Integrand size = 20, antiderivative size = 84 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}+\frac {2 (4 A b-5 a B) \sqrt {a+b x}}{15 a^2 x^{3/2}}-\frac {4 b (4 A b-5 a B) \sqrt {a+b x}}{15 a^3 \sqrt {x}} \] Output:
-2/5*A*(b*x+a)^(1/2)/a/x^(5/2)+2/15*(4*A*b-5*B*a)*(b*x+a)^(1/2)/a^2/x^(3/2 )-4/15*b*(4*A*b-5*B*a)*(b*x+a)^(1/2)/a^3/x^(1/2)
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=-\frac {2 \sqrt {a+b x} \left (3 a^2 A-4 a A b x+5 a^2 B x+8 A b^2 x^2-10 a b B x^2\right )}{15 a^3 x^{5/2}} \] Input:
Integrate[(A + B*x)/(x^(7/2)*Sqrt[a + b*x]),x]
Output:
(-2*Sqrt[a + b*x]*(3*a^2*A - 4*a*A*b*x + 5*a^2*B*x + 8*A*b^2*x^2 - 10*a*b* B*x^2))/(15*a^3*x^(5/2))
Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {87, 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.
\(\displaystyle \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {(4 A b-5 a B) \int \frac {1}{x^{5/2} \sqrt {a+b x}}dx}{5 a}-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {(4 A b-5 a B) \left (-\frac {2 b \int \frac {1}{x^{3/2} \sqrt {a+b x}}dx}{3 a}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right )}{5 a}-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {\left (\frac {4 b \sqrt {a+b x}}{3 a^2 \sqrt {x}}-\frac {2 \sqrt {a+b x}}{3 a x^{3/2}}\right ) (4 A b-5 a B)}{5 a}-\frac {2 A \sqrt {a+b x}}{5 a x^{5/2}}\) |
Input:
Int[(A + B*x)/(x^(7/2)*Sqrt[a + b*x]),x]
Output:
(-2*A*Sqrt[a + b*x])/(5*a*x^(5/2)) - ((4*A*b - 5*a*B)*((-2*Sqrt[a + b*x])/ (3*a*x^(3/2)) + (4*b*Sqrt[a + b*x])/(3*a^2*Sqrt[x])))/(5*a)
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 = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(-\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{2} x^{2}-10 B a b \,x^{2}-4 a A b x +5 B \,a^{2} x +3 a^{2} A \right )}{15 x^{\frac {5}{2}} a^{3}}\) | \(53\) |
default | \(-\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{2} x^{2}-10 B a b \,x^{2}-4 a A b x +5 B \,a^{2} x +3 a^{2} A \right )}{15 x^{\frac {5}{2}} a^{3}}\) | \(53\) |
risch | \(-\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{2} x^{2}-10 B a b \,x^{2}-4 a A b x +5 B \,a^{2} x +3 a^{2} A \right )}{15 x^{\frac {5}{2}} a^{3}}\) | \(53\) |
orering | \(-\frac {2 \sqrt {b x +a}\, \left (8 A \,b^{2} x^{2}-10 B a b \,x^{2}-4 a A b x +5 B \,a^{2} x +3 a^{2} A \right )}{15 x^{\frac {5}{2}} a^{3}}\) | \(53\) |
Input:
int((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/15*(b*x+a)^(1/2)*(8*A*b^2*x^2-10*B*a*b*x^2-4*A*a*b*x+5*B*a^2*x+3*A*a^2) /x^(5/2)/a^3
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.63 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=-\frac {2 \, {\left (3 \, A a^{2} - 2 \, {\left (5 \, B a b - 4 \, A b^{2}\right )} x^{2} + {\left (5 \, B a^{2} - 4 \, A a b\right )} x\right )} \sqrt {b x + a}}{15 \, a^{3} x^{\frac {5}{2}}} \] Input:
integrate((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x, algorithm="fricas")
Output:
-2/15*(3*A*a^2 - 2*(5*B*a*b - 4*A*b^2)*x^2 + (5*B*a^2 - 4*A*a*b)*x)*sqrt(b *x + a)/(a^3*x^(5/2))
Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (82) = 164\).
