Integrand size = 23, antiderivative size = 159 \[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 c}{a (c x)^{5/2} \sqrt {a x^2+b x^3}}-\frac {16 c^3 \sqrt {a x^2+b x^3}}{7 a^2 (c x)^{9/2}}+\frac {96 b c^2 \sqrt {a x^2+b x^3}}{35 a^3 (c x)^{7/2}}-\frac {128 b^2 c \sqrt {a x^2+b x^3}}{35 a^4 (c x)^{5/2}}+\frac {256 b^3 \sqrt {a x^2+b x^3}}{35 a^5 (c x)^{3/2}} \] Output:
2*c/a/(c*x)^(5/2)/(b*x^3+a*x^2)^(1/2)-16/7*c^3*(b*x^3+a*x^2)^(1/2)/a^2/(c* x)^(9/2)+96/35*b*c^2*(b*x^3+a*x^2)^(1/2)/a^3/(c*x)^(7/2)-128/35*b^2*c*(b*x ^3+a*x^2)^(1/2)/a^4/(c*x)^(5/2)+256/35*b^3*(b*x^3+a*x^2)^(1/2)/a^5/(c*x)^( 3/2)
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.43 \[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 c \left (-5 a^4+8 a^3 b x-16 a^2 b^2 x^2+64 a b^3 x^3+128 b^4 x^4\right )}{35 a^5 (c x)^{5/2} \sqrt {x^2 (a+b x)}} \] Input:
Integrate[1/((c*x)^(3/2)*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(2*c*(-5*a^4 + 8*a^3*b*x - 16*a^2*b^2*x^2 + 64*a*b^3*x^3 + 128*b^4*x^4))/( 35*a^5*(c*x)^(5/2)*Sqrt[x^2*(a + b*x)])
Time = 0.59 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1921, 1922, 1922, 1922, 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.
\(\displaystyle \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {8 c^2 \int \frac {1}{(c x)^{7/2} \sqrt {b x^3+a x^2}}dx}{a}+\frac {2 c}{a (c x)^{5/2} \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {8 c^2 \left (-\frac {6 b \int \frac {1}{(c x)^{5/2} \sqrt {b x^3+a x^2}}dx}{7 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{7 a (c x)^{9/2}}\right )}{a}+\frac {2 c}{a (c x)^{5/2} \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {8 c^2 \left (-\frac {6 b \left (-\frac {4 b \int \frac {1}{(c x)^{3/2} \sqrt {b x^3+a x^2}}dx}{5 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{5 a (c x)^{7/2}}\right )}{7 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{7 a (c x)^{9/2}}\right )}{a}+\frac {2 c}{a (c x)^{5/2} \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {8 c^2 \left (-\frac {6 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{\sqrt {c x} \sqrt {b x^3+a x^2}}dx}{3 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{3 a (c x)^{5/2}}\right )}{5 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{5 a (c x)^{7/2}}\right )}{7 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{7 a (c x)^{9/2}}\right )}{a}+\frac {2 c}{a (c x)^{5/2} \sqrt {a x^2+b x^3}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle \frac {8 c^2 \left (-\frac {6 b \left (-\frac {4 b \left (\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 (c x)^{3/2}}-\frac {2 c \sqrt {a x^2+b x^3}}{3 a (c x)^{5/2}}\right )}{5 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{5 a (c x)^{7/2}}\right )}{7 a c}-\frac {2 c \sqrt {a x^2+b x^3}}{7 a (c x)^{9/2}}\right )}{a}+\frac {2 c}{a (c x)^{5/2} \sqrt {a x^2+b x^3}}\) |
Input:
Int[1/((c*x)^(3/2)*(a*x^2 + b*x^3)^(3/2)),x]
Output:
(2*c)/(a*(c*x)^(5/2)*Sqrt[a*x^2 + b*x^3]) + (8*c^2*((-2*c*Sqrt[a*x^2 + b*x ^3])/(7*a*(c*x)^(9/2)) - (6*b*((-2*c*Sqrt[a*x^2 + b*x^3])/(5*a*(c*x)^(7/2) ) - (4*b*((-2*c*Sqrt[a*x^2 + b*x^3])/(3*a*(c*x)^(5/2)) + (4*b*Sqrt[a*x^2 + b*x^3])/(3*a^2*(c*x)^(3/2))))/(5*a*c)))/(7*a*c)))/a
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[((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] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1))) In t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n} , x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
