\(\int \frac {1}{x^7 (a+b x^4)^{5/4}} \, dx\) [1152]

Optimal result

Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac {7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac {7 b^{3/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{a+b x^4}} \]

[Out]

-1/6/a/x^6/(b*x^4+a)^(1/4)+7/12*b/a^2/x^2/(b*x^4+a)^(1/4)+7/4*b^(3/2)*(1+b*x^4/a)^(1/4)*(cos(1/2*arctan(x^2*b^
(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2
^(1/2))/a^(5/2)/(b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {281, 292, 203, 202} \[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\frac {7 b^{3/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{a+b x^4}}+\frac {7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}-\frac {1}{6 a x^6 \sqrt [4]{a+b x^4}} \]

[In]

Int[1/(x^7*(a + b*x^4)^(5/4)),x]

[Out]

-1/6*1/(a*x^6*(a + b*x^4)^(1/4)) + (7*b)/(12*a^2*x^2*(a + b*x^4)^(1/4)) + (7*b^(3/2)*(1 + (b*x^4)/a)^(1/4)*Ell
ipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*a^(5/2)*(a + b*x^4)^(1/4))

Rule 202

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]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + b*
(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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]

Rule 292

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(c*x)^(m + 1)/(a*c*(m + 1)*(a + b*x^2)^(1
/4)), x] - Dist[b*((2*m + 1)/(2*a*c^2*(m + 1))), Int[(c*x)^(m + 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c
}, x] && PosQ[b/a] && IntegerQ[2*m] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^4 \left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 a x^6 \sqrt [4]{a+b x^4}}-\frac {(7 b) \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{12 a} \\ & = -\frac {1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac {7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac {\left (7 b^2\right ) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{8 a^2} \\ & = -\frac {1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac {7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac {\left (7 b^2 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{8 a^3 \sqrt [4]{a+b x^4}} \\ & = -\frac {1}{6 a x^6 \sqrt [4]{a+b x^4}}+\frac {7 b}{12 a^2 x^2 \sqrt [4]{a+b x^4}}+\frac {7 b^{3/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 a^{5/2} \sqrt [4]{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 = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.52 \[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{4},-\frac {1}{2},-\frac {b x^4}{a}\right )}{6 a x^6 \sqrt [4]{a+b x^4}} \]

[In]

Integrate[1/(x^7*(a + b*x^4)^(5/4)),x]

[Out]

-1/6*((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-3/2, 5/4, -1/2, -((b*x^4)/a)])/(a*x^6*(a + b*x^4)^(1/4))

Maple [F]

\[\int \frac {1}{x^{7} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}d x\]

[In]

int(1/x^7/(b*x^4+a)^(5/4),x)

[Out]

int(1/x^7/(b*x^4+a)^(5/4),x)

Fricas [F]

\[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{7}} \,d x } \]

[In]

integrate(1/x^7/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/(b^2*x^15 + 2*a*b*x^11 + a^2*x^7), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.76 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 a^{\frac {5}{4}} x^{6}} \]

[In]

integrate(1/x**7/(b*x**4+a)**(5/4),x)

[Out]

-hyper((-3/2, 5/4), (-1/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(5/4)*x**6)

Maxima [F]

\[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{7}} \,d x } \]

[In]

integrate(1/x^7/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^7), x)

Giac [F]

\[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{7}} \,d x } \]

[In]

integrate(1/x^7/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^7), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^7 \left (a+b x^4\right )^{5/4}} \, dx=\int \frac {1}{x^7\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \]

[In]

int(1/(x^7*(a + b*x^4)^(5/4)),x)

[Out]

int(1/(x^7*(a + b*x^4)^(5/4)), x)