Time = 4.40 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.07 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=- \frac {6 A a^{4} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {4 A a^{3} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {6 A a^{2} b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {24 A a b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {16 A b^{\frac {17}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{15 a^{5} b^{4} x^{2} + 30 a^{4} b^{5} x^{3} + 15 a^{3} b^{6} x^{4}} - \frac {2 B \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{3 a x} + \frac {4 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{2}} \] Input:
integrate((B*x+A)/x**(7/2)/(b*x+a)**(1/2),x)
Output:
-6*A*a**4*b**(9/2)*sqrt(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x** 3 + 15*a**3*b**6*x**4) - 4*A*a**3*b**(11/2)*x*sqrt(a/(b*x) + 1)/(15*a**5*b **4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 6*A*a**2*b**(13/2)*x** 2*sqrt(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6* x**4) - 24*A*a*b**(15/2)*x**3*sqrt(a/(b*x) + 1)/(15*a**5*b**4*x**2 + 30*a* *4*b**5*x**3 + 15*a**3*b**6*x**4) - 16*A*b**(17/2)*x**4*sqrt(a/(b*x) + 1)/ (15*a**5*b**4*x**2 + 30*a**4*b**5*x**3 + 15*a**3*b**6*x**4) - 2*B*sqrt(b)* sqrt(a/(b*x) + 1)/(3*a*x) + 4*B*b**(3/2)*sqrt(a/(b*x) + 1)/(3*a**2)
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.26 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=\frac {4 \, \sqrt {b x^{2} + a x} B b}{3 \, a^{2} x} - \frac {16 \, \sqrt {b x^{2} + a x} A b^{2}}{15 \, a^{3} x} - \frac {2 \, \sqrt {b x^{2} + a x} B}{3 \, a x^{2}} + \frac {8 \, \sqrt {b x^{2} + a x} A b}{15 \, a^{2} x^{2}} - \frac {2 \, \sqrt {b x^{2} + a x} A}{5 \, a x^{3}} \] Input:
integrate((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x, algorithm="maxima")
Output:
4/3*sqrt(b*x^2 + a*x)*B*b/(a^2*x) - 16/15*sqrt(b*x^2 + a*x)*A*b^2/(a^3*x) - 2/3*sqrt(b*x^2 + a*x)*B/(a*x^2) + 8/15*sqrt(b*x^2 + a*x)*A*b/(a^2*x^2) - 2/5*sqrt(b*x^2 + a*x)*A/(a*x^3)
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=\frac {2 \, \sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (5 \, B a b^{4} - 4 \, A b^{5}\right )} {\left (b x + a\right )}}{a^{3}} - \frac {5 \, {\left (5 \, B a^{2} b^{4} - 4 \, A a b^{5}\right )}}{a^{3}}\right )} + \frac {15 \, {\left (B a^{3} b^{4} - A a^{2} b^{5}\right )}}{a^{3}}\right )} b}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}} {\left | b \right |}} \] Input:
integrate((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x, algorithm="giac")
Output:
2/15*sqrt(b*x + a)*((b*x + a)*(2*(5*B*a*b^4 - 4*A*b^5)*(b*x + a)/a^3 - 5*( 5*B*a^2*b^4 - 4*A*a*b^5)/a^3) + 15*(B*a^3*b^4 - A*a^2*b^5)/a^3)*b/(((b*x + a)*b - a*b)^(5/2)*abs(b))
Time = 0.44 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.68 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{5\,a}+\frac {x^2\,\left (16\,A\,b^2-20\,B\,a\,b\right )}{15\,a^3}+\frac {x\,\left (10\,B\,a^2-8\,A\,a\,b\right )}{15\,a^3}\right )}{x^{5/2}} \] Input:
int((A + B*x)/(x^(7/2)*(a + b*x)^(1/2)),x)
Output:
-((a + b*x)^(1/2)*((2*A)/(5*a) + (x^2*(16*A*b^2 - 20*B*a*b))/(15*a^3) + (x *(10*B*a^2 - 8*A*a*b))/(15*a^3)))/x^(5/2)
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x}{x^{7/2} \sqrt {a+b x}} \, dx=\frac {-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a^{2}}{5}-\frac {2 \sqrt {x}\, \sqrt {b x +a}\, a b x}{15}+\frac {4 \sqrt {x}\, \sqrt {b x +a}\, b^{2} x^{2}}{15}-\frac {4 \sqrt {b}\, b^{2} x^{3}}{15}}{a^{2} x^{3}} \] Input:
int((B*x+A)/x^(7/2)/(b*x+a)^(1/2),x)
Output:
(2*( - 3*sqrt(x)*sqrt(a + b*x)*a**2 - sqrt(x)*sqrt(a + b*x)*a*b*x + 2*sqrt (x)*sqrt(a + b*x)*b**2*x**2 - 2*sqrt(b)*b**2*x**3))/(15*a**2*x**3)