Integrand size = 11, antiderivative size = 77 \[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {x}{13 a \left (a+b x^4\right )^{13/4}}+\frac {4 x}{39 a^2 \left (a+b x^4\right )^{9/4}}+\frac {32 x}{195 a^3 \left (a+b x^4\right )^{5/4}}+\frac {128 x}{195 a^4 \sqrt [4]{a+b x^4}} \] Output:
1/13*x/a/(b*x^4+a)^(13/4)+4/39*x/a^2/(b*x^4+a)^(9/4)+32/195*x/a^3/(b*x^4+a )^(5/4)+128/195*x/a^4/(b*x^4+a)^(1/4)
Time = 0.70 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {195 a^3 x+468 a^2 b x^5+416 a b^2 x^9+128 b^3 x^{13}}{195 a^4 \left (a+b x^4\right )^{13/4}} \] Input:
Integrate[(a + b*x^4)^(-17/4),x]
Output:
(195*a^3*x + 468*a^2*b*x^5 + 416*a*b^2*x^9 + 128*b^3*x^13)/(195*a^4*(a + b *x^4)^(13/4))
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {749, 749, 749, 746}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {12 \int \frac {1}{\left (b x^4+a\right )^{13/4}}dx}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {12 \left (\frac {8 \int \frac {1}{\left (b x^4+a\right )^{9/4}}dx}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 749 |
\(\displaystyle \frac {12 \left (\frac {8 \left (\frac {4 \int \frac {1}{\left (b x^4+a\right )^{5/4}}dx}{5 a}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \frac {12 \left (\frac {8 \left (\frac {4 x}{5 a^2 \sqrt [4]{a+b x^4}}+\frac {x}{5 a \left (a+b x^4\right )^{5/4}}\right )}{9 a}+\frac {x}{9 a \left (a+b x^4\right )^{9/4}}\right )}{13 a}+\frac {x}{13 a \left (a+b x^4\right )^{13/4}}\) |
Input:
Int[(a + b*x^4)^(-17/4),x]
Output:
x/(13*a*(a + b*x^4)^(13/4)) + (12*(x/(9*a*(a + b*x^4)^(9/4)) + (8*(x/(5*a* (a + b*x^4)^(5/4)) + (4*x)/(5*a^2*(a + b*x^4)^(1/4))))/(9*a)))/(13*a)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) /a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
Time = 0.54 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(\frac {x \left (128 b^{3} x^{12}+416 a \,b^{2} x^{8}+468 a^{2} b \,x^{4}+195 a^{3}\right )}{195 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(48\) |
trager | \(\frac {x \left (128 b^{3} x^{12}+416 a \,b^{2} x^{8}+468 a^{2} b \,x^{4}+195 a^{3}\right )}{195 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(48\) |
pseudoelliptic | \(\frac {x \left (128 b^{3} x^{12}+416 a \,b^{2} x^{8}+468 a^{2} b \,x^{4}+195 a^{3}\right )}{195 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(48\) |
orering | \(\frac {x \left (128 b^{3} x^{12}+416 a \,b^{2} x^{8}+468 a^{2} b \,x^{4}+195 a^{3}\right )}{195 \left (b \,x^{4}+a \right )^{\frac {13}{4}} a^{4}}\) | \(48\) |
Input:
int(1/(b*x^4+a)^(17/4),x,method=_RETURNVERBOSE)
Output:
1/195*x*(128*b^3*x^12+416*a*b^2*x^8+468*a^2*b*x^4+195*a^3)/(b*x^4+a)^(13/4 )/a^4
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {{\left (128 \, b^{3} x^{13} + 416 \, a b^{2} x^{9} + 468 \, a^{2} b x^{5} + 195 \, a^{3} x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{195 \, {\left (a^{4} b^{4} x^{16} + 4 \, a^{5} b^{3} x^{12} + 6 \, a^{6} b^{2} x^{8} + 4 \, a^{7} b x^{4} + a^{8}\right )}} \] Input:
integrate(1/(b*x^4+a)^(17/4),x, algorithm="fricas")
Output:
1/195*(128*b^3*x^13 + 416*a*b^2*x^9 + 468*a^2*b*x^5 + 195*a^3*x)*(b*x^4 + a)^(3/4)/(a^4*b^4*x^16 + 4*a^5*b^3*x^12 + 6*a^6*b^2*x^8 + 4*a^7*b*x^4 + a^ 8)
Leaf count of result is larger than twice the leaf count of optimal. 1550 vs. \(2 (70) = 140\).
