3.1126 \(\int \frac {x^6}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=80 \[ \frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}+\frac {x^3 \sqrt [4]{a+b x^4}}{4 b} \]

[Out]

1/4*x^3*(b*x^4+a)^(1/4)/b+3/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(7/4)-3/8*a*arctanh(b^(1/4)*x/(b*x^4+a)^(1
/4))/b^(7/4)

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {321, 331, 298, 203, 206} \[ \frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}+\frac {x^3 \sqrt [4]{a+b x^4}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^4)^(3/4),x]

[Out]

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

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 321

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

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx &=\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {(3 a) \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx}{4 b}\\ &=\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {(3 a) \operatorname {Subst}\left (\int \frac {x^2}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4 b}\\ &=\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{3/2}}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{3/2}}\\ &=\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}+\frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 75, normalized size = 0.94 \[ \frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4}+3 a \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^4)^(3/4),x]

[Out]

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

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fricas [B]  time = 0.53, size = 204, normalized size = 2.55 \[ \frac {4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{3} + 12 \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \arctan \left (\frac {b^{5} x \sqrt {\frac {b^{4} x^{2} \sqrt {\frac {a^{4}}{b^{7}}} + \sqrt {b x^{4} + a} a^{2}}{x^{2}}} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{5} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {3}{4}}}{a^{4} x}\right ) - 3 \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (\frac {3 \, {\left (b^{2} x \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )}}{x}\right ) + 3 \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (b^{2} x \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )}}{x}\right )}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/16*(4*(b*x^4 + a)^(1/4)*x^3 + 12*b*(a^4/b^7)^(1/4)*arctan((b^5*x*sqrt((b^4*x^2*sqrt(a^4/b^7) + sqrt(b*x^4 +
a)*a^2)/x^2)*(a^4/b^7)^(3/4) - (b*x^4 + a)^(1/4)*a*b^5*(a^4/b^7)^(3/4))/(a^4*x)) - 3*b*(a^4/b^7)^(1/4)*log(3*(
b^2*x*(a^4/b^7)^(1/4) + (b*x^4 + a)^(1/4)*a)/x) + 3*b*(a^4/b^7)^(1/4)*log(-3*(b^2*x*(a^4/b^7)^(1/4) - (b*x^4 +
a)^(1/4)*a)/x))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^6/(b*x^4 + a)^(3/4), x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^4+a)^(3/4),x)

[Out]

int(x^6/(b*x^4+a)^(3/4),x)

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maxima [A]  time = 3.00, size = 110, normalized size = 1.38 \[ -\frac {3 \, {\left (\frac {2 \, a \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{16 \, b} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{4 \, {\left (b^{2} - \frac {{\left (b x^{4} + a\right )} b}{x^{4}}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a + b*x^4)^(3/4),x)

[Out]

int(x^6/(a + b*x^4)^(3/4), x)

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sympy [C]  time = 1.64, size = 37, normalized size = 0.46 \[ \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**4+a)**(3/4),x)

[Out]

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

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