\(\int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx\) [495]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=-\frac {2 (a+b x)^{3/2}}{5 a x^{5/2}}+\frac {4 b (a+b x)^{3/2}}{15 a^2 x^{3/2}} \]

[Out]

-2/5*(b*x+a)^(3/2)/a/x^(5/2)+4/15*b*(b*x+a)^(3/2)/a^2/x^(3/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {47, 37} \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=\frac {4 b (a+b x)^{3/2}}{15 a^2 x^{3/2}}-\frac {2 (a+b x)^{3/2}}{5 a x^{5/2}} \]

[In]

Int[Sqrt[a + b*x]/x^(7/2),x]

[Out]

(-2*(a + b*x)^(3/2))/(5*a*x^(5/2)) + (4*b*(a + b*x)^(3/2))/(15*a^2*x^(3/2))

Rule 37

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] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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] - Dist[d*(Simplify[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] && NeQ[b*c - a*d, 0] && 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] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (a+b x)^{3/2}}{5 a x^{5/2}}-\frac {(2 b) \int \frac {\sqrt {a+b x}}{x^{5/2}} \, dx}{5 a} \\ & = -\frac {2 (a+b x)^{3/2}}{5 a x^{5/2}}+\frac {4 b (a+b x)^{3/2}}{15 a^2 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (3 a^2+a b x-2 b^2 x^2\right )}{15 a^2 x^{5/2}} \]

[In]

Integrate[Sqrt[a + b*x]/x^(7/2),x]

[Out]

(-2*Sqrt[a + b*x]*(3*a^2 + a*b*x - 2*b^2*x^2))/(15*a^2*x^(5/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.55

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (-2 b x +3 a \right )}{15 x^{\frac {5}{2}} a^{2}}\)	\(24\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (-2 b^{2} x^{2}+a b x +3 a^{2}\right )}{15 x^{\frac {5}{2}} a^{2}}\)	\(34\)
default	\(-\frac {\sqrt {b x +a}}{2 x^{\frac {5}{2}}}-\frac {a \left (-\frac {2 \sqrt {b x +a}}{5 a \,x^{\frac {5}{2}}}-\frac {4 b \left (-\frac {2 \sqrt {b x +a}}{3 a \,x^{\frac {3}{2}}}+\frac {4 b \sqrt {b x +a}}{3 a^{2} \sqrt {x}}\right )}{5 a}\right )}{4}\)	\(71\)

[In]

int((b*x+a)^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*(b*x+a)^(3/2)*(-2*b*x+3*a)/x^(5/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=\frac {2 \, {\left (2 \, b^{2} x^{2} - a b x - 3 \, a^{2}\right )} \sqrt {b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \]

[In]

integrate((b*x+a)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

2/15*(2*b^2*x^2 - a*b*x - 3*a^2)*sqrt(b*x + a)/(a^2*x^(5/2))

Sympy [A] (verification not implemented)

Time = 3.46 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=- \frac {2 \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{5 x^{2}} - \frac {2 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a x} + \frac {4 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}}{15 a^{2}} \]

[In]

integrate((b*x+a)**(1/2)/x**(7/2),x)

[Out]

-2*sqrt(b)*sqrt(a/(b*x) + 1)/(5*x**2) - 2*b**(3/2)*sqrt(a/(b*x) + 1)/(15*a*x) + 4*b**(5/2)*sqrt(a/(b*x) + 1)/(
15*a**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}\right )}}{15 \, a^{2}} \]

[In]

integrate((b*x+a)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

2/15*(5*(b*x + a)^(3/2)*b/x^(3/2) - 3*(b*x + a)^(5/2)/x^(5/2))/a^2

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (b x + a\right )} b^{5}}{a^{2}} - \frac {5 \, b^{5}}{a}\right )} {\left (b x + a\right )}^{\frac {3}{2}} b}{15 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}} {\left | b \right |}} \]

[In]

integrate((b*x+a)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

2/15*(2*(b*x + a)*b^5/a^2 - 5*b^5/a)*(b*x + a)^(3/2)*b/(((b*x + a)*b - a*b)^(5/2)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,b\,x}{15\,a}-\frac {4\,b^2\,x^2}{15\,a^2}+\frac {2}{5}\right )}{x^{5/2}} \]

[In]

int((a + b*x)^(1/2)/x^(7/2),x)

[Out]

-((a + b*x)^(1/2)*((2*b*x)/(15*a) - (4*b^2*x^2)/(15*a^2) + 2/5))/x^(5/2)