3.5.52 \(\int \frac {x^5}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=127 \[ \frac {x^4 \left (a+b x^2\right )}{4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a x^2 \left (a+b x^2\right )}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^2 \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.10, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \begin {gather*} \frac {x^4 \left (a+b x^2\right )}{4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a x^2 \left (a+b x^2\right )}{2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^2 \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.43 \begin {gather*} \frac {\left (a+b x^2\right ) \left (2 a^2 \log \left (a+b x^2\right )+b x^2 \left (b x^2-2 a\right )\right )}{4 b^3 \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 182, normalized size = 1.43 \begin {gather*} -\frac {a^2 \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )}{4 b^4}-\frac {a^2 \left (\sqrt {b^2}-b\right ) \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )}{4 b^4}+\frac {\left (b x^2-3 a\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 b^3}+\frac {2 a x^2-b x^4}{8 b \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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fricas [A]  time = 0.71, size = 33, normalized size = 0.26 \begin {gather*} \frac {b^{2} x^{4} - 2 \, a b x^{2} + 2 \, a^{2} \log \left (b x^{2} + a\right )}{4 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*x^4 - 2*a*b*x^2 + 2*a^2*log(b*x^2 + a))/b^3

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giac [A]  time = 0.18, size = 59, normalized size = 0.46 \begin {gather*} \frac {a^{2} \log \left ({\left | b x^{2} + a \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {b x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) - 2 \, a x^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*log(abs(b*x^2 + a))*sgn(b*x^2 + a)/b^3 + 1/4*(b*x^4*sgn(b*x^2 + a) - 2*a*x^2*sgn(b*x^2 + a))/b^2

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maple [A]  time = 0.01, size = 52, normalized size = 0.41 \begin {gather*} \frac {\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}-2 a b \,x^{2}+2 a^{2} \ln \left (b \,x^{2}+a \right )\right )}{4 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((b*x^2+a)^2)^(1/2),x)

[Out]

1/4*(b*x^2+a)*(b^2*x^4-2*a*b*x^2+2*a^2*ln(b*x^2+a))/((b*x^2+a)^2)^(1/2)/b^3

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maxima [A]  time = 1.39, size = 34, normalized size = 0.27 \begin {gather*} \frac {a^{2} \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {b x^{4} - 2 \, a x^{2}}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*log(b*x^2 + a)/b^3 + 1/4*(b*x^4 - 2*a*x^2)/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{\sqrt {{\left (b\,x^2+a\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((a + b*x^2)^2)^(1/2),x)

[Out]

int(x^5/((a + b*x^2)^2)^(1/2), x)

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sympy [A]  time = 0.20, size = 32, normalized size = 0.25 \begin {gather*} \frac {a^{2} \log {\left (a + b x^{2} \right )}}{2 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{4}}{4 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/((b*x**2+a)**2)**(1/2),x)

[Out]

a**2*log(a + b*x**2)/(2*b**3) - a*x**2/(2*b**2) + x**4/(4*b)

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