Integrand size = 21, antiderivative size = 233 \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\frac {b x}{9 a (b c-a d) \left (a+b x^4\right )^{9/4}}+\frac {b (8 b c-17 a d) x}{45 a^2 (b c-a d)^2 \left (a+b x^4\right )^{5/4}}+\frac {b \left (32 b^2 c^2-100 a b c d+113 a^2 d^2\right ) x}{45 a^3 (b c-a d)^3 \sqrt [4]{a+b x^4}}-\frac {d^3 \arctan \left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}}-\frac {d^3 \text {arctanh}\left (\frac {\sqrt [4]{b c-a d} x}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} (b c-a d)^{13/4}} \] Output:
1/9*b*x/a/(-a*d+b*c)/(b*x^4+a)^(9/4)+1/45*b*(-17*a*d+8*b*c)*x/a^2/(-a*d+b* c)^2/(b*x^4+a)^(5/4)+1/45*b*(113*a^2*d^2-100*a*b*c*d+32*b^2*c^2)*x/a^3/(-a *d+b*c)^3/(b*x^4+a)^(1/4)-1/2*d^3*arctan((-a*d+b*c)^(1/4)*x/c^(1/4)/(b*x^4 +a)^(1/4))/c^(3/4)/(-a*d+b*c)^(13/4)-1/2*d^3*arctanh((-a*d+b*c)^(1/4)*x/c^ (1/4)/(b*x^4+a)^(1/4))/c^(3/4)/(-a*d+b*c)^(13/4)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 15.02 (sec) , antiderivative size = 1172, normalized size of antiderivative = 5.03 \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx =\text {Too large to display} \] Input:
Integrate[1/((a + b*x^4)^(13/4)*(c + d*x^4)),x]
Output:
-1/11475*(-16575*c^5*(b*c - a*d)^2*x^8*(a + b*x^4)^2 - 39780*c^4*d*(b*c - a*d)^2*x^12*(a + b*x^4)^2 - 35360*c^3*d^2*(b*c - a*d)^2*x^16*(a + b*x^4)^2 - 10880*c^2*d^3*(b*c - a*d)^2*x^20*(a + b*x^4)^2 - 29835*c^6*(b*c - a*d)* x^4*(a + b*x^4)^3 - 71604*c^5*d*(b*c - a*d)*x^8*(a + b*x^4)^3 - 63648*c^4* d^2*(b*c - a*d)*x^12*(a + b*x^4)^3 - 19584*c^3*d^3*(b*c - a*d)*x^16*(a + b *x^4)^3 - 149175*c^7*(a + b*x^4)^4 - 358020*c^6*d*x^4*(a + b*x^4)^4 - 3182 40*c^5*d^2*x^8*(a + b*x^4)^4 - 97920*c^4*d^3*x^12*(a + b*x^4)^4 + 149175*c ^7*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 358020*c^6*d*x^4*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ( (b*c - a*d)*x^4)/(c*(a + b*x^4))] + 318240*c^5*d^2*x^8*(a + b*x^4)^4*Hyper geometric2F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 97920*c^4*d ^3*x^12*(a + b*x^4)^4*Hypergeometric2F1[1/4, 1, 5/4, ((b*c - a*d)*x^4)/(c* (a + b*x^4))] + 13620*c^3*(b*c - a*d)^4*x^16*Hypergeometric2F1[2, 17/4, 21 /4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 36900*c^2*d*(b*c - a*d)^4*x^20*Hy pergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 33840* c*d^2*(b*c - a*d)^4*x^24*Hypergeometric2F1[2, 17/4, 21/4, ((b*c - a*d)*x^4 )/(c*(a + b*x^4))] + 10560*d^3*(b*c - a*d)^4*x^28*Hypergeometric2F1[2, 17/ 4, 21/4, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 6480*c^3*(b*c - a*d)^4*x^16* HypergeometricPFQ[{2, 2, 17/4}, {1, 21/4}, ((b*c - a*d)*x^4)/(c*(a + b*x^4 ))] + 18720*c^2*d*(b*c - a*d)^4*x^20*HypergeometricPFQ[{2, 2, 17/4}, {1...
