\(\int \frac {x^{19}}{(a+b x^4)^{5/4}} \, dx\) [1139]

Optimal result

Integrand size = 15, antiderivative size = 99 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^4}{b^5 \sqrt [4]{a+b x^4}}-\frac {4 a^3 \left (a+b x^4\right )^{3/4}}{3 b^5}+\frac {6 a^2 \left (a+b x^4\right )^{7/4}}{7 b^5}-\frac {4 a \left (a+b x^4\right )^{11/4}}{11 b^5}+\frac {\left (a+b x^4\right )^{15/4}}{15 b^5} \]

[Out]

-a^4/b^5/(b*x^4+a)^(1/4)-4/3*a^3*(b*x^4+a)^(3/4)/b^5+6/7*a^2*(b*x^4+a)^(7/4)/b^5-4/11*a*(b*x^4+a)^(11/4)/b^5+1
/15*(b*x^4+a)^(15/4)/b^5

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, 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} \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^4}{b^5 \sqrt [4]{a+b x^4}}-\frac {4 a^3 \left (a+b x^4\right )^{3/4}}{3 b^5}+\frac {6 a^2 \left (a+b x^4\right )^{7/4}}{7 b^5}-\frac {4 a \left (a+b x^4\right )^{11/4}}{11 b^5}+\frac {\left (a+b x^4\right )^{15/4}}{15 b^5} \]

[In]

Int[x^19/(a + b*x^4)^(5/4),x]

[Out]

-(a^4/(b^5*(a + b*x^4)^(1/4))) - (4*a^3*(a + b*x^4)^(3/4))/(3*b^5) + (6*a^2*(a + b*x^4)^(7/4))/(7*b^5) - (4*a*
(a + b*x^4)^(11/4))/(11*b^5) + (a + b*x^4)^(15/4)/(15*b^5)

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^4}{(a+b x)^{5/4}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {a^4}{b^4 (a+b x)^{5/4}}-\frac {4 a^3}{b^4 \sqrt [4]{a+b x}}+\frac {6 a^2 (a+b x)^{3/4}}{b^4}-\frac {4 a (a+b x)^{7/4}}{b^4}+\frac {(a+b x)^{11/4}}{b^4}\right ) \, dx,x,x^4\right ) \\ & = -\frac {a^4}{b^5 \sqrt [4]{a+b x^4}}-\frac {4 a^3 \left (a+b x^4\right )^{3/4}}{3 b^5}+\frac {6 a^2 \left (a+b x^4\right )^{7/4}}{7 b^5}-\frac {4 a \left (a+b x^4\right )^{11/4}}{11 b^5}+\frac {\left (a+b x^4\right )^{15/4}}{15 b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.62 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {-2048 a^4-512 a^3 b x^4+192 a^2 b^2 x^8-112 a b^3 x^{12}+77 b^4 x^{16}}{1155 b^5 \sqrt [4]{a+b x^4}} \]

[In]

Integrate[x^19/(a + b*x^4)^(5/4),x]

[Out]

(-2048*a^4 - 512*a^3*b*x^4 + 192*a^2*b^2*x^8 - 112*a*b^3*x^12 + 77*b^4*x^16)/(1155*b^5*(a + b*x^4)^(1/4))

Maple [A] (verified)

Time = 4.42 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {-77 x^{16} b^{4}+112 a \,b^{3} x^{12}-192 a^{2} b^{2} x^{8}+512 a^{3} b \,x^{4}+2048 a^{4}}{1155 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{5}}\)	\(58\)
trager	\(-\frac {-77 x^{16} b^{4}+112 a \,b^{3} x^{12}-192 a^{2} b^{2} x^{8}+512 a^{3} b \,x^{4}+2048 a^{4}}{1155 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{5}}\)	\(58\)
pseudoelliptic	\(\frac {77 x^{16} b^{4}-112 a \,b^{3} x^{12}+192 a^{2} b^{2} x^{8}-512 a^{3} b \,x^{4}-2048 a^{4}}{1155 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{5}}\)	\(58\)
risch	\(-\frac {\left (-77 b^{3} x^{12}+189 a \,b^{2} x^{8}-381 a^{2} b \,x^{4}+893 a^{3}\right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{1155 b^{5}}-\frac {a^{4}}{b^{5} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\)	\(65\)

