Integrand size = 21, antiderivative size = 274 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\frac {b x \sqrt [4]{a+b x^4}}{2 d}-\frac {\sqrt {a} b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{2 d \left (a+b x^4\right )^{3/4}}-\frac {(b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{2 \sqrt [4]{b} c d}-\frac {(b c-a d) \sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} \operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ),-1\right )}{2 \sqrt [4]{b} c d} \] Output:
1/2*b*x*(b*x^4+a)^(1/4)/d-1/2*a^(1/2)*b^(3/2)*(1+a/b/x^4)^(3/4)*x^3*Invers eJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/d/(b*x^4+a)^(3/4)-1/2*( -a*d+b*c)*(a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)*EllipticPi(b^(1/4)*x/(b*x^4+ a)^(1/4),-(-a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)/b^(1/4)/c/d-1/2*(-a*d+b*c)*( a/(b*x^4+a))^(1/2)*(b*x^4+a)^(1/2)*EllipticPi(b^(1/4)*x/(b*x^4+a)^(1/4),(- a*d+b*c)^(1/2)/b^(1/2)/c^(1/2),I)/b^(1/4)/c/d
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.37 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\frac {x \left (\frac {b (-2 b c+3 a d) x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{c}+\frac {5 \left (-5 a c \left (2 a^2 d+a b d x^4+b^2 x^4 \left (c+d x^4\right )\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b x^4 \left (a+b x^4\right ) \left (c+d x^4\right ) \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}{\left (c+d x^4\right ) \left (-5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+x^4 \left (4 a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )\right )}\right )}{10 d \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(a + b*x^4)^(5/4)/(c + d*x^4),x]
Output:
(x*((b*(-2*b*c + 3*a*d)*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 + (5*(-5*a*c*(2*a^2*d + a*b*d*x^4 + b^2* x^4*(c + d*x^4))*AppellF1[1/4, 3/4, 1, 5/4, -((b*x^4)/a), -((d*x^4)/c)] + b*x^4*(a + b*x^4)*(c + d*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)*(-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)])))))/ (10*d*(a + b*x^4)^(3/4))
Time = 0.78 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {924, 748, 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.
| \(\displaystyle \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx\) |
| \(\Big \downarrow \) 924 |
| \(\displaystyle \frac {b \int \sqrt [4]{b x^4+a}dx}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{d}\) |
| \(\Big \downarrow \) 748 |
| \(\displaystyle \frac {b \left (\frac {1}{2} a \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{2} x \sqrt [4]{a+b x^4}\right )}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{d}\) |
| \(\Big \downarrow \) 768 |
| \(\displaystyle \frac {b \left (\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{2 \left (a+b x^4\right )^{3/4}}+\frac {1}{2} x \sqrt [4]{a+b x^4}\right )}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{d}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{2 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{d}\) |
| \(\Big \downarrow \) 807 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{4 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{d}\) |
| \(\Big \downarrow \) 229 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{b x^4+a}}{d x^4+c}dx}{d}\) |
| \(\Big \downarrow \) 923 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) \int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (c-\frac {(b c-a d) x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{d}\) |
| \(\Big \downarrow \) 925 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) \left (\frac {\int \frac {\sqrt {c}}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 c}+\frac {\int \frac {\sqrt {c}}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 c}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) \left (\frac {\int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\sqrt {c}-\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}+\frac {\int \frac {1}{\sqrt {1-\frac {b x^4}{b x^4+a}} \left (\frac {\sqrt {b c-a d} x^2}{\sqrt {b x^4+a}}+\sqrt {c}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {c}}\right )}{d}\) |
| \(\Big \downarrow \) 1542 |
| \(\displaystyle \frac {b \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )}{d}-\frac {\sqrt {\frac {a}{a+b x^4}} \sqrt {a+b x^4} (b c-a d) \left (\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c}+\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b c-a d}}{\sqrt {b} \sqrt {c}},\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{b x^4+a}}\right ),-1\right )}{2 \sqrt [4]{b} c}\right )}{d}\) |
Input:
Int[(a + b*x^4)^(5/4)/(c + d*x^4),x]
Output:
(b*((x*(a + b*x^4)^(1/4))/2 - (Sqrt[a]*Sqrt[b]*(1 + a/(b*x^4))^(3/4)*x^3*E llipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*(a + b*x^4)^(3/4))))/d - ((b*c - 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]*Sqrt[c]), ArcSin[(b^(1/4)*x) /(a + b*x^4)^(1/4)], -1]/(2*b^(1/4)*c)))/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)^(-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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Simp[a*n*(p/(n*p + 1)) Int[(a + b*x^n)^(p - 1), x], x] /; Fre eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat or[p + 1/n], Denominator[p]])
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[((a_) + (b_.)*(x_)^4)^(5/4)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Simp[b/ d Int[(a + b*x^4)^(1/4), x], x] - Simp[(b*c - a*d)/d Int[(a + b*x^4)^(1 /4)/(c + d*x^4), x], 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[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]
\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{d \,x^{4}+c}d x\]
Input:
int((b*x^4+a)^(5/4)/(d*x^4+c),x)
Output:
int((b*x^4+a)^(5/4)/(d*x^4+c),x)
Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\text {Timed out} \] Input:
integrate((b*x^4+a)^(5/4)/(d*x^4+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\int \frac {\left (a + b x^{4}\right )^{\frac {5}{4}}}{c + d x^{4}}\, dx \] Input:
integrate((b*x**4+a)**(5/4)/(d*x**4+c),x)
Output:
Integral((a + b*x**4)**(5/4)/(c + d*x**4), x)
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{d x^{4} + c} \,d x } \] Input:
integrate((b*x^4+a)^(5/4)/(d*x^4+c),x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(5/4)/(d*x^4 + c), x)
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{d x^{4} + c} \,d x } \] Input:
integrate((b*x^4+a)^(5/4)/(d*x^4+c),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(5/4)/(d*x^4 + c), x)
Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{5/4}}{d\,x^4+c} \,d x \] Input:
int((a + b*x^4)^(5/4)/(c + d*x^4),x)
Output:
int((a + b*x^4)^(5/4)/(c + d*x^4), x)
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{c+d x^4} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} b x +2 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b d \,x^{8}+a d \,x^{4}+b c \,x^{4}+a c}d x \right ) a^{2} d -\left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b d \,x^{8}+a d \,x^{4}+b c \,x^{4}+a c}d x \right ) a b c +3 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{4}}{b d \,x^{8}+a d \,x^{4}+b c \,x^{4}+a c}d x \right ) a b d -2 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{4}}{b d \,x^{8}+a d \,x^{4}+b c \,x^{4}+a c}d x \right ) b^{2} c}{2 d} \] Input:
int((b*x^4+a)^(5/4)/(d*x^4+c),x)
Output:
((a + b*x**4)**(1/4)*b*x + 2*int((a + b*x**4)**(1/4)/(a*c + a*d*x**4 + b*c *x**4 + b*d*x**8),x)*a**2*d - int((a + b*x**4)**(1/4)/(a*c + a*d*x**4 + b* c*x**4 + b*d*x**8),x)*a*b*c + 3*int(((a + b*x**4)**(1/4)*x**4)/(a*c + a*d* x**4 + b*c*x**4 + b*d*x**8),x)*a*b*d - 2*int(((a + b*x**4)**(1/4)*x**4)/(a *c + a*d*x**4 + b*c*x**4 + b*d*x**8),x)*b**2*c)/(2*d)