3.1050 \(\int \frac{\left (a+b x^4\right )^{5/4}}{x} \, dx\)

Optimal. Leaf size=83 \[ -\frac{1}{2} a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac{1}{2} a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )+a \sqrt [4]{a+b x^4}+\frac{1}{5} \left (a+b x^4\right )^{5/4} \]

[Out]

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

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Rubi [A]  time = 0.13528, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4 \[ -\frac{1}{2} a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-\frac{1}{2} a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )+a \sqrt [4]{a+b x^4}+\frac{1}{5} \left (a+b x^4\right )^{5/4} \]

Antiderivative was successfully verified.

[In]  Int[(a + b*x^4)^(5/4)/x,x]

[Out]

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

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Rubi in Sympy [A]  time = 12.868, size = 70, normalized size = 0.84 \[ - \frac{a^{\frac{5}{4}} \operatorname{atan}{\left (\frac{\sqrt [4]{a + b x^{4}}}{\sqrt [4]{a}} \right )}}{2} - \frac{a^{\frac{5}{4}} \operatorname{atanh}{\left (\frac{\sqrt [4]{a + b x^{4}}}{\sqrt [4]{a}} \right )}}{2} + a \sqrt [4]{a + b x^{4}} + \frac{\left (a + b x^{4}\right )^{\frac{5}{4}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x**4+a)**(5/4)/x,x)

[Out]

-a**(5/4)*atan((a + b*x**4)**(1/4)/a**(1/4))/2 - a**(5/4)*atanh((a + b*x**4)**(1
/4)/a**(1/4))/2 + a*(a + b*x**4)**(1/4) + (a + b*x**4)**(5/4)/5

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Mathematica [C]  time = 0.0546093, size = 76, normalized size = 0.92 \[ \frac{3 \left (6 a^2+7 a b x^4+b^2 x^8\right )-5 a^2 \left (\frac{a}{b x^4}+1\right )^{3/4} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};-\frac{a}{b x^4}\right )}{15 \left (a+b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

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

[Out]

(3*(6*a^2 + 7*a*b*x^4 + b^2*x^8) - 5*a^2*(1 + a/(b*x^4))^(3/4)*Hypergeometric2F1
[3/4, 3/4, 7/4, -(a/(b*x^4))])/(15*(a + b*x^4)^(3/4))

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Maple [F]  time = 0.035, size = 0, normalized size = 0. \[ \int{\frac{1}{x} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^4+a)^(5/4)/x,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.281554, size = 163, normalized size = 1.96 \[ \frac{1}{5} \,{\left (b x^{4} + 6 \, a\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}} +{\left (a^{5}\right )}^{\frac{1}{4}} \arctan \left (\frac{{\left (a^{5}\right )}^{\frac{1}{4}}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} a + \sqrt{\sqrt{b x^{4} + a} a^{2} + \sqrt{a^{5}}}}\right ) - \frac{1}{4} \,{\left (a^{5}\right )}^{\frac{1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} a +{\left (a^{5}\right )}^{\frac{1}{4}}\right ) + \frac{1}{4} \,{\left (a^{5}\right )}^{\frac{1}{4}} \log \left ({\left (b x^{4} + a\right )}^{\frac{1}{4}} a -{\left (a^{5}\right )}^{\frac{1}{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/5*(b*x^4 + 6*a)*(b*x^4 + a)^(1/4) + (a^5)^(1/4)*arctan((a^5)^(1/4)/((b*x^4 + a
)^(1/4)*a + sqrt(sqrt(b*x^4 + a)*a^2 + sqrt(a^5)))) - 1/4*(a^5)^(1/4)*log((b*x^4
+ a)^(1/4)*a + (a^5)^(1/4)) + 1/4*(a^5)^(1/4)*log((b*x^4 + a)^(1/4)*a - (a^5)^(
1/4))

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Sympy [A]  time = 7.07708, size = 48, normalized size = 0.58 \[ - \frac{b^{\frac{5}{4}} x^{5} \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{5}{4} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{4}}} \right )}}{4 \Gamma \left (- \frac{1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x**4+a)**(5/4)/x,x)

[Out]

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

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GIAC/XCAS [A]  time = 0.223816, size = 270, normalized size = 3.25 \[ -\frac{1}{4} \, \sqrt{2} \left (-a\right )^{\frac{1}{4}} a \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-a\right )^{\frac{1}{4}} + 2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}\right )}}{2 \, \left (-a\right )^{\frac{1}{4}}}\right ) - \frac{1}{4} \, \sqrt{2} \left (-a\right )^{\frac{1}{4}} a \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-a\right )^{\frac{1}{4}} - 2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}\right )}}{2 \, \left (-a\right )^{\frac{1}{4}}}\right ) - \frac{1}{8} \, \sqrt{2} \left (-a\right )^{\frac{1}{4}} a{\rm ln}\left (\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-a\right )^{\frac{1}{4}} + \sqrt{b x^{4} + a} + \sqrt{-a}\right ) + \frac{1}{8} \, \sqrt{2} \left (-a\right )^{\frac{1}{4}} a{\rm ln}\left (-\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-a\right )^{\frac{1}{4}} + \sqrt{b x^{4} + a} + \sqrt{-a}\right ) + \frac{1}{5} \,{\left (b x^{4} + a\right )}^{\frac{5}{4}} +{\left (b x^{4} + a\right )}^{\frac{1}{4}} a \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/4*sqrt(2)*(-a)^(1/4)*a*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)
^(1/4))/(-a)^(1/4)) - 1/4*sqrt(2)*(-a)^(1/4)*a*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)
^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4)) - 1/8*sqrt(2)*(-a)^(1/4)*a*ln(sqrt(2)*
(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a)) + 1/8*sqrt(2)*(-a)^(1
/4)*a*ln(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a)) + 1
/5*(b*x^4 + a)^(5/4) + (b*x^4 + a)^(1/4)*a