[In]

int(x^19/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-1/1155*(-77*b^4*x^16+112*a*b^3*x^12-192*a^2*b^2*x^8+512*a^3*b*x^4+2048*a^4)/(b*x^4+a)^(1/4)/b^5

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (77 \, b^{4} x^{16} - 112 \, a b^{3} x^{12} + 192 \, a^{2} b^{2} x^{8} - 512 \, a^{3} b x^{4} - 2048 \, a^{4}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{1155 \, {\left (b^{6} x^{4} + a b^{5}\right )}} \]

[In]

integrate(x^19/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/1155*(77*b^4*x^16 - 112*a*b^3*x^12 + 192*a^2*b^2*x^8 - 512*a^3*b*x^4 - 2048*a^4)*(b*x^4 + a)^(3/4)/(b^6*x^4
+ a*b^5)

Sympy [A] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=\begin {cases} - \frac {2048 a^{4}}{1155 b^{5} \sqrt [4]{a + b x^{4}}} - \frac {512 a^{3} x^{4}}{1155 b^{4} \sqrt [4]{a + b x^{4}}} + \frac {64 a^{2} x^{8}}{385 b^{3} \sqrt [4]{a + b x^{4}}} - \frac {16 a x^{12}}{165 b^{2} \sqrt [4]{a + b x^{4}}} + \frac {x^{16}}{15 b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{20}}{20 a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**19/(b*x**4+a)**(5/4),x)

[Out]

Piecewise((-2048*a**4/(1155*b**5*(a + b*x**4)**(1/4)) - 512*a**3*x**4/(1155*b**4*(a + b*x**4)**(1/4)) + 64*a**
2*x**8/(385*b**3*(a + b*x**4)**(1/4)) - 16*a*x**12/(165*b**2*(a + b*x**4)**(1/4)) + x**16/(15*b*(a + b*x**4)**
(1/4)), Ne(b, 0)), (x**20/(20*a**(5/4)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.82 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {15}{4}}}{15 \, b^{5}} - \frac {4 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a}{11 \, b^{5}} + \frac {6 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{7 \, b^{5}} - \frac {4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}}{3 \, b^{5}} - \frac {a^{4}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{5}} \]

[In]

integrate(x^19/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

1/15*(b*x^4 + a)^(15/4)/b^5 - 4/11*(b*x^4 + a)^(11/4)*a/b^5 + 6/7*(b*x^4 + a)^(7/4)*a^2/b^5 - 4/3*(b*x^4 + a)^
(3/4)*a^3/b^5 - a^4/((b*x^4 + a)^(1/4)*b^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {a^{4}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{5}} + \frac {77 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} b^{70} - 420 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a b^{70} + 990 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2} b^{70} - 1540 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3} b^{70}}{1155 \, b^{75}} \]

[In]

integrate(x^19/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

-a^4/((b*x^4 + a)^(1/4)*b^5) + 1/1155*(77*(b*x^4 + a)^(15/4)*b^70 - 420*(b*x^4 + a)^(11/4)*a*b^70 + 990*(b*x^4
+ a)^(7/4)*a^2*b^70 - 1540*(b*x^4 + a)^(3/4)*a^3*b^70)/b^75

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {x^{19}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {420\,a\,{\left (b\,x^4+a\right )}^3+1540\,a^3\,\left (b\,x^4+a\right )-77\,{\left (b\,x^4+a\right )}^4+1155\,a^4-990\,a^2\,{\left (b\,x^4+a\right )}^2}{1155\,b^5\,{\left (b\,x^4+a\right )}^{1/4}} \]

[In]

int(x^19/(a + b*x^4)^(5/4),x)

[Out]

-(420*a*(a + b*x^4)^3 + 1540*a^3*(a + b*x^4) - 77*(a + b*x^4)^4 + 1155*a^4 - 990*a^2*(a + b*x^4)^2)/(1155*b^5*
(a + b*x^4)^(1/4))