Integrand size = 13, antiderivative size = 68 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{3 a^2 x^{3/2}}-\frac {2 b^2}{a^3 \sqrt {x}}-\frac {2 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \] Output:
-2/5/a/x^(5/2)+2/3*b/a^2/x^(3/2)-2*b^2/a^3/x^(1/2)-2*b^(5/2)*arctan(b^(1/2 )*x^(1/2)/a^(1/2))/a^(7/2)
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=-\frac {2 \left (3 a^2-5 a b x+15 b^2 x^2\right )}{15 a^3 x^{5/2}}-\frac {2 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \] Input:
Integrate[1/(x^(7/2)*(a + b*x)),x]
Output:
(-2*(3*a^2 - 5*a*b*x + 15*b^2*x^2))/(15*a^3*x^(5/2)) - (2*b^(5/2)*ArcTan[( Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(7/2)
Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {61, 61, 61, 73, 218}
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.
\(\displaystyle \int \frac {1}{x^{7/2} (a+b x)} \, dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {b \int \frac {1}{x^{5/2} (a+b x)}dx}{a}-\frac {2}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {b \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {b \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {b \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {b \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\) |
Input:
Int[1/(x^(7/2)*(a + b*x)),x]
Output:
-2/(5*a*x^(5/2)) - (b*(-2/(3*a*x^(3/2)) - (b*(-2/(a*Sqrt[x]) - (2*Sqrt[b]* ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a))/a
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {2 \left (15 b^{2} x^{2}-5 a b x +3 a^{2}\right )}{15 a^{3} x^{\frac {5}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\) | \(53\) |
derivativedivides | \(-\frac {2}{5 a \,x^{\frac {5}{2}}}-\frac {2 b^{2}}{a^{3} \sqrt {x}}+\frac {2 b}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\) | \(54\) |
default | \(-\frac {2}{5 a \,x^{\frac {5}{2}}}-\frac {2 b^{2}}{a^{3} \sqrt {x}}+\frac {2 b}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\) | \(54\) |
Input:
int(1/x^(7/2)/(b*x+a),x,method=_RETURNVERBOSE)
Output:
-2/15*(15*b^2*x^2-5*a*b*x+3*a^2)/a^3/x^(5/2)-2*b^3/a^3/(a*b)^(1/2)*arctan( b*x^(1/2)/(a*b)^(1/2))
Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.04 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=\left [\frac {15 \, b^{2} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, -\frac {2 \, {\left (15 \, b^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) + {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \] Input:
integrate(1/x^(7/2)/(b*x+a),x, algorithm="fricas")
Output:
[1/15*(15*b^2*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) - 2*(15*b^2*x^2 - 5*a*b*x + 3*a^2)*sqrt(x))/(a^3*x^3), -2/15*(15*b^2* x^3*sqrt(b/a)*arctan(sqrt(x)*sqrt(b/a)) + (15*b^2*x^2 - 5*a*b*x + 3*a^2)*s qrt(x))/(a^3*x^3)]
Time = 5.36 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.85 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{7 b x^{\frac {7}{2}}} & \text {for}\: a = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} + \frac {2 b}{3 a^{2} x^{\frac {3}{2}}} - \frac {b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} + \frac {b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} - \frac {2 b^{2}}{a^{3} \sqrt {x}} & \text {otherwise} \end {cases} \] Input:
integrate(1/x**(7/2)/(b*x+a),x)
Output:
Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(7*b*x**(7/2)), Eq(a, 0 )), (-2/(5*a*x**(5/2)), Eq(b, 0)), (-2/(5*a*x**(5/2)) + 2*b/(3*a**2*x**(3/ 2)) - b**2*log(sqrt(x) - sqrt(-a/b))/(a**3*sqrt(-a/b)) + b**2*log(sqrt(x) + sqrt(-a/b))/(a**3*sqrt(-a/b)) - 2*b**2/(a**3*sqrt(x)), True))
Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=-\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \] Input:
integrate(1/x^(7/2)/(b*x+a),x, algorithm="maxima")
Output:
-2*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/15*(15*b^2*x^2 - 5* a*b*x + 3*a^2)/(a^3*x^(5/2))
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=-\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \] Input:
integrate(1/x^(7/2)/(b*x+a),x, algorithm="giac")
Output:
-2*b^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/15*(15*b^2*x^2 - 5* a*b*x + 3*a^2)/(a^3*x^(5/2))
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=-\frac {\frac {2}{5\,a}+\frac {2\,b^2\,x^2}{a^3}-\frac {2\,b\,x}{3\,a^2}}{x^{5/2}}-\frac {2\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \] Input:
int(1/(x^(7/2)*(a + b*x)),x)
Output:
- (2/(5*a) + (2*b^2*x^2)/a^3 - (2*b*x)/(3*a^2))/x^(5/2) - (2*b^(5/2)*atan( (b^(1/2)*x^(1/2))/a^(1/2)))/a^(7/2)
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^{7/2} (a+b x)} \, dx=\frac {-2 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{2}-\frac {2 a^{3}}{5}+\frac {2 a^{2} b x}{3}-2 a \,b^{2} x^{2}}{\sqrt {x}\, a^{4} x^{2}} \] Input:
int(1/x^(7/2)/(b*x+a),x)
Output:
(2*( - 15*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**2 *x**2 - 3*a**3 + 5*a**2*b*x - 15*a*b**2*x**2))/(15*sqrt(x)*a**4*x**2)