3.75 \(\int \frac{x^{29/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=159 \[ -\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{3 x^{7/2}}{b^4 \sqrt{a x+b x^3}}+\frac{9 \sqrt{x} \sqrt{a x+b x^3}}{2 b^5}-\frac{9 a \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x+b x^3}}\right )}{2 b^{11/2}}-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[Out]

-x^(25/2)/(7*b*(a*x + b*x^3)^(7/2)) - (9*x^(19/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (3*x^(13/2))/(5*b^3*(a*x + b
*x^3)^(3/2)) - (3*x^(7/2))/(b^4*Sqrt[a*x + b*x^3]) + (9*Sqrt[x]*Sqrt[a*x + b*x^3])/(2*b^5) - (9*a*ArcTanh[(Sqr
t[b]*x^(3/2))/Sqrt[a*x + b*x^3]])/(2*b^(11/2))

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Rubi [A]  time = 0.248376, 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 = {2022, 2024, 2029, 206} \[ -\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{3 x^{7/2}}{b^4 \sqrt{a x+b x^3}}+\frac{9 \sqrt{x} \sqrt{a x+b x^3}}{2 b^5}-\frac{9 a \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x+b x^3}}\right )}{2 b^{11/2}}-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(29/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(25/2)/(7*b*(a*x + b*x^3)^(7/2)) - (9*x^(19/2))/(35*b^2*(a*x + b*x^3)^(5/2)) - (3*x^(13/2))/(5*b^3*(a*x + b
*x^3)^(3/2)) - (3*x^(7/2))/(b^4*Sqrt[a*x + b*x^3]) + (9*Sqrt[x]*Sqrt[a*x + b*x^3])/(2*b^5) - (9*a*ArcTanh[(Sqr
t[b]*x^(3/2))/Sqrt[a*x + b*x^3]])/(2*b^(11/2))

Rule 2022

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*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^{29/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac{9 \int \frac{x^{23/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{7 b}\\ &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}+\frac{9 \int \frac{x^{17/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{5 b^2}\\ &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}+\frac{3 \int \frac{x^{11/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{b^3}\\ &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{3 x^{7/2}}{b^4 \sqrt{a x+b x^3}}+\frac{9 \int \frac{x^{5/2}}{\sqrt{a x+b x^3}} \, dx}{b^4}\\ &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{3 x^{7/2}}{b^4 \sqrt{a x+b x^3}}+\frac{9 \sqrt{x} \sqrt{a x+b x^3}}{2 b^5}-\frac{(9 a) \int \frac{\sqrt{x}}{\sqrt{a x+b x^3}} \, dx}{2 b^5}\\ &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{3 x^{7/2}}{b^4 \sqrt{a x+b x^3}}+\frac{9 \sqrt{x} \sqrt{a x+b x^3}}{2 b^5}-\frac{(9 a) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^{3/2}}{\sqrt{a x+b x^3}}\right )}{2 b^5}\\ &=-\frac{x^{25/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{9 x^{19/2}}{35 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{3 x^{13/2}}{5 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{3 x^{7/2}}{b^4 \sqrt{a x+b x^3}}+\frac{9 \sqrt{x} \sqrt{a x+b x^3}}{2 b^5}-\frac{9 a \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x+b x^3}}\right )}{2 b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.18079, size = 130, normalized size = 0.82 \[ \frac{\sqrt{x} \left (\sqrt{b} x \left (1218 a^2 b^2 x^4+1050 a^3 b x^2+315 a^4+528 a b^3 x^6+35 b^4 x^8\right )-\frac{315 \sqrt{a} \left (a+b x^2\right )^4 \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{70 b^{11/2} \left (a+b x^2\right )^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(29/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(Sqrt[x]*(Sqrt[b]*x*(315*a^4 + 1050*a^3*b*x^2 + 1218*a^2*b^2*x^4 + 528*a*b^3*x^6 + 35*b^4*x^8) - (315*Sqrt[a]*
(a + b*x^2)^4*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(70*b^(11/2)*(a + b*x^2)^3*Sqrt[x*(a + b*x^2
)])

