.45 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88
method | result | size |
pseudoelliptic | \(\frac {\frac {2 x^{2} \left (\frac {3 d x}{5}+c \right ) b^{3}}{3}-\frac {8 x \left (\frac {3 d x}{10}+c \right ) a \,b^{2}}{3}-\frac {16 \left (-\frac {3 d x}{5}+c \right ) a^{2} b}{3}+\frac {32 a^{3} d}{5}}{b^{4} \sqrt {b x +a}}\) | \(58\) |
default | \(-\frac {2 x^{3} \left (b x +a \right ) \left (b c \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \sqrt {b x +a}+a^{\frac {5}{2}} d -a^{\frac {3}{2}} b c \right )}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} b \,a^{\frac {5}{2}}}\) | \(66\) |
Input:
int(x^2*(d*x+c)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
32/5/(b*x+a)^(1/2)*(5/48*x^2*(3/5*d*x+c)*b^3-5/12*x*(3/10*d*x+c)*a*b^2-5/6 *(-3/5*d*x+c)*a^2*b+a^3*d)/b^4
Time = 0.14 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.98 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {{\left (b^{2} c x^{2} + a b c x\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (a b c - a^{2} d\right )}}{a^{2} b^{2} x^{2} + a^{3} b x}, \frac {2 \, {\left ({\left (b^{2} c x^{2} + a b c x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (a b c - a^{2} d\right )}\right )}}{a^{2} b^{2} x^{2} + a^{3} b x}\right ] \] Input:
integrate(x^2*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
[((b^2*c*x^2 + a*b*c*x)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2) *sqrt(a))/x^2) + 2*sqrt(b*x^3 + a*x^2)*(a*b*c - a^2*d))/(a^2*b^2*x^2 + a^3 *b*x), 2*((b^2*c*x^2 + a*b*c*x)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(- a)/(b*x^2 + a*x)) + sqrt(b*x^3 + a*x^2)*(a*b*c - a^2*d))/(a^2*b^2*x^2 + a^ 3*b*x)]
\[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^{2} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**2*(d*x+c)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(x**2*(c + d*x)/(x**2*(a + b*x))**(3/2), x)
\[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )} x^{2}}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^2*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x + c)*x^2/(b*x^3 + a*x^2)^(3/2), x)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.61 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, c \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (\sqrt {a} b c \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + \sqrt {-a} b c - \sqrt {-a} a d\right )} \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{\frac {3}{2}} b} + \frac {2 \, {\left (b c - a d\right )}}{\sqrt {b x + a} a b \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^2*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
2*c*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a*sgn(x)) - 2*(sqrt(a)*b*c*ar ctan(sqrt(a)/sqrt(-a)) + sqrt(-a)*b*c - sqrt(-a)*a*d)*sgn(x)/(sqrt(-a)*a^( 3/2)*b) + 2*(b*c - a*d)/(sqrt(b*x + a)*a*b*sgn(x))
Timed out. \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {x^2\,\left (c+d\,x\right )}{{\left (b\,x^3+a\,x^2\right )}^{3/2}} \,d x \] Input:
int((x^2*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x)
Output:
int((x^2*(c + d*x))/(a*x^2 + b*x^3)^(3/2), x)
Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b c -\sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b c -2 a^{2} d +2 a b c}{\sqrt {b x +a}\, a^{2} b} \] Input:
int(x^2*(d*x+c)/(b*x^3+a*x^2)^(3/2),x)
Output:
(sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b*c - sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b*c - 2*a**2*d + 2*a*b*c)/(sqrt(a + b*x )*a**2*b)