Integrand size = 21, antiderivative size = 308 \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}-\frac {\sqrt {a} b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{4 c (b c-a d) \left (a+b x^4\right )^{3/4}}+\frac {(2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)}+\frac {(2 b c-3 a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{8 \sqrt [4]{b} c^2 (b c-a d)} \]
1/4*x*(b*x^4+a)^(1/4)/c/(d*x^4+c)-1/4*b^(3/2)*(1+a/b/x^4)^(3/4)*x^3*(cos(1 /2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2 )))*EllipticF(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)/c/(-a* d+b*c)/(b*x^4+a)^(3/4)+1/8*(-3*a*d+2*b*c)*EllipticPi(b^(1/4)*x/(b*x^4+a)^( 1/4),-(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1 /2)/b^(1/4)/c^2/(-a*d+b*c)+1/8*(-3*a*d+2*b*c)*EllipticPi(b^(1/4)*x/(b*x^4+ a)^(1/4),(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)*(a/(b*x^4+a))^(1/2)*(b*x^4+a) ^(1/2)/b^(1/4)/c^2/(-a*d+b*c)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.22 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\frac {x \left (\frac {2 b x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c^2}+\frac {5 \left (\frac {a+b x^4}{c}-\frac {15 a^2 \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{-5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+x^4 \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}\right )}{c+d x^4}\right )}{20 \left (a+b x^4\right )^{3/4}} \]
(x*((2*b*x^4*(1 + (b*x^4)/a)^(3/4)*AppellF1[5/4, 3/4, 1, 9/4, -((b*x^4)/a) , -((d*x^4)/c)])/c^2 + (5*((a + b*x^4)/c - (15*a^2*AppellF1[1/4, 3/4, 1, 5 /4, -((b*x^4)/a), -((d*x^4)/c)])/(-5*a*c*AppellF1[1/4, 3/4, 1, 5/4, -((b*x ^4)/a), -((d*x^4)/c)] + x^4*(4*a*d*AppellF1[5/4, 3/4, 2, 9/4, -((b*x^4)/a) , -((d*x^4)/c)] + 3*b*c*AppellF1[5/4, 7/4, 1, 9/4, -((b*x^4)/a), -((d*x^4) /c)]))))/(c + d*x^4)))/(20*(a + b*x^4)^(3/4))
Time = 0.48 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.87, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {929, 25, 404, 768, 858, 807, 229, 923, 925, 27, 1542}
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.
\(\displaystyle \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 929 |
\(\displaystyle \frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}-\frac {\int -\frac {2 b x^4+3 a}{\left (b x^4+a\right )^{3/4} \left (d x^4+c\right )}dx}{4 c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {2 b x^4+3 a}{\left (b x^4+a\right )^{3/4} \left (d x^4+c\right )}dx}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 404 |
\(\displaystyle \frac {\frac {a b \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{b c-a d}+\frac {(2 b c-3 a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{b c-a d}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {\frac {(2 b c-3 a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{b c-a d}+\frac {a b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {\frac {(2 b c-3 a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{b c-a d}-\frac {a b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {\frac {(2 b c-3 a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{b c-a d}-\frac {a b x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{2 \left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {\frac {(2 b c-3 a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{b c-a d}-\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 923 |
\(\displaystyle \frac {\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-3 a d) \int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (c-\frac {(b c-a d) x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{b c-a d}-\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 925 |
\(\displaystyle \frac {\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-3 a d) \left (\frac {\int \frac {\sqrt {c}}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 c}+\frac {\int \frac {\sqrt {c}}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 c}\right )}{b c-a d}-\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-3 a d) \left (\frac {\int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{b c-a d}-\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
\(\Big \downarrow \) 1542 |
\(\displaystyle \frac {\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (2 b c-3 a d) \left (\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c}+\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c}\right )}{b c-a d}-\frac {\sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{\left (a+b x^4\right )^{3/4} (b c-a d)}}{4 c}+\frac {x \sqrt [4]{a+b x^4}}{4 c \left (c+d x^4\right )}\) |
(x*(a + b*x^4)^(1/4))/(4*c*(c + d*x^4)) + (-((Sqrt[a]*b^(3/2)*(1 + a/(b*x^ 4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/((b*c - a*d)* (a + b*x^4)^(3/4))) + ((2*b*c - 3*a*d)*Sqrt[a/(a + b*x^4)]*Sqrt[a + b*x^4] *(EllipticPi[-(Sqrt[b*c - a*d]/(Sqrt[b]*Sqrt[c])), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1]/(2*b^(1/4)*c) + EllipticPi[Sqrt[b*c - a*d]/(Sqrt[b]*Sq rt[c]), ArcSin[(b^(1/4)*x)/(a + b*x^4)^(1/4)], -1]/(2*b^(1/4)*c)))/(b*c - a*d))/(4*c)
3.3.13.3.1 Defintions of rubi rules used
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)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((e_) + (f_.)*(x_)^4)/(((a_) + (b_.)*(x_)^4)^(3/4)*((c_) + (d_.)*(x_)^4 )), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^4)^(3/4), x] , x] - Simp[(d*e - c*f)/(b*c - a*d) Int[(a + b*x^4)^(1/4)/(c + d*x^4), x] , x] /; FreeQ[{a, b, c, d, e, f}, x]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 /4)/(a + b*x^4)^(3/4)) Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ [{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Int[((a_) + (b_.)*(x_)^4)^(1/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Simp[Sq rt[a + b*x^4]*Sqrt[a/(a + b*x^4)] Subst[Int[1/(Sqrt[1 - b*x^4]*(c - (b*c - a*d)*x^4)), x], x, x/(a + b*x^4)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Simp[ 1/(2*c) Int[1/(Sqrt[a + b*x^4]*(1 - Rt[-d/c, 2]*x^2)), x], x] + Simp[1/(2 *c) Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*n*(p + 1))), x] + Simp[1 /(a*n*(p + 1)) Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(n*(p + 1) + 1) + d*(n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ {q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x ], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{\left (d \,x^{4}+c \right )^{2}}d x\]
Timed out. \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {\sqrt [4]{a + b x^{4}}}{\left (c + d x^{4}\right )^{2}}\, dx \]
\[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [4]{a+b x^4}}{\left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{1/4}}{{\left (d\,x^4+c\right )}^2} \,d x \]