\(\int \frac {(a+c x^4)^{3/2}}{x^6} \, dx\) [802]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [F]
   Sympy [C] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 15, antiderivative size = 252 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=-\frac {6 c \sqrt {a+c x^4}}{5 x}+\frac {12 c^{3/2} x \sqrt {a+c x^4}}{5 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}-\frac {12 \sqrt [4]{a} c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+c x^4}}+\frac {6 \sqrt [4]{a} c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+c x^4}} \]

[Out]

-1/5*(c*x^4+a)^(3/2)/x^5-6/5*c*(c*x^4+a)^(1/2)/x+12/5*c^(3/2)*x*(c*x^4+a)^(1/2)/(a^(1/2)+x^2*c^(1/2))-12/5*a^(
1/4)*c^(5/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arcta
n(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/(c*x^4+a)^(
1/2)+6/5*a^(1/4)*c^(5/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF
(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)
/(c*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {283, 311, 226, 1210} \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\frac {6 \sqrt [4]{a} c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 \sqrt {a+c x^4}}-\frac {12 \sqrt [4]{a} c^{5/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+c x^4}}+\frac {12 c^{3/2} x \sqrt {a+c x^4}}{5 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {6 c \sqrt {a+c x^4}}{5 x}-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5} \]

[In]

Int[(a + c*x^4)^(3/2)/x^6,x]

[Out]

(-6*c*Sqrt[a + c*x^4])/(5*x) + (12*c^(3/2)*x*Sqrt[a + c*x^4])/(5*(Sqrt[a] + Sqrt[c]*x^2)) - (a + c*x^4)^(3/2)/
(5*x^5) - (12*a^(1/4)*c^(5/4)*(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])/(5*Sqrt[a + c*x^4]) + (6*a^(1/4)*c^(5/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a +
c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(5*Sqrt[a + c*x^4])

Rule 226

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]

Rule 283

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] - Dist[b*n*(p/(c^n*(m + 1))), Int[(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] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
 x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1210

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
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}+\frac {1}{5} (6 c) \int \frac {\sqrt {a+c x^4}}{x^2} \, dx \\ & = -\frac {6 c \sqrt {a+c x^4}}{5 x}-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}+\frac {1}{5} \left (12 c^2\right ) \int \frac {x^2}{\sqrt {a+c x^4}} \, dx \\ & = -\frac {6 c \sqrt {a+c x^4}}{5 x}-\frac {\left (a+c x^4\right )^{3/2}}{5 x^5}+\frac {1}{5} \left (12 \sqrt {a} c^{3/2}\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx-\frac {1}{5} \left (12 \sqrt {a} c^{3/2}\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx \\ & = -\frac {6 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 {\left (a+c x^4\right )^{3/2}}{5 x^5}-\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 \tan ^{-1}\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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a+c x^4}} \\ \end{align*}

Mathematica [C] (verified)

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 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=-\frac {a \sqrt {a+c x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{4},-\frac {c x^4}{a}\right )}{5 x^5 \sqrt {1+\frac {c x^4}{a}}} \]

[In]

Integrate[(a + c*x^4)^(3/2)/x^6,x]

[Out]

-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])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.51 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.48

method result size
risch \(-\frac {\sqrt {x^{4} c +a}\, \left (7 x^{4} c +a \right )}{5 x^{5}}+\frac {12 i c^{\frac {3}{2}} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\) \(120\)
default \(-\frac {a \sqrt {x^{4} c +a}}{5 x^{5}}-\frac {7 c \sqrt {x^{4} c +a}}{5 x}+\frac {12 i c^{\frac {3}{2}} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\) \(128\)
elliptic \(-\frac {a \sqrt {x^{4} c +a}}{5 x^{5}}-\frac {7 c \sqrt {x^{4} c +a}}{5 x}+\frac {12 i c^{\frac {3}{2}} \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\) \(128\)

[In]

int((c*x^4+a)^(3/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

-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))

Fricas [F]

\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \]

[In]

integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="fricas")

[Out]

integral((c*x^4 + a)^(3/2)/x^6, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.18 \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\frac {a^{\frac {3}{2}} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} \]

[In]

integrate((c*x**4+a)**(3/2)/x**6,x)

[Out]

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))

Maxima [F]

\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \]

[In]

integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="maxima")

[Out]

integrate((c*x^4 + a)^(3/2)/x^6, x)

Giac [F]

\[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \]

[In]

integrate((c*x^4+a)^(3/2)/x^6,x, algorithm="giac")

[Out]

integrate((c*x^4 + a)^(3/2)/x^6, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+c x^4\right )^{3/2}}{x^6} \, dx=\int \frac {{\left (c\,x^4+a\right )}^{3/2}}{x^6} \,d x \]

[In]

int((a + c*x^4)^(3/2)/x^6,x)

[Out]

int((a + c*x^4)^(3/2)/x^6, x)