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

Optimal. Leaf size=152 \[ \frac{256 b \sqrt{a x+b x^3}}{21 a^6 x^{3/2}}-\frac{128 \sqrt{a x+b x^3}}{21 a^5 x^{7/2}}+\frac{32}{7 a^4 x^{5/2} \sqrt{a x+b x^3}}+\frac{16}{21 a^3 x^{3/2} \left (a x+b x^3\right )^{3/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}} \]

[Out]

Sqrt[x]/(7*a*(a*x + b*x^3)^(7/2)) + 2/(7*a^2*Sqrt[x]*(a*x + b*x^3)^(5/2)) + 16/(21*a^3*x^(3/2)*(a*x + b*x^3)^(
3/2)) + 32/(7*a^4*x^(5/2)*Sqrt[a*x + b*x^3]) - (128*Sqrt[a*x + b*x^3])/(21*a^5*x^(7/2)) + (256*b*Sqrt[a*x + b*
x^3])/(21*a^6*x^(3/2))

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Rubi [A]  time = 0.233374, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ \frac{256 b \sqrt{a x+b x^3}}{21 a^6 x^{3/2}}-\frac{128 \sqrt{a x+b x^3}}{21 a^5 x^{7/2}}+\frac{32}{7 a^4 x^{5/2} \sqrt{a x+b x^3}}+\frac{16}{21 a^3 x^{3/2} \left (a x+b x^3\right )^{3/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a*x + b*x^3)^(9/2),x]

[Out]

Sqrt[x]/(7*a*(a*x + b*x^3)^(7/2)) + 2/(7*a^2*Sqrt[x]*(a*x + b*x^3)^(5/2)) + 16/(21*a^3*x^(3/2)*(a*x + b*x^3)^(
3/2)) + 32/(7*a^4*x^(5/2)*Sqrt[a*x + b*x^3]) - (128*Sqrt[a*x + b*x^3])/(21*a^5*x^(7/2)) + (256*b*Sqrt[a*x + b*
x^3])/(21*a^6*x^(3/2))

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])

Rule 2016

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*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

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])

Rubi steps

\begin{align*} \int \frac{\sqrt{x}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{10 \int \frac{1}{\sqrt{x} \left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{16 \int \frac{1}{x^{3/2} \left (a x+b x^3\right )^{5/2}} \, dx}{7 a^2}\\ &=\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{16}{21 a^3 x^{3/2} \left (a x+b x^3\right )^{3/2}}+\frac{32 \int \frac{1}{x^{5/2} \left (a x+b x^3\right )^{3/2}} \, dx}{7 a^3}\\ &=\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{16}{21 a^3 x^{3/2} \left (a x+b x^3\right )^{3/2}}+\frac{32}{7 a^4 x^{5/2} \sqrt{a x+b x^3}}+\frac{128 \int \frac{1}{x^{7/2} \sqrt{a x+b x^3}} \, dx}{7 a^4}\\ &=\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{16}{21 a^3 x^{3/2} \left (a x+b x^3\right )^{3/2}}+\frac{32}{7 a^4 x^{5/2} \sqrt{a x+b x^3}}-\frac{128 \sqrt{a x+b x^3}}{21 a^5 x^{7/2}}-\frac{(256 b) \int \frac{1}{x^{3/2} \sqrt{a x+b x^3}} \, dx}{21 a^5}\\ &=\frac{\sqrt{x}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{2}{7 a^2 \sqrt{x} \left (a x+b x^3\right )^{5/2}}+\frac{16}{21 a^3 x^{3/2} \left (a x+b x^3\right )^{3/2}}+\frac{32}{7 a^4 x^{5/2} \sqrt{a x+b x^3}}-\frac{128 \sqrt{a x+b x^3}}{21 a^5 x^{7/2}}+\frac{256 b \sqrt{a x+b x^3}}{21 a^6 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.024839, size = 88, normalized size = 0.58 \[ \frac{\sqrt{x \left (a+b x^2\right )} \left (1120 a^2 b^3 x^6+560 a^3 b^2 x^4+70 a^4 b x^2-7 a^5+896 a b^4 x^8+256 b^5 x^{10}\right )}{21 a^6 x^{7/2} \left (a+b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(a + b*x^2)]*(-7*a^5 + 70*a^4*b*x^2 + 560*a^3*b^2*x^4 + 1120*a^2*b^3*x^6 + 896*a*b^4*x^8 + 256*b^5*x^1
0))/(21*a^6*x^(7/2)*(a + b*x^2)^4)

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Maple [A]  time = 0.006, size = 81, normalized size = 0.5 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( -256\,{b}^{5}{x}^{10}-896\,{b}^{4}{x}^{8}a-1120\,{b}^{3}{x}^{6}{a}^{2}-560\,{b}^{2}{x}^{4}{a}^{3}-70\,b{x}^{2}{a}^{4}+7\,{a}^{5} \right ) }{21\,{a}^{6}}{x}^{{\frac{3}{2}}} \left ( b{x}^{3}+ax \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x^3+a*x)^(9/2),x)

[Out]

-1/21*x^(3/2)*(b*x^2+a)*(-256*b^5*x^10-896*a*b^4*x^8-1120*a^2*b^3*x^6-560*a^3*b^2*x^4-70*a^4*b*x^2+7*a^5)/a^6/
(b*x^3+a*x)^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{{\left (b x^{3} + a x\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.85057, size = 265, normalized size = 1.74 \begin{align*} \frac{{\left (256 \, b^{5} x^{10} + 896 \, a b^{4} x^{8} + 1120 \, a^{2} b^{3} x^{6} + 560 \, a^{3} b^{2} x^{4} + 70 \, a^{4} b x^{2} - 7 \, a^{5}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{21 \,{\left (a^{6} b^{4} x^{12} + 4 \, a^{7} b^{3} x^{10} + 6 \, a^{8} b^{2} x^{8} + 4 \, a^{9} b x^{6} + a^{10} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

1/21*(256*b^5*x^10 + 896*a*b^4*x^8 + 1120*a^2*b^3*x^6 + 560*a^3*b^2*x^4 + 70*a^4*b*x^2 - 7*a^5)*sqrt(b*x^3 + a
*x)*sqrt(x)/(a^6*b^4*x^12 + 4*a^7*b^3*x^10 + 6*a^8*b^2*x^8 + 4*a^9*b*x^6 + a^10*x^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.38026, size = 116, normalized size = 0.76 \begin{align*} \frac{{\left ({\left (x^{2}{\left (\frac{158 \, b^{5} x^{2}}{a^{6}} + \frac{511 \, b^{4}}{a^{5}}\right )} + \frac{560 \, b^{3}}{a^{4}}\right )} x^{2} + \frac{210 \, b^{2}}{a^{3}}\right )} x}{21 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} - \frac{{\left (b + \frac{a}{x^{2}}\right )}^{\frac{3}{2}} - 15 \, \sqrt{b + \frac{a}{x^{2}}} b}{3 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/21*((x^2*(158*b^5*x^2/a^6 + 511*b^4/a^5) + 560*b^3/a^4)*x^2 + 210*b^2/a^3)*x/(b*x^2 + a)^(7/2) - 1/3*((b + a
/x^2)^(3/2) - 15*sqrt(b + a/x^2)*b)/a^6