Integrand size = 18, antiderivative size = 147 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=-\frac {5 (7 A b-3 a B)}{12 a^3 b x^{3/2}}+\frac {5 (7 A b-3 a B)}{4 a^4 \sqrt {x}}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}+\frac {7 A b-3 a B}{4 a^2 b x^{3/2} (a+b x)}+\frac {5 \sqrt {b} (7 A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}} \]
-5/12*(7*A*b-3*B*a)/a^3/b/x^(3/2)+1/2*(A*b-B*a)/a/b/x^(3/2)/(b*x+a)^2+1/4* (7*A*b-3*B*a)/a^2/b/x^(3/2)/(b*x+a)+5/4*(7*A*b-3*B*a)*arctan(b^(1/2)*x^(1/ 2)/a^(1/2))*b^(1/2)/a^(9/2)+5/4*(7*A*b-3*B*a)/a^4/x^(1/2)
Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=\frac {105 A b^3 x^3+a^2 b x (56 A-75 B x)+5 a b^2 x^2 (35 A-9 B x)-8 a^3 (A+3 B x)}{12 a^4 x^{3/2} (a+b x)^2}+\frac {5 \sqrt {b} (7 A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}} \]
(105*A*b^3*x^3 + a^2*b*x*(56*A - 75*B*x) + 5*a*b^2*x^2*(35*A - 9*B*x) - 8* a^3*(A + 3*B*x))/(12*a^4*x^(3/2)*(a + b*x)^2) + (5*Sqrt[b]*(7*A*b - 3*a*B) *ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(9/2))
Time = 0.22 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 52, 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.
| \(\displaystyle \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(7 A b-3 a B) \int \frac {1}{x^{5/2} (a+b x)^2}dx}{4 a b}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(7 A b-3 a B) \left (\frac {5 \int \frac {1}{x^{5/2} (a+b x)}dx}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(7 A b-3 a B) \left (\frac {5 \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(7 A b-3 a B) \left (\frac {5 \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(7 A b-3 a B) \left (\frac {5 \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(7 A b-3 a B) \left (\frac {5 \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a}+\frac {1}{a x^{3/2} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b x^{3/2} (a+b x)^2}\) |
(A*b - a*B)/(2*a*b*x^(3/2)*(a + b*x)^2) + ((7*A*b - 3*a*B)*(1/(a*x^(3/2)*( a + b*x)) + (5*(-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))/(2*a)))/(4*a*b)
3.4.69.3.1 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 = 1.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.67
| method | result | size |
| risch | \(-\frac {2 \left (-9 A b x +3 B a x +A a \right )}{3 a^{4} x^{\frac {3}{2}}}+\frac {b \left (\frac {2 \left (\frac {11}{8} b^{2} A -\frac {7}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (13 A b -9 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {5 \left (7 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{a^{4}}\) | \(98\) |
| derivativedivides | \(\frac {2 b \left (\frac {\left (\frac {11}{8} b^{2} A -\frac {7}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (13 A b -9 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {5 \left (7 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 \left (-3 A b +B a \right )}{a^{4} \sqrt {x}}\) | \(101\) |
| default | \(\frac {2 b \left (\frac {\left (\frac {11}{8} b^{2} A -\frac {7}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (13 A b -9 B a \right ) \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {5 \left (7 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}-\frac {2 A}{3 a^{3} x^{\frac {3}{2}}}-\frac {2 \left (-3 A b +B a \right )}{a^{4} \sqrt {x}}\) | \(101\) |
-2/3*(-9*A*b*x+3*B*a*x+A*a)/a^4/x^(3/2)+1/a^4*b*(2*((11/8*b^2*A-7/8*a*b*B) *x^(3/2)+1/8*a*(13*A*b-9*B*a)*x^(1/2))/(b*x+a)^2+5/4*(7*A*b-3*B*a)/(a*b)^( 1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.24 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.59 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=\left [-\frac {15 \, {\left ({\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + 2 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3} + {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (8 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, \frac {15 \, {\left ({\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} + 2 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{3} + {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (8 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \]
[-1/24*(15*((3*B*a*b^2 - 7*A*b^3)*x^4 + 2*(3*B*a^2*b - 7*A*a*b^2)*x^3 + (3 *B*a^3 - 7*A*a^2*b)*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a) /(b*x + a)) + 2*(8*A*a^3 + 15*(3*B*a*b^2 - 7*A*b^3)*x^3 + 25*(3*B*a^2*b - 7*A*a*b^2)*x^2 + 8*(3*B*a^3 - 7*A*a^2*b)*x)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5* b*x^3 + a^6*x^2), 1/12*(15*((3*B*a*b^2 - 7*A*b^3)*x^4 + 2*(3*B*a^2*b - 7*A *a*b^2)*x^3 + (3*B*a^3 - 7*A*a^2*b)*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*s qrt(x))) - (8*A*a^3 + 15*(3*B*a*b^2 - 7*A*b^3)*x^3 + 25*(3*B*a^2*b - 7*A*a *b^2)*x^2 + 8*(3*B*a^3 - 7*A*a^2*b)*x)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)]
Leaf count of result is larger than twice the leaf count of optimal. 1703 vs. \(2 (139) = 278\).
