3.1125 \(\int \frac {x^{10}}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=106 \[ -\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b} \]

[Out]

-7/32*a*x^3*(b*x^4+a)^(1/4)/b^2+1/8*x^7*(b*x^4+a)^(1/4)/b-21/64*a^2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(11/4)
+21/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(11/4)

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Rubi [A]  time = 0.04, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(a + b*x^4)^(3/4),x]

[Out]

(-7*a*x^3*(a + b*x^4)^(1/4))/(32*b^2) + (x^7*(a + b*x^4)^(1/4))/(8*b) - (21*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)
^(1/4)])/(64*b^(11/4)) + (21*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(11/4))

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^{10}}{\left (a+b x^4\right )^{3/4}} \, dx &=\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {(7 a) \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx}{8 b}\\ &=-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}+\frac {\left (21 a^2\right ) \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx}{32 b^2}\\ &=-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}+\frac {\left (21 a^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{32 b^2}\\ &=-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}+\frac {\left (21 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/2}}-\frac {\left (21 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{5/2}}\\ &=-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.84 \[ \frac {-21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (4 b x^4-7 a\right )}{64 b^{11/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(a + b*x^4)^(3/4),x]

[Out]

(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(-7*a + 4*b*x^4) - 21*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + 21*a^2*ArcT
anh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(11/4))

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fricas [B]  time = 1.01, size = 227, normalized size = 2.14 \[ -\frac {84 \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b^{8} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}} - b^{8} x \sqrt {\frac {b^{6} x^{2} \sqrt {\frac {a^{8}}{b^{11}}} + \sqrt {b x^{4} + a} a^{4}}{x^{2}}} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}}}{a^{8} x}\right ) - 21 \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (\frac {21 \, {\left (b^{3} x \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) + 21 \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {21 \, {\left (b^{3} x \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) - 4 \, {\left (4 \, b x^{7} - 7 \, a x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

-1/128*(84*b^2*(a^8/b^11)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*a^2*b^8*(a^8/b^11)^(3/4) - b^8*x*sqrt((b^6*x^2*sqrt
(a^8/b^11) + sqrt(b*x^4 + a)*a^4)/x^2)*(a^8/b^11)^(3/4))/(a^8*x)) - 21*b^2*(a^8/b^11)^(1/4)*log(21*(b^3*x*(a^8
/b^11)^(1/4) + (b*x^4 + a)^(1/4)*a^2)/x) + 21*b^2*(a^8/b^11)^(1/4)*log(-21*(b^3*x*(a^8/b^11)^(1/4) - (b*x^4 +
a)^(1/4)*a^2)/x) - 4*(4*b*x^7 - 7*a*x^3)*(b*x^4 + a)^(1/4))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(x^10/(b*x^4 + a)^(3/4), x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(b*x^4+a)^(3/4),x)

[Out]

int(x^10/(b*x^4+a)^(3/4),x)

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maxima [A]  time = 3.00, size = 155, normalized size = 1.46 \[ \frac {\frac {11 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b}{x} - \frac {7 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}{x^{5}}}{32 \, {\left (b^{4} - \frac {2 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}}\right )}} + \frac {21 \, {\left (\frac {2 \, a^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{2} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

1/32*(11*(b*x^4 + a)^(1/4)*a^2*b/x - 7*(b*x^4 + a)^(5/4)*a^2/x^5)/(b^4 - 2*(b*x^4 + a)*b^3/x^4 + (b*x^4 + a)^2
*b^2/x^8) + 21/128*(2*a^2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - a^2*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^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{10}}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(a + b*x^4)^(3/4),x)

[Out]

int(x^10/(a + b*x^4)^(3/4), x)

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sympy [C]  time = 2.35, size = 37, normalized size = 0.35 \[ \frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(b*x**4+a)**(3/4),x)

[Out]

x**11*gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/4)*gamma(15/4))

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