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

Optimal. Leaf size=159 \[ \frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3}{a^4 x^{3/2} \sqrt{a x+b x^3}}-\frac{9 \sqrt{a x+b x^3}}{2 a^5 x^{5/2}}+\frac{9 b \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b x^3}}\right )}{2 a^{11/2}}+\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

[Out]

x^(3/2)/(7*a*(a*x + b*x^3)^(7/2)) + (9*Sqrt[x])/(35*a^2*(a*x + b*x^3)^(5/2)) + 3/(5*a^3*Sqrt[x]*(a*x + b*x^3)^
(3/2)) + 3/(a^4*x^(3/2)*Sqrt[a*x + b*x^3]) - (9*Sqrt[a*x + b*x^3])/(2*a^5*x^(5/2)) + (9*b*ArcTanh[(Sqrt[a]*Sqr
t[x])/Sqrt[a*x + b*x^3]])/(2*a^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.240548, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2023, 2025, 2029, 206} \[ \frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3}{a^4 x^{3/2} \sqrt{a x+b x^3}}-\frac{9 \sqrt{a x+b x^3}}{2 a^5 x^{5/2}}+\frac{9 b \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b x^3}}\right )}{2 a^{11/2}}+\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a*x + b*x^3)^(9/2),x]

[Out]

x^(3/2)/(7*a*(a*x + b*x^3)^(7/2)) + (9*Sqrt[x])/(35*a^2*(a*x + b*x^3)^(5/2)) + 3/(5*a^3*Sqrt[x]*(a*x + b*x^3)^
(3/2)) + 3/(a^4*x^(3/2)*Sqrt[a*x + b*x^3]) - (9*Sqrt[a*x + b*x^3])/(2*a^5*x^(5/2)) + (9*b*ArcTanh[(Sqrt[a]*Sqr
t[x])/Sqrt[a*x + b*x^3]])/(2*a^(11/2))

Rule 2023

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, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2025

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{3/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \int \frac{\sqrt{x}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 a}\\ &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{9 \int \frac{1}{\sqrt{x} \left (a x+b x^3\right )^{5/2}} \, dx}{5 a^2}\\ &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3 \int \frac{1}{x^{3/2} \left (a x+b x^3\right )^{3/2}} \, dx}{a^3}\\ &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3}{a^4 x^{3/2} \sqrt{a x+b x^3}}+\frac{9 \int \frac{1}{x^{5/2} \sqrt{a x+b x^3}} \, dx}{a^4}\\ &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3}{a^4 x^{3/2} \sqrt{a x+b x^3}}-\frac{9 \sqrt{a x+b x^3}}{2 a^5 x^{5/2}}-\frac{(9 b) \int \frac{1}{\sqrt{x} \sqrt{a x+b x^3}} \, dx}{2 a^5}\\ &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3}{a^4 x^{3/2} \sqrt{a x+b x^3}}-\frac{9 \sqrt{a x+b x^3}}{2 a^5 x^{5/2}}+\frac{(9 b) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a x+b x^3}}\right )}{2 a^5}\\ &=\frac{x^{3/2}}{7 a \left (a x+b x^3\right )^{7/2}}+\frac{9 \sqrt{x}}{35 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{3}{5 a^3 \sqrt{x} \left (a x+b x^3\right )^{3/2}}+\frac{3}{a^4 x^{3/2} \sqrt{a x+b x^3}}-\frac{9 \sqrt{a x+b x^3}}{2 a^5 x^{5/2}}+\frac{9 b \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b x^3}}\right )}{2 a^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0196505, size = 44, normalized size = 0.28 \[ -\frac{b x^{7/2} \, _2F_1\left (-\frac{7}{2},2;-\frac{5}{2};\frac{b x^2}{a}+1\right )}{7 a^2 \left (x \left (a+b x^2\right )\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^(7/2)*Hypergeometric2F1[-7/2, 2, -5/2, 1 + (b*x^2)/a])/(7*a^2*(x*(a + b*x^2))^(7/2))

