\(\int \frac {x^{11}}{(a+b x^4)^{5/4}} \, dx\) [1141]

Optimal result

Integrand size = 15, antiderivative size = 57 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^2}{b^3 \sqrt [4]{a+b x^4}}-\frac {2 a \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac {\left (a+b x^4\right )^{7/4}}{7 b^3} \]

[Out]

-a^2/b^3/(b*x^4+a)^(1/4)-2/3*a*(b*x^4+a)^(3/4)/b^3+1/7*(b*x^4+a)^(7/4)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^2}{b^3 \sqrt [4]{a+b x^4}}-\frac {2 a \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac {\left (a+b x^4\right )^{7/4}}{7 b^3} \]

[In]

Int[x^11/(a + b*x^4)^(5/4),x]

[Out]

-(a^2/(b^3*(a + b*x^4)^(1/4))) - (2*a*(a + b*x^4)^(3/4))/(3*b^3) + (a + b*x^4)^(7/4)/(7*b^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {x^2}{(a+b x)^{5/4}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {a^2}{b^2 (a+b x)^{5/4}}-\frac {2 a}{b^2 \sqrt [4]{a+b x}}+\frac {(a+b x)^{3/4}}{b^2}\right ) \, dx,x,x^4\right ) \\ & = -\frac {a^2}{b^3 \sqrt [4]{a+b x^4}}-\frac {2 a \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac {\left (a+b x^4\right )^{7/4}}{7 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.68 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {-32 a^2-8 a b x^4+3 b^2 x^8}{21 b^3 \sqrt [4]{a+b x^4}} \]

[In]

Integrate[x^11/(a + b*x^4)^(5/4),x]

[Out]

(-32*a^2 - 8*a*b*x^4 + 3*b^2*x^8)/(21*b^3*(a + b*x^4)^(1/4))

Maple [A] (verified)

Time = 4.45 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.63

method	result	size
gosper	\(-\frac {-3 b^{2} x^{8}+8 a b \,x^{4}+32 a^{2}}{21 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3}}\)	\(36\)
trager	\(-\frac {-3 b^{2} x^{8}+8 a b \,x^{4}+32 a^{2}}{21 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3}}\)	\(36\)
pseudoelliptic	\(\frac {3 b^{2} x^{8}-8 a b \,x^{4}-32 a^{2}}{21 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3}}\)	\(36\)
risch	\(-\frac {\left (-3 b \,x^{4}+11 a \right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{21 b^{3}}-\frac {a^{2}}{b^{3} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(43\)

[In]

int(x^11/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-1/21*(-3*b^2*x^8+8*a*b*x^4+32*a^2)/(b*x^4+a)^(1/4)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (3 \, b^{2} x^{8} - 8 \, a b x^{4} - 32 \, a^{2}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{21 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \]

[In]

integrate(x^11/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/21*(3*b^2*x^8 - 8*a*b*x^4 - 32*a^2)*(b*x^4 + a)^(3/4)/(b^4*x^4 + a*b^3)

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=\begin {cases} - \frac {32 a^{2}}{21 b^{3} \sqrt [4]{a + b x^{4}}} - \frac {8 a x^{4}}{21 b^{2} \sqrt [4]{a + b x^{4}}} + \frac {x^{8}}{7 b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{12}}{12 a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**11/(b*x**4+a)**(5/4),x)

[Out]

Piecewise((-32*a**2/(21*b**3*(a + b*x**4)**(1/4)) - 8*a*x**4/(21*b**2*(a + b*x**4)**(1/4)) + x**8/(7*b*(a + b*
x**4)**(1/4)), Ne(b, 0)), (x**12/(12*a**(5/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {7}{4}}}{7 \, b^{3}} - \frac {2 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a}{3 \, b^{3}} - \frac {a^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}} \]

[In]

integrate(x^11/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

1/7*(b*x^4 + a)^(7/4)/b^3 - 2/3*(b*x^4 + a)^(3/4)*a/b^3 - a^2/((b*x^4 + a)^(1/4)*b^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{3}} + \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{18} - 14 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{18}}{21 \, b^{21}} \]

[In]

integrate(x^11/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

-a^2/((b*x^4 + a)^(1/4)*b^3) + 1/21*(3*(b*x^4 + a)^(7/4)*b^18 - 14*(b*x^4 + a)^(3/4)*a*b^18)/b^21

Mupad [B] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.68 \[ \int \frac {x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {\frac {2\,a\,\left (b\,x^4+a\right )}{3}-\frac {{\left (b\,x^4+a\right )}^2}{7}+a^2}{b^3\,{\left (b\,x^4+a\right )}^{1/4}} \]

[In]

int(x^11/(a + b*x^4)^(5/4),x)

[Out]

-((2*a*(a + b*x^4))/3 - (a + b*x^4)^2/7 + a^2)/(b^3*(a + b*x^4)^(1/4))