Integrand size = 20, antiderivative size = 144 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}-\frac {2 (8 A b-5 a B)}{15 a^2 x^{3/2} (a+b x)^{3/2}}-\frac {4 (8 A b-5 a B)}{5 a^3 x^{3/2} \sqrt {a+b x}}+\frac {16 (8 A b-5 a B) \sqrt {a+b x}}{15 a^4 x^{3/2}}-\frac {32 b (8 A b-5 a B) \sqrt {a+b x}}{15 a^5 \sqrt {x}} \]
-2/5*A/a/x^(5/2)/(b*x+a)^(3/2)-2/15*(8*A*b-5*B*a)/a^2/x^(3/2)/(b*x+a)^(3/2 )-4/5*(8*A*b-5*B*a)/a^3/x^(3/2)/(b*x+a)^(1/2)+16/15*(8*A*b-5*B*a)*(b*x+a)^ (1/2)/a^4/x^(3/2)-32/15*b*(8*A*b-5*B*a)*(b*x+a)^(1/2)/a^5/x^(1/2)
Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=-\frac {2 \left (128 A b^4 x^4+24 a^2 b^2 x^2 (2 A-5 B x)+16 a b^3 x^3 (12 A-5 B x)+a^4 (3 A+5 B x)-2 a^3 b x (4 A+15 B x)\right )}{15 a^5 x^{5/2} (a+b x)^{3/2}} \]
(-2*(128*A*b^4*x^4 + 24*a^2*b^2*x^2*(2*A - 5*B*x) + 16*a*b^3*x^3*(12*A - 5 *B*x) + a^4*(3*A + 5*B*x) - 2*a^3*b*x*(4*A + 15*B*x)))/(15*a^5*x^(5/2)*(a + b*x)^(3/2))
Time = 0.20 (sec) , antiderivative size = 135, 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.
\(\displaystyle \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {(8 A b-5 a B) \int \frac {1}{x^{5/2} (a+b x)^{5/2}}dx}{5 a}-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(8 A b-5 a B) \left (\frac {2 \int \frac {1}{x^{5/2} (a+b x)^{3/2}}dx}{a}+\frac {2}{3 a x^{3/2} (a+b x)^{3/2}}\right )}{5 a}-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(8 A b-5 a B) \left (\frac {2 \left (\frac {4 \int \frac {1}{x^{5/2} \sqrt {a+b x}}dx}{a}+\frac {2}{a x^{3/2} \sqrt {a+b x}}\right )}{a}+\frac {2}{3 a x^{3/2} (a+b x)^{3/2}}\right )}{5 a}-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(8 A b-5 a B) \left (\frac {2 \left (\frac {4 \left (-\frac {2 b \int \frac {1}{x^{3/2} \sqrt {a+b x}}dx}{3 a}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{a}+\frac {2}{a x^{3/2} \sqrt {a+b x}}\right )}{a}+\frac {2}{3 a x^{3/2} (a+b x)^{3/2}}\right )}{5 a}-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (\frac {2 \left (\frac {4 \left (\frac {4 b \sqrt {a+b x}}{3 a^2 \sqrt {x}}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{a}+\frac {2}{a x^{3/2} \sqrt {a+b x}}\right )}{a}+\frac {2}{3 a x^{3/2} (a+b x)^{3/2}}\right ) (8 A b-5 a B)}{5 a}-\frac {2 A}{5 a x^{5/2} (a+b x)^{3/2}}\) |
(-2*A)/(5*a*x^(5/2)*(a + b*x)^(3/2)) - ((8*A*b - 5*a*B)*(2/(3*a*x^(3/2)*(a + b*x)^(3/2)) + (2*(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))/a))/(5*a)
3.6.44.3.1 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 = 0.53 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.67
method | result | size |
risch | \(-\frac {2 \sqrt {b x +a}\, \left (73 A \,b^{2} x^{2}-40 B a b \,x^{2}-14 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 a^{5} x^{\frac {5}{2}}}-\frac {2 b^{2} \left (11 A \,b^{2} x -8 B a b x +12 a b A -9 a^{2} B \right ) \sqrt {x}}{3 \left (b x +a \right )^{\frac {3}{2}} a^{5}}\) | \(97\) |
gosper | \(-\frac {2 \left (128 A \,b^{4} x^{4}-80 B a \,b^{3} x^{4}+192 A a \,b^{3} x^{3}-120 B \,a^{2} b^{2} x^{3}+48 A \,a^{2} b^{2} x^{2}-30 B \,a^{3} b \,x^{2}-8 A \,a^{3} b x +5 B \,a^{4} x +3 A \,a^{4}\right )}{15 x^{\frac {5}{2}} \left (b x +a \right )^{\frac {3}{2}} a^{5}}\) | \(101\) |
default | \(-\frac {2 \left (128 A \,b^{4} x^{4}-80 B a \,b^{3} x^{4}+192 A a \,b^{3} x^{3}-120 B \,a^{2} b^{2} x^{3}+48 A \,a^{2} b^{2} x^{2}-30 B \,a^{3} b \,x^{2}-8 A \,a^{3} b x +5 B \,a^{4} x +3 A \,a^{4}\right )}{15 x^{\frac {5}{2}} \left (b x +a \right )^{\frac {3}{2}} a^{5}}\) | \(101\) |
-2/15*(b*x+a)^(1/2)*(73*A*b^2*x^2-40*B*a*b*x^2-14*A*a*b*x+5*B*a^2*x+3*A*a^ 2)/a^5/x^(5/2)-2/3*b^2*(11*A*b^2*x-8*B*a*b*x+12*A*a*b-9*B*a^2)*x^(1/2)/(b* x+a)^(3/2)/a^5
