3.1.44 \(\int \frac {x^{17/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=51 \[ \frac {2 x^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

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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^{15/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(17/2)/(7*a*(a*x + b*x^3)^(7/2)) + (2*x^(15/2))/(35*a^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^{17/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {2 \int \frac {x^{15/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac {x^{17/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {2 x^{15/2}}{35 a^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 {x^{9/2} \sqrt {x \left (a+b x^2\right )} \left (7 a+2 b x^2\right )}{35 a^2 \left (a+b x^2\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(9/2)*Sqrt[x*(a + b*x^2)]*(7*a + 2*b*x^2))/(35*a^2*(a + b*x^2)^4)

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IntegrateAlgebraic [A]  time = 0.11, size = 42, normalized size = 0.82 \begin {gather*} \frac {x^{19/2} \left (a+b x^2\right ) \left (7 a+2 b x^2\right )}{35 a^2 \left (x \left (a+b x^2\right )\right )^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(17/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(19/2)*(a + b*x^2)*(7*a + 2*b*x^2))/(35*a^2*(x*(a + b*x^2))^(9/2))

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fricas [A]  time = 0.43, size = 76, normalized size = 1.49 \begin {gather*} \frac {{\left (2 \, b x^{6} + 7 \, a x^{4}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{35 \, {\left (a^{2} b^{4} x^{8} + 4 \, a^{3} b^{3} x^{6} + 6 \, a^{4} b^{2} x^{4} + 4 \, a^{5} b x^{2} + a^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 29, normalized size = 0.57 \begin {gather*} \frac {x^{5} {\left (\frac {2 \, b x^{2}}{a^{2}} + \frac {7}{a}\right )}}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

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

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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 (2 b \,x^{2}+7 a \right ) x^{\frac {19}{2}}}{35 \left (b \,x^{3}+a x \right )^{\frac {9}{2}} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/35*(b*x^2+a)*x^(19/2)*(2*b*x^2+7*a)/a^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 {17}{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^(17/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(17/2)/(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^{17/2}}{{\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^(17/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(17/2)/(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**(17/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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