Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=\frac {b x^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 a x^2}-\frac {\sqrt {b} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt [4]{a+b x^4}} \] Output:
1/2*b*x^2/a/(b*x^4+a)^(1/4)-1/2*(b*x^4+a)^(3/4)/a/x^2-1/2*b^(1/2)*(1+b*x^4 /a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x^2/a^(1/2))),2^(1/2))/a^(1/2)/ (b*x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {b x^4}{a}\right )}{2 x^2 \sqrt [4]{a+b x^4}} \] Input:
Integrate[1/(x^3*(a + b*x^4)^(1/4)),x]
Output:
-1/2*((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-1/2, 1/4, 1/2, -((b*x^4)/a) ])/(x^2*(a + b*x^4)^(1/4))
Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {807, 264, 227, 225, 212}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \sqrt [4]{b x^4+a}}dx^2\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{2} \left (\frac {b \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2}{2 a}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {1}{2} \left (\frac {b \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {1}{2} \left (\frac {b \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {1}{2} \left (\frac {b \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{2 a \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{a x^2}\right )\) |
Input:
Int[1/(x^3*(a + b*x^4)^(1/4)),x]
Output:
(-((a + b*x^4)^(3/4)/(a*x^2)) + (b*(1 + (b*x^4)/a)^(1/4)*((2*x^2)/(1 + (b* x^4)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/ Sqrt[b]))/(2*a*(a + b*x^4)^(1/4)))/2
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
\[\int \frac {1}{x^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]
Input:
int(1/x^3/(b*x^4+a)^(1/4),x)
Output:
int(1/x^3/(b*x^4+a)^(1/4),x)
\[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)/(b*x^7 + a*x^3), x)
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.30 \[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} x^{2}} \] Input:
integrate(1/x**3/(b*x**4+a)**(1/4),x)
Output:
-hyper((-1/2, 1/4), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(1/4)*x**2)
\[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^4 + a)^(1/4)*x^3), x)
\[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{3}} \,d x } \] Input:
integrate(1/x^3/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate(1/((b*x^4 + a)^(1/4)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=\int \frac {1}{x^3\,{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:
int(1/(x^3*(a + b*x^4)^(1/4)),x)
Output:
int(1/(x^3*(a + b*x^4)^(1/4)), x)
\[ \int \frac {1}{x^3 \sqrt [4]{a+b x^4}} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{3}}d x \] Input:
int(1/x^3/(b*x^4+a)^(1/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*x**3),x)