3.12.67 \(\int \frac {(-b+a x^4) \sqrt [4]{b+a x^4}}{x^2} \, dx\)

Optimal. Leaf size=86 \[ \frac {\sqrt [4]{a x^4+b} \left (a x^4+4 b\right )}{4 x}+\frac {3}{8} \sqrt [4]{a} b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-\frac {3}{8} \sqrt [4]{a} b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 93, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {453, 279, 331, 298, 203, 206} \begin {gather*} \frac {\left (a x^4+b\right )^{5/4}}{x}+\frac {3}{8} \sqrt [4]{a} b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-\frac {3}{8} \sqrt [4]{a} b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-\frac {3}{4} a x^3 \sqrt [4]{a x^4+b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rule 453

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

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 61, normalized size = 0.71 \begin {gather*} \frac {\sqrt [4]{a x^4+b} \left (-\frac {a x^4 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {a x^4}{b}\right )}{\sqrt [4]{\frac {a x^4}{b}+1}}+a x^4+b\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*x^4)^(1/4)*(b + a*x^4 - (a*x^4*Hypergeometric2F1[-1/4, 3/4, 7/4, -((a*x^4)/b)])/(1 + (a*x^4)/b)^(1/4))
)/x

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IntegrateAlgebraic [A]  time = 0.35, size = 86, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{b+a x^4} \left (4 b+a x^4\right )}{4 x}+\frac {3}{8} \sqrt [4]{a} b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )-\frac {3}{8} \sqrt [4]{a} b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  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((a*x^4-b)*(a*x^4+b)^(1/4)/x^2,x, algorithm="fricas")

[Out]

Timed out

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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 {1}{4}} {\left (a x^{4} - b\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4+b)^(1/4)/x^2,x, algorithm="giac")

[Out]

integrate((a*x^4 + b)^(1/4)*(a*x^4 - b)/x^2, x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \right ) \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)*(a*x^4+b)^(1/4)/x^2,x)

[Out]

int((a*x^4-b)*(a*x^4+b)^(1/4)/x^2,x)

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maxima [B]  time = 0.42, size = 190, normalized size = 2.21 \begin {gather*} \frac {1}{16} \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}} - \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a - \frac {a x^{4} + b}{x^{4}}\right )} x}\right )} a - \frac {1}{4} \, {\left (2 \, a^{\frac {1}{4}} \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) - a^{\frac {1}{4}} \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right ) - \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4+b)^(1/4)/x^2,x, algorithm="maxima")

[Out]

1/16*(2*b*arctan((a*x^4 + b)^(1/4)/(a^(1/4)*x))/a^(3/4) - b*log(-(a^(1/4) - (a*x^4 + b)^(1/4)/x)/(a^(1/4) + (a
*x^4 + b)^(1/4)/x))/a^(3/4) - 4*(a*x^4 + b)^(1/4)*b/((a - (a*x^4 + b)/x^4)*x))*a - 1/4*(2*a^(1/4)*arctan((a*x^
4 + b)^(1/4)/(a^(1/4)*x)) - a^(1/4)*log(-(a^(1/4) - (a*x^4 + b)^(1/4)/x)/(a^(1/4) + (a*x^4 + b)^(1/4)/x)) - 4*
(a*x^4 + b)^(1/4)/x)*b

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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 )}^{1/4}\,\left (b-a\,x^4\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((b + a*x^4)^(1/4)*(b - a*x^4))/x^2,x)

[Out]

-int(((b + a*x^4)^(1/4)*(b - a*x^4))/x^2, x)

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sympy [C]  time = 3.16, size = 83, normalized size = 0.97 \begin {gather*} \frac {a \sqrt [4]{b} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {b^{\frac {5}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4-b)*(a*x**4+b)**(1/4)/x**2,x)

[Out]

a*b**(1/4)*x**3*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), a*x**4*exp_polar(I*pi)/b)/(4*gamma(7/4)) - b**(5/4)*gamm
a(-1/4)*hyper((-1/4, -1/4), (3/4,), a*x**4*exp_polar(I*pi)/b)/(4*x*gamma(3/4))

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