Integrand size = 20, antiderivative size = 147 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=-\frac {2 A}{7 a x^{7/2} \sqrt {a+b x}}-\frac {2 (8 A b-7 a B)}{7 a^2 x^{5/2} \sqrt {a+b x}}+\frac {12 (8 A b-7 a B) \sqrt {a+b x}}{35 a^3 x^{5/2}}-\frac {16 b (8 A b-7 a B) \sqrt {a+b x}}{35 a^4 x^{3/2}}+\frac {32 b^2 (8 A b-7 a B) \sqrt {a+b x}}{35 a^5 \sqrt {x}} \]
-2/7*A/a/x^(7/2)/(b*x+a)^(1/2)-2/7*(8*A*b-7*B*a)/a^2/x^(5/2)/(b*x+a)^(1/2) +12/35*(8*A*b-7*B*a)*(b*x+a)^(1/2)/a^3/x^(5/2)-16/35*b*(8*A*b-7*B*a)*(b*x+ a)^(1/2)/a^4/x^(3/2)+32/35*b^2*(8*A*b-7*B*a)*(b*x+a)^(1/2)/a^5/x^(1/2)
Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=-\frac {2 \left (-128 A b^4 x^4+16 a b^3 x^3 (-4 A+7 B x)+8 a^2 b^2 x^2 (2 A+7 B x)-2 a^3 b x (4 A+7 B x)+a^4 (5 A+7 B x)\right )}{35 a^5 x^{7/2} \sqrt {a+b x}} \]
(-2*(-128*A*b^4*x^4 + 16*a*b^3*x^3*(-4*A + 7*B*x) + 8*a^2*b^2*x^2*(2*A + 7 *B*x) - 2*a^3*b*x*(4*A + 7*B*x) + a^4*(5*A + 7*B*x)))/(35*a^5*x^(7/2)*Sqrt [a + b*x])
Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {87, 55, 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.
| \(\displaystyle \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {(8 A b-7 a B) \int \frac {1}{x^{7/2} (a+b x)^{3/2}}dx}{7 a}-\frac {2 A}{7 a x^{7/2} \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {(8 A b-7 a B) \left (\frac {6 \int \frac {1}{x^{7/2} \sqrt {a+b x}}dx}{a}+\frac {2}{a x^{5/2} \sqrt {a+b x}}\right )}{7 a}-\frac {2 A}{7 a x^{7/2} \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {(8 A b-7 a B) \left (\frac {6 \left (-\frac {4 b \int \frac {1}{x^{5/2} \sqrt {a+b x}}dx}{5 a}-\frac {2 \sqrt {a+b x}}{5 a x^{5/2}}\right )}{a}+\frac {2}{a x^{5/2} \sqrt {a+b x}}\right )}{7 a}-\frac {2 A}{7 a x^{7/2} \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {(8 A b-7 a B) \left (\frac {6 \left (-\frac {4 b \left (-\frac {2 b \int \frac {1}{x^{3/2} \sqrt {a+b x}}dx}{3 a}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{5 a}-\frac {2 \sqrt {a+b x}}{5 a x^{5/2}}\right )}{a}+\frac {2}{a x^{5/2} \sqrt {a+b x}}\right )}{7 a}-\frac {2 A}{7 a x^{7/2} \sqrt {a+b x}}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle -\frac {\left (\frac {6 \left (-\frac {4 b \left (\frac {4 b \sqrt {a+b x}}{3 a^2 \sqrt {x}}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{5 a}-\frac {2 \sqrt {a+b x}}{5 a x^{5/2}}\right )}{a}+\frac {2}{a x^{5/2} \sqrt {a+b x}}\right ) (8 A b-7 a B)}{7 a}-\frac {2 A}{7 a x^{7/2} \sqrt {a+b x}}\) |
(-2*A)/(7*a*x^(7/2)*Sqrt[a + b*x]) - ((8*A*b - 7*a*B)*(2/(a*x^(5/2)*Sqrt[a + b*x]) + (6*((-2*Sqrt[a + b*x])/(5*a*x^(5/2)) - (4*b*((-2*Sqrt[a + b*x]) /(3*a*x^(3/2)) + (4*b*Sqrt[a + b*x])/(3*a^2*Sqrt[x])))/(5*a)))/a))/(7*a)
3.6.34.3.1 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 = 1.47 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.69
| method | result | size |
| gosper | \(-\frac {2 \left (-128 A \,b^{4} x^{4}+112 B a \,b^{3} x^{4}-64 A a \,b^{3} x^{3}+56 B \,a^{2} b^{2} x^{3}+16 A \,a^{2} b^{2} x^{2}-14 B \,a^{3} b \,x^{2}-8 A \,a^{3} b x +7 B \,a^{4} x +5 A \,a^{4}\right )}{35 x^{\frac {7}{2}} \sqrt {b x +a}\, a^{5}}\) | \(101\) |
| default | \(-\frac {2 \left (-128 A \,b^{4} x^{4}+112 B a \,b^{3} x^{4}-64 A a \,b^{3} x^{3}+56 B \,a^{2} b^{2} x^{3}+16 A \,a^{2} b^{2} x^{2}-14 B \,a^{3} b \,x^{2}-8 A \,a^{3} b x +7 B \,a^{4} x +5 A \,a^{4}\right )}{35 x^{\frac {7}{2}} \sqrt {b x +a}\, a^{5}}\) | \(101\) |
