Integrand size = 19, antiderivative size = 905 \[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx =\text {Too large to display} \] Output:
1/4*(2+2^(1/2))^(1/2)*(c^(1/2)*d+a^(1/2)*e)*x^3*((-a^(1/2))^(1/2)*(1+c^(1/ 4)*x^2/(-a^(1/2))^(1/2))^2/c^(1/4)/x^2)^(1/2)*((c*x^8+a)/a^(1/2)/c^(1/2)/x ^4)^(1/2)*EllipticE(1/2*(-(-a^(1/2))^(1/2)*(2^(1/2)-2*c^(1/4)*x^2/(-a^(1/2 ))^(1/2)-2^(1/2)*c^(1/2)*x^4/a^(1/2))/c^(1/4)/x^2)^(1/2),(-2+2*2^(1/2))^(1 /2))/(-a^(1/2))^(1/2)/c^(1/4)/(1+c^(1/4)*x^2/(-a^(1/2))^(1/2))/(c*x^8+a)^( 1/2)-1/4*(2+2^(1/2))^(1/2)*(c^(1/2)*d+a^(1/2)*e)*x^3*(-(-a^(1/2))^(1/2)*(1 -c^(1/4)*x^2/(-a^(1/2))^(1/2))^2/c^(1/4)/x^2)^(1/2)*((c*x^8+a)/a^(1/2)/c^( 1/2)/x^4)^(1/2)*EllipticE(1/2*((-a^(1/2))^(1/2)*(2^(1/2)+2*c^(1/4)*x^2/(-a ^(1/2))^(1/2)-2^(1/2)*c^(1/2)*x^4/a^(1/2))/c^(1/4)/x^2)^(1/2),(-2+2*2^(1/2 ))^(1/2))/(-a^(1/2))^(1/2)/c^(1/4)/(1-c^(1/4)*x^2/(-a^(1/2))^(1/2))/(c*x^8 +a)^(1/2)+1/4*(2+2^(1/2))^(1/2)*(c^(1/2)*d-a^(1/2)*e)*x^3*((a^(1/4)+c^(1/4 )*x^2)^2/a^(1/4)/c^(1/4)/x^2)^(1/2)*(-(c*x^8+a)/a^(1/2)/c^(1/2)/x^4)^(1/2) *EllipticE(1/2*(-a^(1/4)*(2^(1/2)-2*c^(1/4)*x^2/a^(1/4)+2^(1/2)*c^(1/2)*x^ 4/a^(1/2))/c^(1/4)/x^2)^(1/2),(-2+2*2^(1/2))^(1/2))/c^(1/4)/(a^(1/4)+c^(1/ 4)*x^2)/(c*x^8+a)^(1/2)-1/4*(2+2^(1/2))^(1/2)*(c^(1/2)*d-a^(1/2)*e)*x^3*(- (a^(1/4)-c^(1/4)*x^2)^2/a^(1/4)/c^(1/4)/x^2)^(1/2)*(-(c*x^8+a)/a^(1/2)/c^( 1/2)/x^4)^(1/2)*EllipticE(1/2*(a^(1/4)*(2^(1/2)+2*c^(1/4)*x^2/a^(1/4)+2^(1 /2)*c^(1/2)*x^4/a^(1/2))/c^(1/4)/x^2)^(1/2),(-2+2*2^(1/2))^(1/2))/c^(1/4)/ (a^(1/4)-c^(1/4)*x^2)/(c*x^8+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.09 \[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\frac {\sqrt {1+\frac {c x^8}{a}} \left (5 d x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},-\frac {c x^8}{a}\right )+e x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {c x^8}{a}\right )\right )}{5 \sqrt {a+c x^8}} \] Input:
Integrate[(d + e*x^4)/Sqrt[a + c*x^8],x]
Output:
(Sqrt[1 + (c*x^8)/a]*(5*d*x*Hypergeometric2F1[1/8, 1/2, 9/8, -((c*x^8)/a)] + e*x^5*Hypergeometric2F1[1/2, 5/8, 13/8, -((c*x^8)/a)]))/(5*Sqrt[a + c*x ^8])
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.49 (sec) , antiderivative size = 433, normalized size of antiderivative = 0.48, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1763, 2009}
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.
