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

Optimal. Leaf size=77 \[ \frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b} \]

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Rubi [A]  time = 0.04, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {459, 331, 298, 203, 206} \begin {gather*} \frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(b + a*x^4)^(1/4))/4 + (7*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3/4)) - (7*b*ArcTanh[(a^(1/4)*x)
/(b + a*x^4)^(1/4)])/(8*a^(3/4))

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 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 459

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx &=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {1}{4} (7 b) \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx\\ &=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {1}{4} (7 b) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )\\ &=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 \sqrt {a}}+\frac {(7 b) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 \sqrt {a}}\\ &=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 77, normalized size = 1.00 \begin {gather*} \frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(b + a*x^4)^(1/4))/4 + (7*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3/4)) - (7*b*ArcTanh[(a^(1/4)*x)
/(b + a*x^4)^(1/4)])/(8*a^(3/4))

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IntegrateAlgebraic [A]  time = 0.51, size = 77, normalized size = 1.00 \begin {gather*} \frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(b + a*x^4)^(1/4))/4 + (7*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3/4)) - (7*b*ArcTanh[(a^(1/4)*x)
/(b + a*x^4)^(1/4)])/(8*a^(3/4))

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fricas [B]  time = 0.50, size = 192, normalized size = 2.49 \begin {gather*} \frac {1}{4} \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} x^{3} + \frac {7}{4} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {a^{2} \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}} x \sqrt {\frac {a^{2} \sqrt {\frac {b^{4}}{a^{3}}} x^{2} + \sqrt {a x^{4} + b} b^{2}}{x^{2}}} - {\left (a x^{4} + b\right )}^{\frac {1}{4}} a^{2} b \left (\frac {b^{4}}{a^{3}}\right )^{\frac {3}{4}}}{b^{4} x}\right ) - \frac {7}{16} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + \frac {7}{16} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*x^4 + b)^(1/4)*x^3 + 7/4*(b^4/a^3)^(1/4)*arctan((a^2*(b^4/a^3)^(3/4)*x*sqrt((a^2*sqrt(b^4/a^3)*x^2 + sq
rt(a*x^4 + b)*b^2)/x^2) - (a*x^4 + b)^(1/4)*a^2*b*(b^4/a^3)^(3/4))/(b^4*x)) - 7/16*(b^4/a^3)^(1/4)*log(7*(a*(b
^4/a^3)^(1/4)*x + (a*x^4 + b)^(1/4)*b)/x) + 7/16*(b^4/a^3)^(1/4)*log(-7*(a*(b^4/a^3)^(1/4)*x - (a*x^4 + b)^(1/
4)*b)/x)

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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 )} x^{2}}{{\left (a x^{4} + b\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 185, normalized size = 2.40 \begin {gather*} -\frac {1}{4} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {\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}}}\right )} - \frac {1}{16} \, a {\left (\frac {3 \, {\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}}}\right )}}{a} + \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 2.74, size = 78, normalized size = 1.01 \begin {gather*} \frac {a x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} - \frac {\sqrt [4]{b} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**7*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*x**4*exp_polar(I*pi)/b)/(4*b**(3/4)*gamma(11/4)) - b**(1/4)*x**
3*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*x**4*exp_polar(I*pi)/b)/(4*gamma(7/4))

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