Time = 2.77 (sec) , antiderivative size = 1550, normalized size of antiderivative = 20.13 \[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\text {Too large to display} \] Input:
integrate(1/(b*x**4+a)**(17/4),x)
Output:
585*a**14*x*gamma(1/4)/(256*a**(73/4)*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 1536*a**(69/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 3840*a**(65/4)*b **2*x**8*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 5120*a**(61/4)*b**3*x**12*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 3840*a**(57/4)*b**4*x**16*(1 + b*x**4/a)* *(1/4)*gamma(17/4) + 1536*a**(53/4)*b**5*x**20*(1 + b*x**4/a)**(1/4)*gamma (17/4) + 256*a**(49/4)*b**6*x**24*(1 + b*x**4/a)**(1/4)*gamma(17/4)) + 315 9*a**13*b*x**5*gamma(1/4)/(256*a**(73/4)*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 1536*a**(69/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 3840*a**(65/4 )*b**2*x**8*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 5120*a**(61/4)*b**3*x**12* (1 + b*x**4/a)**(1/4)*gamma(17/4) + 3840*a**(57/4)*b**4*x**16*(1 + b*x**4/ a)**(1/4)*gamma(17/4) + 1536*a**(53/4)*b**5*x**20*(1 + b*x**4/a)**(1/4)*ga mma(17/4) + 256*a**(49/4)*b**6*x**24*(1 + b*x**4/a)**(1/4)*gamma(17/4)) + 7215*a**12*b**2*x**9*gamma(1/4)/(256*a**(73/4)*(1 + b*x**4/a)**(1/4)*gamma (17/4) + 1536*a**(69/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 3840*a* *(65/4)*b**2*x**8*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 5120*a**(61/4)*b**3* x**12*(1 + b*x**4/a)**(1/4)*gamma(17/4) + 3840*a**(57/4)*b**4*x**16*(1 + b *x**4/a)**(1/4)*gamma(17/4) + 1536*a**(53/4)*b**5*x**20*(1 + b*x**4/a)**(1 /4)*gamma(17/4) + 256*a**(49/4)*b**6*x**24*(1 + b*x**4/a)**(1/4)*gamma(17/ 4)) + 8925*a**11*b**3*x**13*gamma(1/4)/(256*a**(73/4)*(1 + b*x**4/a)**(1/4 )*gamma(17/4) + 1536*a**(69/4)*b*x**4*(1 + b*x**4/a)**(1/4)*gamma(17/4)...
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=-\frac {{\left (15 \, b^{3} - \frac {65 \, {\left (b x^{4} + a\right )} b^{2}}{x^{4}} + \frac {117 \, {\left (b x^{4} + a\right )}^{2} b}{x^{8}} - \frac {195 \, {\left (b x^{4} + a\right )}^{3}}{x^{12}}\right )} x^{13}}{195 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{4}} \] Input:
integrate(1/(b*x^4+a)^(17/4),x, algorithm="maxima")
Output:
-1/195*(15*b^3 - 65*(b*x^4 + a)*b^2/x^4 + 117*(b*x^4 + a)^2*b/x^8 - 195*(b *x^4 + a)^3/x^12)*x^13/((b*x^4 + a)^(13/4)*a^4)
\[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {17}{4}}} \,d x } \] Input:
integrate(1/(b*x^4+a)^(17/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(-17/4), x)
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\frac {128\,x}{195\,a^4\,{\left (b\,x^4+a\right )}^{1/4}}+\frac {32\,x}{195\,a^3\,{\left (b\,x^4+a\right )}^{5/4}}+\frac {4\,x}{39\,a^2\,{\left (b\,x^4+a\right )}^{9/4}}+\frac {x}{13\,a\,{\left (b\,x^4+a\right )}^{13/4}} \] Input:
int(1/(a + b*x^4)^(17/4),x)
Output:
(128*x)/(195*a^4*(a + b*x^4)^(1/4)) + (32*x)/(195*a^3*(a + b*x^4)^(5/4)) + (4*x)/(39*a^2*(a + b*x^4)^(9/4)) + x/(13*a*(a + b*x^4)^(13/4))
\[ \int \frac {1}{\left (a+b x^4\right )^{17/4}} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{4}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} b \,x^{4}+6 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b^{2} x^{8}+4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{3} x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4} x^{16}}d x \] Input:
int(1/(b*x^4+a)^(17/4),x)
Output:
int(1/((a + b*x**4)**(1/4)*a**4 + 4*(a + b*x**4)**(1/4)*a**3*b*x**4 + 6*(a + b*x**4)**(1/4)*a**2*b**2*x**8 + 4*(a + b*x**4)**(1/4)*a*b**3*x**12 + (a + b*x**4)**(1/4)*b**4*x**16),x)