3.6.76 \(\int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=92 \[ \frac {32 b^3 \sqrt {a+b x}}{35 a^4 \sqrt {x}}-\frac {16 b^2 \sqrt {a+b x}}{35 a^3 x^{3/2}}+\frac {12 b \sqrt {a+b x}}{35 a^2 x^{5/2}}-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {45, 37} \begin {gather*} -\frac {16 b^2 \sqrt {a+b x}}{35 a^3 x^{3/2}}+\frac {32 b^3 \sqrt {a+b x}}{35 a^4 \sqrt {x}}+\frac {12 b \sqrt {a+b x}}{35 a^2 x^{5/2}}-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(9/2)*Sqrt[a + b*x]),x]

[Out]

(-2*Sqrt[a + b*x])/(7*a*x^(7/2)) + (12*b*Sqrt[a + b*x])/(35*a^2*x^(5/2)) - (16*b^2*Sqrt[a + b*x])/(35*a^3*x^(3
/2)) + (32*b^3*Sqrt[a + b*x])/(35*a^4*Sqrt[x])

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 45

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*} \int \frac {1}{x^{9/2} \sqrt {a+b x}} \, dx &=-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}-\frac {(6 b) \int \frac {1}{x^{7/2} \sqrt {a+b x}} \, dx}{7 a}\\ &=-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}+\frac {12 b \sqrt {a+b x}}{35 a^2 x^{5/2}}+\frac {\left (24 b^2\right ) \int \frac {1}{x^{5/2} \sqrt {a+b x}} \, dx}{35 a^2}\\ &=-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}+\frac {12 b \sqrt {a+b x}}{35 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a+b x}}{35 a^3 x^{3/2}}-\frac {\left (16 b^3\right ) \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{35 a^3}\\ &=-\frac {2 \sqrt {a+b x}}{7 a x^{7/2}}+\frac {12 b \sqrt {a+b x}}{35 a^2 x^{5/2}}-\frac {16 b^2 \sqrt {a+b x}}{35 a^3 x^{3/2}}+\frac {32 b^3 \sqrt {a+b x}}{35 a^4 \sqrt {x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 51, normalized size = 0.55 \begin {gather*} -\frac {2 \sqrt {a+b x} \left (5 a^3-6 a^2 b x+8 a b^2 x^2-16 b^3 x^3\right )}{35 a^4 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(9/2)*Sqrt[a + b*x]),x]

[Out]

(-2*Sqrt[a + b*x]*(5*a^3 - 6*a^2*b*x + 8*a*b^2*x^2 - 16*b^3*x^3))/(35*a^4*x^(7/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.10, size = 51, normalized size = 0.55 \begin {gather*} \frac {2 \sqrt {a+b x} \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right )}{35 a^4 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(9/2)*Sqrt[a + b*x]),x]

[Out]

(2*Sqrt[a + b*x]*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3))/(35*a^4*x^(7/2))

________________________________________________________________________________________

fricas [A]  time = 1.21, size = 45, normalized size = 0.49 \begin {gather*} \frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x + a}}{35 \, a^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x + a)/(a^4*x^(7/2))

________________________________________________________________________________________

giac [A]  time = 1.41, size = 82, normalized size = 0.89 \begin {gather*} -\frac {2 \, {\left (\frac {35 \, b^{7}}{a} - 2 \, {\left (\frac {35 \, b^{7}}{a^{2}} + 4 \, {\left (\frac {2 \, {\left (b x + a\right )} b^{7}}{a^{4}} - \frac {7 \, b^{7}}{a^{3}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a} b}{35 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 46, normalized size = 0.50 \begin {gather*} -\frac {2 \sqrt {b x +a}\, \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 a^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(b*x+a)^(1/2)*(-16*b^3*x^3+8*a*b^2*x^2-6*a^2*b*x+5*a^3)/x^(7/2)/a^4

________________________________________________________________________________________

maxima [A]  time = 1.30, size = 61, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (\frac {35 \, \sqrt {b x + a} b^{3}}{\sqrt {x}} - \frac {35 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} + \frac {21 \, {\left (b x + a\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} - \frac {5 \, {\left (b x + a\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}\right )}}{35 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(35*sqrt(b*x + a)*b^3/sqrt(x) - 35*(b*x + a)^(3/2)*b^2/x^(3/2) + 21*(b*x + a)^(5/2)*b/x^(5/2) - 5*(b*x +
a)^(7/2)/x^(7/2))/a^4

________________________________________________________________________________________

mupad [B]  time = 0.38, size = 47, normalized size = 0.51 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2}{7\,a}+\frac {16\,b^2\,x^2}{35\,a^3}-\frac {32\,b^3\,x^3}{35\,a^4}-\frac {12\,b\,x}{35\,a^2}\right )}{x^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(9/2)*(a + b*x)^(1/2)),x)

[Out]

-((a + b*x)^(1/2)*(2/(7*a) + (16*b^2*x^2)/(35*a^3) - (32*b^3*x^3)/(35*a^4) - (12*b*x)/(35*a^2)))/x^(7/2)

________________________________________________________________________________________

sympy [B]  time = 16.14, size = 488, normalized size = 5.30 \begin {gather*} - \frac {10 a^{6} b^{\frac {19}{2}} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} - \frac {18 a^{5} b^{\frac {21}{2}} x \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} - \frac {10 a^{4} b^{\frac {23}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {10 a^{3} b^{\frac {25}{2}} x^{3} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {60 a^{2} b^{\frac {27}{2}} x^{4} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {80 a b^{\frac {29}{2}} x^{5} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} + \frac {32 b^{\frac {31}{2}} x^{6} \sqrt {\frac {a}{b x} + 1}}{35 a^{7} b^{9} x^{3} + 105 a^{6} b^{10} x^{4} + 105 a^{5} b^{11} x^{5} + 35 a^{4} b^{12} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*a**6*b**(19/2)*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*
b**12*x**6) - 18*a**5*b**(21/2)*x*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*
x**5 + 35*a**4*b**12*x**6) - 10*a**4*b**(23/2)*x**2*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4
+ 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6) + 10*a**3*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 1
05*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6) + 60*a**2*b**(27/2)*x**4*sqrt(a/(b*x) + 1)/(35*
a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6) + 80*a*b**(29/2)*x**5*sqrt(a/
(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6) + 32*b**(31/2)
*x**6*sqrt(a/(b*x) + 1)/(35*a**7*b**9*x**3 + 105*a**6*b**10*x**4 + 105*a**5*b**11*x**5 + 35*a**4*b**12*x**6)

________________________________________________________________________________________