3.15.76 \(\int \frac {(b+a x^4)^{3/4}}{x^4 (2 b+a x^4)} \, dx\)

Optimal. Leaf size=104 \[ \frac {a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{4\ 2^{3/4} b}+\frac {a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{a x^4+b}}\right )}{4\ 2^{3/4} b}-\frac {\left (a x^4+b\right )^{3/4}}{6 b x^3} \]

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Rubi [C]  time = 0.07, antiderivative size = 46, normalized size of antiderivative = 0.44, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {511, 510} \begin {gather*} -\frac {\left (a x^4+b\right )^{3/4} \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\frac {a x^4}{2 \left (a x^4+b\right )}\right )}{6 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x]

[Out]

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

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx &=\frac {\left (b+a x^4\right )^{3/4} \int \frac {\left (1+\frac {a x^4}{b}\right )^{3/4}}{x^4 \left (2 b+a x^4\right )} \, dx}{\left (1+\frac {a x^4}{b}\right )^{3/4}}\\ &=-\frac {\left (b+a x^4\right )^{3/4} \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\frac {a x^4}{2 \left (b+a x^4\right )}\right )}{6 b x^3}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 77, normalized size = 0.74 \begin {gather*} -\frac {\left (a x^4+b\right )^{3/4} \left (\frac {a x^4}{b}+2\right )^{3/4} \, _2F_1\left (-\frac {3}{4},-\frac {3}{4};\frac {1}{4};-\frac {a x^4}{a x^4+2 b}\right )}{6 b x^3 \left (\frac {2 a x^4}{b}+2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.35, size = 104, normalized size = 1.00 \begin {gather*} -\frac {\left (b+a x^4\right )^{3/4}}{6 b x^3}+\frac {a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} b}+\frac {a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{2} \sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x]

[Out]

-1/6*(b + a*x^4)^(3/4)/(b*x^3) + (a^(3/4)*ArcTan[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))])/(4*2^(3/4)*b) + (a^
(3/4)*ArcTanh[(a^(1/4)*x)/(2^(1/4)*(b + a*x^4)^(1/4))])/(4*2^(3/4)*b)

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fricas [B]  time = 63.20, size = 474, normalized size = 4.56 \begin {gather*} -\frac {12 \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {4 \, {\left (\left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a^{4} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} + 4 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} - 2 \, \sqrt {\frac {1}{2}} {\left (\left (\frac {1}{8}\right )^{\frac {1}{4}} \sqrt {a x^{4} + b} a^{2} b x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} + \left (\frac {1}{8}\right )^{\frac {3}{4}} {\left (3 \, a b^{3} x^{4} + 2 \, b^{4}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}}\right )} \sqrt {\sqrt {\frac {1}{2}} a^{2} b^{2} \sqrt {\frac {a^{3}}{b^{4}}}}\right )}}{a^{5} x^{4} + 2 \, a^{4} b}\right ) - 3 \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {a^{3}}{b^{4}}} + 8 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} b^{3} x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} x + \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{4} + 2 \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) + 3 \, \left (\frac {1}{8}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b\right )}^{\frac {1}{4}} a b^{2} x^{3} \sqrt {\frac {a^{3}}{b^{4}}} - 8 \, \left (\frac {1}{8}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} b^{3} x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} a^{2} x - \left (\frac {1}{8}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{4} + 2 \, a b^{2}\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{2 \, {\left (a x^{4} + 2 \, b\right )}}\right ) + 8 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}}}{48 \, b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x, algorithm="fricas")

[Out]

-1/48*(12*(1/8)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*arctan(-4*((1/8)^(1/4)*(a*x^4 + b)^(1/4)*a^4*b*x^3*(a^3/b^4)^(1/4)
+ 4*(1/8)^(3/4)*(a*x^4 + b)^(3/4)*a^2*b^3*x*(a^3/b^4)^(3/4) - 2*sqrt(1/2)*((1/8)^(1/4)*sqrt(a*x^4 + b)*a^2*b*
x^2*(a^3/b^4)^(1/4) + (1/8)^(3/4)*(3*a*b^3*x^4 + 2*b^4)*(a^3/b^4)^(3/4))*sqrt(sqrt(1/2)*a^2*b^2*sqrt(a^3/b^4))
)/(a^5*x^4 + 2*a^4*b)) - 3*(1/8)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(1/2*(2*sqrt(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*
sqrt(a^3/b^4) + 8*(1/8)^(3/4)*sqrt(a*x^4 + b)*b^3*x^2*(a^3/b^4)^(3/4) + 2*(a*x^4 + b)^(3/4)*a^2*x + (1/8)^(1/4
)*(3*a^2*b*x^4 + 2*a*b^2)*(a^3/b^4)^(1/4))/(a*x^4 + 2*b)) + 3*(1/8)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log(1/2*(2*sqr
t(1/2)*(a*x^4 + b)^(1/4)*a*b^2*x^3*sqrt(a^3/b^4) - 8*(1/8)^(3/4)*sqrt(a*x^4 + b)*b^3*x^2*(a^3/b^4)^(3/4) + 2*(
a*x^4 + b)^(3/4)*a^2*x - (1/8)^(1/4)*(3*a^2*b*x^4 + 2*a*b^2)*(a^3/b^4)^(1/4))/(a*x^4 + 2*b)) + 8*(a*x^4 + b)^(
3/4))/(b*x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{{\left (a x^{4} + 2 \, b\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x, algorithm="giac")

[Out]

integrate((a*x^4 + b)^(3/4)/((a*x^4 + 2*b)*x^4), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}+b \right )^{\frac {3}{4}}}{x^{4} \left (a \,x^{4}+2 b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x)

[Out]

int((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{{\left (a x^{4} + 2 \, b\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4+b)^(3/4)/x^4/(a*x^4+2*b),x, algorithm="maxima")

[Out]

integrate((a*x^4 + b)^(3/4)/((a*x^4 + 2*b)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x^4+b\right )}^{3/4}}{x^4\,\left (a\,x^4+2\,b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)),x)

[Out]

int((b + a*x^4)^(3/4)/(x^4*(2*b + a*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x^{4} + b\right )^{\frac {3}{4}}}{x^{4} \left (a x^{4} + 2 b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4+b)**(3/4)/x**4/(a*x**4+2*b),x)

[Out]

Integral((a*x**4 + b)**(3/4)/(x**4*(a*x**4 + 2*b)), x)

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