3.1065 \(\int \frac{(a+b x^4)^{5/4}}{x^2} \, dx\)

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*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

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Rubi [A]  time = 0.0357187, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {277, 279, 331, 298, 203, 206} \[ \frac{5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac{\left (a+b x^4\right )^{5/4}}{x}-\frac{5}{8} a \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac{5}{8} a \sqrt [4]{b} \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) \]

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 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 279

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 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]

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 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])

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) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.0088159, size = 50, normalized size = 0.53 \[ -\frac{a \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac{5}{4},-\frac{1}{4};\frac{3}{4};-\frac{b x^4}{a}\right )}{x \sqrt [4]{\frac{b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^2,x]

[Out]

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

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

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]  time = 2.64722, size = 41, normalized size = 0.44 \begin{align*} \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{align*}

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 [B]  time = 1.14041, size = 309, normalized size = 3.29 \begin{align*} \frac{1}{32} \,{\left (\frac{8 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b x^{3}}{a} + 10 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right ) + 10 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right ) + 5 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right ) - 5 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right ) - \frac{32 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*(b*x^4 + a)^(1/4)*b*x^3/a + 10*sqrt(2)*(-b)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 +
a)^(1/4)/x)/(-b)^(1/4)) + 10*sqrt(2)*(-b)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(1/4)/
x)/(-b)^(1/4)) + 5*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) + sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/
x^2) - 5*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) - sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2) - 32*
(b*x^4 + a)^(1/4)/x)*a