3.1.41 \(\int \frac {x^{23/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2015, 2014} \begin {gather*} -\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}}-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(19/2)/(7*b*(a*x + b*x^3)^(7/2)) - (6*x^(13/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (8*x^(7/2))/(35*b^3*(a*x + b
*x^3)^(3/2)) - (16*Sqrt[x])/(35*b^4*Sqrt[a*x + b*x^3])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2015

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \frac {x^{23/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {6 \int \frac {x^{17/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac {24 \int \frac {x^{11/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 b^2}\\ &=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}+\frac {16 \int \frac {x^{5/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{35 b^3}\\ &=-\frac {x^{19/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {6 x^{13/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {8 x^{7/2}}{35 b^3 \left (a x+b x^3\right )^{3/2}}-\frac {16 \sqrt {x}}{35 b^4 \sqrt {a x+b x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 66, normalized size = 0.65 \begin {gather*} -\frac {\sqrt {x} \left (16 a^3+56 a^2 b x^2+70 a b^2 x^4+35 b^3 x^6\right )}{35 b^4 \left (a+b x^2\right )^3 \sqrt {x \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/35*(Sqrt[x]*(16*a^3 + 56*a^2*b*x^2 + 70*a*b^2*x^4 + 35*b^3*x^6))/(b^4*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.11, size = 64, normalized size = 0.63 \begin {gather*} \frac {x^{9/2} \left (a+b x^2\right ) \left (-16 a^3-56 a^2 b x^2-70 a b^2 x^4-35 b^3 x^6\right )}{35 b^4 \left (x \left (a+b x^2\right )\right )^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(23/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(9/2)*(a + b*x^2)*(-16*a^3 - 56*a^2*b*x^2 - 70*a*b^2*x^4 - 35*b^3*x^6))/(35*b^4*(x*(a + b*x^2))^(9/2))

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 97, normalized size = 0.96 \begin {gather*} -\frac {{\left (35 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} + 16 \, a^{3}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (b^{8} x^{9} + 4 \, a b^{7} x^{7} + 6 \, a^{2} b^{6} x^{5} + 4 \, a^{3} b^{5} x^{3} + a^{4} b^{4} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

-1/35*(35*b^3*x^6 + 70*a*b^2*x^4 + 56*a^2*b*x^2 + 16*a^3)*sqrt(b*x^3 + a*x)*sqrt(x)/(b^8*x^9 + 4*a*b^7*x^7 + 6
*a^2*b^6*x^5 + 4*a^3*b^5*x^3 + a^4*b^4*x)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 64, normalized size = 0.63 \begin {gather*} \frac {16}{35 \, \sqrt {a} b^{4}} - \frac {35 \, {\left (b x^{2} + a\right )}^{3} - 35 \, {\left (b x^{2} + a\right )}^{2} a + 21 \, {\left (b x^{2} + a\right )} a^{2} - 5 \, a^{3}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

16/35/(sqrt(a)*b^4) - 1/35*(35*(b*x^2 + a)^3 - 35*(b*x^2 + a)^2*a + 21*(b*x^2 + a)*a^2 - 5*a^3)/((b*x^2 + a)^(
7/2)*b^4)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 59, normalized size = 0.58 \begin {gather*} -\frac {\left (b \,x^{2}+a \right ) \left (35 b^{3} x^{6}+70 a \,b^{2} x^{4}+56 a^{2} b \,x^{2}+16 a^{3}\right ) x^{\frac {9}{2}}}{35 \left (b \,x^{3}+a x \right )^{\frac {9}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(23/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/35*(b*x^2+a)*(35*b^3*x^6+70*a*b^2*x^4+56*a^2*b*x^2+16*a^3)*x^(9/2)/b^4/(b*x^3+a*x)^(9/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {23}{2}}}{{\left (b x^{3} + a x\right )}^{\frac {9}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(23/2)/(b*x^3 + a*x)^(9/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{23/2}}{{\left (b\,x^3+a\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(23/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(23/2)/(a*x + b*x^3)^(9/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(23/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________