3.1.83 \(\int \frac {x^{13/2}}{(a x+b x^3)^{9/2}} \, dx\) [83]

Optimal. Leaf size=76 \[ \frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {4 x^{11/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 x^{9/2}}{105 a^3 \left (a x+b x^3\right )^{3/2}} \]

[Out]

1/7*x^(13/2)/a/(b*x^3+a*x)^(7/2)+4/35*x^(11/2)/a^2/(b*x^3+a*x)^(5/2)+8/105*x^(9/2)/a^3/(b*x^3+a*x)^(3/2)

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Rubi [A]
time = 0.08, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2040, 2039} \begin {gather*} \frac {8 x^{9/2}}{105 a^3 \left (a x+b x^3\right )^{3/2}}+\frac {4 x^{11/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(13/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(13/2)/(7*a*(a*x + b*x^3)^(7/2)) + (4*x^(11/2))/(35*a^2*(a*x + b*x^3)^(5/2)) + (8*x^(9/2))/(105*a^3*(a*x + b
*x^3)^(3/2))

Rule 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2040

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^{13/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {4 \int \frac {x^{11/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {4 x^{11/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 \int \frac {x^{9/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac {x^{13/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac {4 x^{11/2}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac {8 x^{9/2}}{105 a^3 \left (a x+b x^3\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 56, normalized size = 0.74 \begin {gather*} \frac {x^{9/2} \left (a+b x^2\right ) \left (35 a^2 x^3+28 a b x^5+8 b^2 x^7\right )}{105 a^3 \left (x \left (a+b x^2\right )\right )^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(13/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(x^(9/2)*(a + b*x^2)*(35*a^2*x^3 + 28*a*b*x^5 + 8*b^2*x^7))/(105*a^3*(x*(a + b*x^2))^(9/2))

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Maple [A]
time = 0.36, size = 50, normalized size = 0.66

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) x^{\frac {15}{2}} \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 a^{3} \left (b \,x^{3}+a x \right )^{\frac {9}{2}}}\)	\(48\)
default	\(\frac {x^{\frac {5}{2}} \sqrt {x \left (b \,x^{2}+a \right )}\, \left (8 b^{2} x^{4}+28 a b \,x^{2}+35 a^{2}\right )}{105 a^{3} \left (b \,x^{2}+a \right )^{4}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(b*x^3+a*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/105*x^(5/2)*(x*(b*x^2+a))^(1/2)*(8*b^2*x^4+28*a*b*x^2+35*a^2)/a^3/(b*x^2+a)^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(13/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]
time = 1.02, size = 87, normalized size = 1.14 \begin {gather*} \frac {{\left (8 \, b^{2} x^{6} + 28 \, a b x^{4} + 35 \, a^{2} x^{2}\right )} \sqrt {b x^{3} + a x} \sqrt {x}}{105 \, {\left (a^{3} b^{4} x^{8} + 4 \, a^{4} b^{3} x^{6} + 6 \, a^{5} b^{2} x^{4} + 4 \, a^{6} b x^{2} + a^{7}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

1/105*(8*b^2*x^6 + 28*a*b*x^4 + 35*a^2*x^2)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^3*b^4*x^8 + 4*a^4*b^3*x^6 + 6*a^5*b^2
*x^4 + 4*a^6*b*x^2 + a^7)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(13/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3655 deep

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Giac [A]
time = 0.99, size = 43, normalized size = 0.57 \begin {gather*} \frac {{\left (4 \, x^{2} {\left (\frac {2 \, b^{2} x^{2}}{a^{3}} + \frac {7 \, b}{a^{2}}\right )} + \frac {35}{a}\right )} x^{3}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/105*(4*x^2*(2*b^2*x^2/a^3 + 7*b/a^2) + 35/a)*x^3/(b*x^2 + a)^(7/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{13/2}}{{\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^(13/2)/(a*x + b*x^3)^(9/2),x)

[Out]

int(x^(13/2)/(a*x + b*x^3)^(9/2), x)

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