Integrand size = 21, antiderivative size = 116 \[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}+\frac {12 b \sqrt {a x^2+b x^3}}{35 a^2 x^{7/2}}-\frac {16 b^2 \sqrt {a x^2+b x^3}}{35 a^3 x^{5/2}}+\frac {32 b^3 \sqrt {a x^2+b x^3}}{35 a^4 x^{3/2}} \] Output:
-2/7*(b*x^3+a*x^2)^(1/2)/a/x^(9/2)+12/35*b*(b*x^3+a*x^2)^(1/2)/a^2/x^(7/2) -16/35*b^2*(b*x^3+a*x^2)^(1/2)/a^3/x^(5/2)+32/35*b^3*(b*x^3+a*x^2)^(1/2)/a ^4/x^(3/2)
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {x^2 (a+b x)} \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right )}{35 a^4 x^{9/2}} \] Input:
Integrate[1/(x^(7/2)*Sqrt[a*x^2 + b*x^3]),x]
Output:
(2*Sqrt[x^2*(a + b*x)]*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3))/(3 5*a^4*x^(9/2))
Time = 0.45 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {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.
\(\displaystyle \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {6 b \int \frac {1}{x^{5/2} \sqrt {b x^3+a x^2}}dx}{7 a}-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \int \frac {1}{x^{3/2} \sqrt {b x^3+a x^2}}dx}{5 a}-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}\right )}{7 a}-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{\sqrt {x} \sqrt {b x^3+a x^2}}dx}{3 a}-\frac {2 \sqrt {a x^2+b x^3}}{3 a x^{5/2}}\right )}{5 a}-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}\right )}{7 a}-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}\) |
\(\Big \downarrow \) 1920 |
\(\displaystyle -\frac {6 b \left (-\frac {4 b \left (\frac {4 b \sqrt {a x^2+b x^3}}{3 a^2 x^{3/2}}-\frac {2 \sqrt {a x^2+b x^3}}{3 a x^{5/2}}\right )}{5 a}-\frac {2 \sqrt {a x^2+b x^3}}{5 a x^{7/2}}\right )}{7 a}-\frac {2 \sqrt {a x^2+b x^3}}{7 a x^{9/2}}\) |
Input:
Int[1/(x^(7/2)*Sqrt[a*x^2 + b*x^3]),x]
Output:
(-2*Sqrt[a*x^2 + b*x^3])/(7*a*x^(9/2)) - (6*b*((-2*Sqrt[a*x^2 + b*x^3])/(5 *a*x^(7/2)) - (4*b*((-2*Sqrt[a*x^2 + b*x^3])/(3*a*x^(5/2)) + (4*b*Sqrt[a*x ^2 + b*x^3])/(3*a^2*x^(3/2))))/(5*a)))/(7*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*(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 = 0.41 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47
method | result | size |
risch | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 \sqrt {x^{2} \left (b x +a \right )}\, x^{\frac {5}{2}} a^{4}}\) | \(55\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{\frac {5}{2}} a^{4} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(57\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{\frac {5}{2}} a^{4} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(57\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{\frac {5}{2}} a^{4} \sqrt {b \,x^{3}+a \,x^{2}}}\) | \(57\) |
Input:
int(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/35/(x^2*(b*x+a))^(1/2)/x^(5/2)*(b*x+a)*(-16*b^3*x^3+8*a*b^2*x^2-6*a^2*b *x+5*a^3)/a^4
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{35 \, a^{4} x^{\frac {9}{2}}} \] Input:
integrate(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x^3 + a*x^2)/(a ^4*x^(9/2))
\[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{\frac {7}{2}} \sqrt {x^{2} \left (a + b x\right )}}\, dx \] Input:
integrate(1/x**(7/2)/(b*x**3+a*x**2)**(1/2),x)
Output:
Integral(1/(x**(7/2)*sqrt(x**2*(a + b*x))), x)
\[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}} x^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(b*x^3 + a*x^2)*x^(7/2)), x)
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\frac {64 \, {\left (35 \, {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 21 \, a {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 7 \, a^{2} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{3}\right )} b^{\frac {7}{2}}}{35 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{7} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
64/35*(35*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^6 - 21*a*(sqrt(b)*sqrt(x) - sq rt(b*x + a))^4 + 7*a^2*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^3)*b^(7/2)/ (((sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a)^7*sgn(x))
Timed out. \[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^{7/2}\,\sqrt {b\,x^3+a\,x^2}} \,d x \] Input:
int(1/(x^(7/2)*(a*x^2 + b*x^3)^(1/2)),x)
Output:
int(1/(x^(7/2)*(a*x^2 + b*x^3)^(1/2)), x)
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x^{7/2} \sqrt {a x^2+b x^3}} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{3}}{7}+\frac {12 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b x}{35}-\frac {16 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x^{2}}{35}+\frac {32 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{3}}{35}-\frac {32 \sqrt {b}\, b^{3} x^{4}}{35}}{a^{4} x^{4}} \] Input:
int(1/x^(7/2)/(b*x^3+a*x^2)^(1/2),x)
Output:
(2*( - 5*sqrt(x)*sqrt(a + b*x)*a**3 + 6*sqrt(x)*sqrt(a + b*x)*a**2*b*x - 8 *sqrt(x)*sqrt(a + b*x)*a*b**2*x**2 + 16*sqrt(x)*sqrt(a + b*x)*b**3*x**3 - 16*sqrt(b)*b**3*x**4))/(35*a**4*x**4)