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 234, normalized size = 1.5 \begin{align*}{\frac{1}{70\, \left ( b{x}^{2}+a \right ) ^{4}}\sqrt{x \left ( b{x}^{2}+a \right ) } \left ( 315\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{8}{b}^{4}\sqrt{b{x}^{2}+a}-315\,\sqrt{a}{x}^{8}{b}^{4}+945\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{6}a{b}^{3}\sqrt{b{x}^{2}+a}-1050\,{a}^{3/2}{x}^{6}{b}^{3}+945\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{4}{a}^{2}{b}^{2}\sqrt{b{x}^{2}+a}-1218\,{a}^{5/2}{x}^{4}{b}^{2}+315\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{2}{a}^{3}b\sqrt{b{x}^{2}+a}-528\,{a}^{7/2}{x}^{2}b-35\,{a}^{9/2} \right ){a}^{-{\frac{11}{2}}}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*(x*(b*x^2+a))^(1/2)/a^(11/2)*(315*ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a)/x)*x^8*b^4*(b*x^2+a)^(1/2)-315*a^(1/2)
*x^8*b^4+945*ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a)/x)*x^6*a*b^3*(b*x^2+a)^(1/2)-1050*a^(3/2)*x^6*b^3+945*ln(2*(a^(1
/2)*(b*x^2+a)^(1/2)+a)/x)*x^4*a^2*b^2*(b*x^2+a)^(1/2)-1218*a^(5/2)*x^4*b^2+315*ln(2*(a^(1/2)*(b*x^2+a)^(1/2)+a
)/x)*x^2*a^3*b*(b*x^2+a)^(1/2)-528*a^(7/2)*x^2*b-35*a^(9/2))/x^(5/2)/(b*x^2+a)^4

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.3772, size = 884, normalized size = 5.56 \begin{align*} \left [\frac{315 \,{\left (b^{5} x^{11} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{7} + 4 \, a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )} \sqrt{a} \log \left (\frac{b x^{3} + 2 \, a x + 2 \, \sqrt{b x^{3} + a x} \sqrt{a} \sqrt{x}}{x^{3}}\right ) - 2 \,{\left (315 \, a b^{4} x^{8} + 1050 \, a^{2} b^{3} x^{6} + 1218 \, a^{3} b^{2} x^{4} + 528 \, a^{4} b x^{2} + 35 \, a^{5}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{140 \,{\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}}, -\frac{315 \,{\left (b^{5} x^{11} + 4 \, a b^{4} x^{9} + 6 \, a^{2} b^{3} x^{7} + 4 \, a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a x} \sqrt{-a}}{a \sqrt{x}}\right ) +{\left (315 \, a b^{4} x^{8} + 1050 \, a^{2} b^{3} x^{6} + 1218 \, a^{3} b^{2} x^{4} + 528 \, a^{4} b x^{2} + 35 \, a^{5}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{70 \,{\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/140*(315*(b^5*x^11 + 4*a*b^4*x^9 + 6*a^2*b^3*x^7 + 4*a^3*b^2*x^5 + a^4*b*x^3)*sqrt(a)*log((b*x^3 + 2*a*x +
2*sqrt(b*x^3 + a*x)*sqrt(a)*sqrt(x))/x^3) - 2*(315*a*b^4*x^8 + 1050*a^2*b^3*x^6 + 1218*a^3*b^2*x^4 + 528*a^4*b
*x^2 + 35*a^5)*sqrt(b*x^3 + a*x)*sqrt(x))/(a^6*b^4*x^11 + 4*a^7*b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x
^3), -1/70*(315*(b^5*x^11 + 4*a*b^4*x^9 + 6*a^2*b^3*x^7 + 4*a^3*b^2*x^5 + a^4*b*x^3)*sqrt(-a)*arctan(sqrt(b*x^
3 + a*x)*sqrt(-a)/(a*sqrt(x))) + (315*a*b^4*x^8 + 1050*a^2*b^3*x^6 + 1218*a^3*b^2*x^4 + 528*a^4*b*x^2 + 35*a^5
)*sqrt(b*x^3 + a*x)*sqrt(x))/(a^6*b^4*x^11 + 4*a^7*b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x^3)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.32919, size = 142, normalized size = 0.89 \begin{align*} -\frac{1}{70} \, b{\left (\frac{315 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{5}} + \frac{2 \,{\left (140 \,{\left (b x^{2} + a\right )}^{3} + 35 \,{\left (b x^{2} + a\right )}^{2} a + 14 \,{\left (b x^{2} + a\right )} a^{2} + 5 \, a^{3}\right )}}{{\left (b x^{2} + a\right )}^{\frac{7}{2}} a^{5}} + \frac{35 \, \sqrt{b x^{2} + a}}{a^{5} b x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*b*(315*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^5) + 2*(140*(b*x^2 + a)^3 + 35*(b*x^2 + a)^2*a + 14*
(b*x^2 + a)*a^2 + 5*a^3)/((b*x^2 + a)^(7/2)*a^5) + 35*sqrt(b*x^2 + a)/(a^5*b*x^2))