3.2.56 \(\int \frac {x^7}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=81 \[ -\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{5/2}}+\frac {3 \sqrt {b x^2+c x^4}}{2 c^2}-\frac {x^4}{c \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.11, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2018, 668, 640, 620, 206} \begin {gather*} \frac {3 \sqrt {b x^2+c x^4}}{2 c^2}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{5/2}}-\frac {x^4}{c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 76, normalized size = 0.94 \begin {gather*} \frac {x \left (\sqrt {c} x \left (3 b+c x^2\right )-3 b^{3/2} \sqrt {\frac {c x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )\right )}{2 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(Sqrt[c]*x*(3*b + c*x^2) - 3*b^(3/2)*Sqrt[1 + (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(2*c^(5/2)*Sqrt[x^2
*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.39, size = 88, normalized size = 1.09 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (3 b+c x^2\right )}{2 c^2 \left (b+c x^2\right )}+\frac {3 b \log \left (-2 c^{5/2} \sqrt {b x^2+c x^4}+b c^2+2 c^3 x^2\right )}{4 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.09, size = 180, normalized size = 2.22 \begin {gather*} \left [\frac {3 \, {\left (b c x^{2} + b^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (c^{2} x^{2} + 3 \, b c\right )}}{4 \, {\left (c^{4} x^{2} + b c^{3}\right )}}, \frac {3 \, {\left (b c x^{2} + b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (c^{2} x^{2} + 3 \, b c\right )}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 99, normalized size = 1.22 \begin {gather*} \frac {3 \, b \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{4 \, c^{\frac {5}{2}}} + \frac {b^{2}}{{\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} c + b \sqrt {c}\right )} c^{2}} + \frac {\sqrt {c x^{4} + b x^{2}}}{2 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

3/4*b*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*sqrt(c) - b))/c^(5/2) + b^2/(((sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2))*c + b*sqrt(c))*c^2) + 1/2*sqrt(c*x^4 + b*x^2)/c^2

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maple [A]  time = 0.01, size = 73, normalized size = 0.90 \begin {gather*} \frac {\left (c \,x^{2}+b \right ) \left (c^{\frac {5}{2}} x^{3}+3 b \,c^{\frac {3}{2}} x -3 \sqrt {c \,x^{2}+b}\, b c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )\right ) x^{3}}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(c*x^4+b*x^2)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.51, size = 77, normalized size = 0.95 \begin {gather*} \frac {x^{4}}{2 \, \sqrt {c x^{4} + b x^{2}} c} + \frac {3 \, b x^{2}}{2 \, \sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {3 \, b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{4 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*x^4/(sqrt(c*x^4 + b*x^2)*c) + 3/2*b*x^2/(sqrt(c*x^4 + b*x^2)*c^2) - 3/4*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 +
b*x^2)*sqrt(c))/c^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^7}{{\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^7/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^7/(b*x^2 + c*x^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\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**7/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**7/(x**2*(b + c*x**2))**(3/2), x)

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