Integrand size = 15, antiderivative size = 80 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=-\frac {a^3 \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {3 a^2 \left (a+b x^4\right )^{11/4}}{11 b^4}-\frac {a \left (a+b x^4\right )^{15/4}}{5 b^4}+\frac {\left (a+b x^4\right )^{19/4}}{19 b^4} \] Output:
-1/7*a^3*(b*x^4+a)^(7/4)/b^4+3/11*a^2*(b*x^4+a)^(11/4)/b^4-1/5*a*(b*x^4+a) ^(15/4)/b^4+1/19*(b*x^4+a)^(19/4)/b^4
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=\frac {\left (a+b x^4\right )^{7/4} \left (-128 a^3+224 a^2 b x^4-308 a b^2 x^8+385 b^3 x^{12}\right )}{7315 b^4} \] Input:
Integrate[x^15*(a + b*x^4)^(3/4),x]
Output:
((a + b*x^4)^(7/4)*(-128*a^3 + 224*a^2*b*x^4 - 308*a*b^2*x^8 + 385*b^3*x^1 2))/(7315*b^4)
Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {798, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^{15} \left (a+b x^4\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int x^{12} \left (b x^4+a\right )^{3/4}dx^4\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{4} \int \left (\frac {\left (b x^4+a\right )^{15/4}}{b^3}-\frac {3 a \left (b x^4+a\right )^{11/4}}{b^3}+\frac {3 a^2 \left (b x^4+a\right )^{7/4}}{b^3}-\frac {a^3 \left (b x^4+a\right )^{3/4}}{b^3}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-\frac {4 a^3 \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {12 a^2 \left (a+b x^4\right )^{11/4}}{11 b^4}+\frac {4 \left (a+b x^4\right )^{19/4}}{19 b^4}-\frac {4 a \left (a+b x^4\right )^{15/4}}{5 b^4}\right )\) |
Input:
Int[x^15*(a + b*x^4)^(3/4),x]
Output:
((-4*a^3*(a + b*x^4)^(7/4))/(7*b^4) + (12*a^2*(a + b*x^4)^(11/4))/(11*b^4) - (4*a*(a + b*x^4)^(15/4))/(5*b^4) + (4*(a + b*x^4)^(19/4))/(19*b^4))/4
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] && 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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Time = 0.54 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59
method | result | size |
gosper | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (-385 b^{3} x^{12}+308 a \,b^{2} x^{8}-224 a^{2} b \,x^{4}+128 a^{3}\right )}{7315 b^{4}}\) | \(47\) |
pseudoelliptic | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (-385 b^{3} x^{12}+308 a \,b^{2} x^{8}-224 a^{2} b \,x^{4}+128 a^{3}\right )}{7315 b^{4}}\) | \(47\) |
orering | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (-385 b^{3} x^{12}+308 a \,b^{2} x^{8}-224 a^{2} b \,x^{4}+128 a^{3}\right )}{7315 b^{4}}\) | \(47\) |
trager | \(-\frac {\left (-385 x^{16} b^{4}-77 a \,b^{3} x^{12}+84 a^{2} b^{2} x^{8}-96 a^{3} b \,x^{4}+128 a^{4}\right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{7315 b^{4}}\) | \(58\) |
risch | \(-\frac {\left (-385 x^{16} b^{4}-77 a \,b^{3} x^{12}+84 a^{2} b^{2} x^{8}-96 a^{3} b \,x^{4}+128 a^{4}\right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{7315 b^{4}}\) | \(58\) |
Input:
int(x^15*(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
-1/7315*(b*x^4+a)^(7/4)*(-385*b^3*x^12+308*a*b^2*x^8-224*a^2*b*x^4+128*a^3 )/b^4
