3.11.65 \(\int \frac {(a+b x^4)^{5/4}}{x^2} \, dx\) [1065]

Optimal. Leaf size=94 \[ \frac {5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{x}-\frac {5}{8} a \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {5}{8} a \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) \]

[Out]

5/4*b*x^3*(b*x^4+a)^(1/4)-(b*x^4+a)^(5/4)/x-5/8*a*b^(1/4)*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))+5/8*a*b^(1/4)*arct
anh(b^(1/4)*x/(b*x^4+a)^(1/4))

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Rubi [A]
time = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {283, 285, 338, 304, 209, 212} \begin {gather*} -\frac {5}{8} a \sqrt [4]{b} \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{5/4}}{x}+\frac {5}{8} a \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {5}{4} b x^3 \sqrt [4]{a+b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^2,x]

[Out]

(5*b*x^3*(a + b*x^4)^(1/4))/4 - (a + b*x^4)^(5/4)/x - (5*a*b^(1/4)*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/8 +
(5*a*b^(1/4)*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/8

Rule 209

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

Rule 212

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

Rule 283

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 285

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

Rule 304

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 338

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^2} \, dx &=-\frac {\left (a+b x^4\right )^{5/4}}{x}+(5 b) \int x^2 \sqrt [4]{a+b x^4} \, dx\\ &=\frac {5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{x}+\frac {1}{4} (5 a b) \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{x}+\frac {1}{4} (5 a b) \text {Subst}\left (\int \frac {x^2}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac {5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{x}+\frac {1}{8} \left (5 a \sqrt {b}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )-\frac {1}{8} \left (5 a \sqrt {b}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac {5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{x}-\frac {5}{8} a \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {5}{8} a \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 86, normalized size = 0.91 \begin {gather*} \frac {\left (-4 a+b x^4\right ) \sqrt [4]{a+b x^4}}{4 x}-\frac {5}{8} a \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {5}{8} a \sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^2,x]

[Out]

((-4*a + b*x^4)*(a + b*x^4)^(1/4))/(4*x) - (5*a*b^(1/4)*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/8 + (5*a*b^(1/4
)*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/8

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(5/4)/x^2,x)

[Out]

int((b*x^4+a)^(5/4)/x^2,x)

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Maxima [A]
time = 0.50, size = 119, normalized size = 1.27 \begin {gather*} \frac {5}{16} \, {\left (2 \, b^{\frac {1}{4}} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) - b^{\frac {1}{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 )\right )} a - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a b}{4 \, {\left (b - \frac {b x^{4} + a}{x^{4}}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^2,x, algorithm="maxima")

[Out]

5/16*(2*b^(1/4)*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x)) - b^(1/4)*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4)
+ (b*x^4 + a)^(1/4)/x)))*a - (b*x^4 + a)^(1/4)*a/x - 1/4*(b*x^4 + a)^(1/4)*a*b/((b - (b*x^4 + a)/x^4)*x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [C] Result contains complex when optimal does not.
time = 0.97, size = 41, normalized size = 0.44 \begin {gather*} \frac {a^{\frac {5}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(5/4)/x**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(5/4)/x^2,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)/x^2, x)

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Mupad [B]
time = 1.53, size = 40, normalized size = 0.43 \begin {gather*} -\frac {{\left (b\,x^4+a\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {b\,x^4}{a}\right )}{x\,{\left (\frac {b\,x^4}{a}+1\right )}^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(5/4)/x^2,x)

[Out]

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

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