3.77 \(\int \frac{x^{25/2}}{(a x+b x^3)^{9/2}} \, dx\)

Optimal. Leaf size=130 \[ -\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{3/2}}{b^4 \sqrt{a x+b x^3}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x+b x^3}}\right )}{b^{9/2}}-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

[Out]

-x^(21/2)/(7*b*(a*x + b*x^3)^(7/2)) - x^(15/2)/(5*b^2*(a*x + b*x^3)^(5/2)) - x^(9/2)/(3*b^3*(a*x + b*x^3)^(3/2
)) - x^(3/2)/(b^4*Sqrt[a*x + b*x^3]) + ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x + b*x^3]]/b^(9/2)

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Rubi [A]  time = 0.205549, antiderivative size = 130, 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 = {2022, 2029, 206} \[ -\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{3/2}}{b^4 \sqrt{a x+b x^3}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x+b x^3}}\right )}{b^{9/2}}-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(25/2)/(a*x + b*x^3)^(9/2),x]

[Out]

-x^(21/2)/(7*b*(a*x + b*x^3)^(7/2)) - x^(15/2)/(5*b^2*(a*x + b*x^3)^(5/2)) - x^(9/2)/(3*b^3*(a*x + b*x^3)^(3/2
)) - x^(3/2)/(b^4*Sqrt[a*x + b*x^3]) + ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a*x + b*x^3]]/b^(9/2)

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 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^{25/2}}{\left (a x+b x^3\right )^{9/2}} \, dx &=-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}+\frac{\int \frac{x^{19/2}}{\left (a x+b x^3\right )^{7/2}} \, dx}{b}\\ &=-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}+\frac{\int \frac{x^{13/2}}{\left (a x+b x^3\right )^{5/2}} \, dx}{b^2}\\ &=-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}+\frac{\int \frac{x^{7/2}}{\left (a x+b x^3\right )^{3/2}} \, dx}{b^3}\\ &=-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{3/2}}{b^4 \sqrt{a x+b x^3}}+\frac{\int \frac{\sqrt{x}}{\sqrt{a x+b x^3}} \, dx}{b^4}\\ &=-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{3/2}}{b^4 \sqrt{a x+b x^3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^{3/2}}{\sqrt{a x+b x^3}}\right )}{b^4}\\ &=-\frac{x^{21/2}}{7 b \left (a x+b x^3\right )^{7/2}}-\frac{x^{15/2}}{5 b^2 \left (a x+b x^3\right )^{5/2}}-\frac{x^{9/2}}{3 b^3 \left (a x+b x^3\right )^{3/2}}-\frac{x^{3/2}}{b^4 \sqrt{a x+b x^3}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x+b x^3}}\right )}{b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.252861, size = 120, normalized size = 0.92 \[ \frac{\sqrt{x} \left (105 \sqrt{a} \left (a+b x^2\right )^3 \sqrt{\frac{b x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )-\sqrt{b} x \left (350 a^2 b x^2+105 a^3+406 a b^2 x^4+176 b^3 x^6\right )\right )}{105 b^{9/2} \left (a+b x^2\right )^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(25/2)/(a*x + b*x^3)^(9/2),x]

[Out]

(Sqrt[x]*(-(Sqrt[b]*x*(105*a^3 + 350*a^2*b*x^2 + 406*a*b^2*x^4 + 176*b^3*x^6)) + 105*Sqrt[a]*(a + b*x^2)^3*Sqr
t[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]]))/(105*b^(9/2)*(a + b*x^2)^3*Sqrt[x*(a + b*x^2)])

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Maple [A]  time = 0.022, size = 198, normalized size = 1.5 \begin{align*}{\frac{1}{105\, \left ( b{x}^{2}+a \right ) ^{4}}\sqrt{x \left ( b{x}^{2}+a \right ) } \left ( 105\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){x}^{6}{b}^{3}\sqrt{b{x}^{2}+a}-176\,{x}^{7}{b}^{7/2}+315\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){x}^{4}a{b}^{2}\sqrt{b{x}^{2}+a}-406\,{b}^{5/2}{x}^{5}a+315\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){x}^{2}{a}^{2}b\sqrt{b{x}^{2}+a}-350\,{b}^{3/2}{x}^{3}{a}^{2}+105\,\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){a}^{3}\sqrt{b{x}^{2}+a}-105\,\sqrt{b}x{a}^{3} \right ){b}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(25/2)/(b*x^3+a*x)^(9/2),x)

[Out]

1/105*(x*(b*x^2+a))^(1/2)/b^(9/2)*(105*ln(x*b^(1/2)+(b*x^2+a)^(1/2))*x^6*b^3*(b*x^2+a)^(1/2)-176*x^7*b^(7/2)+3
15*ln(x*b^(1/2)+(b*x^2+a)^(1/2))*x^4*a*b^2*(b*x^2+a)^(1/2)-406*b^(5/2)*x^5*a+315*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
*x^2*a^2*b*(b*x^2+a)^(1/2)-350*b^(3/2)*x^3*a^2+105*ln(x*b^(1/2)+(b*x^2+a)^(1/2))*a^3*(b*x^2+a)^(1/2)-105*b^(1/
2)*x*a^3)/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{25}{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^(25/2)/(b*x^3+a*x)^(9/2),x, algorithm="maxima")

[Out]

integrate(x^(25/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.47245, size = 783, normalized size = 6.02 \begin{align*} \left [\frac{105 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{b} \log \left (2 \, b x^{2} + 2 \, \sqrt{b x^{3} + a x} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (176 \, b^{4} x^{6} + 406 \, a b^{3} x^{4} + 350 \, a^{2} b^{2} x^{2} + 105 \, a^{3} b\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{210 \,{\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, -\frac{105 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{3} + a x} \sqrt{-b}}{b x^{\frac{3}{2}}}\right ) +{\left (176 \, b^{4} x^{6} + 406 \, a b^{3} x^{4} + 350 \, a^{2} b^{2} x^{2} + 105 \, a^{3} b\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{105 \,{\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(25/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

[1/210*(105*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(b)*log(2*b*x^2 + 2*sqrt(b*x^3 + a
*x)*sqrt(b)*sqrt(x) + a) - 2*(176*b^4*x^6 + 406*a*b^3*x^4 + 350*a^2*b^2*x^2 + 105*a^3*b)*sqrt(b*x^3 + a*x)*sqr
t(x))/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5), -1/105*(105*(b^4*x^8 + 4*a*b^3*x^6 +
6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(-b)*arctan(sqrt(b*x^3 + a*x)*sqrt(-b)/(b*x^(3/2))) + (176*b^4*x^6 + 40
6*a*b^3*x^4 + 350*a^2*b^2*x^2 + 105*a^3*b)*sqrt(b*x^3 + a*x)*sqrt(x))/(b^9*x^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 +
4*a^3*b^6*x^2 + a^4*b^5)]

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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**(25/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.40829, size = 105, normalized size = 0.81 \begin{align*} -\frac{{\left (2 \,{\left (x^{2}{\left (\frac{88 \, x^{2}}{b} + \frac{203 \, a}{b^{2}}\right )} + \frac{175 \, a^{2}}{b^{3}}\right )} x^{2} + \frac{105 \, a^{3}}{b^{4}}\right )} x}{105 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} - \frac{\log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{b^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(25/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

-1/105*(2*(x^2*(88*x^2/b + 203*a/b^2) + 175*a^2/b^3)*x^2 + 105*a^3/b^4)*x/(b*x^2 + a)^(7/2) - log(abs(-sqrt(b)
*x + sqrt(b*x^2 + a)))/b^(9/2)