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Maple [A]  time = 0.04, size = 212, normalized size = 1.3 \begin{align*} -{\frac{1}{70\, \left ( b{x}^{2}+a \right ) ^{4}}\sqrt{x \left ( b{x}^{2}+a \right ) } \left ( -35\,{x}^{9}{b}^{9/2}+315\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){x}^{6}a{b}^{3}\sqrt{b{x}^{2}+a}-528\,{b}^{7/2}{x}^{7}a+945\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){x}^{4}{a}^{2}{b}^{2}\sqrt{b{x}^{2}+a}-1218\,{b}^{5/2}{x}^{5}{a}^{2}+945\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){x}^{2}{a}^{3}b\sqrt{b{x}^{2}+a}-1050\,{b}^{3/2}{x}^{3}{a}^{3}+315\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){a}^{4}\sqrt{b{x}^{2}+a}-315\,\sqrt{b}x{a}^{4} \right ){b}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(29/2)/(b*x^3+a*x)^(9/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{29}{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^(29/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(29/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.58916, size = 845, normalized size = 5.31 \begin{align*} \left [\frac{315 \,{\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \sqrt{b} \log \left (2 \, b x^{2} - 2 \, \sqrt{b x^{3} + a x} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (35 \, b^{5} x^{8} + 528 \, a b^{4} x^{6} + 1218 \, a^{2} b^{3} x^{4} + 1050 \, a^{3} b^{2} x^{2} + 315 \, a^{4} b\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{140 \,{\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}}, \frac{315 \,{\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{3} + a x} \sqrt{-b}}{b x^{\frac{3}{2}}}\right ) +{\left (35 \, b^{5} x^{8} + 528 \, a b^{4} x^{6} + 1218 \, a^{2} b^{3} x^{4} + 1050 \, a^{3} b^{2} x^{2} + 315 \, a^{4} b\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{70 \,{\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(29/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

[1/140*(315*(a*b^4*x^8 + 4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4*b*x^2 + a^5)*sqrt(b)*log(2*b*x^2 - 2*sqrt(b*x^3
+ a*x)*sqrt(b)*sqrt(x) + a) + 2*(35*b^5*x^8 + 528*a*b^4*x^6 + 1218*a^2*b^3*x^4 + 1050*a^3*b^2*x^2 + 315*a^4*b
)*sqrt(b*x^3 + a*x)*sqrt(x))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4*a^3*b^7*x^2 + a^4*b^6), 1/70*(315*(a*
b^4*x^8 + 4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4*b*x^2 + a^5)*sqrt(-b)*arctan(sqrt(b*x^3 + a*x)*sqrt(-b)/(b*x^(
3/2))) + (35*b^5*x^8 + 528*a*b^4*x^6 + 1218*a^2*b^3*x^4 + 1050*a^3*b^2*x^2 + 315*a^4*b)*sqrt(b*x^3 + a*x)*sqrt
(x))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4*a^3*b^7*x^2 + a^4*b^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(x**(29/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.32596, size = 123, normalized size = 0.77 \begin{align*} \frac{{\left ({\left ({\left (x^{2}{\left (\frac{35 \, x^{2}}{b} + \frac{528 \, a}{b^{2}}\right )} + \frac{1218 \, a^{2}}{b^{3}}\right )} x^{2} + \frac{1050 \, a^{3}}{b^{4}}\right )} x^{2} + \frac{315 \, a^{4}}{b^{5}}\right )} x}{70 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} + \frac{9 \, a \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(29/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

1/70*(((x^2*(35*x^2/b + 528*a/b^2) + 1218*a^2/b^3)*x^2 + 1050*a^3/b^4)*x^2 + 315*a^4/b^5)*x/(b*x^2 + a)^(7/2)
+ 9/2*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(11/2)