3.91 \(\int \frac{1}{x^{3/2} (a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=180 \[ -\frac{1024 b^2 \sqrt{a x+b x^3}}{35 a^7 x^{3/2}}+\frac{512 b \sqrt{a x+b x^3}}{35 a^6 x^{7/2}}-\frac{384 \sqrt{a x+b x^3}}{35 a^5 x^{11/2}}+\frac{64}{7 a^4 x^{9/2} \sqrt{a x+b x^3}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}} \]

[Out]

1/(7*a*x^(3/2)*(a*x + b*x^3)^(7/2)) + 12/(35*a^2*x^(5/2)*(a*x + b*x^3)^(5/2)) + 8/(7*a^3*x^(7/2)*(a*x + b*x^3)
^(3/2)) + 64/(7*a^4*x^(9/2)*Sqrt[a*x + b*x^3]) - (384*Sqrt[a*x + b*x^3])/(35*a^5*x^(11/2)) + (512*b*Sqrt[a*x +
b*x^3])/(35*a^6*x^(7/2)) - (1024*b^2*Sqrt[a*x + b*x^3])/(35*a^7*x^(3/2))

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Rubi [A]  time = 0.284987, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ -\frac{1024 b^2 \sqrt{a x+b x^3}}{35 a^7 x^{3/2}}+\frac{512 b \sqrt{a x+b x^3}}{35 a^6 x^{7/2}}-\frac{384 \sqrt{a x+b x^3}}{35 a^5 x^{11/2}}+\frac{64}{7 a^4 x^{9/2} \sqrt{a x+b x^3}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a*x + b*x^3)^(9/2)),x]

[Out]

1/(7*a*x^(3/2)*(a*x + b*x^3)^(7/2)) + 12/(35*a^2*x^(5/2)*(a*x + b*x^3)^(5/2)) + 8/(7*a^3*x^(7/2)*(a*x + b*x^3)
^(3/2)) + 64/(7*a^4*x^(9/2)*Sqrt[a*x + b*x^3]) - (384*Sqrt[a*x + b*x^3])/(35*a^5*x^(11/2)) + (512*b*Sqrt[a*x +
b*x^3])/(35*a^6*x^(7/2)) - (1024*b^2*Sqrt[a*x + b*x^3])/(35*a^7*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{1}{x^{3/2} \left (a x+b x^3\right )^{9/2}} \, dx &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12 \int \frac{1}{x^{5/2} \left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{24 \int \frac{1}{x^{7/2} \left (a x+b x^3\right )^{5/2}} \, dx}{7 a^2}\\ &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{64 \int \frac{1}{x^{9/2} \left (a x+b x^3\right )^{3/2}} \, dx}{7 a^3}\\ &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{64}{7 a^4 x^{9/2} \sqrt{a x+b x^3}}+\frac{384 \int \frac{1}{x^{11/2} \sqrt{a x+b x^3}} \, dx}{7 a^4}\\ &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{64}{7 a^4 x^{9/2} \sqrt{a x+b x^3}}-\frac{384 \sqrt{a x+b x^3}}{35 a^5 x^{11/2}}-\frac{(1536 b) \int \frac{1}{x^{7/2} \sqrt{a x+b x^3}} \, dx}{35 a^5}\\ &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{64}{7 a^4 x^{9/2} \sqrt{a x+b x^3}}-\frac{384 \sqrt{a x+b x^3}}{35 a^5 x^{11/2}}+\frac{512 b \sqrt{a x+b x^3}}{35 a^6 x^{7/2}}+\frac{\left (1024 b^2\right ) \int \frac{1}{x^{3/2} \sqrt{a x+b x^3}} \, dx}{35 a^6}\\ &=\frac{1}{7 a x^{3/2} \left (a x+b x^3\right )^{7/2}}+\frac{12}{35 a^2 x^{5/2} \left (a x+b x^3\right )^{5/2}}+\frac{8}{7 a^3 x^{7/2} \left (a x+b x^3\right )^{3/2}}+\frac{64}{7 a^4 x^{9/2} \sqrt{a x+b x^3}}-\frac{384 \sqrt{a x+b x^3}}{35 a^5 x^{11/2}}+\frac{512 b \sqrt{a x+b x^3}}{35 a^6 x^{7/2}}-\frac{1024 b^2 \sqrt{a x+b x^3}}{35 a^7 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0310446, size = 99, normalized size = 0.55 \[ -\frac{\sqrt{x \left (a+b x^2\right )} \left (4480 a^2 b^4 x^8+2240 a^3 b^3 x^6+280 a^4 b^2 x^4-28 a^5 b x^2+7 a^6+3584 a b^5 x^{10}+1024 b^6 x^{12}\right )}{35 a^7 x^{11/2} \left (a+b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a*x + b*x^3)^(9/2)),x]

