3.13.49 \(\int \frac {x^{10}}{(a-b x^4)^{3/4}} \, dx\) [1249]

Optimal. Leaf size=266 \[ -\frac {7 a x^3 \sqrt [4]{a-b x^4}}{32 b^2}-\frac {x^7 \sqrt [4]{a-b x^4}}{8 b}-\frac {21 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}}-\frac {21 a^2 \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}} \]

[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+21/128*a^2*arctan(-1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^
(1/4))/b^(11/4)*2^(1/2)+21/128*a^2*arctan(1+b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4))/b^(11/4)*2^(1/2)+21/256*a^2*ln
(1-b^(1/4)*x*2^(1/2)/(-b*x^4+a)^(1/4)+x^2*b^(1/2)/(-b*x^4+a)^(1/2))/b^(11/4)*2^(1/2)-21/256*a^2*ln(1+b^(1/4)*x
*2^(1/2)/(-b*x^4+a)^(1/4)+x^2*b^(1/2)/(-b*x^4+a)^(1/2))/b^(11/4)*2^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {327, 338, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {21 a^2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+1\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{128 \sqrt {2} b^{11/4}}-\frac {21 a^2 \log \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+1\right )}{128 \sqrt {2} 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} \end {gather*}

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[1 - (Sqrt[2]*b^(1/4)*x)
/(a - b*x^4)^(1/4)])/(64*Sqrt[2]*b^(11/4)) + (21*a^2*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(64*Sq
rt[2]*b^(11/4)) + (21*a^2*Log[1 + (Sqrt[b]*x^2)/Sqrt[a - b*x^4] - (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(128
*Sqrt[2]*b^(11/4)) - (21*a^2*Log[1 + (Sqrt[b]*x^2)/Sqrt[a - b*x^4] + (Sqrt[2]*b^(1/4)*x)/(a - b*x^4)^(1/4)])/(
128*Sqrt[2]*b^(11/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 327

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 338

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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 ) \text {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 ) \text {Subst}\left (\int \frac {1-\sqrt {b} x^2}{1+b x^4} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{64 b^{5/2}}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1+\sqrt {b} x^2}{1+b x^4} \, 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 {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {b}}-\frac {\sqrt {2} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{128 b^3}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {b}}+\frac {\sqrt {2} x}{\sqrt [4]{b}}+x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{128 b^3}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{b}}+2 x}{-\frac {1}{\sqrt {b}}-\frac {\sqrt {2} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{b}}-2 x}{-\frac {1}{\sqrt {b}}+\frac {\sqrt {2} x}{\sqrt [4]{b}}-x^2} \, dx,x,\frac {x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} 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}+\frac {21 a^2 \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}}-\frac {21 a^2 \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}}+\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} b^{11/4}}-\frac {\left (21 a^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} 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}-\frac {21 a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{64 \sqrt {2} b^{11/4}}+\frac {21 a^2 \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}}-\frac {21 a^2 \log \left (1+\frac {\sqrt {b} x^2}{\sqrt {a-b x^4}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a-b x^4}}\right )}{128 \sqrt {2} b^{11/4}}\\ \end {align*}

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Mathematica [A]
time = 0.54, size = 161, normalized size = 0.61 \begin {gather*} \frac {-4 b^{3/4} x^3 \sqrt [4]{a-b x^4} \left (7 a+4 b x^4\right )+21 \sqrt {2} a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x \sqrt [4]{a-b x^4}}{-\sqrt {b} x^2+\sqrt {a-b x^4}}\right )-21 \sqrt {2} a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2+\sqrt {a-b x^4}}{\sqrt {2} \sqrt [4]{b} x \sqrt [4]{a-b x^4}}\right )}{128 b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^10/(a - b*x^4)^(3/4),x]

[Out]

(-4*b^(3/4)*x^3*(a - b*x^4)^(1/4)*(7*a + 4*b*x^4) + 21*Sqrt[2]*a^2*ArcTan[(Sqrt[2]*b^(1/4)*x*(a - b*x^4)^(1/4)
)/(-(Sqrt[b]*x^2) + Sqrt[a - b*x^4])] - 21*Sqrt[2]*a^2*ArcTanh[(Sqrt[b]*x^2 + Sqrt[a - b*x^4])/(Sqrt[2]*b^(1/4
)*x*(a - b*x^4)^(1/4))])/(128*b^(11/4))

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Maple [F]
time = 0.01, 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 = 0.52, size = 274, normalized size = 1.03 \begin {gather*} -\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 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{x}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {b} + \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (\sqrt {b} - \frac {\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{\frac {1}{4}}}{x} + \frac {\sqrt {-b x^{4} + a}}{x^{2}}\right )}{b^{\frac {3}{4}}}\right )}}{256 \, b^{2}} \end {gather*}

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) - 21/256*(2*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^
(3/4) + 2*sqrt(2)*a^2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(-b*x^4 + a)^(1/4)/x)/b^(1/4))/b^(3/4) + sqrt(2
)*a^2*log(sqrt(b) + sqrt(2)*(-b*x^4 + a)^(1/4)*b^(1/4)/x + sqrt(-b*x^4 + a)/x^2)/b^(3/4) - sqrt(2)*a^2*log(sqr
t(b) - sqrt(2)*(-b*x^4 + a)^(1/4)*b^(1/4)/x + sqrt(-b*x^4 + a)/x^2)/b^(3/4))/b^2

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Fricas [A]
time = 0.38, size = 240, normalized size = 0.90 \begin {gather*} -\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}} \end {gather*}

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*s
qrt(-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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Sympy [C] Result contains complex when optimal does not.
time = 1.60, size = 39, normalized size = 0.15 \begin {gather*} \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^{2 i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {15}{4}\right )} \end {gather*}

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(2*I*pi)/a)/(4*a**(3/4)*gamma(15/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{10}}{{\left (a-b\,x^4\right )}^{3/4}} \,d x \end {gather*}

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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