\(\int (a^2+2 a b x^2+b^2 x^4)^{3/4} \, dx\) [1]

Optimal result

Integrand size = 22, antiderivative size = 128 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\frac {1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}+\frac {3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac {3 \sqrt {a} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1103, 201, 221} \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\frac {3 \sqrt {a} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \text {arcsinh}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/2}}+\frac {3 a x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}{8 \left (a+b x^2\right )}+\frac {1}{4} x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \]

[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 201

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 221

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 1103

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*} \text {integral}& = \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] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.69 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\frac {\left (\left (a+b x^2\right )^2\right )^{3/4} \left (\sqrt {b} x \sqrt {a+b x^2} \left (5 a+2 b x^2\right )-3 a^2 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{8 \sqrt {b} \left (a+b x^2\right )^{3/2}} \]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.60

method	result	size
risch	\(\frac {x \left (2 b \,x^{2}+5 a \right ) \left (b \,x^{2}+a \right )}{8 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}+\frac {3 a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {b \,x^{2}+a}}{8 \sqrt {b}\, {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {1}{4}}}\)	\(77\)

[In]

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

[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(b^(1/2)*x+(b*x^2+a)^(1/2))/b^(1/2)/((b*x^2+a)^2)^
(1/4)*(b*x^2+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.38 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} {\left (2 \, b^{2} x^{3} + 5 \, a b x\right )}}{16 \, b}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} \sqrt {-b} x}{b x^{2} + a}\right ) - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} {\left (2 \, b^{2} x^{3} + 5 \, a b x\right )}}{8 \, b}\right ] \]

[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]

Sympy [F]

\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {3}{4}}\, dx \]

[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)

Maxima [F]

\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {3}{4}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.46 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=-\frac {1}{8} \, {\left (2 \, b x^{2} + 5 \, a\right )} \sqrt {-b x^{2} - a} x - \frac {3 \, a^{2} \log \left ({\left | -\sqrt {-b} x + \sqrt {-b x^{2} - a} \right |}\right )}{8 \, \sqrt {-b}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4} \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4} \,d x \]

[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)