Time = 0.23 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (3 \, A a^{4} - 16 \, {\left (5 \, B a b^{3} - 8 \, A b^{4}\right )} x^{4} - 24 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} - 6 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{2} + {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \]
-2/15*(3*A*a^4 - 16*(5*B*a*b^3 - 8*A*b^4)*x^4 - 24*(5*B*a^2*b^2 - 8*A*a*b^ 3)*x^3 - 6*(5*B*a^3*b - 8*A*a^2*b^2)*x^2 + (5*B*a^4 - 8*A*a^3*b)*x)*sqrt(b *x + a)*sqrt(x)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)
Timed out. \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=-\frac {4 \, B b x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} + \frac {32 \, B b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {32 \, A b^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, A b^{3} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {2 \, B}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} + \frac {16 \, B b}{3 \, \sqrt {b x^{2} + a x} a^{3}} + \frac {16 \, A b}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, A b^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {2 \, A}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} \]
-4/3*B*b*x/((b*x^2 + a*x)^(3/2)*a^2) + 32/3*B*b^2*x/(sqrt(b*x^2 + a*x)*a^4 ) + 32/15*A*b^2*x/((b*x^2 + a*x)^(3/2)*a^3) - 256/15*A*b^3*x/(sqrt(b*x^2 + a*x)*a^5) - 2/3*B/((b*x^2 + a*x)^(3/2)*a) + 16/3*B*b/(sqrt(b*x^2 + a*x)*a ^3) + 16/15*A*b/((b*x^2 + a*x)^(3/2)*a^2) - 128/15*A*b^2/(sqrt(b*x^2 + a*x )*a^4) - 2/5*A/((b*x^2 + a*x)^(3/2)*a*x)
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (114) = 228\).
Time = 0.42 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.37 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=\frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {{\left (40 \, B a^{8} b^{7} - 73 \, A a^{7} b^{8}\right )} {\left (b x + a\right )}}{a^{12} b^{2} {\left | b \right |}} - \frac {5 \, {\left (17 \, B a^{9} b^{7} - 32 \, A a^{8} b^{8}\right )}}{a^{12} b^{2} {\left | b \right |}}\right )} + \frac {45 \, {\left (B a^{10} b^{7} - 2 \, A a^{9} b^{8}\right )}}{a^{12} b^{2} {\left | b \right |}}\right )}}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}} + \frac {4 \, {\left (6 \, B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {7}{2}} + 18 \, B a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {9}{2}} - 9 \, A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {9}{2}} + 8 \, B a^{3} b^{\frac {11}{2}} - 24 \, A a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {11}{2}} - 11 \, A a^{2} b^{\frac {13}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{4} {\left | b \right |}} \]
2/15*sqrt(b*x + a)*((b*x + a)*((40*B*a^8*b^7 - 73*A*a^7*b^8)*(b*x + a)/(a^ 12*b^2*abs(b)) - 5*(17*B*a^9*b^7 - 32*A*a^8*b^8)/(a^12*b^2*abs(b))) + 45*( B*a^10*b^7 - 2*A*a^9*b^8)/(a^12*b^2*abs(b)))/((b*x + a)*b - a*b)^(5/2) + 4 /3*(6*B*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(7/2) + 18 *B*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(9/2) - 9*A*( sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(9/2) + 8*B*a^3*b^(11 /2) - 24*A*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(11/2) - 11*A*a^2*b^(13/2))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*a^4*abs(b))
Time = 1.12 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{x^{7/2} (a+b x)^{5/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5\,a\,b^2}+\frac {16\,x^3\,\left (8\,A\,b-5\,B\,a\right )}{5\,a^4}+\frac {4\,x^2\,\left (8\,A\,b-5\,B\,a\right )}{5\,a^3\,b}+\frac {x^4\,\left (256\,A\,b^4-160\,B\,a\,b^3\right )}{15\,a^5\,b^2}+\frac {x\,\left (10\,B\,a^4-16\,A\,a^3\,b\right )}{15\,a^5\,b^2}\right )}{x^{9/2}+\frac {2\,a\,x^{7/2}}{b}+\frac {a^2\,x^{5/2}}{b^2}} \]