Integrand size = 21, antiderivative size = 180 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\frac {b x}{5 a (b c-a d) \left (a+b x^4\right )^{5/4}}+\frac {b (4 b c-9 a d) x}{5 a^2 (b c-a d)^2 \sqrt [4]{a+b x^4}}+\frac {d^2 \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{9/4}}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{9/4}} \] Output:
1/5*b*x/a/(-a*d+b*c)/(b*x^4+a)^(5/4)+1/5*b*(-9*a*d+4*b*c)*x/a^2/(-a*d+b*c) ^2/(b*x^4+a)^(1/4)+1/2*d^2*arctan((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4+a)^(1/ 4))/c^(3/4)/(-a*d+b*c)^(9/4)+1/2*d^2*arctanh((-a*d+b*c)^(1/4)*x/c^(1/4)/(b *x^4+a)^(1/4))/c^(3/4)/(-a*d+b*c)^(9/4)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 11.98 (sec) , antiderivative size = 621, normalized size of antiderivative = 3.45 \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\frac {-585 c^4 (b c-a d) x^4 \left (a+b x^4\right )^2-936 c^3 d (b c-a d) x^8 \left (a+b x^4\right )^2-416 c^2 d^2 (b c-a d) x^{12} \left (a+b x^4\right )^2-2925 c^5 \left (a+b x^4\right )^3-4680 c^4 d x^4 \left (a+b x^4\right )^3-2080 c^3 d^2 x^8 \left (a+b x^4\right )^3+2925 c^5 \left (a+b x^4\right )^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+4680 c^4 d x^4 \left (a+b x^4\right )^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+2080 c^3 d^2 x^8 \left (a+b x^4\right )^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+280 c^2 (b c-a d)^3 x^{12} \operatorname {Hypergeometric2F1}\left (2,\frac {13}{4},\frac {17}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+520 c d (b c-a d)^3 x^{16} \operatorname {Hypergeometric2F1}\left (2,\frac {13}{4},\frac {17}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+240 d^2 (b c-a d)^3 x^{20} \operatorname {Hypergeometric2F1}\left (2,\frac {13}{4},\frac {17}{4},\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+80 c^2 (b c-a d)^3 x^{12} \, _3F_2\left (2,2,\frac {13}{4};1,\frac {17}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+160 c d (b c-a d)^3 x^{16} \, _3F_2\left (2,2,\frac {13}{4};1,\frac {17}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )+80 d^2 (b c-a d)^3 x^{20} \, _3F_2\left (2,2,\frac {13}{4};1,\frac {17}{4};\frac {(b c-a d) x^4}{c \left (a+b x^4\right )}\right )}{325 c^4 (b c-a d)^2 x^7 \left (a+b x^4\right )^{13/4}} \] Input:
Integrate[1/((a + b*x^4)^(9/4)*(c + d*x^4)),x]
Output:
(-585*c^4*(b*c - a*d)*x^4*(a + b*x^4)^2 - 936*c^3*d*(b*c - a*d)*x^8*(a + b *x^4)^2 - 416*c^2*d^2*(b*c - a*d)*x^12*(a + b*x^4)^2 - 2925*c^5*(a + b*x^4 )^3 - 4680*c^4*d*x^4*(a + b*x^4)^3 - 2080*c^3*d^2*x^8*(a + b*x^4)^3 + 2925 *c^5*(a + b*x^4)^3*Hypergeometric2F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 4680*c^4*d*x^4*(a + b*x^4)^3*Hypergeometric2F1[1/4, 1, 5/4, ( (b*c - a*d)*x^4)/(c*(a + b*x^4))] + 2080*c^3*d^2*x^8*(a + b*x^4)^3*Hyperge ometric2F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 280*c^2*(b*c - a*d)^3*x^12*Hypergeometric2F1[2, 13/4, 17/4, ((b*c - a*d)*x^4)/(c*(a + b *x^4))] + 520*c*d*(b*c - a*d)^3*x^16*Hypergeometric2F1[2, 13/4, 17/4, ((b* c - a*d)*x^4)/(c*(a + b*x^4))] + 240*d^2*(b*c - a*d)^3*x^20*Hypergeometric 2F1[2, 13/4, 17/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 80*c^2*(b*c - a*d) ^3*x^12*HypergeometricPFQ[{2, 2, 13/4}, {1, 17/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 160*c*d*(b*c - a*d)^3*x^16*HypergeometricPFQ[{2, 2, 13/4}, { 1, 17/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 80*d^2*(b*c - a*d)^3*x^20*H ypergeometricPFQ[{2, 2, 13/4}, {1, 17/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4) )])/(325*c^4*(b*c - a*d)^2*x^7*(a + b*x^4)^(13/4))
Time = 0.61 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {931, 25, 1024, 27, 902, 756, 218, 221}
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.
