Integrand size = 20, antiderivative size = 114 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}-\frac {2 (6 A b-5 a B)}{5 a^2 x^{3/2} \sqrt {a+b x}}+\frac {8 (6 A b-5 a B) \sqrt {a+b x}}{15 a^3 x^{3/2}}-\frac {16 b (6 A b-5 a B) \sqrt {a+b x}}{15 a^4 \sqrt {x}} \]
-2/5*A/a/x^(5/2)/(b*x+a)^(1/2)-2/5*(6*A*b-5*B*a)/a^2/x^(3/2)/(b*x+a)^(1/2) +8/15*(6*A*b-5*B*a)*(b*x+a)^(1/2)/a^3/x^(3/2)-16/15*b*(6*A*b-5*B*a)*(b*x+a )^(1/2)/a^4/x^(1/2)
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=-\frac {2 \left (48 A b^3 x^3+8 a b^2 x^2 (3 A-5 B x)+a^3 (3 A+5 B x)-2 a^2 b x (3 A+10 B x)\right )}{15 a^4 x^{5/2} \sqrt {a+b x}} \]
(-2*(48*A*b^3*x^3 + 8*a*b^2*x^2*(3*A - 5*B*x) + a^3*(3*A + 5*B*x) - 2*a^2* b*x*(3*A + 10*B*x)))/(15*a^4*x^(5/2)*Sqrt[a + b*x])
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {87, 55, 55, 48}
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.
\(\displaystyle \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {(6 A b-5 a B) \int \frac {1}{x^{5/2} (a+b x)^{3/2}}dx}{5 a}-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(6 A b-5 a B) \left (\frac {4 \int \frac {1}{x^{5/2} \sqrt {a+b x}}dx}{a}+\frac {2}{a x^{3/2} \sqrt {a+b x}}\right )}{5 a}-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(6 A b-5 a B) \left (\frac {4 \left (-\frac {2 b \int \frac {1}{x^{3/2} \sqrt {a+b x}}dx}{3 a}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{a}+\frac {2}{a x^{3/2} \sqrt {a+b x}}\right )}{5 a}-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (\frac {4 \left (\frac {4 b \sqrt {a+b x}}{3 a^2 \sqrt {x}}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{a}+\frac {2}{a x^{3/2} \sqrt {a+b x}}\right ) (6 A b-5 a B)}{5 a}-\frac {2 A}{5 a x^{5/2} \sqrt {a+b x}}\) |
(-2*A)/(5*a*x^(5/2)*Sqrt[a + b*x]) - ((6*A*b - 5*a*B)*(2/(a*x^(3/2)*Sqrt[a + b*x]) + (4*((-2*Sqrt[a + b*x])/(3*a*x^(3/2)) + (4*b*Sqrt[a + b*x])/(3*a ^2*Sqrt[x])))/a))/(5*a)
3.6.33.3.1 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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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]))))
Time = 0.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(-\frac {2 \left (48 A \,b^{3} x^{3}-40 B a \,b^{2} x^{3}+24 a A \,b^{2} x^{2}-20 B \,a^{2} b \,x^{2}-6 a^{2} A b x +5 a^{3} B x +3 a^{3} A \right )}{15 x^{\frac {5}{2}} \sqrt {b x +a}\, a^{4}}\) | \(77\) |
default | \(-\frac {2 \left (48 A \,b^{3} x^{3}-40 B a \,b^{2} x^{3}+24 a A \,b^{2} x^{2}-20 B \,a^{2} b \,x^{2}-6 a^{2} A b x +5 a^{3} B x +3 a^{3} A \right )}{15 x^{\frac {5}{2}} \sqrt {b x +a}\, a^{4}}\) | \(77\) |
risch | \(-\frac {2 \sqrt {b x +a}\, \left (33 A \,b^{2} x^{2}-25 B a b \,x^{2}-9 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 a^{4} x^{\frac {5}{2}}}-\frac {2 b^{2} \left (A b -B a \right ) \sqrt {x}}{a^{4} \sqrt {b x +a}}\) | \(80\) |
-2/15*(48*A*b^3*x^3-40*B*a*b^2*x^3+24*A*a*b^2*x^2-20*B*a^2*b*x^2-6*A*a^2*b *x+5*B*a^3*x+3*A*a^3)/x^(5/2)/(b*x+a)^(1/2)/a^4
Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=-\frac {2 \, {\left (3 \, A a^{3} - 8 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} - 4 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x^{2} + {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \]
-2/15*(3*A*a^3 - 8*(5*B*a*b^2 - 6*A*b^3)*x^3 - 4*(5*B*a^2*b - 6*A*a*b^2)*x ^2 + (5*B*a^3 - 6*A*a^2*b)*x)*sqrt(b*x + a)*sqrt(x)/(a^4*b*x^4 + a^5*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (110) = 220\).
