3.273 \(\int \frac{x^9}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=109 \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{7/2}}+\frac{5 x^2 \sqrt{b x^2+c x^4}}{4 c^2}-\frac{15 b \sqrt{b x^2+c x^4}}{8 c^3}-\frac{x^6}{c \sqrt{b x^2+c x^4}} \]

[Out]

-(x^6/(c*Sqrt[b*x^2 + c*x^4])) - (15*b*Sqrt[b*x^2 + c*x^4])/(8*c^3) + (5*x^2*Sqrt[b*x^2 + c*x^4])/(4*c^2) + (1
5*b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*c^(7/2))

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Rubi [A]  time = 0.128751, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2018, 668, 670, 640, 620, 206} \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{7/2}}+\frac{5 x^2 \sqrt{b x^2+c x^4}}{4 c^2}-\frac{15 b \sqrt{b x^2+c x^4}}{8 c^3}-\frac{x^6}{c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x^6/(c*Sqrt[b*x^2 + c*x^4])) - (15*b*Sqrt[b*x^2 + c*x^4])/(8*c^3) + (5*x^2*Sqrt[b*x^2 + c*x^4])/(4*c^2) + (1
5*b^2*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*c^(7/2))

Rule 2018

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

Rule 668

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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^9}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{x^6}{c \sqrt{b x^2+c x^4}}+\frac{5 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{2 c}\\ &=-\frac{x^6}{c \sqrt{b x^2+c x^4}}+\frac{5 x^2 \sqrt{b x^2+c x^4}}{4 c^2}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{8 c^2}\\ &=-\frac{x^6}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{8 c^3}+\frac{5 x^2 \sqrt{b x^2+c x^4}}{4 c^2}+\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{16 c^3}\\ &=-\frac{x^6}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{8 c^3}+\frac{5 x^2 \sqrt{b x^2+c x^4}}{4 c^2}+\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^3}\\ &=-\frac{x^6}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{8 c^3}+\frac{5 x^2 \sqrt{b x^2+c x^4}}{4 c^2}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0512658, size = 88, normalized size = 0.81 \[ \frac{x \left (\sqrt{c} x \left (-15 b^2-5 b c x^2+2 c^2 x^4\right )+15 b^{5/2} \sqrt{\frac{c x^2}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{8 c^{7/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(Sqrt[c]*x*(-15*b^2 - 5*b*c*x^2 + 2*c^2*x^4) + 15*b^(5/2)*Sqrt[1 + (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]])
)/(8*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.051, size = 87, normalized size = 0.8 \begin{align*}{\frac{{x}^{3} \left ( c{x}^{2}+b \right ) }{8} \left ( 2\,{x}^{5}{c}^{7/2}-5\,{c}^{5/2}{x}^{3}b-15\,{c}^{3/2}x{b}^{2}+15\,\sqrt{c{x}^{2}+b}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{2}c \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/8*x^3*(c*x^2+b)*(2*x^5*c^(7/2)-5*c^(5/2)*x^3*b-15*c^(3/2)*x*b^2+15*(c*x^2+b)^(1/2)*ln(x*c^(1/2)+(c*x^2+b)^(1
/2))*b^2*c)/(c*x^4+b*x^2)^(3/2)/c^(9/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60523, size = 451, normalized size = 4.14 \begin{align*} \left [\frac{15 \,{\left (b^{2} c x^{2} + b^{3}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) + 2 \,{\left (2 \, c^{3} x^{4} - 5 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt{c x^{4} + b x^{2}}}{16 \,{\left (c^{5} x^{2} + b c^{4}\right )}}, -\frac{15 \,{\left (b^{2} c x^{2} + b^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) -{\left (2 \, c^{3} x^{4} - 5 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt{c x^{4} + b x^{2}}}{8 \,{\left (c^{5} x^{2} + b c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*(b^2*c*x^2 + b^3)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(2*c^3*x^4 - 5*b*c^2
*x^2 - 15*b^2*c)*sqrt(c*x^4 + b*x^2))/(c^5*x^2 + b*c^4), -1/8*(15*(b^2*c*x^2 + b^3)*sqrt(-c)*arctan(sqrt(c*x^4
+ b*x^2)*sqrt(-c)/(c*x^2 + b)) - (2*c^3*x^4 - 5*b*c^2*x^2 - 15*b^2*c)*sqrt(c*x^4 + b*x^2))/(c^5*x^2 + b*c^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**9/(x**2*(b + c*x**2))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9}}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^9/(c*x^4 + b*x^2)^(3/2), x)