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*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Time = 0.40 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.45
method | result | size |
gosper | \(-\frac {2 x \left (b x +a \right ) \left (-128 b^{4} x^{4}-64 a \,b^{3} x^{3}+16 a^{2} b^{2} x^{2}-8 a^{3} b x +5 a^{4}\right )}{35 a^{5} \left (c x \right )^{\frac {3}{2}} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(71\) |
orering | \(-\frac {2 x \left (b x +a \right ) \left (-128 b^{4} x^{4}-64 a \,b^{3} x^{3}+16 a^{2} b^{2} x^{2}-8 a^{3} b x +5 a^{4}\right )}{35 a^{5} \left (c x \right )^{\frac {3}{2}} \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\) | \(71\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-128 b^{4} x^{4}-64 a \,b^{3} x^{3}+16 a^{2} b^{2} x^{2}-8 a^{3} b x +5 a^{4}\right )}{35 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \sqrt {c x}\, c \,a^{5}}\) | \(73\) |
risch | \(-\frac {2 \left (b x +a \right ) \left (-93 b^{3} x^{3}+29 a \,b^{2} x^{2}-13 a^{2} b x +5 a^{3}\right )}{35 a^{5} x^{2} c \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {c x}}+\frac {2 b^{4} x^{2}}{a^{5} c \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {c x}}\) | \(94\) |
Input:
int(1/(c*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2/35*x*(b*x+a)*(-128*b^4*x^4-64*a*b^3*x^3+16*a^2*b^2*x^2-8*a^3*b*x+5*a^4) /a^5/(c*x)^(3/2)/(b*x^3+a*x^2)^(3/2)
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \, {\left (128 \, b^{4} x^{4} + 64 \, a b^{3} x^{3} - 16 \, a^{2} b^{2} x^{2} + 8 \, a^{3} b x - 5 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}} \sqrt {c x}}{35 \, {\left (a^{5} b c^{2} x^{6} + a^{6} c^{2} x^{5}\right )}} \] Input:
integrate(1/(c*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")
Output:
2/35*(128*b^4*x^4 + 64*a*b^3*x^3 - 16*a^2*b^2*x^2 + 8*a^3*b*x - 5*a^4)*sqr t(b*x^3 + a*x^2)*sqrt(c*x)/(a^5*b*c^2*x^6 + a^6*c^2*x^5)
\[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (c x\right )^{\frac {3}{2}} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(c*x)**(3/2)/(b*x**3+a*x**2)**(3/2),x)
Output:
Integral(1/((c*x)**(3/2)*(x**2*(a + b*x))**(3/2)), x)
\[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} \left (c x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*x^3 + a*x^2)^(3/2)*(c*x)^(3/2)), x)
Timed out. \[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(c*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x, algorithm="giac")
Output:
Timed out
Time = 9.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {-10\,a^4+16\,a^3\,b\,x-32\,a^2\,b^2\,x^2+128\,a\,b^3\,x^3+256\,b^4\,x^4}{35\,a^5\,c\,x^2\,\sqrt {c\,x}\,\sqrt {b\,x^3+a\,x^2}} \] Input:
int(1/((c*x)^(3/2)*(a*x^2 + b*x^3)^(3/2)),x)
Output:
(256*b^4*x^4 - 10*a^4 + 128*a*b^3*x^3 - 32*a^2*b^2*x^2 + 16*a^3*b*x)/(35*a ^5*c*x^2*(c*x)^(1/2)*(a*x^2 + b*x^3)^(1/2))
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(c x)^{3/2} \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2 \sqrt {c}\, \left (-128 \sqrt {b}\, \sqrt {b x +a}\, b^{3} x^{4}-5 \sqrt {x}\, a^{4}+8 \sqrt {x}\, a^{3} b x -16 \sqrt {x}\, a^{2} b^{2} x^{2}+64 \sqrt {x}\, a \,b^{3} x^{3}+128 \sqrt {x}\, b^{4} x^{4}\right )}{35 \sqrt {b x +a}\, a^{5} c^{2} x^{4}} \] Input:
int(1/(c*x)^(3/2)/(b*x^3+a*x^2)^(3/2),x)
Output:
(2*sqrt(c)*( - 128*sqrt(b)*sqrt(a + b*x)*b**3*x**4 - 5*sqrt(x)*a**4 + 8*sq rt(x)*a**3*b*x - 16*sqrt(x)*a**2*b**2*x**2 + 64*sqrt(x)*a*b**3*x**3 + 128* sqrt(x)*b**4*x**4))/(35*sqrt(a + b*x)*a**5*c**2*x**4)