Integrand size = 18, antiderivative size = 126 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=-\frac {3 (5 A b-a B)}{4 a^3 b \sqrt {x}}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}+\frac {5 A b-a B}{4 a^2 b \sqrt {x} (a+b x)}-\frac {3 (5 A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}} \]
-3/4*(5*A*b-B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(7/2)/b^(1/2)-3/4*(5*A* b-B*a)/a^3/b/x^(1/2)+1/2*(A*b-B*a)/a/b/(b*x+a)^2/x^(1/2)+1/4*(5*A*b-B*a)/a ^2/b/(b*x+a)/x^(1/2)
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=\frac {-15 A b^2 x^2+a b x (-25 A+3 B x)+a^2 (-8 A+5 B x)}{4 a^3 \sqrt {x} (a+b x)^2}+\frac {3 (-5 A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}} \]
(-15*A*b^2*x^2 + a*b*x*(-25*A + 3*B*x) + a^2*(-8*A + 5*B*x))/(4*a^3*Sqrt[x ]*(a + b*x)^2) + (3*(-5*A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a ^(7/2)*Sqrt[b])
Time = 0.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {87, 52, 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.
\(\displaystyle \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(5 A b-a B) \int \frac {1}{x^{3/2} (a+b x)^2}dx}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(5 A b-a B) \left (\frac {3 \int \frac {1}{x^{3/2} (a+b x)}dx}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(5 A b-a B) \left (\frac {3 \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(5 A b-a B) \left (\frac {3 \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(5 A b-a B) \left (\frac {3 \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\) |
(A*b - a*B)/(2*a*b*Sqrt[x]*(a + b*x)^2) + ((5*A*b - a*B)*(1/(a*Sqrt[x]*(a + b*x)) + (3*(-2/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a] ])/a^(3/2)))/(2*a)))/(4*a*b)
3.4.68.3.1 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.49 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(-\frac {2 \left (\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (9 A b -5 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {2 A}{a^{3} \sqrt {x}}\) | \(84\) |
default | \(-\frac {2 \left (\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (9 A b -5 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}-\frac {2 A}{a^{3} \sqrt {x}}\) | \(84\) |
risch | \(-\frac {2 A}{a^{3} \sqrt {x}}-\frac {\frac {2 \left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (9 A b -5 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}}{a^{3}}\) | \(85\) |
-2/a^3*(((7/8*b^2*A-3/8*a*b*B)*x^(3/2)+1/8*a*(9*A*b-5*B*a)*x^(1/2))/(b*x+a )^2+3/8*(5*A*b-B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))-2*A/a^3/x^( 1/2)
Time = 0.24 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.63 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=\left [\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a + 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}, -\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}\right ] \]
[1/8*(3*((B*a*b^2 - 5*A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x - a + 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) - 2*(8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)*x^2 - 5*(B*a^3*b - 5*A*a^2*b^2)* x)*sqrt(x))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x), -1/4*(3*((B*a*b^2 - 5 *A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(a* b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)* x^2 - 5*(B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x)]
Leaf count of result is larger than twice the leaf count of optimal. 1598 vs. \(2 (116) = 232\).
Time = 16.76 (sec) , antiderivative size = 1598, normalized size of antiderivative = 12.68 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=\text {Too large to display} \]
Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0) ), ((-2*A/sqrt(x) + 2*B*sqrt(x))/a**3, Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 2* B/(5*x**(5/2)))/b**3, Eq(a, 0)), (-15*A*a**2*b*sqrt(x)*log(sqrt(x) - sqrt( -a/b))/(8*a**5*b*sqrt(x)*sqrt(-a/b) + 16*a**4*b**2*x**(3/2)*sqrt(-a/b) + 8 *a**3*b**3*x**(5/2)*sqrt(-a/b)) + 15*A*a**2*b*sqrt(x)*log(sqrt(x) + sqrt(- a/b))/(8*a**5*b*sqrt(x)*sqrt(-a/b) + 16*a**4*b**2*x**(3/2)*sqrt(-a/b) + 8* a**3*b**3*x**(5/2)*sqrt(-a/b)) - 16*A*a**2*b*sqrt(-a/b)/(8*a**5*b*sqrt(x)* sqrt(-a/b) + 16*a**4*b**2*x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt( -a/b)) - 30*A*a*b**2*x**(3/2)*log(sqrt(x) - sqrt(-a/b))/(8*a**5*b*sqrt(x)* sqrt(-a/b) + 16*a**4*b**2*x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt( -a/b)) + 30*A*a*b**2*x**(3/2)*log(sqrt(x) + sqrt(-a/b))/(8*a**5*b*sqrt(x)* sqrt(-a/b) + 16*a**4*b**2*x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt( -a/b)) - 50*A*a*b**2*x*sqrt(-a/b)/(8*a**5*b*sqrt(x)*sqrt(-a/b) + 16*a**4*b **2*x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt(-a/b)) - 15*A*b**3*x** (5/2)*log(sqrt(x) - sqrt(-a/b))/(8*a**5*b*sqrt(x)*sqrt(-a/b) + 16*a**4*b** 2*x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt(-a/b)) + 15*A*b**3*x**(5 /2)*log(sqrt(x) + sqrt(-a/b))/(8*a**5*b*sqrt(x)*sqrt(-a/b) + 16*a**4*b**2* x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt(-a/b)) - 30*A*b**3*x**2*sq rt(-a/b)/(8*a**5*b*sqrt(x)*sqrt(-a/b) + 16*a**4*b**2*x**(3/2)*sqrt(-a/b) + 8*a**3*b**3*x**(5/2)*sqrt(-a/b)) + 3*B*a**3*sqrt(x)*log(sqrt(x) - sqrt...
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=-\frac {8 \, A a^{2} - 3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{2} - 5 \, {\left (B a^{2} - 5 \, A a b\right )} x}{4 \, {\left (a^{3} b^{2} x^{\frac {5}{2}} + 2 \, a^{4} b x^{\frac {3}{2}} + a^{5} \sqrt {x}\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} \]
-1/4*(8*A*a^2 - 3*(B*a*b - 5*A*b^2)*x^2 - 5*(B*a^2 - 5*A*a*b)*x)/(a^3*b^2* x^(5/2) + 2*a^4*b*x^(3/2) + a^5*sqrt(x)) + 3/4*(B*a - 5*A*b)*arctan(b*sqrt (x)/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=\frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} - \frac {2 \, A}{a^{3} \sqrt {x}} + \frac {3 \, B a b x^{\frac {3}{2}} - 7 \, A b^{2} x^{\frac {3}{2}} + 5 \, B a^{2} \sqrt {x} - 9 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3}} \]
3/4*(B*a - 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2*A/(a^3*s qrt(x)) + 1/4*(3*B*a*b*x^(3/2) - 7*A*b^2*x^(3/2) + 5*B*a^2*sqrt(x) - 9*A*a *b*sqrt(x))/((b*x + a)^2*a^3)
Time = 0.55 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{3/2} (a+b x)^3} \, dx=-\frac {\frac {2\,A}{a}+\frac {5\,x\,\left (5\,A\,b-B\,a\right )}{4\,a^2}+\frac {3\,b\,x^2\,\left (5\,A\,b-B\,a\right )}{4\,a^3}}{a^2\,\sqrt {x}+b^2\,x^{5/2}+2\,a\,b\,x^{3/2}}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,\sqrt {x}\,\left (5\,A\,b-B\,a\right )}{\sqrt {a}\,\left (15\,A\,b-3\,B\,a\right )}\right )\,\left (5\,A\,b-B\,a\right )}{4\,a^{7/2}\,\sqrt {b}} \]