3.1066 \(\int \frac {(a+b x^4)^{5/4}}{x^6} \, dx\)

Optimal. Leaf size=92 \[ -\frac {1}{2} b^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{2} b^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac {b \sqrt [4]{a+b x^4}}{x}-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5} \]

[Out]

-b*(b*x^4+a)^(1/4)/x-1/5*(b*x^4+a)^(5/4)/x^5-1/2*b^(5/4)*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))+1/2*b^(5/4)*arctanh
(b^(1/4)*x/(b*x^4+a)^(1/4))

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Rubi [A]  time = 0.03, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {277, 331, 298, 203, 206} \[ -\frac {1}{2} b^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{2} b^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac {b \sqrt [4]{a+b x^4}}{x}-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^6,x]

[Out]

-((b*(a + b*x^4)^(1/4))/x) - (a + b*x^4)^(5/4)/(5*x^5) - (b^(5/4)*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/2 + (
b^(5/4)*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/2

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 277

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^4\right )^{5/4}}{x^6} \, dx &=-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5}+b \int \frac {\sqrt [4]{a+b x^4}}{x^2} \, dx\\ &=-\frac {b \sqrt [4]{a+b x^4}}{x}-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5}+b^2 \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=-\frac {b \sqrt [4]{a+b x^4}}{x}-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5}+b^2 \operatorname {Subst}\left (\int \frac {x^2}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )\\ &=-\frac {b \sqrt [4]{a+b x^4}}{x}-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5}+\frac {1}{2} b^{3/2} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )-\frac {1}{2} b^{3/2} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )\\ &=-\frac {b \sqrt [4]{a+b x^4}}{x}-\frac {\left (a+b x^4\right )^{5/4}}{5 x^5}-\frac {1}{2} b^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{2} b^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.57 \[ -\frac {a \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac {5}{4},-\frac {5}{4};-\frac {1}{4};-\frac {b x^4}{a}\right )}{5 x^5 \sqrt [4]{\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^6,x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^6,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^6,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)/x^6, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(5/4)/x^6,x)

[Out]

int((b*x^4+a)^(5/4)/x^6,x)

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maxima [A]  time = 2.95, size = 97, normalized size = 1.05 \[ \frac {1}{2} \, b^{\frac {5}{4}} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) - \frac {1}{4} \, b^{\frac {5}{4}} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right ) - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^6,x, algorithm="maxima")

[Out]

1/2*b^(5/4)*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x)) - 1/4*b^(5/4)*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4)
+ (b*x^4 + a)^(1/4)/x)) - (b*x^4 + a)^(1/4)*b/x - 1/5*(b*x^4 + a)^(5/4)/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^4+a\right )}^{5/4}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(5/4)/x^6,x)

[Out]

int((a + b*x^4)^(5/4)/x^6, x)

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sympy [C]  time = 4.83, size = 46, normalized size = 0.50 \[ \frac {a^{\frac {5}{4}} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(5/4)/x**6,x)

[Out]

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

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