\(\int x^{15} (a+b x^4)^{5/4} \, dx\) [1048]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=-\frac {a^3 \left (a+b x^4\right )^{9/4}}{9 b^4}+\frac {3 a^2 \left (a+b x^4\right )^{13/4}}{13 b^4}-\frac {3 a \left (a+b x^4\right )^{17/4}}{17 b^4}+\frac {\left (a+b x^4\right )^{21/4}}{21 b^4} \]

[Out]

-1/9*a^3*(b*x^4+a)^(9/4)/b^4+3/13*a^2*(b*x^4+a)^(13/4)/b^4-3/17*a*(b*x^4+a)^(17/4)/b^4+1/21*(b*x^4+a)^(21/4)/b
^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, 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} \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=-\frac {a^3 \left (a+b x^4\right )^{9/4}}{9 b^4}+\frac {3 a^2 \left (a+b x^4\right )^{13/4}}{13 b^4}+\frac {\left (a+b x^4\right )^{21/4}}{21 b^4}-\frac {3 a \left (a+b x^4\right )^{17/4}}{17 b^4} \]

[In]

Int[x^15*(a + b*x^4)^(5/4),x]

[Out]

-1/9*(a^3*(a + b*x^4)^(9/4))/b^4 + (3*a^2*(a + b*x^4)^(13/4))/(13*b^4) - (3*a*(a + b*x^4)^(17/4))/(17*b^4) + (
a + b*x^4)^(21/4)/(21*b^4)

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 x^3 (a+b x)^{5/4} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (-\frac {a^3 (a+b x)^{5/4}}{b^3}+\frac {3 a^2 (a+b x)^{9/4}}{b^3}-\frac {3 a (a+b x)^{13/4}}{b^3}+\frac {(a+b x)^{17/4}}{b^3}\right ) \, dx,x,x^4\right ) \\ & = -\frac {a^3 \left (a+b x^4\right )^{9/4}}{9 b^4}+\frac {3 a^2 \left (a+b x^4\right )^{13/4}}{13 b^4}-\frac {3 a \left (a+b x^4\right )^{17/4}}{17 b^4}+\frac {\left (a+b x^4\right )^{21/4}}{21 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=\frac {\left (a+b x^4\right )^{9/4} \left (-128 a^3+288 a^2 b x^4-468 a b^2 x^8+663 b^3 x^{12}\right )}{13923 b^4} \]

[In]

Integrate[x^15*(a + b*x^4)^(5/4),x]

[Out]

((a + b*x^4)^(9/4)*(-128*a^3 + 288*a^2*b*x^4 - 468*a*b^2*x^8 + 663*b^3*x^12))/(13923*b^4)

Maple [A] (verified)

Time = 4.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (-663 b^{3} x^{12}+468 a \,b^{2} x^{8}-288 a^{2} b \,x^{4}+128 a^{3}\right )}{13923 b^{4}}\)	\(47\)
pseudoelliptic	\(-\frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (-663 b^{3} x^{12}+468 a \,b^{2} x^{8}-288 a^{2} b \,x^{4}+128 a^{3}\right )}{13923 b^{4}}\)	\(47\)
trager	\(-\frac {\left (-663 b^{5} x^{20}-858 a \,b^{4} x^{16}-15 a^{2} b^{3} x^{12}+20 a^{3} b^{2} x^{8}-32 a^{4} b \,x^{4}+128 a^{5}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{13923 b^{4}}\)	\(69\)
risch	\(-\frac {\left (-663 b^{5} x^{20}-858 a \,b^{4} x^{16}-15 a^{2} b^{3} x^{12}+20 a^{3} b^{2} x^{8}-32 a^{4} b \,x^{4}+128 a^{5}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{13923 b^{4}}\)	\(69\)

[In]

int(x^15*(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-1/13923*(b*x^4+a)^(9/4)*(-663*b^3*x^12+468*a*b^2*x^8-288*a^2*b*x^4+128*a^3)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=\frac {{\left (663 \, b^{5} x^{20} + 858 \, a b^{4} x^{16} + 15 \, a^{2} b^{3} x^{12} - 20 \, a^{3} b^{2} x^{8} + 32 \, a^{4} b x^{4} - 128 \, a^{5}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{13923 \, b^{4}} \]