| \(\displaystyle \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx\) |
| \(\Big \downarrow \) 931 |
| \(\displaystyle \frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}-\frac {\int -\frac {4 b d x^4+4 b c-5 a d}{\left (b x^4+a\right )^{5/4} \left (d x^4+c\right )}dx}{5 a (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {4 b d x^4+4 b c-5 a d}{\left (b x^4+a\right )^{5/4} \left (d x^4+c\right )}dx}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
| \(\Big \downarrow \) 1024 |
| \(\displaystyle \frac {\frac {b x (4 b c-9 a d)}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {\int -\frac {5 a^2 d^2}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{a (b c-a d)}}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5 a d^2 \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{b c-a d}+\frac {b x (4 b c-9 a d)}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \frac {\frac {5 a d^2 \int \frac {1}{c-\frac {(b c-a d) x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{b c-a d}+\frac {b x (4 b c-9 a d)}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {\frac {5 a d^2 \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{b c-a d}+\frac {b x (4 b c-9 a d)}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {5 a d^2 \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{b c-a d}+\frac {b x (4 b c-9 a d)}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {5 a d^2 \left (\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{b c-a d}+\frac {b x (4 b c-9 a d)}{a \sqrt [4]{a+b x^4} (b c-a d)}}{5 a (b c-a d)}+\frac {b x}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}\) |
Input:
Int[1/((a + b*x^4)^(9/4)*(c + d*x^4)),x]
Output:
(b*x)/(5*a*(b*c - a*d)*(a + b*x^4)^(5/4)) + ((b*(4*b*c - 9*a*d)*x)/(a*(b*c - a*d)*(a + b*x^4)^(1/4)) + (5*a*d^2*(ArcTan[((b*c - a*d)^(1/4)*x)/(c^(1/ 4)*(a + b*x^4)^(1/4))]/(2*c^(3/4)*(b*c - a*d)^(1/4)) + ArcTanh[((b*c - a*d )^(1/4)*x)/(c^(1/4)*(a + b*x^4)^(1/4))]/(2*c^(3/4)*(b*c - a*d)^(1/4))))/(b *c - a*d))/(5*a*(b*c - a*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b , c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Simp[1/(a*n*(p + 1)*(b*c - a*d)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, n, p, q, x]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f _.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*( p + 1)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b *c - a*d)*(p + 1) + d*(b*e - a*f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; Fr eeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(299\) vs. \(2(148)=296\).
Time = 1.62 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.67
| method | result | size |
| pseudoelliptic | \(\frac {\frac {a^{2} d^{2} \left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (2 \arctan \left (-\frac {\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}+1\right )-2 \arctan \left (\frac {\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}+1\right )-\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )\right ) \sqrt {2}}{2}-8 \left (d \,a^{2}-\frac {b \left (-\frac {9 d \,x^{4}}{5}+c \right ) a}{2}-\frac {2 b^{2} c \,x^{4}}{5}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} c b x}{4 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {5}{4}} \left (a d -b c \right )^{2} c \,a^{2}}\) | \(300\) |
Input:
int(1/(b*x^4+a)^(9/4)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
1/4/((a*d-b*c)/c)^(1/4)/(b*x^4+a)^(5/4)*(1/2*a^2*d^2*(b*x^4+a)^(5/4)*(2*ar ctan(-2^(1/2)/((a*d-b*c)/c)^(1/4)*(b*x^4+a)^(1/4)/x+1)-2*arctan(2^(1/2)/(( a*d-b*c)/c)^(1/4)*(b*x^4+a)^(1/4)/x+1)-ln((-((a*d-b*c)/c)^(1/4)*(b*x^4+a)^ (1/4)*2^(1/2)*x+((a*d-b*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2))/(((a*d-b*c)/c)^(1 /4)*(b*x^4+a)^(1/4)*2^(1/2)*x+((a*d-b*c)/c)^(1/2)*x^2+(b*x^4+a)^(1/2))))*2 ^(1/2)-8*(d*a^2-1/2*b*(-9/5*d*x^4+c)*a-2/5*b^2*c*x^4)*((a*d-b*c)/c)^(1/4)* c*b*x)/(a*d-b*c)^2/c/a^2
Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^4+a)^(9/4)/(d*x^4+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (a + b x^{4}\right )^{\frac {9}{4}} \left (c + d x^{4}\right )}\, dx \] Input:
integrate(1/(b*x**4+a)**(9/4)/(d*x**4+c),x)
Output:
Integral(1/((a + b*x**4)**(9/4)*(c + d*x**4)), x)
\[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {9}{4}} {\left (d x^{4} + c\right )}} \,d x } \] Input:
integrate(1/(b*x^4+a)^(9/4)/(d*x^4+c),x, algorithm="maxima")
Output:
integrate(1/((b*x^4 + a)^(9/4)*(d*x^4 + c)), x)
\[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {9}{4}} {\left (d x^{4} + c\right )}} \,d x } \] Input:
integrate(1/(b*x^4+a)^(9/4)/(d*x^4+c),x, algorithm="giac")
Output:
integrate(1/((b*x^4 + a)^(9/4)*(d*x^4 + c)), x)
Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{9/4}\,\left (d\,x^4+c\right )} \,d x \] Input:
int(1/((a + b*x^4)^(9/4)*(c + d*x^4)),x)
Output:
int(1/((a + b*x^4)^(9/4)*(c + d*x^4)), x)
\[ \int \frac {1}{\left (a+b x^4\right )^{9/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} c +\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} d \,x^{4}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b c \,x^{4}+2 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a b d \,x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} c \,x^{8}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{2} d \,x^{12}}d x \] Input:
int(1/(b*x^4+a)^(9/4)/(d*x^4+c),x)
Output:
int(1/((a + b*x**4)**(1/4)*a**2*c + (a + b*x**4)**(1/4)*a**2*d*x**4 + 2*(a + b*x**4)**(1/4)*a*b*c*x**4 + 2*(a + b*x**4)**(1/4)*a*b*d*x**8 + (a + b*x **4)**(1/4)*b**2*c*x**8 + (a + b*x**4)**(1/4)*b**2*d*x**12),x)