3.3.61 \(\int \frac {x^5}{\sqrt {b x^2+c x^4}} \, dx\) [261]

Optimal. Leaf size=86 \[ -\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{5/2}} \]

[Out]

3/8*b^2*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(5/2)-3/8*b*(c*x^4+b*x^2)^(1/2)/c^2+1/4*x^2*(c*x^4+b*x^2)^(
1/2)/c

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Rubi [A]
time = 0.06, antiderivative size = 86, 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 = {2043, 684, 654, 634, 212} \begin {gather*} \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{5/2}}-\frac {3 b \sqrt {b x^2+c x^4}}{8 c^2}+\frac {x^2 \sqrt {b x^2+c x^4}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

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 654

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 684

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 2043

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

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Mathematica [A]
time = 0.06, size = 91, normalized size = 1.06 \begin {gather*} \frac {x \left (\sqrt {c} x \left (-3 b^2-b c x^2+2 c^2 x^4\right )-3 b^2 \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{8 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[b*x^2 + c*x^4],x]

[Out]

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

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Maple [A]
time = 0.10, size = 85, normalized size = 0.99

method	result	size
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \left (2 x^{3} \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}}-3 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b x +3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{2} c \right )}{8 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {7}{2}}}\)	\(85\)
risch	\(-\frac {x^{2} \left (-2 c \,x^{2}+3 b \right ) \left (c \,x^{2}+b \right )}{8 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {3 b^{2} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{8 c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 76, normalized size = 0.88 \begin {gather*} \frac {\sqrt {c x^{4} + b x^{2}} x^{2}}{4 \, c} + \frac {3 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{16 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{8 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 145, normalized size = 1.69 \begin {gather*} \left [\frac {3 \, b^{2} \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 (2 \, c^{2} x^{2} - 3 \, b c\right )}}{16 \, c^{3}}, -\frac {3 \, b^{2} \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 (2 \, c^{2} x^{2} - 3 \, b c\right )}}{8 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*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)*(2*c^2*x^2 - 3*
b*c))/c^3, -1/8*(3*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)*(2*c^2*
x^2 - 3*b*c))/c^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\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/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 3.20, size = 79, normalized size = 0.92 \begin {gather*} \frac {1}{8} \, \sqrt {c x^{2} + b} x {\left (\frac {2 \, x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {3 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {5}{2}}} - \frac {3 \, b^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/8*sqrt(c*x^2 + b)*x*(2*x^2/(c*sgn(x)) - 3*b/(c^2*sgn(x))) + 3/16*b^2*log(abs(b))*sgn(x)/c^(5/2) - 3/8*b^2*lo
g(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(5/2)*sgn(x))

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

[Out]

int(x^5/(b*x^2 + c*x^4)^(1/2), x)

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