[In]

integrate(x^15*(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/13923*(663*b^5*x^20 + 858*a*b^4*x^16 + 15*a^2*b^3*x^12 - 20*a^3*b^2*x^8 + 32*a^4*b*x^4 - 128*a^5)*(b*x^4 + a
)^(1/4)/b^4

Sympy [A] (verification not implemented)

Time = 1.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.68 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=\begin {cases} - \frac {128 a^{5} \sqrt [4]{a + b x^{4}}}{13923 b^{4}} + \frac {32 a^{4} x^{4} \sqrt [4]{a + b x^{4}}}{13923 b^{3}} - \frac {20 a^{3} x^{8} \sqrt [4]{a + b x^{4}}}{13923 b^{2}} + \frac {5 a^{2} x^{12} \sqrt [4]{a + b x^{4}}}{4641 b} + \frac {22 a x^{16} \sqrt [4]{a + b x^{4}}}{357} + \frac {b x^{20} \sqrt [4]{a + b x^{4}}}{21} & \text {for}\: b \neq 0 \\\frac {a^{\frac {5}{4}} x^{16}}{16} & \text {otherwise} \end {cases} \]

[In]

integrate(x**15*(b*x**4+a)**(5/4),x)

[Out]

Piecewise((-128*a**5*(a + b*x**4)**(1/4)/(13923*b**4) + 32*a**4*x**4*(a + b*x**4)**(1/4)/(13923*b**3) - 20*a**
3*x**8*(a + b*x**4)**(1/4)/(13923*b**2) + 5*a**2*x**12*(a + b*x**4)**(1/4)/(4641*b) + 22*a*x**16*(a + b*x**4)*
*(1/4)/357 + b*x**20*(a + b*x**4)**(1/4)/21, Ne(b, 0)), (a**(5/4)*x**16/16, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {21}{4}}}{21 \, b^{4}} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {17}{4}} a}{17 \, b^{4}} + \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{2}}{13 \, b^{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3}}{9 \, b^{4}} \]

[In]

integrate(x^15*(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

1/21*(b*x^4 + a)^(21/4)/b^4 - 3/17*(b*x^4 + a)^(17/4)*a/b^4 + 3/13*(b*x^4 + a)^(13/4)*a^2/b^4 - 1/9*(b*x^4 + a
)^(9/4)*a^3/b^4

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx=\frac {663 \, {\left (b x^{4} + a\right )}^{\frac {21}{4}} - 2457 \, {\left (b x^{4} + a\right )}^{\frac {17}{4}} a + 3213 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{2} - 1547 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3}}{13923 \, b^{4}} \]

[In]

integrate(x^15*(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

1/13923*(663*(b*x^4 + a)^(21/4) - 2457*(b*x^4 + a)^(17/4)*a + 3213*(b*x^4 + a)^(13/4)*a^2 - 1547*(b*x^4 + a)^(
9/4)*a^3)/b^4

Mupad [B] (verification not implemented)

Time = 5.61 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int x^{15} \left (a+b x^4\right )^{5/4} \, dx={\left (b\,x^4+a\right )}^{1/4}\,\left (\frac {22\,a\,x^{16}}{357}+\frac {b\,x^{20}}{21}-\frac {128\,a^5}{13923\,b^4}+\frac {32\,a^4\,x^4}{13923\,b^3}-\frac {20\,a^3\,x^8}{13923\,b^2}+\frac {5\,a^2\,x^{12}}{4641\,b}\right ) \]

[In]

int(x^15*(a + b*x^4)^(5/4),x)

[Out]

(a + b*x^4)^(1/4)*((22*a*x^16)/357 + (b*x^20)/21 - (128*a^5)/(13923*b^4) + (32*a^4*x^4)/(13923*b^3) - (20*a^3*
x^8)/(13923*b^2) + (5*a^2*x^12)/(4641*b))