[Out]

-(Sqrt[x*(a + b*x^2)]*(7*a^6 - 28*a^5*b*x^2 + 280*a^4*b^2*x^4 + 2240*a^3*b^3*x^6 + 4480*a^2*b^4*x^8 + 3584*a*b
^5*x^10 + 1024*b^6*x^12))/(35*a^7*x^(11/2)*(a + b*x^2)^4)

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Maple [A]  time = 0.007, size = 92, normalized size = 0.5 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 1024\,{b}^{6}{x}^{12}+3584\,{b}^{5}{x}^{10}a+4480\,{b}^{4}{x}^{8}{a}^{2}+2240\,{b}^{3}{x}^{6}{a}^{3}+280\,{b}^{2}{x}^{4}{a}^{4}-28\,b{x}^{2}{a}^{5}+7\,{a}^{6} \right ) }{35\,{a}^{7}}{\frac{1}{\sqrt{x}}} \left ( b{x}^{3}+ax \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(b*x^2+a)*(1024*b^6*x^12+3584*a*b^5*x^10+4480*a^2*b^4*x^8+2240*a^3*b^3*x^6+280*a^4*b^2*x^4-28*a^5*b*x^2+
7*a^6)/x^(1/2)/a^7/(b*x^3+a*x)^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a x\right )}^{\frac{9}{2}} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.28488, size = 298, normalized size = 1.66 \begin{align*} -\frac{{\left (1024 \, b^{6} x^{12} + 3584 \, a b^{5} x^{10} + 4480 \, a^{2} b^{4} x^{8} + 2240 \, a^{3} b^{3} x^{6} + 280 \, a^{4} b^{2} x^{4} - 28 \, a^{5} b x^{2} + 7 \, a^{6}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{35 \,{\left (a^{7} b^{4} x^{14} + 4 \, a^{8} b^{3} x^{12} + 6 \, a^{9} b^{2} x^{10} + 4 \, a^{10} b x^{8} + a^{11} x^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(1024*b^6*x^12 + 3584*a*b^5*x^10 + 4480*a^2*b^4*x^8 + 2240*a^3*b^3*x^6 + 280*a^4*b^2*x^4 - 28*a^5*b*x^2
+ 7*a^6)*sqrt(b*x^3 + a*x)*sqrt(x)/(a^7*b^4*x^14 + 4*a^8*b^3*x^12 + 6*a^9*b^2*x^10 + 4*a^10*b*x^8 + a^11*x^6)

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

[Out]

Timed out

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Giac [A]  time = 1.40183, size = 136, normalized size = 0.76 \begin{align*} -\frac{{\left ({\left (2 \, x^{2}{\left (\frac{281 \, b^{6} x^{2}}{a^{7}} + \frac{896 \, b^{5}}{a^{6}}\right )} + \frac{1925 \, b^{4}}{a^{5}}\right )} x^{2} + \frac{700 \, b^{3}}{a^{4}}\right )} x}{35 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} - \frac{{\left (b + \frac{a}{x^{2}}\right )}^{\frac{5}{2}} - 10 \,{\left (b + \frac{a}{x^{2}}\right )}^{\frac{3}{2}} b + 75 \, \sqrt{b + \frac{a}{x^{2}}} b^{2}}{5 \, a^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*((2*x^2*(281*b^6*x^2/a^7 + 896*b^5/a^6) + 1925*b^4/a^5)*x^2 + 700*b^3/a^4)*x/(b*x^2 + a)^(7/2) - 1/5*((b
+ a/x^2)^(5/2) - 10*(b + a/x^2)^(3/2)*b + 75*sqrt(b + a/x^2)*b^2)/a^7