3.1 \(\int (a^2+2 a b x^2+b^2 x^4)^{3/4} \, dx\)

Optimal. Leaf size=128 \[ \frac{3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac{1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac{3 \sqrt{a} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{b} \left (\frac{b x^2}{a}+1\right )^{3/2}} \]

[Out]

(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))/4 + (3*a*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))/(8*(a + b*x^2)) + (3*Sqrt[
a]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[b]*(1 + (b*x^2)/a)^(3/2))

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Rubi [A]  time = 0.0308191, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1089, 195, 215} \[ \frac{3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac{1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac{3 \sqrt{a} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{b} \left (\frac{b x^2}{a}+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4),x]

[Out]

(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))/4 + (3*a*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))/(8*(a + b*x^2)) + (3*Sqrt[
a]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[b]*(1 + (b*x^2)/a)^(3/2))

Rule 1089

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 + c*x^4)^FracPart[p]
)/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2
- 4*a*c, 0] &&  !IntegerQ[2*p]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx &=\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \int \left (1+\frac{b x^2}{a}\right )^{3/2} \, dx}{\left (1+\frac{b x^2}{a}\right )^{3/2}}\\ &=\frac{1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac{\left (3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}\right ) \int \sqrt{1+\frac{b x^2}{a}} \, dx}{4 \left (1+\frac{b x^2}{a}\right )^{3/2}}\\ &=\frac{1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac{3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac{\left (3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}\right ) \int \frac{1}{\sqrt{1+\frac{b x^2}{a}}} \, dx}{8 \left (1+\frac{b x^2}{a}\right )^{3/2}}\\ &=\frac{1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac{3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac{3 \sqrt{a} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{b} \left (1+\frac{b x^2}{a}\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0640802, size = 97, normalized size = 0.76 \[ \frac{\left (\left (a+b x^2\right )^2\right )^{3/4} \left (3 a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )+\sqrt{b} x \left (5 a+2 b x^2\right ) \sqrt{\frac{b x^2}{a}+1}\right )}{8 \sqrt{b} \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4),x]

[Out]

(((a + b*x^2)^2)^(3/4)*(Sqrt[b]*x*(5*a + 2*b*x^2)*Sqrt[1 + (b*x^2)/a] + 3*a^(3/2)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]]
))/(8*Sqrt[b]*(a + b*x^2)*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.199, size = 77, normalized size = 0.6 \begin{align*}{\frac{x \left ( 2\,b{x}^{2}+5\,a \right ) \left ( b{x}^{2}+a \right ) }{8}{\frac{1}{\sqrt [4]{ \left ( b{x}^{2}+a \right ) ^{2}}}}}+{\frac{3\,{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) \sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt [4]{ \left ( b{x}^{2}+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/4),x)

[Out]

1/8*x*(2*b*x^2+5*a)*(b*x^2+a)/((b*x^2+a)^2)^(1/4)+3/8*a^2*ln(x*b^(1/2)+(b*x^2+a)^(1/2))/b^(1/2)/((b*x^2+a)^2)^
(1/4)*(b*x^2+a)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{3}{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/4),x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(3/4), x)

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Fricas [A]  time = 1.41979, size = 408, normalized size = 3.19 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{1}{4}} \sqrt{b} x - a\right ) + 2 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{1}{4}}{\left (2 \, b^{2} x^{3} + 5 \, a b x\right )}}{16 \, b}, -\frac{3 \, a^{2} \sqrt{-b} \arctan \left (\frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{1}{4}} \sqrt{-b} x}{b x^{2} + a}\right ) -{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{1}{4}}{\left (2 \, b^{2} x^{3} + 5 \, a b x\right )}}{8 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/4),x, algorithm="fricas")

[Out]

[1/16*(3*a^2*sqrt(b)*log(-2*b*x^2 - 2*(b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)*sqrt(b)*x - a) + 2*(b^2*x^4 + 2*a*b*x^
2 + a^2)^(1/4)*(2*b^2*x^3 + 5*a*b*x))/b, -1/8*(3*a^2*sqrt(-b)*arctan((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)*sqrt(-b
)*x/(b*x^2 + a)) - (b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)*(2*b^2*x^3 + 5*a*b*x))/b]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac{3}{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/4),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(3/4), x)

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Giac [A]  time = 1.21555, size = 117, normalized size = 0.91 \begin{align*} -\frac{1}{8} \,{\left (\frac{x^{4}{\left (\frac{5 \, \sqrt{-b x^{2} - a}{\left (b + \frac{a}{x^{2}}\right )}{\left | x \right |}}{x^{2}} - \frac{3 \, \sqrt{-b x^{2} - a} b{\left | x \right |}}{x^{2}}\right )}}{a^{2}} + \frac{3 \, \arctan \left (\frac{\sqrt{-b x^{2} - a}{\left | x \right |}}{\sqrt{b} x^{2}}\right )}{\sqrt{b}}\right )} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/4),x, algorithm="giac")

[Out]

-1/8*(x^4*(5*sqrt(-b*x^2 - a)*(b + a/x^2)*abs(x)/x^2 - 3*sqrt(-b*x^2 - a)*b*abs(x)/x^2)/a^2 + 3*arctan(sqrt(-b
*x^2 - a)*abs(x)/(sqrt(b)*x^2))/sqrt(b))*a^2