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

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

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Rubi [A]  time = 0.03, antiderivative size = 128, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.08, size = 97, normalized size = 0.76 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 6.81, size = 85, normalized size = 0.66 \begin {gather*} \frac {\left (\left (a+b x^2\right )^2\right )^{3/4} \left (\frac {1}{8} \sqrt {a+b x^2} \left (5 a x+2 b x^3\right )-\frac {3 a^2 \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{8 \sqrt {b}}\right )}{\left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 177, normalized size = 1.38 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.44, size = 87, normalized size = 0.68 \begin {gather*} -\frac {\frac {3 \, a^{3} \arctan \left (\frac {\sqrt {-\frac {b x^{2} + a}{x^{2}}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {{\left (5 \, a^{3} {\left (b + \frac {a}{x^{2}}\right )} \sqrt {-\frac {b x^{2} + a}{x^{2}}} - 3 \, a^{3} b \sqrt {-\frac {b x^{2} + a}{x^{2}}}\right )} x^{4}}{a^{2}}}{8 \, a} \end {gather*}

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

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maple [A]  time = 0.03, size = 77, normalized size = 0.60 \begin {gather*} \frac {3 \sqrt {b \,x^{2}+a}\, a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {1}{4}} \sqrt {b}}+\frac {\left (2 b \,x^{2}+5 a \right ) \left (b \,x^{2}+a \right ) x}{8 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {1}{4}}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {3}{4}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {3}{4}}\, dx \end {gather*}

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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