| risch | \(-\frac {2 \sqrt {b x +a}\, \left (-93 A \,b^{3} x^{3}+77 B a \,b^{2} x^{3}+29 a A \,b^{2} x^{2}-21 B \,a^{2} b \,x^{2}-13 a^{2} A b x +7 a^{3} B x +5 a^{3} A \right )}{35 a^{5} x^{\frac {7}{2}}}+\frac {2 b^{3} \left (A b -B a \right ) \sqrt {x}}{a^{5} \sqrt {b x +a}}\) | \(104\) |
-2/35*(-128*A*b^4*x^4+112*B*a*b^3*x^4-64*A*a*b^3*x^3+56*B*a^2*b^2*x^3+16*A *a^2*b^2*x^2-14*B*a^3*b*x^2-8*A*a^3*b*x+7*B*a^4*x+5*A*a^4)/x^(7/2)/(b*x+a) ^(1/2)/a^5
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=-\frac {2 \, {\left (5 \, A a^{4} + 16 \, {\left (7 \, B a b^{3} - 8 \, A b^{4}\right )} x^{4} + 8 \, {\left (7 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} - 2 \, {\left (7 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{2} + {\left (7 \, B a^{4} - 8 \, A a^{3} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{35 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}} \]
-2/35*(5*A*a^4 + 16*(7*B*a*b^3 - 8*A*b^4)*x^4 + 8*(7*B*a^2*b^2 - 8*A*a*b^3 )*x^3 - 2*(7*B*a^3*b - 8*A*a^2*b^2)*x^2 + (7*B*a^4 - 8*A*a^3*b)*x)*sqrt(b* x + a)*sqrt(x)/(a^5*b*x^5 + a^6*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 1008 vs. \(2 (144) = 288\).
Time = 80.04 (sec) , antiderivative size = 1008, normalized size of antiderivative = 6.86 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=A \left (- \frac {10 a^{7} b^{\frac {33}{2}} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} - \frac {14 a^{6} b^{\frac {35}{2}} x \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} - \frac {14 a^{5} b^{\frac {37}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} + \frac {70 a^{4} b^{\frac {39}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} + \frac {560 a^{3} b^{\frac {41}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} + \frac {1120 a^{2} b^{\frac {43}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} + \frac {896 a b^{\frac {45}{2}} x^{6} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}} + \frac {256 b^{\frac {47}{2}} x^{7} \sqrt {\frac {a}{b x} + 1}}{35 a^{9} b^{16} x^{3} + 140 a^{8} b^{17} x^{4} + 210 a^{7} b^{18} x^{5} + 140 a^{6} b^{19} x^{6} + 35 a^{5} b^{20} x^{7}}\right ) + B \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 ) \]
A*(-10*a**7*b**(33/2)*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 140*a**8*b** 17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**7) - 14*a**6*b**(35/2)*x*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 140*a**8*b** 17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**7) - 14*a**5*b**(37/2)*x**2*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 140*a**8* b**17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x** 7) + 70*a**4*b**(39/2)*x**3*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 140*a* *8*b**17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a**5*b**20* x**7) + 560*a**3*b**(41/2)*x**4*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 14 0*a**8*b**17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a**5*b* *20*x**7) + 1120*a**2*b**(43/2)*x**5*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 140*a**8*b**17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a* *5*b**20*x**7) + 896*a*b**(45/2)*x**6*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x** 3 + 140*a**8*b**17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a **5*b**20*x**7) + 256*b**(47/2)*x**7*sqrt(a/(b*x) + 1)/(35*a**9*b**16*x**3 + 140*a**8*b**17*x**4 + 210*a**7*b**18*x**5 + 140*a**6*b**19*x**6 + 35*a* *5*b**20*x**7)) + B*(-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*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) - 60*a**2*b**(25/2)*x**3*sqrt(a...
