3.2.47 \(\int \frac {x^5 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=112 \[ -\frac {(3 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{5/2}}+\frac {\sqrt {b x^2+c x^4} (3 b B-2 A c)}{2 b c^2}-\frac {x^4 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.24, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2034, 788, 640, 620, 206} \begin {gather*} \frac {\sqrt {b x^2+c x^4} (3 b B-2 A c)}{2 b c^2}-\frac {(3 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{5/2}}-\frac {x^4 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^4)/(b*c*Sqrt[b*x^2 + c*x^4])) + ((3*b*B - 2*A*c)*Sqrt[b*x^2 + c*x^4])/(2*b*c^2) - ((3*b*B - 2
*A*c)*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 788

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

Rule 2034

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

Rubi steps

\begin {align*} \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) x^4}{b c \sqrt {b x^2+c x^4}}+\frac {1}{2} \left (-\frac {2 A}{b}+\frac {3 B}{c}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {(b B-A c) x^4}{b c \sqrt {b x^2+c x^4}}+\frac {(3 b B-2 A c) \sqrt {b x^2+c x^4}}{2 b c^2}-\frac {(3 b B-2 A c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{4 c^2}\\ &=-\frac {(b B-A c) x^4}{b c \sqrt {b x^2+c x^4}}+\frac {(3 b B-2 A c) \sqrt {b x^2+c x^4}}{2 b c^2}-\frac {(3 b B-2 A c) \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 {(b B-A c) x^4}{b c \sqrt {b x^2+c x^4}}+\frac {(3 b B-2 A c) \sqrt {b x^2+c x^4}}{2 b c^2}-\frac {(3 b B-2 A c) \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 = 0.17, size = 91, normalized size = 0.81 \begin {gather*} \frac {x \left (\sqrt {c} x \left (-2 A c+3 b B+B c x^2\right )-\sqrt {b} \sqrt {\frac {c x^2}{b}+1} (3 b B-2 A c) \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^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(Sqrt[c]*x*(3*b*B - 2*A*c + B*c*x^2) - Sqrt[b]*(3*b*B - 2*A*c)*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 = 0.54, size = 102, normalized size = 0.91 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-2 A c+3 b B+B c x^2\right )}{2 c^2 \left (b+c x^2\right )}+\frac {(3 b B-2 A c) \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^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

((3*b*B - 2*A*c + B*c*x^2)*Sqrt[b*x^2 + c*x^4])/(2*c^2*(b + c*x^2)) + ((3*b*B - 2*A*c)*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 = 0.43, size = 230, normalized size = 2.05 \begin {gather*} \left [-\frac {{\left (3 \, B b^{2} - 2 \, A b c + {\left (3 \, B b c - 2 \, A c^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (B c^{2} x^{2} + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{4 \, {\left (c^{4} x^{2} + b c^{3}\right )}}, \frac {{\left (3 \, B b^{2} - 2 \, A b c + {\left (3 \, B b c - 2 \, A c^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (B c^{2} x^{2} + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{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^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 116, normalized size = 1.04 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2}} B}{2 \, c^{2}} + \frac {{\left (3 \, B b - 2 \, A c\right )} \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 b^{2} - A b c}{{\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} c + b \sqrt {c}\right )} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 115, normalized size = 1.03 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-B \,c^{\frac {5}{2}} x^{3}+2 A \,c^{\frac {5}{2}} x -3 B b \,c^{\frac {3}{2}} x -2 \sqrt {c \,x^{2}+b}\, A \,c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 \sqrt {c \,x^{2}+b}\, 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^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/2*x^3*(c*x^2+b)*(-B*c^(5/2)*x^3+2*A*c^(5/2)*x-3*B*c^(3/2)*x*b-2*A*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*(c*x^2+b)^(
1/2)*c^2+3*B*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 = 138, normalized size = 1.23 \begin {gather*} \frac {1}{4} \, {\left (\frac {2 \, x^{4}}{\sqrt {c x^{4} + b x^{2}} c} + \frac {6 \, b x^{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 )}{c^{\frac {5}{2}}}\right )} B - \frac {1}{2} \, A {\left (\frac {2 \, x^{2}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/4*(2*x^4/(sqrt(c*x^4 + b*x^2)*c) + 6*b*x^2/(sqrt(c*x^4 + b*x^2)*c^2) - 3*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 +
b*x^2)*sqrt(c))/c^(5/2))*B - 1/2*A*(2*x^2/(sqrt(c*x^4 + b*x^2)*c) - log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sq
rt(c))/c^(3/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,\left (B\,x^2+A\right )}{{\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^5*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int((x^5*(A + B*x^2))/(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^{5} \left (A + B x^{2}\right )}{\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**5*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**5*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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