\(\int \frac {x^{12}}{(a+b x^4)^{5/4}} \, dx\) [1154]

Optimal result

Integrand size = 15, antiderivative size = 123 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {x^9}{b \sqrt [4]{a+b x^4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}+\frac {45 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}}+\frac {45 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}} \]

[Out]

-x^9/b/(b*x^4+a)^(1/4)-45/32*a*x*(b*x^4+a)^(3/4)/b^3+9/8*x^5*(b*x^4+a)^(3/4)/b^2+45/64*a^2*arctan(b^(1/4)*x/(b
*x^4+a)^(1/4))/b^(13/4)+45/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(13/4)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {294, 327, 246, 218, 212, 209} \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {45 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}}+\frac {45 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}-\frac {x^9}{b \sqrt [4]{a+b x^4}} \]

[In]

Int[x^12/(a + b*x^4)^(5/4),x]

[Out]

-(x^9/(b*(a + b*x^4)^(1/4))) - (45*a*x*(a + b*x^4)^(3/4))/(32*b^3) + (9*x^5*(a + b*x^4)^(3/4))/(8*b^2) + (45*a
^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(13/4)) + (45*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*
b^(13/4))

Rule 209

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

Rule 212

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

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^9}{b \sqrt [4]{a+b x^4}}+\frac {9 \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx}{b} \\ & = -\frac {x^9}{b \sqrt [4]{a+b x^4}}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}-\frac {(45 a) \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx}{8 b^2} \\ & = -\frac {x^9}{b \sqrt [4]{a+b x^4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}+\frac {\left (45 a^2\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{32 b^3} \\ & = -\frac {x^9}{b \sqrt [4]{a+b x^4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}+\frac {\left (45 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{32 b^3} \\ & = -\frac {x^9}{b \sqrt [4]{a+b x^4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}+\frac {\left (45 a^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b^3}+\frac {\left (45 a^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{64 b^3} \\ & = -\frac {x^9}{b \sqrt [4]{a+b x^4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}+\frac {45 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}}+\frac {45 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (-45 a^2-9 a b x^4+4 b^2 x^8\right )}{\sqrt [4]{a+b x^4}}+45 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+45 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}} \]

[In]

Integrate[x^12/(a + b*x^4)^(5/4),x]

[Out]

((2*b^(1/4)*x*(-45*a^2 - 9*a*b*x^4 + 4*b^2*x^8))/(a + b*x^4)^(1/4) + 45*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/
4)] + 45*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(13/4))

Maple [A] (verified)

Time = 4.90 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(-\frac {45 \left (-\frac {8 b^{\frac {9}{4}} x^{9}}{45}+\frac {2 a \,b^{\frac {5}{4}} x^{5}}{5}+\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2} \left (b \,x^{4}+a \right )^{\frac {1}{4}}-\frac {\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{2}+2 a^{2} x \,b^{\frac {1}{4}}\right )}{64 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {13}{4}}}\)	\(125\)

[In]

int(x^12/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-45/64/(b*x^4+a)^(1/4)*(-8/45*b^(9/4)*x^9+2/5*a*b^(5/4)*x^5+arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^2*(b*x^4+a)^
(1/4)-1/2*ln((-b^(1/4)*x-(b*x^4+a)^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))*a^2*(b*x^4+a)^(1/4)+2*a^2*x*b^(1/4))/b^
(13/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.37 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {45 \, {\left (b^{4} x^{4} + a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (\frac {91125 \, {\left (b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 45 \, {\left (b^{4} x^{4} + a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-\frac {91125 \, {\left (b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 45 \, {\left (-i \, b^{4} x^{4} - i \, a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-\frac {91125 \, {\left (i \, b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 45 \, {\left (i \, b^{4} x^{4} + i \, a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-\frac {91125 \, {\left (-i \, b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) + 4 \, {\left (4 \, b^{2} x^{9} - 9 \, a b x^{5} - 45 \, a^{2} x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \]

[In]

integrate(x^12/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/128*(45*(b^4*x^4 + a*b^3)*(a^8/b^13)^(1/4)*log(91125*(b^10*x*(a^8/b^13)^(3/4) + (b*x^4 + a)^(1/4)*a^6)/x) -
45*(b^4*x^4 + a*b^3)*(a^8/b^13)^(1/4)*log(-91125*(b^10*x*(a^8/b^13)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) - 45*(-I
*b^4*x^4 - I*a*b^3)*(a^8/b^13)^(1/4)*log(-91125*(I*b^10*x*(a^8/b^13)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) - 45*(I
*b^4*x^4 + I*a*b^3)*(a^8/b^13)^(1/4)*log(-91125*(-I*b^10*x*(a^8/b^13)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) + 4*(4
*b^2*x^9 - 9*a*b*x^5 - 45*a^2*x)*(b*x^4 + a)^(3/4))/(b^4*x^4 + a*b^3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.43 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.30 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {17}{4}\right )} \]

[In]

integrate(x**12/(b*x**4+a)**(5/4),x)

[Out]

x**13*gamma(13/4)*hyper((5/4, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(17/4))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.40 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {32 \, a^{2} b^{2} - \frac {81 \, {\left (b x^{4} + a\right )} a^{2} b}{x^{4}} + \frac {45 \, {\left (b x^{4} + a\right )}^{2} a^{2}}{x^{8}}}{32 \, {\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{5}}{x} - \frac {2 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{4}}{x^{5}} + \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} b^{3}}{x^{9}}\right )}} - \frac {45 \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{128 \, b^{3}} \]

[In]

integrate(x^12/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

-1/32*(32*a^2*b^2 - 81*(b*x^4 + a)*a^2*b/x^4 + 45*(b*x^4 + a)^2*a^2/x^8)/((b*x^4 + a)^(1/4)*b^5/x - 2*(b*x^4 +
a)^(5/4)*b^4/x^5 + (b*x^4 + a)^(9/4)*b^3/x^9) - 45/128*a^2*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) +
log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1/4))/b^3

Giac [F]

\[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(x^12/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^12/(b*x^4 + a)^(5/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^{12}}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \]

[In]

int(x^12/(a + b*x^4)^(5/4),x)

[Out]

int(x^12/(a + b*x^4)^(5/4), x)