3.2.31 \(\int \frac {x^5 (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=139 \[ -\frac {b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}+\frac {b \sqrt {b x^2+c x^4} (5 b B-6 A c)}{16 c^3}-\frac {x^2 \sqrt {b x^2+c x^4} (5 b B-6 A c)}{24 c^2}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c} \]

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Rubi [A]  time = 0.27, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2034, 794, 670, 640, 620, 206} \begin {gather*} -\frac {b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}-\frac {x^2 \sqrt {b x^2+c x^4} (5 b B-6 A c)}{24 c^2}+\frac {b \sqrt {b x^2+c x^4} (5 b B-6 A c)}{16 c^3}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(5*b*B - 6*A*c)*Sqrt[b*x^2 + c*x^4])/(16*c^3) - ((5*b*B - 6*A*c)*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^2) + (B*x^4
*Sqrt[b*x^2 + c*x^4])/(6*c) - (b^2*(5*b*B - 6*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*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 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 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 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 )}{\sqrt {b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c}+\frac {\left (2 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{6 c}\\ &=-\frac {(5 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c}+\frac {(b (5 b B-6 A c)) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=\frac {b (5 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^3}-\frac {(5 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {\left (b^2 (5 b B-6 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{32 c^3}\\ &=\frac {b (5 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^3}-\frac {(5 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {\left (b^2 (5 b B-6 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^3}\\ &=\frac {b (5 b B-6 A c) \sqrt {b x^2+c x^4}}{16 c^3}-\frac {(5 b B-6 A c) x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {B x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 123, normalized size = 0.88 \begin {gather*} \frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (-2 b c \left (9 A+5 B x^2\right )+4 c^2 x^2 \left (3 A+2 B x^2\right )+15 b^2 B\right )-3 b^2 \sqrt {b+c x^2} (5 b B-6 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b+c x^2}}\right )\right )}{48 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Sqrt[c]*x*(b + c*x^2)*(15*b^2*B + 4*c^2*x^2*(3*A + 2*B*x^2) - 2*b*c*(9*A + 5*B*x^2)) - 3*b^2*(5*b*B - 6*A*
c)*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/Sqrt[b + c*x^2]]))/(48*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.58, size = 115, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-18 A b c+12 A c^2 x^2+15 b^2 B-10 b B c x^2+8 B c^2 x^4\right )}{48 c^3}+\frac {\left (5 b^3 B-6 A b^2 c\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{32 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(15*b^2*B - 18*A*b*c - 10*b*B*c*x^2 + 12*A*c^2*x^2 + 8*B*c^2*x^4))/(48*c^3) + ((5*b^3*B -
6*A*b^2*c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(32*c^(7/2))

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fricas [A]  time = 0.45, size = 226, normalized size = 1.63 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, B c^{3} x^{4} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{4}}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(8*B*c^3*x^4 + 1
5*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4, 1/48*(3*(5*B*b^3 - 6*A*b^2*c)*
sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*B*c^3*x^4 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*
c^2 - 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4]

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giac [A]  time = 0.25, size = 119, normalized size = 0.86 \begin {gather*} \frac {1}{48} \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, {\left (\frac {4 \, B x^{2}}{c} - \frac {5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x^{2} + \frac {3 \, {\left (5 \, B b^{2} - 6 \, A b c\right )}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{32 \, c^{\frac {7}{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)^(1/2),x, algorithm="giac")

[Out]

1/48*sqrt(c*x^4 + b*x^2)*(2*(4*B*x^2/c - (5*B*b*c - 6*A*c^2)/c^3)*x^2 + 3*(5*B*b^2 - 6*A*b*c)/c^3) + 1/32*(5*B
*b^3 - 6*A*b^2*c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*sqrt(c) - b))/c^(7/2)

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maple [A]  time = 0.06, size = 169, normalized size = 1.22 \begin {gather*} \frac {\sqrt {c \,x^{2}+b}\, \left (8 \sqrt {c \,x^{2}+b}\, B \,c^{\frac {7}{2}} x^{5}+12 \sqrt {c \,x^{2}+b}\, A \,c^{\frac {7}{2}} x^{3}-10 \sqrt {c \,x^{2}+b}\, B b \,c^{\frac {5}{2}} x^{3}+18 A \,b^{2} c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 B \,b^{3} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-18 \sqrt {c \,x^{2}+b}\, A b \,c^{\frac {5}{2}} x +15 \sqrt {c \,x^{2}+b}\, B \,b^{2} c^{\frac {3}{2}} x \right ) x}{48 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {9}{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)^(1/2),x)

[Out]

1/48*x*(c*x^2+b)^(1/2)*(8*B*(c*x^2+b)^(1/2)*c^(7/2)*x^5+12*A*(c*x^2+b)^(1/2)*c^(7/2)*x^3-10*B*(c*x^2+b)^(1/2)*
c^(5/2)*x^3*b-18*A*(c*x^2+b)^(1/2)*c^(5/2)*x*b+15*B*(c*x^2+b)^(1/2)*c^(3/2)*x*b^2+18*A*ln(c^(1/2)*x+(c*x^2+b)^
(1/2))*b^2*c^2-15*B*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3*c)/(c*x^4+b*x^2)^(1/2)/c^(9/2)

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maxima [A]  time = 1.45, size = 183, normalized size = 1.32 \begin {gather*} \frac {1}{16} \, {\left (\frac {4 \, \sqrt {c x^{4} + b x^{2}} x^{2}}{c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{c^{2}}\right )} A + \frac {1}{96} \, {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} x^{4}}{c} - \frac {20 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c^{2}} - \frac {15 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{3}}\right )} B \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)^(1/2),x, algorithm="maxima")

[Out]

1/16*(4*sqrt(c*x^4 + b*x^2)*x^2/c + 3*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2) - 6*sqrt(c*
x^4 + b*x^2)*b/c^2)*A + 1/96*(16*sqrt(c*x^4 + b*x^2)*x^4/c - 20*sqrt(c*x^4 + b*x^2)*b*x^2/c^2 - 15*b^3*log(2*c
*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 30*sqrt(c*x^4 + b*x^2)*b^2/c^3)*B

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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 )}{\sqrt {c\,x^4+b\,x^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)^(1/2),x)

[Out]

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4)^(1/2), 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 )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, 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)**(1/2),x)

[Out]

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

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