Time = 30.11 (sec) , antiderivative size = 573, normalized size of antiderivative = 5.03 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=A \left (- \frac {2 a^{5} b^{\frac {19}{2}} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {10 a^{3} b^{\frac {23}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {60 a^{2} b^{\frac {25}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {80 a b^{\frac {27}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac {32 b^{\frac {29}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}}\right ) + B \left (- \frac {2 a^{3} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {6 a^{2} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {24 a b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}} + \frac {16 b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} x + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{3}}\right ) \]
A*(-2*a**5*b**(19/2)*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x **3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 10*a**3*b**(23/2)*x**2*sqr t(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 60*a**2*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(5*a**7*b **9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 80*a*b**(27/2)*x**4*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x* *3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 32*b**(29/2)*x**5*sqrt(a/(b *x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a **4*b**12*x**5)) + B*(-2*a**3*b**(9/2)*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 6*a**2*b**(11/2)*x*sqrt(a/(b*x) + 1 )/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3) + 24*a*b**(13/2)*x **2*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3 ) + 16*b**(15/2)*x**3*sqrt(a/(b*x) + 1)/(3*a**5*b**4*x + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**3))
Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=\frac {16 \, B b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {32 \, A b^{3} x}{5 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {8 \, B b}{3 \, \sqrt {b x^{2} + a x} a^{2}} - \frac {16 \, A b^{2}}{5 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {2 \, B}{3 \, \sqrt {b x^{2} + a x} a x} + \frac {4 \, A b}{5 \, \sqrt {b x^{2} + a x} a^{2} x} - \frac {2 \, A}{5 \, \sqrt {b x^{2} + a x} a x^{2}} \]
16/3*B*b^2*x/(sqrt(b*x^2 + a*x)*a^3) - 32/5*A*b^3*x/(sqrt(b*x^2 + a*x)*a^4 ) + 8/3*B*b/(sqrt(b*x^2 + a*x)*a^2) - 16/5*A*b^2/(sqrt(b*x^2 + a*x)*a^3) - 2/3*B/(sqrt(b*x^2 + a*x)*a*x) + 4/5*A*b/(sqrt(b*x^2 + a*x)*a^2*x) - 2/5*A /(sqrt(b*x^2 + a*x)*a*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (90) = 180\).
Time = 0.34 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.23 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=\frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {{\left (25 \, B a^{6} b^{7} - 33 \, A a^{5} b^{8}\right )} {\left (b x + a\right )}}{a^{9} b^{2} {\left | b \right |}} - \frac {5 \, {\left (11 \, B a^{7} b^{7} - 15 \, A a^{6} b^{8}\right )}}{a^{9} b^{2} {\left | b \right |}}\right )} + \frac {15 \, {\left (2 \, B a^{8} b^{7} - 3 \, A a^{7} b^{8}\right )}}{a^{9} b^{2} {\left | b \right |}}\right )}}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}} + \frac {4 \, {\left (B^{2} a^{2} b^{7} - 2 \, A B a b^{8} + A^{2} b^{9}\right )}}{{\left (B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {7}{2}} + B a^{2} b^{\frac {9}{2}} - A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {9}{2}} - A a b^{\frac {11}{2}}\right )} a^{3} {\left | b \right |}} \]
2/15*sqrt(b*x + a)*((b*x + a)*((25*B*a^6*b^7 - 33*A*a^5*b^8)*(b*x + a)/(a^ 9*b^2*abs(b)) - 5*(11*B*a^7*b^7 - 15*A*a^6*b^8)/(a^9*b^2*abs(b))) + 15*(2* B*a^8*b^7 - 3*A*a^7*b^8)/(a^9*b^2*abs(b)))/((b*x + a)*b - a*b)^(5/2) + 4*( B^2*a^2*b^7 - 2*A*B*a*b^8 + A^2*b^9)/((B*a*(sqrt(b*x + a)*sqrt(b) - sqrt(( b*x + a)*b - a*b))^2*b^(7/2) + B*a^2*b^(9/2) - A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(9/2) - A*a*b^(11/2))*a^3*abs(b))
Time = 0.97 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{3/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5\,a\,b}+\frac {8\,x^2\,\left (6\,A\,b-5\,B\,a\right )}{15\,a^3}+\frac {x^3\,\left (96\,A\,b^3-80\,B\,a\,b^2\right )}{15\,a^4\,b}+\frac {x\,\left (10\,B\,a^3-12\,A\,a^2\,b\right )}{15\,a^4\,b}\right )}{x^{7/2}+\frac {a\,x^{5/2}}{b}} \]