3.86 \(\int \frac{x^{7/2}}{(a x+b x^3)^{9/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.202778, 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 = {2023, 2029, 206} \[ \frac{x^{5/2}}{5 a^2 \left (a x+b x^3\right )^{5/2}}+\frac{x^{3/2}}{3 a^3 \left (a x+b x^3\right )^{3/2}}+\frac{\sqrt{x}}{a^4 \sqrt{a x+b x^3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b x^3}}\right )}{a^{9/2}}+\frac{x^{7/2}}{7 a \left (a x+b x^3\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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

Mathematica [C]  time = 0.0117814, size = 43, normalized size = 0.33 \[ \frac{x^{7/2} \, _2F_1\left (-\frac{7}{2},1;-\frac{5}{2};\frac{b x^2}{a}+1\right )}{7 a \left (x \left (a+b x^2\right )\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a*x + b*x^3)^(9/2),x]

[Out]

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

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Maple [B]  time = 0.012, size = 217, normalized size = 1.7 \begin{align*} -{\frac{1}{105\, \left ( b{x}^{2}+a \right ) ^{4}}\sqrt{x \left ( b{x}^{2}+a \right ) } \left ( 105\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{6}{b}^{3}\sqrt{b{x}^{2}+a}-105\,\sqrt{a}{x}^{6}{b}^{3}+315\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{4}a{b}^{2}\sqrt{b{x}^{2}+a}-350\,{a}^{3/2}{x}^{4}{b}^{2}+315\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){x}^{2}{a}^{2}b\sqrt{b{x}^{2}+a}-406\,{a}^{5/2}{x}^{2}b+105\,\ln \left ( 2\,{\frac{\sqrt{a}\sqrt{b{x}^{2}+a}+a}{x}} \right ){a}^{3}\sqrt{b{x}^{2}+a}-176\,{a}^{7/2} \right ){a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^3+a*x)^(9/2),x)

[Out]

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

[Out]

integrate(x^(7/2)/(b*x^3 + a*x)^(9/2), x)

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Fricas [A]  time = 1.29511, size = 803, normalized size = 6.18 \begin{align*} \left [\frac{105 \,{\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\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 (105 \, a b^{3} x^{6} + 350 \, a^{2} b^{2} x^{4} + 406 \, a^{3} b x^{2} + 176 \, a^{4}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{210 \,{\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}, \frac{105 \,{\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a x} \sqrt{-a}}{a \sqrt{x}}\right ) +{\left (105 \, a b^{3} x^{6} + 350 \, a^{2} b^{2} x^{4} + 406 \, a^{3} b x^{2} + 176 \, a^{4}\right )} \sqrt{b x^{3} + a x} \sqrt{x}}{105 \,{\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^3+a*x)^(9/2),x, algorithm="fricas")

[Out]

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

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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**(7/2)/(b*x**3+a*x)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.28796, size = 154, normalized size = 1.18 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{4}} - \frac{105 \, \sqrt{a} \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) + 176 \, \sqrt{-a}}{105 \, \sqrt{-a} a^{\frac{9}{2}}} + \frac{105 \,{\left (b x^{2} + a\right )}^{3} + 35 \,{\left (b x^{2} + a\right )}^{2} a + 21 \,{\left (b x^{2} + a\right )} a^{2} + 15 \, a^{3}}{105 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^3+a*x)^(9/2),x, algorithm="giac")

[Out]

arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^4) - 1/105*(105*sqrt(a)*arctan(sqrt(a)/sqrt(-a)) + 176*sqrt(-a))/
(sqrt(-a)*a^(9/2)) + 1/105*(105*(b*x^2 + a)^3 + 35*(b*x^2 + a)^2*a + 21*(b*x^2 + a)*a^2 + 15*a^3)/((b*x^2 + a)
^(7/2)*a^4)