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

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

[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])

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Rubi [A]  time = 0.101596, antiderivative size = 127, normalized size of antiderivative = 1., 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} \[ \frac{x^4 \left (a+b x^2\right )}{4 b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{a x^2 \left (a+b x^2\right )}{2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{a^2 \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

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 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])

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 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])

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*}

Mathematica [A]  time = 0.0240429, size = 55, normalized size = 0.43 \[ \frac{\left (a+b x^2\right ) \left (2 a^2 \log \left (a+b x^2\right )+b x^2 \left (b x^2-2 a\right )\right )}{4 b^3 \sqrt{\left (a+b x^2\right )^2}} \]

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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Maple [A]  time = 0.211, size = 52, normalized size = 0.4 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) \left ({b}^{2}{x}^{4}-2\,ab{x}^{2}+2\,{a}^{2}\ln \left ( b{x}^{2}+a \right ) \right ) }{4\,{b}^{3}}{\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

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 = 0.999292, size = 62, normalized size = 0.49 \begin{align*} \frac{x^{4}}{4 \, \sqrt{b^{2}}} - \frac{a b x^{2}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{a^{2} b^{2} \log \left (x^{2} + \frac{a}{b}\right )}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}} \end{align*}

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]

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

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Fricas [A]  time = 1.22679, size = 73, normalized size = 0.57 \begin{align*} \frac{b^{2} x^{4} - 2 \, a b x^{2} + 2 \, a^{2} \log \left (b x^{2} + a\right )}{4 \, b^{3}} \end{align*}

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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Sympy [A]  time = 0.327823, size = 32, normalized size = 0.25 \begin{align*} \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{align*}

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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Giac [A]  time = 1.13911, size = 80, normalized size = 0.63 \begin{align*} \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{align*}

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