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.71 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=\frac {{\left (385 \, b^{4} x^{16} + 77 \, a b^{3} x^{12} - 84 \, a^{2} b^{2} x^{8} + 96 \, a^{3} b x^{4} - 128 \, a^{4}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{7315 \, b^{4}} \] Input:
integrate(x^15*(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
1/7315*(385*b^4*x^16 + 77*a*b^3*x^12 - 84*a^2*b^2*x^8 + 96*a^3*b*x^4 - 128 *a^4)*(b*x^4 + a)^(3/4)/b^4
Time = 0.90 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=\begin {cases} - \frac {128 a^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{7315 b^{4}} + \frac {96 a^{3} x^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{7315 b^{3}} - \frac {12 a^{2} x^{8} \left (a + b x^{4}\right )^{\frac {3}{4}}}{1045 b^{2}} + \frac {a x^{12} \left (a + b x^{4}\right )^{\frac {3}{4}}}{95 b} + \frac {x^{16} \left (a + b x^{4}\right )^{\frac {3}{4}}}{19} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{4}} x^{16}}{16} & \text {otherwise} \end {cases} \] Input:
integrate(x**15*(b*x**4+a)**(3/4),x)
Output:
Piecewise((-128*a**4*(a + b*x**4)**(3/4)/(7315*b**4) + 96*a**3*x**4*(a + b *x**4)**(3/4)/(7315*b**3) - 12*a**2*x**8*(a + b*x**4)**(3/4)/(1045*b**2) + a*x**12*(a + b*x**4)**(3/4)/(95*b) + x**16*(a + b*x**4)**(3/4)/19, Ne(b, 0)), (a**(3/4)*x**16/16, True))
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {19}{4}}}{19 \, b^{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {15}{4}} a}{5 \, b^{4}} + \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a^{2}}{11 \, b^{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{3}}{7 \, b^{4}} \] Input:
integrate(x^15*(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
1/19*(b*x^4 + a)^(19/4)/b^4 - 1/5*(b*x^4 + a)^(15/4)*a/b^4 + 3/11*(b*x^4 + a)^(11/4)*a^2/b^4 - 1/7*(b*x^4 + a)^(7/4)*a^3/b^4
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (64) = 128\).
Time = 0.12 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.68 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=\frac {\frac {19 \, {\left (77 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}\right )} a}{b^{3}} + \frac {1155 \, {\left (b x^{4} + a\right )}^{\frac {19}{4}} - 5852 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} a + 11970 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a^{2} - 12540 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{3} + 7315 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{4}}{b^{3}}}{21945 \, b} \] Input:
integrate(x^15*(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
1/21945*(19*(77*(b*x^4 + a)^(15/4) - 315*(b*x^4 + a)^(11/4)*a + 495*(b*x^4 + a)^(7/4)*a^2 - 385*(b*x^4 + a)^(3/4)*a^3)*a/b^3 + (1155*(b*x^4 + a)^(19 /4) - 5852*(b*x^4 + a)^(15/4)*a + 11970*(b*x^4 + a)^(11/4)*a^2 - 12540*(b* x^4 + a)^(7/4)*a^3 + 7315*(b*x^4 + a)^(3/4)*a^4)/b^3)/b