Time = 0.19 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=-\frac {32 \, B b^{3} x}{5 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {256 \, A b^{4} x}{35 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {16 \, B b^{2}}{5 \, \sqrt {b x^{2} + a x} a^{3}} + \frac {128 \, A b^{3}}{35 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {4 \, B b}{5 \, \sqrt {b x^{2} + a x} a^{2} x} - \frac {32 \, A b^{2}}{35 \, \sqrt {b x^{2} + a x} a^{3} x} - \frac {2 \, B}{5 \, \sqrt {b x^{2} + a x} a x^{2}} + \frac {16 \, A b}{35 \, \sqrt {b x^{2} + a x} a^{2} x^{2}} - \frac {2 \, A}{7 \, \sqrt {b x^{2} + a x} a x^{3}} \]
-32/5*B*b^3*x/(sqrt(b*x^2 + a*x)*a^4) + 256/35*A*b^4*x/(sqrt(b*x^2 + a*x)* a^5) - 16/5*B*b^2/(sqrt(b*x^2 + a*x)*a^3) + 128/35*A*b^3/(sqrt(b*x^2 + a*x )*a^4) + 4/5*B*b/(sqrt(b*x^2 + a*x)*a^2*x) - 32/35*A*b^2/(sqrt(b*x^2 + a*x )*a^3*x) - 2/5*B/(sqrt(b*x^2 + a*x)*a*x^2) + 16/35*A*b/(sqrt(b*x^2 + a*x)* a^2*x^2) - 2/7*A/(sqrt(b*x^2 + a*x)*a*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (117) = 234\).
Time = 0.35 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.99 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=-\frac {2 \, {\left ({\left (b x + a\right )} {\left ({\left (b x + a\right )} {\left (\frac {{\left (77 \, B a^{10} b^{9} {\left | b \right |} - 93 \, A a^{9} b^{10} {\left | b \right |}\right )} {\left (b x + a\right )}}{a^{14} b^{4}} - \frac {28 \, {\left (9 \, B a^{11} b^{9} {\left | b \right |} - 11 \, A a^{10} b^{10} {\left | b \right |}\right )}}{a^{14} b^{4}}\right )} + \frac {70 \, {\left (4 \, B a^{12} b^{9} {\left | b \right |} - 5 \, A a^{11} b^{10} {\left | b \right |}\right )}}{a^{14} b^{4}}\right )} - \frac {35 \, {\left (3 \, B a^{13} b^{9} {\left | b \right |} - 4 \, A a^{12} b^{10} {\left | b \right |}\right )}}{a^{14} b^{4}}\right )} \sqrt {b x + a}}{35 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}}} - \frac {4 \, {\left (B^{2} a^{2} b^{9} - 2 \, A B a b^{10} + A^{2} b^{11}\right )}}{{\left (B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {9}{2}} + B a^{2} b^{\frac {11}{2}} - A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {11}{2}} - A a b^{\frac {13}{2}}\right )} a^{4} {\left | b \right |}} \]
-2/35*((b*x + a)*((b*x + a)*((77*B*a^10*b^9*abs(b) - 93*A*a^9*b^10*abs(b)) *(b*x + a)/(a^14*b^4) - 28*(9*B*a^11*b^9*abs(b) - 11*A*a^10*b^10*abs(b))/( a^14*b^4)) + 70*(4*B*a^12*b^9*abs(b) - 5*A*a^11*b^10*abs(b))/(a^14*b^4)) - 35*(3*B*a^13*b^9*abs(b) - 4*A*a^12*b^10*abs(b))/(a^14*b^4))*sqrt(b*x + a) /((b*x + a)*b - a*b)^(7/2) - 4*(B^2*a^2*b^9 - 2*A*B*a*b^10 + A^2*b^11)/((B *a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(9/2) + B*a^2*b^( 11/2) - A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(11/2) - A *a*b^(13/2))*a^4*abs(b))
Time = 0.92 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{3/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{7\,a\,b}+\frac {4\,x^2\,\left (8\,A\,b-7\,B\,a\right )}{35\,a^3}-\frac {x^4\,\left (256\,A\,b^4-224\,B\,a\,b^3\right )}{35\,a^5\,b}-\frac {16\,b\,x^3\,\left (8\,A\,b-7\,B\,a\right )}{35\,a^4}+\frac {x\,\left (14\,B\,a^4-16\,A\,a^3\,b\right )}{35\,a^5\,b}\right )}{x^{9/2}+\frac {a\,x^{7/2}}{b}} \]