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

Optimal. Leaf size=124 \[ \frac {2 a^{3/4} c^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt {a+c x^4}}+\frac {2}{3} c x \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {277, 195, 220} \[ \frac {2 a^{3/4} c^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt {a+c x^4}}-\frac {\left (a+c x^4\right )^{3/2}}{3 x^3}+\frac {2}{3} c x \sqrt {a+c x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^4,x]

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.42 \[ -\frac {a \sqrt {a+c x^4} \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {c x^4}{a}\right )}{3 x^3 \sqrt {\frac {c x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^4,x]

[Out]

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

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^4 + a)^(3/2)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + a)^(3/2)/x^4, x)

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maple [C]  time = 0.01, size = 102, normalized size = 0.82 \[ \frac {4 \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a c \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {\sqrt {c \,x^{4}+a}\, c x}{3}-\frac {\sqrt {c \,x^{4}+a}\, a}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(c*x^4+a)^(1/2)/x^3+1/3*c*x*(c*x^4+a)^(1/2)+4/3*a*c/(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 \[ \int \frac {{\left (c x^{4} + a\right )}^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + a)^(3/2)/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(3/2)/x^4,x)

[Out]

int((a + c*x^4)^(3/2)/x^4, x)

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sympy [C]  time = 2.12, size = 42, normalized size = 0.34 \[ \frac {a^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*gamma(-3/4)*hyper((-3/2, -3/4), (1/4,), c*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4))

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