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.69 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx={\left (b\,x^4+a\right )}^{3/4}\,\left (\frac {x^{16}}{19}-\frac {128\,a^4}{7315\,b^4}+\frac {a\,x^{12}}{95\,b}+\frac {96\,a^3\,x^4}{7315\,b^3}-\frac {12\,a^2\,x^8}{1045\,b^2}\right ) \] Input:
int(x^15*(a + b*x^4)^(3/4),x)
Output:
(a + b*x^4)^(3/4)*(x^16/19 - (128*a^4)/(7315*b^4) + (a*x^12)/(95*b) + (96* a^3*x^4)/(7315*b^3) - (12*a^2*x^8)/(1045*b^2))
Time = 0.26 (sec) , antiderivative size = 461, normalized size of antiderivative = 5.76 \[ \int x^{15} \left (a+b x^4\right )^{3/4} \, dx=\frac {\sqrt {\sqrt {b}\, \sqrt {b \,x^{4}+a}\, x^{2}+a +b \,x^{4}}\, \left (-1152 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{8} x^{2}-14496 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{7} b \,x^{6}-44532 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{6} b^{2} x^{10}-41643 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{5} b^{3} x^{14}-1055 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{4} b^{4} x^{18}+55656 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{3} b^{5} x^{22}+189168 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a^{2} b^{6} x^{26}+241472 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a \,b^{7} x^{30}+98560 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, b^{8} x^{34}-128 a^{9}-5152 a^{8} b \,x^{4}-31988 a^{7} b^{2} x^{8}-64551 a^{6} b^{3} x^{12}-44042 a^{5} b^{4} x^{16}+14369 a^{4} b^{5} x^{20}+126216 a^{3} b^{6} x^{24}+297584 a^{2} b^{7} x^{28}+290752 a \,b^{8} x^{32}+98560 b^{9} x^{36}\right )}{7315 \sqrt {\sqrt {b \,x^{4}+a}+\sqrt {b}\, x^{2}}\, b^{4} \left (\sqrt {b \,x^{4}+a}\, a^{4}+40 \sqrt {b \,x^{4}+a}\, a^{3} b \,x^{4}+240 \sqrt {b \,x^{4}+a}\, a^{2} b^{2} x^{8}+448 \sqrt {b \,x^{4}+a}\, a \,b^{3} x^{12}+256 \sqrt {b \,x^{4}+a}\, b^{4} x^{16}+9 \sqrt {b}\, a^{4} x^{2}+120 \sqrt {b}\, a^{3} b \,x^{6}+432 \sqrt {b}\, a^{2} b^{2} x^{10}+576 \sqrt {b}\, a \,b^{3} x^{14}+256 \sqrt {b}\, b^{4} x^{18}\right )} \] Input:
int(x^15*(b*x^4+a)^(3/4),x)
Output:
(sqrt(sqrt(b)*sqrt(a + b*x**4)*x**2 + a + b*x**4)*( - 1152*sqrt(b)*sqrt(a + b*x**4)*a**8*x**2 - 14496*sqrt(b)*sqrt(a + b*x**4)*a**7*b*x**6 - 44532*s qrt(b)*sqrt(a + b*x**4)*a**6*b**2*x**10 - 41643*sqrt(b)*sqrt(a + b*x**4)*a **5*b**3*x**14 - 1055*sqrt(b)*sqrt(a + b*x**4)*a**4*b**4*x**18 + 55656*sqr t(b)*sqrt(a + b*x**4)*a**3*b**5*x**22 + 189168*sqrt(b)*sqrt(a + b*x**4)*a* *2*b**6*x**26 + 241472*sqrt(b)*sqrt(a + b*x**4)*a*b**7*x**30 + 98560*sqrt( b)*sqrt(a + b*x**4)*b**8*x**34 - 128*a**9 - 5152*a**8*b*x**4 - 31988*a**7* b**2*x**8 - 64551*a**6*b**3*x**12 - 44042*a**5*b**4*x**16 + 14369*a**4*b** 5*x**20 + 126216*a**3*b**6*x**24 + 297584*a**2*b**7*x**28 + 290752*a*b**8* x**32 + 98560*b**9*x**36))/(7315*sqrt(sqrt(a + b*x**4) + sqrt(b)*x**2)*b** 4*(sqrt(a + b*x**4)*a**4 + 40*sqrt(a + b*x**4)*a**3*b*x**4 + 240*sqrt(a + b*x**4)*a**2*b**2*x**8 + 448*sqrt(a + b*x**4)*a*b**3*x**12 + 256*sqrt(a + b*x**4)*b**4*x**16 + 9*sqrt(b)*a**4*x**2 + 120*sqrt(b)*a**3*b*x**6 + 432*s qrt(b)*a**2*b**2*x**10 + 576*sqrt(b)*a*b**3*x**14 + 256*sqrt(b)*b**4*x**18 ))