3.1.45 \(\int \frac {x^{15/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=51 \[ -\frac {2 x^{5/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, 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 {2 x^{5/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac {x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(15/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(11/2)/(7*b*(a*x + b*x^3)^(7/2)) - (2*x^(5/2))/(35*b^2*(a*x + b*x^3)^(5/2))

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^{15/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac {x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac {2 \int \frac {x^{9/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac {x^{11/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac {2 x^{5/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.86 \begin {gather*} -\frac {\sqrt {x} \left (2 a+7 b x^2\right )}{35 b^2 \left (a+b x^2\right )^3 \sqrt {x \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(15/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-1/35*(Sqrt[x]*(2*a + 7*b*x^2))/(b^2*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

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IntegrateAlgebraic [A]  time = 0.63, size = 35, normalized size = 0.69 \begin {gather*} -\frac {x^{7/2} \left (2 a+7 b x^2\right )}{35 b^2 \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(15/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 75, normalized size = 1.47 \begin {gather*} -\frac {\sqrt {b x^{3} + a x} {\left (7 \, b x^{2} + 2 \, a\right )} \sqrt {x}}{35 \, {\left (b^{6} x^{9} + 4 \, a b^{5} x^{7} + 6 \, a^{2} b^{4} x^{5} + 4 \, a^{3} b^{3} x^{3} + a^{4} b^{2} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 33, normalized size = 0.65 \begin {gather*} -\frac {7 \, b x^{2} + 2 \, a}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} + \frac {2}{35 \, a^{\frac {5}{2}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/35*(7*b*x^2 + 2*a)/((b*x^2 + a)^(7/2)*b^2) + 2/35/(a^(5/2)*b^2)

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maple [A]  time = 0.06, size = 37, normalized size = 0.73 \begin {gather*} -\frac {\left (b \,x^{2}+a \right ) \left (7 b \,x^{2}+2 a \right ) x^{\frac {9}{2}}}{35 \left (b \,x^{3}+a x \right )^{\frac {9}{2}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/35*(b*x^2+a)*(7*b*x^2+2*a)*x^(9/2)/b^2/(b*x^3+a*x)^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {15}{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^(15/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(15/2)/(b*x^3 + a*x)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{15/2}}{{\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^(15/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(15/2)/(a*x + b*x^3)^(9/2), x)

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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**(15/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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