Time = 0.84 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.20, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {931, 25, 1024, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx\) |
| \(\Big \downarrow \) 931 |
| \(\displaystyle \frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}-\frac {\int -\frac {8 b d x^4+8 b c-9 a d}{\left (b x^4+a\right )^{9/4} \left (d x^4+c\right )}dx}{9 a (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {8 b d x^4+8 b c-9 a d}{\left (b x^4+a\right )^{9/4} \left (d x^4+c\right )}dx}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 1024 |
| \(\displaystyle \frac {\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}-\frac {\int -\frac {4 b d (8 b c-17 a d) x^4+32 b^2 c^2+45 a^2 d^2-68 a b c d}{\left (b x^4+a\right )^{5/4} \left (d x^4+c\right )}dx}{5 a (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {4 b d (8 b c-17 a d) x^4+32 b^2 c^2+45 a^2 d^2-68 a b c d}{\left (b x^4+a\right )^{5/4} \left (d x^4+c\right )}dx}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 1024 |
| \(\displaystyle \frac {\frac {\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {\int \frac {45 a^3 d^3}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{a (b c-a d)}}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {45 a^2 d^3 \int \frac {1}{\sqrt [4]{b x^4+a} \left (d x^4+c\right )}dx}{b c-a d}}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \frac {\frac {\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {45 a^2 d^3 \int \frac {1}{c-\frac {(b c-a d) x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{b c-a d}}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {\frac {\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {45 a^2 d^3 \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{b c-a d}}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {45 a^2 d^3 \left (\frac {\int \frac {1}{\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{b c-a d}}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {b x \left (113 a^2 d^2-100 a b c d+32 b^2 c^2\right )}{a \sqrt [4]{a+b x^4} (b c-a d)}-\frac {45 a^2 d^3 \left (\frac {\arctan \left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}+\frac {\text {arctanh}\left (\frac {x \sqrt [4]{b c-a d}}{\sqrt [4]{c} \sqrt [4]{a+b x^4}}\right )}{2 c^{3/4} \sqrt [4]{b c-a d}}\right )}{b c-a d}}{5 a (b c-a d)}+\frac {b x (8 b c-17 a d)}{5 a \left (a+b x^4\right )^{5/4} (b c-a d)}}{9 a (b c-a d)}+\frac {b x}{9 a \left (a+b x^4\right )^{9/4} (b c-a d)}\) |
Input:
Int[1/((a + b*x^4)^(13/4)*(c + d*x^4)),x]
Output:
(b*x)/(9*a*(b*c - a*d)*(a + b*x^4)^(9/4)) + ((b*(8*b*c - 17*a*d)*x)/(5*a*( b*c - a*d)*(a + b*x^4)^(5/4)) + ((b*(32*b^2*c^2 - 100*a*b*c*d + 113*a^2*d^ 2)*x)/(a*(b*c - a*d)*(a + b*x^4)^(1/4)) - (45*a^2*d^3*(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)) + Arc Tanh[((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)))/(9*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]
Time = 1.74 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.50
| method | result | size |
| pseudoelliptic | \(-\frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}} a^{3} d^{3} \left (\ln \left (\frac {-\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a d -b c}{c}}\, x^{2}+\sqrt {b \,x^{4}+a}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} x}+1\right )\right ) \sqrt {2}+24 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} c \left (a^{4} d^{2}-d b \left (-\frac {9 d \,x^{4}}{5}+c \right ) a^{3}+\frac {b^{2} \left (\frac {113}{45} d^{2} x^{8}-5 c d \,x^{4}+c^{2}\right ) a^{2}}{3}+\frac {8 c \,b^{3} x^{4} \left (-\frac {25 d \,x^{4}}{18}+c \right ) a}{15}+\frac {32 b^{4} c^{2} x^{8}}{135}\right ) b x}{8 \left (\frac {a d -b c}{c}\right )^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (a d -b c \right )^{3} c \,a^{3}}\) | \(349\) |
Input:
int(1/(b*x^4+a)^(13/4)/(d*x^4+c),x,method=_RETURNVERBOSE)
Output:
-1/8/((a*d-b*c)/c)^(1/4)/(b*x^4+a)^(9/4)*((b*x^4+a)^(9/4)*a^3*d^3*(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*arctan(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))*2^(1 /2)+24*((a*d-b*c)/c)^(1/4)*c*(a^4*d^2-d*b*(-9/5*d*x^4+c)*a^3+1/3*b^2*(113/ 45*d^2*x^8-5*c*d*x^4+c^2)*a^2+8/15*c*b^3*x^4*(-25/18*d*x^4+c)*a+32/135*b^4 *c^2*x^8)*b*x)/(a*d-b*c)^3/c/a^3
Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (a + b x^{4}\right )^{\frac {13}{4}} \left (c + d x^{4}\right )}\, dx \] Input:
integrate(1/(b*x**4+a)**(13/4)/(d*x**4+c),x)
Output:
Integral(1/((a + b*x**4)**(13/4)*(c + d*x**4)), x)
\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} {\left (d x^{4} + c\right )}} \,d x } \] Input:
integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="maxima")
Output:
integrate(1/((b*x^4 + a)^(13/4)*(d*x^4 + c)), x)
\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int { \frac {1}{{\left (b x^{4} + a\right )}^{\frac {13}{4}} {\left (d x^{4} + c\right )}} \,d x } \] Input:
integrate(1/(b*x^4+a)^(13/4)/(d*x^4+c),x, algorithm="giac")
Output:
integrate(1/((b*x^4 + a)^(13/4)*(d*x^4 + c)), x)
Timed out. \[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{{\left (b\,x^4+a\right )}^{13/4}\,\left (d\,x^4+c\right )} \,d x \] Input:
int(1/((a + b*x^4)^(13/4)*(c + d*x^4)),x)
Output:
int(1/((a + b*x^4)^(13/4)*(c + d*x^4)), x)
\[ \int \frac {1}{\left (a+b x^4\right )^{13/4} \left (c+d x^4\right )} \, dx=\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} c +\left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{3} d \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b c \,x^{4}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{2} b d \,x^{8}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} c \,x^{8}+3 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a \,b^{2} d \,x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} c \,x^{12}+\left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{3} d \,x^{16}}d x \] Input:
int(1/(b*x^4+a)^(13/4)/(d*x^4+c),x)
Output:
int(1/((a + b*x**4)**(1/4)*a**3*c + (a + b*x**4)**(1/4)*a**3*d*x**4 + 3*(a + b*x**4)**(1/4)*a**2*b*c*x**4 + 3*(a + b*x**4)**(1/4)*a**2*b*d*x**8 + 3* (a + b*x**4)**(1/4)*a*b**2*c*x**8 + 3*(a + b*x**4)**(1/4)*a*b**2*d*x**12 + (a + b*x**4)**(1/4)*b**3*c*x**12 + (a + b*x**4)**(1/4)*b**3*d*x**16),x)