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

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 109, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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])

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 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 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 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]

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 = 0.05, size = 88, normalized size = 0.81 \begin {gather*} \frac {x \left (15 b^{5/2} \sqrt {\frac {c x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \left (-15 b^2-5 b c x^2+2 c^2 x^4\right )\right )}{8 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

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)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.45, size = 102, normalized size = 0.94 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-15 b^2-5 b c x^2+2 c^2 x^4\right )}{8 c^3 \left (b+c x^2\right )}-\frac {15 b^2 \log \left (-2 c^{7/2} \sqrt {b x^2+c x^4}+b c^3+2 c^4 x^2\right )}{16 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-15*b^2 - 5*b*c*x^2 + 2*c^2*x^4))/(8*c^3*(b + c*x^2)) - (15*b^2*Log[b*c^3 + 2*c^4*x^2 -
2*c^(7/2)*Sqrt[b*x^2 + c*x^4]])/(16*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 209, normalized size = 1.92 \begin {gather*} \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 {gather*}

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)]

________________________________________________________________________________________

giac [A]  time = 0.27, size = 114, normalized size = 1.05 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{4} + b x^{2}} {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {7 \, b}{c^{3}}\right )} - \frac {15 \, b^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {7}{2}}} - \frac {b^{3}}{{\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} c + b \sqrt {c}\right )} c^{3}} \end {gather*}

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]

1/8*sqrt(c*x^4 + b*x^2)*(2*x^2/c^2 - 7*b/c^3) - 15/16*b^2*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*sqrt(
c) - b))/c^(7/2) - b^3/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*c + b*sqrt(c))*c^3)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 87, normalized size = 0.80 \begin {gather*} \frac {\left (c \,x^{2}+b \right ) \left (2 c^{\frac {7}{2}} x^{5}-5 b \,c^{\frac {5}{2}} x^{3}-15 b^{2} c^{\frac {3}{2}} x +15 \sqrt {c \,x^{2}+b}\, b^{2} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )\right ) x^{3}}{8 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {9}{2}}} \end {gather*}

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*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(c*x^2+b)^(
1/2)*b^2*c)/(c*x^4+b*x^2)^(3/2)/c^(9/2)

________________________________________________________________________________________

maxima [A]  time = 1.48, size = 103, normalized size = 0.94 \begin {gather*} \frac {x^{6}}{4 \, \sqrt {c x^{4} + b x^{2}} c} - \frac {5 \, b x^{4}}{8 \, \sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {15 \, b^{2} x^{2}}{8 \, \sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} \end {gather*}

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]

1/4*x^6/(sqrt(c*x^4 + b*x^2)*c) - 5/8*b*x^4/(sqrt(c*x^4 + b*x^2)*c^2) - 15/8*b^2*x^2/(sqrt(c*x^4 + b*x^2)*c^3)
+ 15/16*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^9}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^9/(b*x^2 + c*x^4)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{9}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

________________________________________________________________________________________