Integrand size = 28, antiderivative size = 66 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 c e \sqrt {e x}}{a \sqrt {a x^2+b x^3}}-\frac {2 (2 b c-a d) (e x)^{3/2}}{a^2 \sqrt {a x^2+b x^3}} \] Output:
-2*c*e*(e*x)^(1/2)/a/(b*x^3+a*x^2)^(1/2)-2*(-a*d+2*b*c)*(e*x)^(3/2)/a^2/(b *x^3+a*x^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.61 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 e \sqrt {e x} (-a c-2 b c x+a d x)}{a^2 \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(2*e*Sqrt[e*x]*(-(a*c) - 2*b*c*x + a*d*x))/(a^2*Sqrt[x^2*(a + b*x)])
Time = 0.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1944, 1920}
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.
\(\displaystyle \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle -\frac {(2 b c-a d) \int \frac {(e x)^{5/2}}{\left (b x^3+a x^2\right )^{3/2}}dx}{a e}-\frac {2 c e \sqrt {e x}}{a \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle -\frac {2 (e x)^{3/2} (2 b c-a d)}{a^2 \sqrt {a x^2+b x^3}}-\frac {2 c e \sqrt {e x}}{a \sqrt {a x^2+b x^3}}\) |
Input:
Int[((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x]
Output:
(-2*c*e*Sqrt[e*x])/(a*Sqrt[a*x^2 + b*x^3]) - (2*(2*b*c - a*d)*(e*x)^(3/2)) /(a^2*Sqrt[a*x^2 + b*x^3])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[ n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 0.41 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.67
method | result | size |
gosper | \(-\frac {2 x \left (b x +a \right ) \left (-a d x +2 c b x +a c \right ) \left (e x \right )^{\frac {3}{2}}}{a^{2} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(44\) |
orering | \(-\frac {2 x \left (b x +a \right ) \left (-a d x +2 c b x +a c \right ) \left (e x \right )^{\frac {3}{2}}}{a^{2} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(44\) |
default | \(-\frac {2 x^{2} \left (b x +a \right ) \left (-a d x +2 c b x +a c \right ) \sqrt {e x}\, e}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{2}}\) | \(47\) |
risch | \(-\frac {2 c \left (b x +a \right ) e^{2} x}{a^{2} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {2 \left (a d -b c \right ) x^{2} e^{2}}{a^{2} \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\) | \(68\) |
Input:
int((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2*x*(b*x+a)*(-a*d*x+2*b*c*x+a*c)*(e*x)^(3/2)/a^2/(b*x^3+a*x^2)^(3/2)
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {b x^{3} + a x^{2}} {\left (a c e + {\left (2 \, b c - a d\right )} e x\right )} \sqrt {e x}}{a^{2} b x^{3} + a^{3} x^{2}} \] Input:
integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
-2*sqrt(b*x^3 + a*x^2)*(a*c*e + (2*b*c - a*d)*e*x)*sqrt(e*x)/(a^2*b*x^3 + a^3*x^2)
\[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {\left (e x\right )^{\frac {3}{2}} \left (c + d x\right )}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((e*x)**(3/2)*(d*x+c)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral((e*x)**(3/2)*(c + d*x)/(x**2*(a + b*x))**(3/2), x)
\[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate((d*x + c)*(e*x)^(3/2)/(b*x^3 + a*x^2)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.68 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-2 \, e {\left (\frac {2 \, \sqrt {b e} c e^{3}}{{\left (a e^{2} - {\left (\sqrt {b e} \sqrt {e x} - \sqrt {b e^{2} x + a e^{2}}\right )}^{2}\right )} a {\left | e \right |} \mathrm {sgn}\left (x\right )} + \frac {{\left (b c e^{2} - a d e^{2}\right )} \sqrt {e x}}{\sqrt {b e^{2} x + a e^{2}} a^{2} {\left | e \right |} \mathrm {sgn}\left (x\right )}\right )} \] Input:
integrate((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
-2*e*(2*sqrt(b*e)*c*e^3/((a*e^2 - (sqrt(b*e)*sqrt(e*x) - sqrt(b*e^2*x + a* e^2))^2)*a*abs(e)*sgn(x)) + (b*c*e^2 - a*d*e^2)*sqrt(e*x)/(sqrt(b*e^2*x + a*e^2)*a^2*abs(e)*sgn(x)))
Time = 9.73 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {2\,c\,e\,\sqrt {e\,x}}{a\,b}-\frac {2\,e\,x\,\sqrt {e\,x}\,\left (a\,d-2\,b\,c\right )}{a^2\,b}\right )}{x^3+\frac {a\,x^2}{b}} \] Input:
int(((e*x)^(3/2)*(c + d*x))/(a*x^2 + b*x^3)^(3/2),x)
Output:
-((a*x^2 + b*x^3)^(1/2)*((2*c*e*(e*x)^(1/2))/(a*b) - (2*e*x*(e*x)^(1/2)*(a *d - 2*b*c))/(a^2*b)))/(x^3 + (a*x^2)/b)
Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08 \[ \int \frac {(e x)^{3/2} (c+d x)}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {e}\, e \left (\sqrt {b}\, \sqrt {b x +a}\, a d x -2 \sqrt {b}\, \sqrt {b x +a}\, b c x -\sqrt {x}\, a b c +\sqrt {x}\, a b d x -2 \sqrt {x}\, b^{2} c x \right )}{\sqrt {b x +a}\, a^{2} b x} \] Input:
int((e*x)^(3/2)*(d*x+c)/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*sqrt(e)*e*(sqrt(b)*sqrt(a + b*x)*a*d*x - 2*sqrt(b)*sqrt(a + b*x)*b*c*x - sqrt(x)*a*b*c + sqrt(x)*a*b*d*x - 2*sqrt(x)*b**2*c*x))/(sqrt(a + b*x)*a* *2*b*x)