\(\displaystyle \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx\) |
\(\Big \downarrow \) 1763 |
\(\displaystyle \int \left (\frac {d}{\sqrt {a+c x^8}}+\frac {e x^4}{\sqrt {a+c x^8}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [4]{c} d x^3 \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{c} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{c} x^2}} \sqrt {-\frac {a+c x^8}{\sqrt {a} \sqrt {c} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\frac {\sqrt {2} \sqrt {c} x^4}{\sqrt {a}}-\frac {2 \sqrt [4]{c} x^2}{\sqrt [4]{a}}+\sqrt {2}\right )}{\sqrt [4]{c} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}+\sqrt [4]{c} x^2\right ) \sqrt {a+c x^8}}-\frac {\sqrt [4]{c} d x^3 \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{c} x^2\right )^2}{\sqrt [4]{a} \sqrt [4]{c} x^2}} \sqrt {-\frac {a+c x^8}{\sqrt {a} \sqrt {c} x^4}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\frac {\sqrt {2} \sqrt {c} x^4}{\sqrt {a}}+\frac {2 \sqrt [4]{c} x^2}{\sqrt [4]{a}}+\sqrt {2}\right )}{\sqrt [4]{c} x^2}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} \left (\sqrt [4]{a}-\sqrt [4]{c} x^2\right ) \sqrt {a+c x^8}}+\frac {e x^5 \sqrt {\frac {c x^8}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {c x^8}{a}\right )}{5 \sqrt {a+c x^8}}\) |
Input:
Int[(d + e*x^4)/Sqrt[a + c*x^8],x]
Output:
(c^(1/4)*d*x^3*Sqrt[(a^(1/4) + c^(1/4)*x^2)^2/(a^(1/4)*c^(1/4)*x^2)]*Sqrt[ -((a + c*x^8)/(Sqrt[a]*Sqrt[c]*x^4))]*EllipticF[ArcSin[Sqrt[-((a^(1/4)*(Sq rt[2] - (2*c^(1/4)*x^2)/a^(1/4) + (Sqrt[2]*Sqrt[c]*x^4)/Sqrt[a]))/(c^(1/4) *x^2))]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*(a^(1/4) + c^(1/4)*x^2 )*Sqrt[a + c*x^8]) - (c^(1/4)*d*x^3*Sqrt[-((a^(1/4) - c^(1/4)*x^2)^2/(a^(1 /4)*c^(1/4)*x^2))]*Sqrt[-((a + c*x^8)/(Sqrt[a]*Sqrt[c]*x^4))]*EllipticF[Ar cSin[Sqrt[(a^(1/4)*(Sqrt[2] + (2*c^(1/4)*x^2)/a^(1/4) + (Sqrt[2]*Sqrt[c]*x ^4)/Sqrt[a]))/(c^(1/4)*x^2)]/2], -2*(1 - Sqrt[2])])/(2*Sqrt[2 + Sqrt[2]]*( a^(1/4) - c^(1/4)*x^2)*Sqrt[a + c*x^8]) + (e*x^5*Sqrt[1 + (c*x^8)/a]*Hyper geometric2F1[1/2, 5/8, 13/8, -((c*x^8)/a)])/(5*Sqrt[a + c*x^8])
Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> I nt[ExpandIntegrand[(d + e*x^n)*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n]
\[\int \frac {x^{4} e +d}{\sqrt {c \,x^{8}+a}}d x\]
Input:
int((e*x^4+d)/(c*x^8+a)^(1/2),x)
Output:
int((e*x^4+d)/(c*x^8+a)^(1/2),x)
\[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {c x^{8} + a}} \,d x } \] Input:
integrate((e*x^4+d)/(c*x^8+a)^(1/2),x, algorithm="fricas")
Output:
integral((e*x^4 + d)/sqrt(c*x^8 + a), x)
Result contains complex when optimal does not.
Time = 1.00 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.09 \[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\frac {d x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, \frac {1}{2} \\ \frac {9}{8} \end {matrix}\middle | {\frac {c x^{8} e^{i \pi }}{a}} \right )}}{8 \sqrt {a} \Gamma \left (\frac {9}{8}\right )} + \frac {e x^{5} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{8} \\ \frac {13}{8} \end {matrix}\middle | {\frac {c x^{8} e^{i \pi }}{a}} \right )}}{8 \sqrt {a} \Gamma \left (\frac {13}{8}\right )} \] Input:
integrate((e*x**4+d)/(c*x**8+a)**(1/2),x)
Output:
d*x*gamma(1/8)*hyper((1/8, 1/2), (9/8,), c*x**8*exp_polar(I*pi)/a)/(8*sqrt (a)*gamma(9/8)) + e*x**5*gamma(5/8)*hyper((1/2, 5/8), (13/8,), c*x**8*exp_ polar(I*pi)/a)/(8*sqrt(a)*gamma(13/8))
\[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {c x^{8} + a}} \,d x } \] Input:
integrate((e*x^4+d)/(c*x^8+a)^(1/2),x, algorithm="maxima")
Output:
integrate((e*x^4 + d)/sqrt(c*x^8 + a), x)
\[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {c x^{8} + a}} \,d x } \] Input:
integrate((e*x^4+d)/(c*x^8+a)^(1/2),x, algorithm="giac")
Output:
integrate((e*x^4 + d)/sqrt(c*x^8 + a), x)
Timed out. \[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\int \frac {e\,x^4+d}{\sqrt {c\,x^8+a}} \,d x \] Input:
int((d + e*x^4)/(a + c*x^8)^(1/2),x)
Output:
int((d + e*x^4)/(a + c*x^8)^(1/2), x)
\[ \int \frac {d+e x^4}{\sqrt {a+c x^8}} \, dx=\left (\int \frac {\sqrt {c \,x^{8}+a}}{c \,x^{8}+a}d x \right ) d +\left (\int \frac {\sqrt {c \,x^{8}+a}\, x^{4}}{c \,x^{8}+a}d x \right ) e \] Input:
int((e*x^4+d)/(c*x^8+a)^(1/2),x)
Output:
int(sqrt(a + c*x**8)/(a + c*x**8),x)*d + int((sqrt(a + c*x**8)*x**4)/(a + c*x**8),x)*e