Time = 38.49 (sec) , antiderivative size = 1703, normalized size of antiderivative = 11.59 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=\text {Too large to display} \]
Piecewise((zoo*(-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2))), Eq(a, 0) & Eq(b, 0) ), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/a**3, Eq(b, 0)), ((-2*A/(9*x**(9/2)) - 2*B/(7*x**(7/2)))/b**3, Eq(a, 0)), (-16*A*a**3*sqrt(-a/b)/(24*a**6*x**( 3/2)*sqrt(-a/b) + 48*a**5*b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sq rt(-a/b)) + 105*A*a**2*b*x**(3/2)*log(sqrt(x) - sqrt(-a/b))/(24*a**6*x**(3 /2)*sqrt(-a/b) + 48*a**5*b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqr t(-a/b)) - 105*A*a**2*b*x**(3/2)*log(sqrt(x) + sqrt(-a/b))/(24*a**6*x**(3/ 2)*sqrt(-a/b) + 48*a**5*b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqrt (-a/b)) + 112*A*a**2*b*x*sqrt(-a/b)/(24*a**6*x**(3/2)*sqrt(-a/b) + 48*a**5 *b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqrt(-a/b)) + 210*A*a*b**2* x**(5/2)*log(sqrt(x) - sqrt(-a/b))/(24*a**6*x**(3/2)*sqrt(-a/b) + 48*a**5* b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqrt(-a/b)) - 210*A*a*b**2*x **(5/2)*log(sqrt(x) + sqrt(-a/b))/(24*a**6*x**(3/2)*sqrt(-a/b) + 48*a**5*b *x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqrt(-a/b)) + 350*A*a*b**2*x* *2*sqrt(-a/b)/(24*a**6*x**(3/2)*sqrt(-a/b) + 48*a**5*b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqrt(-a/b)) + 105*A*b**3*x**(7/2)*log(sqrt(x) - s qrt(-a/b))/(24*a**6*x**(3/2)*sqrt(-a/b) + 48*a**5*b*x**(5/2)*sqrt(-a/b) + 24*a**4*b**2*x**(7/2)*sqrt(-a/b)) - 105*A*b**3*x**(7/2)*log(sqrt(x) + sqrt (-a/b))/(24*a**6*x**(3/2)*sqrt(-a/b) + 48*a**5*b*x**(5/2)*sqrt(-a/b) + 24* a**4*b**2*x**(7/2)*sqrt(-a/b)) + 210*A*b**3*x**3*sqrt(-a/b)/(24*a**6*x*...
Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=-\frac {8 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{3} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} + 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} - \frac {5 \, {\left (3 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} \]
-1/12*(8*A*a^3 + 15*(3*B*a*b^2 - 7*A*b^3)*x^3 + 25*(3*B*a^2*b - 7*A*a*b^2) *x^2 + 8*(3*B*a^3 - 7*A*a^2*b)*x)/(a^4*b^2*x^(7/2) + 2*a^5*b*x^(5/2) + a^6 *x^(3/2)) - 5/4*(3*B*a*b - 7*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b) *a^4)
Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=-\frac {5 \, {\left (3 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} - \frac {2 \, {\left (3 \, B a x - 9 \, A b x + A a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} - \frac {7 \, B a b^{2} x^{\frac {3}{2}} - 11 \, A b^{3} x^{\frac {3}{2}} + 9 \, B a^{2} b \sqrt {x} - 13 \, A a b^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{4}} \]
-5/4*(3*B*a*b - 7*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 2/3 *(3*B*a*x - 9*A*b*x + A*a)/(a^4*x^(3/2)) - 1/4*(7*B*a*b^2*x^(3/2) - 11*A*b ^3*x^(3/2) + 9*B*a^2*b*sqrt(x) - 13*A*a*b^2*sqrt(x))/((b*x + a)^2*a^4)
Time = 0.55 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x}{x^{5/2} (a+b x)^3} \, dx=\frac {\frac {2\,x\,\left (7\,A\,b-3\,B\,a\right )}{3\,a^2}-\frac {2\,A}{3\,a}+\frac {5\,b^2\,x^3\,\left (7\,A\,b-3\,B\,a\right )}{4\,a^4}+\frac {25\,b\,x^2\,\left (7\,A\,b-3\,B\,a\right )}{12\,a^3}}{a^2\,x^{3/2}+b^2\,x^{7/2}+2\,a\,b\,x^{5/2}}+\frac {5\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (7\,A\,b-3\,B\,a\right )}{4\,a^{9/2}} \]