3.799 \(\int \frac {(a+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=126 \[ \frac {2 c^{7/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{a} \sqrt {a+c x^4}}-\frac {\left (a+c x^4\right )^{3/2}}{7 x^7}-\frac {2 c \sqrt {a+c x^4}}{7 x^3} \]

[Out]

-1/7*(c*x^4+a)^(3/2)/x^7-2/7*c*(c*x^4+a)^(1/2)/x^3+2/7*c^(7/4)*(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)/a^(1/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {277, 220} \[ \frac {2 c^{7/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{a} \sqrt {a+c x^4}}-\frac {2 c \sqrt {a+c x^4}}{7 x^3}-\frac {\left (a+c x^4\right )^{3/2}}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^8,x]

[Out]

(-2*c*Sqrt[a + c*x^4])/(7*x^3) - (a + c*x^4)^(3/2)/(7*x^7) + (2*c^(7/4)*(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])/(7*a^(1/4)*Sqrt[a + c*x^4])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 277

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]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^4\right )^{3/2}}{x^8} \, dx &=-\frac {\left (a+c x^4\right )^{3/2}}{7 x^7}+\frac {1}{7} (6 c) \int \frac {\sqrt {a+c x^4}}{x^4} \, dx\\ &=-\frac {2 c \sqrt {a+c x^4}}{7 x^3}-\frac {\left (a+c x^4\right )^{3/2}}{7 x^7}+\frac {1}{7} \left (4 c^2\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx\\ &=-\frac {2 c \sqrt {a+c x^4}}{7 x^3}-\frac {\left (a+c x^4\right )^{3/2}}{7 x^7}+\frac {2 c^{7/4} \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 )}{7 \sqrt [4]{a} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.41 \[ -\frac {a \sqrt {a+c x^4} \, _2F_1\left (-\frac {7}{4},-\frac {3}{2};-\frac {3}{4};-\frac {c x^4}{a}\right )}{7 x^7 \sqrt {\frac {c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^8,x]

[Out]

-1/7*(a*Sqrt[a + c*x^4]*Hypergeometric2F1[-7/4, -3/2, -3/4, -((c*x^4)/a)])/(x^7*Sqrt[1 + (c*x^4)/a])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^8,x, algorithm="fricas")

[Out]

integral((c*x^4 + a)^(3/2)/x^8, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^8,x, algorithm="giac")

[Out]

integrate((c*x^4 + a)^(3/2)/x^8, x)

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maple [C]  time = 0.01, size = 105, normalized size = 0.83 \[ \frac {4 \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, c^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{7 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}-\frac {3 \sqrt {c \,x^{4}+a}\, c}{7 x^{3}}-\frac {\sqrt {c \,x^{4}+a}\, a}{7 x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(3/2)/x^8,x)

[Out]

-1/7*a*(c*x^4+a)^(1/2)/x^7-3/7*c*(c*x^4+a)^(1/2)/x^3+4/7*c^2/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2
+1)^(1/2)*(I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{8}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^8,x, algorithm="maxima")

[Out]

integrate((c*x^4 + a)^(3/2)/x^8, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+a\right )}^{3/2}}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(3/2)/x^8,x)

[Out]

int((a + c*x^4)^(3/2)/x^8, x)

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sympy [C]  time = 2.47, size = 46, normalized size = 0.37 \[ \frac {a^{\frac {3}{2}} \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(3/2)/x**8,x)

[Out]

a**(3/2)*gamma(-7/4)*hyper((-7/4, -3/2), (-3/4,), c*x**4*exp_polar(I*pi)/a)/(4*x**7*gamma(-3/4))

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