Integrand size = 15, antiderivative size = 253 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=-\frac {a \sqrt {a+c x^4}}{5 x^5}-\frac {7 c \sqrt {a+c x^4}}{5 x}+\frac {12 c^{3/2} x \sqrt {a+c x^4}}{5 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {12 \sqrt [4]{a} c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+c x^4}}+\frac {6 \sqrt [4]{a} c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+c x^4}} \] Output:
-1/5*a*(c*x^4+a)^(1/2)/x^5-7/5*c*(c*x^4+a)^(1/2)/x+12*c^(3/2)*x*(c*x^4+a)^ (1/2)/(5*a^(1/2)+5*c^(1/2)*x^2)-12/5*a^(1/4)*c^(5/4)*(a^(1/2)+c^(1/2)*x^2) *((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan(c^(1/4)* x/a^(1/4))),1/2*2^(1/2))/(c*x^4+a)^(1/2)+6/5*a^(1/4)*c^(5/4)*(a^(1/2)+c^(1 /2)*x^2)*((c*x^4+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arcta n(c^(1/4)*x/a^(1/4)),1/2*2^(1/2))/(c*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.21 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=-\frac {a \sqrt {a+c x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {c x^4}{a}\right )}{5 x^5 \sqrt {1+\frac {c x^4}{a}}} \] Input:
Integrate[(a + c*x^4)^(3/2)/x^6,x]
Output:
-1/5*(a*Sqrt[a + c*x^4]*Hypergeometric2F1[-3/2, -5/4, -1/4, -((c*x^4)/a)]) /(x^5*Sqrt[1 + (c*x^4)/a])
Time = 0.55 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {809, 809, 834, 27, 761, 1510}
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.
| \(\displaystyle \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx\) |
| \(\Big \downarrow \) 809 |
| \(\displaystyle \frac {6}{5} c \int \frac {\sqrt {c x^4+a}}{x^2}dx-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 809 |
| \(\displaystyle \frac {6}{5} c \left (2 c \int \frac {x^2}{\sqrt {c x^4+a}}dx-\frac {\sqrt {a+c x^4}}{x}\right )-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {6}{5} c \left (2 c \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+a}}dx}{\sqrt {c}}\right )-\frac {\sqrt {a+c x^4}}{x}\right )-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {6}{5} c \left (2 c \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+a}}dx}{\sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}\right )-\frac {\sqrt {a+c x^4}}{x}\right )-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {6}{5} c \left (2 c \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}\right )-\frac {\sqrt {a+c x^4}}{x}\right )-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {6}{5} c \left (2 c \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {a+c x^4}}-\frac {x \sqrt {a+c x^4}}{\sqrt {a}+\sqrt {c} x^2}}{\sqrt {c}}\right )-\frac {\sqrt {a+c x^4}}{x}\right )-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}\) |
Input:
Int[(a + c*x^4)^(3/2)/x^6,x]
Output:
-1/5*(a + c*x^4)^(3/2)/x^5 + (6*c*(-(Sqrt[a + c*x^4]/x) + 2*c*(-((-((x*Sqr t[a + c*x^4])/(Sqrt[a] + Sqrt[c]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)* Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x) /a^(1/4)], 1/2])/(c^(1/4)*Sqrt[a + c*x^4]))/Sqrt[c]) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcT an[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*c^(3/4)*Sqrt[a + c*x^4]))))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[a, b, c, n, m, p, x]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Result contains complex when optimal does not.
Time = 0.84 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.47
| method | result | size |
| risch | \(-\frac {\sqrt {c \,x^{4}+a}\, \left (7 c \,x^{4}+a \right )}{5 x^{5}}+\frac {12 i c^{\frac {3}{2}} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(120\) |
| default | \(-\frac {a \sqrt {c \,x^{4}+a}}{5 x^{5}}-\frac {7 c \sqrt {c \,x^{4}+a}}{5 x}+\frac {12 i c^{\frac {3}{2}} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(128\) |
| elliptic | \(-\frac {a \sqrt {c \,x^{4}+a}}{5 x^{5}}-\frac {7 c \sqrt {c \,x^{4}+a}}{5 x}+\frac {12 i c^{\frac {3}{2}} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\) | \(128\) |
Input:
int((c*x^4+a)^(3/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/5*(c*x^4+a)^(1/2)*(7*c*x^4+a)/x^5+12/5*I*c^(3/2)*a^(1/2)/(I/a^(1/2)*c^( 1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2 )/(c*x^4+a)^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I /a^(1/2)*c^(1/2))^(1/2),I))
\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="fricas")
Output:
integral((c*x^4 + a)^(3/2)/x^6, x)
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.18 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \] Input:
integrate((c*x**4+a)**(3/2)/x**6,x)
Output:
a**(3/2)*gamma(-5/4)*hyper((-3/2, -5/4), (-1/4,), c*x**4*exp_polar(I*pi)/a )/(4*x**5*gamma(-1/4))
\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="maxima")
Output:
integrate((c*x^4 + a)^(3/2)/x^6, x)
\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:
integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="giac")
Output:
integrate((c*x^4 + a)^(3/2)/x^6, x)
Timed out. \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int \frac {{\left (c\,x^4+a\right )}^{3/2}}{x^6} \,d x \] Input:
int((a + c*x^4)^(3/2)/x^6,x)
Output:
int((a + c*x^4)^(3/2)/x^6, x)
\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\frac {-\sqrt {c \,x^{4}+a}\, a +\sqrt {c \,x^{4}+a}\, c \,x^{4}-4 \left (\int \frac {\sqrt {c \,x^{4}+a}}{c \,x^{10}+a \,x^{6}}d x \right ) a^{2} x^{5}}{x^{5}} \] Input:
int((c*x^4+a)^(3/2)/x^6,x)
Output:
( - sqrt(a + c*x**4)*a + sqrt(a + c*x**4)*c*x**4 - 4*int(sqrt(a + c*x**4)/ (a*x**6 + c*x**10),x)*a**2*x**5)/x**5