\(\int (a+c x^4)^{3/2} \, dx\) [797]

Optimal result

Integrand size = 11, antiderivative size = 122 \[ \int \left (a+c x^4\right )^{3/2} \, dx=\frac {2}{7} a x \sqrt {a+c x^4}+\frac {1}{7} x \left (a+c x^4\right )^{3/2}+\frac {2 a^{7/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{7 \sqrt [4]{c} \sqrt {a+c x^4}} \]

[Out]

1/7*x*(c*x^4+a)^(3/2)+2/7*a*x*(c*x^4+a)^(1/2)+2/7*a^(7/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arc
tan(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^(1/4)/(c*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {201, 226} \[ \int \left (a+c x^4\right )^{3/2} \, dx=\frac {2 a^{7/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{7 \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {2}{7} a x \sqrt {a+c x^4}+\frac {1}{7} x \left (a+c x^4\right )^{3/2} \]

[In]

Int[(a + c*x^4)^(3/2),x]

[Out]

(2*a*x*Sqrt[a + c*x^4])/7 + (x*(a + c*x^4)^(3/2))/7 + (2*a^(7/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqr
t[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(7*c^(1/4)*Sqrt[a + c*x^4])

Rule 201

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{7} x \left (a+c x^4\right )^{3/2}+\frac {1}{7} (6 a) \int \sqrt {a+c x^4} \, dx \\ & = \frac {2}{7} a x \sqrt {a+c x^4}+\frac {1}{7} x \left (a+c x^4\right )^{3/2}+\frac {1}{7} \left (4 a^2\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx \\ & = \frac {2}{7} a x \sqrt {a+c x^4}+\frac {1}{7} x \left (a+c x^4\right )^{3/2}+\frac {2 a^{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]{c} \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 = 4.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.39 \[ \int \left (a+c x^4\right )^{3/2} \, dx=\frac {a x \sqrt {a+c x^4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^4}{a}\right )}{\sqrt {1+\frac {c x^4}{a}}} \]

[In]

Integrate[(a + c*x^4)^(3/2),x]

[Out]

(a*x*Sqrt[a + c*x^4]*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^4)/a)])/Sqrt[1 + (c*x^4)/a]

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {x \left (x^{4} c +3 a \right ) \sqrt {x^{4} c +a}}{7}+\frac {4 a^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\)	\(96\)
default	\(\frac {x^{5} c \sqrt {x^{4} c +a}}{7}+\frac {3 a x \sqrt {x^{4} c +a}}{7}+\frac {4 a^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\)	\(103\)
elliptic	\(\frac {x^{5} c \sqrt {x^{4} c +a}}{7}+\frac {3 a x \sqrt {x^{4} c +a}}{7}+\frac {4 a^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{7 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {x^{4} c +a}}\)	\(103\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.42 \[ \int \left (a+c x^4\right )^{3/2} \, dx=\frac {4}{7} \, a \sqrt {c} \left (-\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \frac {1}{7} \, {\left (c x^{5} + 3 \, a x\right )} \sqrt {c x^{4} + a} \]

[In]

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

[Out]

4/7*a*sqrt(c)*(-a/c)^(3/4)*elliptic_f(arcsin((-a/c)^(1/4)/x), -1) + 1/7*(c*x^5 + 3*a*x)*sqrt(c*x^4 + a)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.30 \[ \int \left (a+c x^4\right )^{3/2} \, dx=\frac {a^{\frac {3}{2}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

a**(3/2)*x*gamma(1/4)*hyper((-3/2, 1/4), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4))

Maxima [F]

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

[In]

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

[Out]

integrate((c*x^4 + a)^(3/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((c*x^4 + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 5.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.30 \[ \int \left (a+c x^4\right )^{3/2} \, dx=\frac {x\,{\left (c\,x^4+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {c\,x^4}{a}\right )}{{\left (\frac {c\,x^4}{a}+1\right )}^{3/2}} \]

[In]

int((a + c*x^4)^(3/2),x)

[Out]

(x*(a + c*x^4)^(3/2)*hypergeom([-3/2, 1/4], 5/4, -(c*x^4)/a))/((c*x^4)/a + 1)^(3/2)