\(\int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx\) [496]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {8 b (a+b x)^{3/2}}{35 a^2 x^{5/2}}-\frac {16 b^2 (a+b x)^{3/2}}{105 a^3 x^{3/2}} \]

[Out]

-2/7*(b*x+a)^(3/2)/a/x^(7/2)+8/35*b*(b*x+a)^(3/2)/a^2/x^(5/2)-16/105*b^2*(b*x+a)^(3/2)/a^3/x^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {47, 37} \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {16 b^2 (a+b x)^{3/2}}{105 a^3 x^{3/2}}+\frac {8 b (a+b x)^{3/2}}{35 a^2 x^{5/2}}-\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}} \]

[In]

Int[Sqrt[a + b*x]/x^(9/2),x]

[Out]

(-2*(a + b*x)^(3/2))/(7*a*x^(7/2)) + (8*b*(a + b*x)^(3/2))/(35*a^2*x^(5/2)) - (16*b^2*(a + b*x)^(3/2))/(105*a^
3*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}}{7 a x^{7/2}}-\frac {(4 b) \int \frac {\sqrt {a+b x}}{x^{7/2}} \, dx}{7 a} \\ & = -\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {8 b (a+b x)^{3/2}}{35 a^2 x^{5/2}}+\frac {\left (8 b^2\right ) \int \frac {\sqrt {a+b x}}{x^{5/2}} \, dx}{35 a^2} \\ & = -\frac {2 (a+b x)^{3/2}}{7 a x^{7/2}}+\frac {8 b (a+b x)^{3/2}}{35 a^2 x^{5/2}}-\frac {16 b^2 (a+b x)^{3/2}}{105 a^3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (15 a^3+3 a^2 b x-4 a b^2 x^2+8 b^3 x^3\right )}{105 a^3 x^{7/2}} \]

[In]

Integrate[Sqrt[a + b*x]/x^(9/2),x]

[Out]

(-2*Sqrt[a + b*x]*(15*a^3 + 3*a^2*b*x - 4*a*b^2*x^2 + 8*b^3*x^3))/(105*a^3*x^(7/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51

method	result	size
gosper	\(-\frac {2 \left (b x +a \right )^{\frac {3}{2}} \left (8 b^{2} x^{2}-12 a b x +15 a^{2}\right )}{105 x^{\frac {7}{2}} a^{3}}\)	\(35\)
risch	\(-\frac {2 \sqrt {b x +a}\, \left (8 b^{3} x^{3}-4 a \,b^{2} x^{2}+3 a^{2} b x +15 a^{3}\right )}{105 x^{\frac {7}{2}} a^{3}}\)	\(46\)
default	\(-\frac {\sqrt {b x +a}}{3 x^{\frac {7}{2}}}-\frac {a \left (-\frac {2 \sqrt {b x +a}}{7 a \,x^{\frac {7}{2}}}-\frac {6 b \left (-\frac {2 \sqrt {b x +a}}{5 a \,x^{\frac {5}{2}}}-\frac {4 b \left (-\frac {2 \sqrt {b x +a}}{3 a \,x^{\frac {3}{2}}}+\frac {4 b \sqrt {b x +a}}{3 a^{2} \sqrt {x}}\right )}{5 a}\right )}{7 a}\right )}{6}\)	\(93\)

[In]

int((b*x+a)^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)

[Out]

-2/105*(b*x+a)^(3/2)*(8*b^2*x^2-12*a*b*x+15*a^2)/x^(7/2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} - 4 \, a b^{2} x^{2} + 3 \, a^{2} b x + 15 \, a^{3}\right )} \sqrt {b x + a}}{105 \, a^{3} x^{\frac {7}{2}}} \]

[In]

integrate((b*x+a)^(1/2)/x^(9/2),x, algorithm="fricas")

[Out]

-2/105*(8*b^3*x^3 - 4*a*b^2*x^2 + 3*a^2*b*x + 15*a^3)*sqrt(b*x + a)/(a^3*x^(7/2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (63) = 126\).

Time = 9.99 (sec) , antiderivative size = 347, normalized size of antiderivative = 5.10 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=- \frac {30 a^{5} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {66 a^{4} b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {34 a^{3} b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {6 a^{2} b^{\frac {15}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {24 a b^{\frac {17}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} - \frac {16 b^{\frac {19}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{105 a^{5} b^{4} x^{3} + 210 a^{4} b^{5} x^{4} + 105 a^{3} b^{6} x^{5}} \]

[In]

integrate((b*x+a)**(1/2)/x**(9/2),x)

[Out]

-30*a**5*b**(9/2)*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 66*a**4*b
**(11/2)*x*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 34*a**3*b**(13/2
)*x**2*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 6*a**2*b**(15/2)*x**
3*sqrt(a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 24*a*b**(17/2)*x**4*sqrt(
a/(b*x) + 1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5) - 16*b**(19/2)*x**5*sqrt(a/(b*x) +
1)/(105*a**5*b**4*x**3 + 210*a**4*b**5*x**4 + 105*a**3*b**6*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {2 \, {\left (\frac {35 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} - \frac {42 \, {\left (b x + a\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}\right )}}{105 \, a^{3}} \]

[In]

integrate((b*x+a)^(1/2)/x^(9/2),x, algorithm="maxima")

[Out]

-2/105*(35*(b*x + a)^(3/2)*b^2/x^(3/2) - 42*(b*x + a)^(5/2)*b/x^(5/2) + 15*(b*x + a)^(7/2)/x^(7/2))/a^3

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {2 \, {\left (\frac {35 \, b^{7}}{a} + 4 \, {\left (\frac {2 \, {\left (b x + a\right )} b^{7}}{a^{3}} - \frac {7 \, b^{7}}{a^{2}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {3}{2}} b}{105 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} {\left | b \right |}} \]

[In]

integrate((b*x+a)^(1/2)/x^(9/2),x, algorithm="giac")

[Out]

-2/105*(35*b^7/a + 4*(2*(b*x + a)*b^7/a^3 - 7*b^7/a^2)*(b*x + a))*(b*x + a)^(3/2)*b/(((b*x + a)*b - a*b)^(7/2)
*abs(b))

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {a+b x}}{x^{9/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {16\,b^3\,x^3}{105\,a^3}-\frac {8\,b^2\,x^2}{105\,a^2}+\frac {2\,b\,x}{35\,a}+\frac {2}{7}\right )}{x^{7/2}} \]

[In]

int((a + b*x)^(1/2)/x^(9/2),x)

[Out]

-((a + b*x)^(1/2)*((16*b^3*x^3)/(105*a^3) - (8*b^2*x^2)/(105*a^2) + (2*b*x)/(35*a) + 2/7))/x^(7/2)