\(\int \frac {1}{x^4 (a+b x^4)^{3/2}} \, dx\) [865]

Optimal result

Integrand size = 15, antiderivative size = 131 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {1}{2 a x^3 \sqrt {a+b x^4}}-\frac {5 \sqrt {a+b x^4}}{6 a^2 x^3}-\frac {5 b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+b x^4}} \]

[Out]

1/2/a/x^3/(b*x^4+a)^(1/2)-5/6*(b*x^4+a)^(1/2)/a^2/x^3-5/12*b^(3/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/
cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))
*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(9/4)/(b*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {296, 331, 226} \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {5 b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+b x^4}}-\frac {5 \sqrt {a+b x^4}}{6 a^2 x^3}+\frac {1}{2 a x^3 \sqrt {a+b x^4}} \]

[In]

Int[1/(x^4*(a + b*x^4)^(3/2)),x]

[Out]

1/(2*a*x^3*Sqrt[a + b*x^4]) - (5*Sqrt[a + b*x^4])/(6*a^2*x^3) - (5*b^(3/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b
*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(12*a^(9/4)*Sqrt[a + b*x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 a x^3 \sqrt {a+b x^4}}+\frac {5 \int \frac {1}{x^4 \sqrt {a+b x^4}} \, dx}{2 a} \\ & = \frac {1}{2 a x^3 \sqrt {a+b x^4}}-\frac {5 \sqrt {a+b x^4}}{6 a^2 x^3}-\frac {(5 b) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{6 a^2} \\ & = \frac {1}{2 a x^3 \sqrt {a+b x^4}}-\frac {5 \sqrt {a+b x^4}}{6 a^2 x^3}-\frac {5 b^{3/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{12 a^{9/4} \sqrt {a+b x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {1}{4},-\frac {b x^4}{a}\right )}{3 a x^3 \sqrt {a+b x^4}} \]

[In]

Integrate[1/(x^4*(a + b*x^4)^(3/2)),x]

[Out]

-1/3*(Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[-3/4, 3/2, 1/4, -((b*x^4)/a)])/(a*x^3*Sqrt[a + b*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {b x}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {5 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(113\)
elliptic	\(-\frac {b x}{2 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {5 b \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{6 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(113\)
risch	\(-\frac {\sqrt {b \,x^{4}+a}}{3 a^{2} x^{3}}-\frac {b \left (4 a \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{3 a^{2}}\)	\(217\)

[In]

int(1/x^4/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*b/a^2*x/((x^4+a/b)*b)^(1/2)-1/3*(b*x^4+a)^(1/2)/a^2/x^3-5/6*b/a^2/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*
b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {5 \, {\left (b x^{7} + a x^{3}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (5 \, b x^{4} + 2 \, a\right )} \sqrt {b x^{4} + a}}{6 \, {\left (a^{2} b x^{7} + a^{3} x^{3}\right )}} \]

[In]

integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

1/6*(5*(b*x^7 + a*x^3)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - (5*b*x^4 + 2*a)*sqrt(b*x^
4 + a))/(a^2*b*x^7 + a^3*x^3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{3} \Gamma \left (\frac {1}{4}\right )} \]

[In]

integrate(1/x**4/(b*x**4+a)**(3/2),x)

[Out]

gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*x**3*gamma(1/4))

Maxima [F]

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

[In]

integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^4), x)

Giac [F]

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

[In]

integrate(1/x^4/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^4\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \]

[In]

int(1/(x^4*(a + b*x^4)^(3/2)),x)

[Out]

int(1/(x^4*(a + b*x^4)^(3/2)), x)