Integrand size = 15, antiderivative size = 91 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\frac {\frac {15 b^2}{4 a^3}-\frac {1}{2 a x^2}+\frac {5 b}{4 a^2 x}}{\sqrt {a+b x}}+\frac {15 b^2 \log \left (\frac {-\sqrt {a}+\sqrt {a+b x}}{\sqrt {a}+\sqrt {a+b x}}\right )}{8 a^{5/2}} \]
(-1/2/a/x^2+5/4*b/a^2/x+15/4*b^2/a^3)/(b*x+a)^(1/2)+15/8*b^2/a^(5/2)*ln((( b*x+a)^(1/2)-a^(1/2))/((b*x+a)^(1/2)+a^(1/2)))
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=-\frac {(a+b x) \left (\sqrt {a} \left (2 a^2-5 a b x-15 b^2 x^2\right )+15 b^2 x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 a^{7/2} x^2 \sqrt {(a+b x)^3}} \]
-1/4*((a + b*x)*(Sqrt[a]*(2*a^2 - 5*a*b*x - 15*b^2*x^2) + 15*b^2*x^2*Sqrt[ a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(7/2)*x^2*Sqrt[(a + b*x)^3])
Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2008, 52, 52, 61, 73, 221}
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.
\(\displaystyle \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx\) |
\(\Big \downarrow \) 2008 |
\(\displaystyle \frac {(a+b x)^{3/2} \int \frac {1}{x^3 (a+b x)^{3/2}}dx}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(a+b x)^{3/2} \left (-\frac {5 b \int \frac {1}{x^2 (a+b x)^{3/2}}dx}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\right )}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(a+b x)^{3/2} \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x (a+b x)^{3/2}}dx}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\right )}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(a+b x)^{3/2} \left (-\frac {5 b \left (-\frac {3 b \left (\frac {\int \frac {1}{x \sqrt {a+b x}}dx}{a}+\frac {2}{a \sqrt {a+b x}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\right )}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x)^{3/2} \left (-\frac {5 b \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a b}+\frac {2}{a \sqrt {a+b x}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\right )}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b x)^{3/2} \left (-\frac {5 b \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\right )}{\sqrt {(a+b x)^3}}\) |
((a + b*x)^(3/2)*(-1/2*1/(a*x^2*Sqrt[a + b*x]) - (5*b*(-(1/(a*x*Sqrt[a + b *x])) - (3*b*(2/(a*Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^( 3/2)))/(2*a)))/(4*a)))/Sqrt[(a + b*x)^3]
3.2.50.3.1 Defintions of rubi rules used
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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p) Int[u*(a + b*x)^(Expon[Px, x]*p), x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; !IntegerQ[p] && PolyQ[Px, x ] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\frac {\left (b x +a \right ) \left (15 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-5 a^{\frac {3}{2}} b x -15 b^{2} x^{2} \sqrt {a}+2 a^{\frac {5}{2}}\right )}{4 \sqrt {\left (b x +a \right )^{3}}\, a^{\frac {7}{2}} x^{2}}\) | \(74\) |
risch | \(-\frac {\left (b x +a \right )^{2} \left (-7 b x +2 a \right )}{4 a^{3} x^{2} \sqrt {\left (b x +a \right )^{3}}}+\frac {b^{2} \left (\frac {16}{\sqrt {b x +a}}-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \left (b x +a \right )^{\frac {3}{2}}}{8 a^{3} \sqrt {\left (b x +a \right )^{3}}}\) | \(85\) |
-1/4*(b*x+a)*(15*(b*x+a)^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*b^2*x^2-5*a^ (3/2)*b*x-15*b^2*x^2*a^(1/2)+2*a^(5/2))/((b*x+a)^3)^(1/2)/a^(7/2)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (72) = 144\).
Time = 0.26 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.81 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\left [\frac {15 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b^{2} x^{2} + 3 \, a b x + 2 \, a^{2} - 2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} \sqrt {a}}{b x^{2} + a x}\right ) + 2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )}}{8 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {15 \, {\left (b^{4} x^{4} + 2 \, a b^{3} x^{3} + a^{2} b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} \sqrt {-a}}{a b x + a^{2}}\right ) + \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )}}{4 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \]
[1/8*(15*(b^4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(a)*log((b^2*x^2 + 3*a* b*x + 2*a^2 - 2*sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(a))/(b* x^2 + a*x)) + 2*sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(15*a*b^2*x^ 2 + 5*a^2*b*x - 2*a^3))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), 1/4*(15*(b^ 4*x^4 + 2*a*b^3*x^3 + a^2*b^2*x^2)*sqrt(-a)*arctan(sqrt(b^3*x^3 + 3*a*b^2* x^2 + 3*a^2*b*x + a^3)*sqrt(-a)/(a*b*x + a^2)) + sqrt(b^3*x^3 + 3*a*b^2*x^ 2 + 3*a^2*b*x + a^3)*(15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3))/(a^4*b^2*x^4 + 2* a^5*b*x^3 + a^6*x^2)]
\[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\int \frac {1}{x^{3} \sqrt {\left (a + b x\right )^{3}}}\, dx \]
\[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\int { \frac {1}{\sqrt {{\left (b x + a\right )}^{3}} x^{3}} \,d x } \]
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]
15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2*b^2/(sqrt(b*x + a)*a^3) + 1/4*(7*(b*x + a)^(3/2)*b^2 - 9*sqrt(b*x + a)*a*b^2)/(a^3*b^2*x^ 2)
Timed out. \[ \int \frac {1}{x^3 \sqrt {(a+b x)^3}} \, dx=\int \frac {1}{x^3\,\sqrt {{\left (a+b\,x\right )}^3}} \,d x \]