3.2948 \(\int \sqrt {a+b (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=306 \[ \frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a x \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [3]{b} \sqrt {c x^2} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}+\frac {2}{5} x \sqrt {a+b \left (c x^2\right )^{3/2}} \]

[Out]

2/5*x*(a+b*(c*x^2)^(3/2))^(1/2)+2/5*3^(3/4)*a*x*EllipticF((a^(1/3)*(1-3^(1/2))+b^(1/3)*(c*x^2)^(1/2))/(a^(1/3)
*(1+3^(1/2))+b^(1/3)*(c*x^2)^(1/2)),I*3^(1/2)+2*I)*(a^(1/3)+b^(1/3)*(c*x^2)^(1/2))*(1/2*6^(1/2)+1/2*2^(1/2))*(
(a^(2/3)+b^(2/3)*c*x^2-a^(1/3)*b^(1/3)*(c*x^2)^(1/2))/(a^(1/3)*(1+3^(1/2))+b^(1/3)*(c*x^2)^(1/2))^2)^(1/2)/b^(
1/3)/(c*x^2)^(1/2)/(a+b*(c*x^2)^(3/2))^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*(c*x^2)^(1/2))/(a^(1/3)*(1+3^(1/2))+b^(
1/3)*(c*x^2)^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {254, 195, 218} \[ \frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a x \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [3]{b} \sqrt {c x^2} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}+\frac {2}{5} x \sqrt {a+b \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(2*x*Sqrt[a + b*(c*x^2)^(3/2)])/5 + (2*3^(3/4)*Sqrt[2 + Sqrt[3]]*a*x*(a^(1/3) + b^(1/3)*Sqrt[c*x^2])*Sqrt[(a^(
2/3) + b^(2/3)*c*x^2 - a^(1/3)*b^(1/3)*Sqrt[c*x^2])/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])^2]*EllipticF
[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])], -7 - 4*S
qrt[3]])/(5*b^(1/3)*Sqrt[c*x^2]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*Sqrt[c*x^2]))/((1 + Sqrt[3])*a^(1/3) + b^(1/3
)*Sqrt[c*x^2])^2]*Sqrt[a + b*(c*x^2)^(3/2)])

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 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 254

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> Dist[x/(c*x^q)^(1/q), Subst[Int[(a + b*x^(n*q))
^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, n, p, q}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

Rubi steps

\begin {align*} \int \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx &=\frac {x \operatorname {Subst}\left (\int \sqrt {a+b x^3} \, dx,x,\sqrt {c x^2}\right )}{\sqrt {c x^2}}\\ &=\frac {2}{5} x \sqrt {a+b \left (c x^2\right )^{3/2}}+\frac {(3 a x) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\sqrt {c x^2}\right )}{5 \sqrt {c x^2}}\\ &=\frac {2}{5} x \sqrt {a+b \left (c x^2\right )^{3/2}}+\frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a x \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}+b^{2/3} c x^2-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [3]{b} \sqrt {c x^2} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 64, normalized size = 0.21 \[ \frac {x \sqrt {a+b \left (c x^2\right )^{3/2}} \, _2F_1\left (-\frac {1}{2},\frac {1}{3};\frac {4}{3};-\frac {b \left (c x^2\right )^{3/2}}{a}\right )}{\sqrt {\frac {b \left (c x^2\right )^{3/2}}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\sqrt {c x^{2}} b c x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(3/2))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(sqrt(c*x^2)*b*c*x^2 + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(3/2))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt((c*x^2)^(3/2)*b + a), x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \sqrt {a +\left (c \,x^{2}\right )^{\frac {3}{2}} b}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+(c*x^2)^(3/2)*b)^(1/2),x)

[Out]

int((a+(c*x^2)^(3/2)*b)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(3/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt((c*x^2)^(3/2)*b + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c*x^2)^(3/2))^(1/2),x)

[Out]

int((a + b*(c*x^2)^(3/2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \left (c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x**2)**(3/2))**(1/2),x)

[Out]

Integral(sqrt(a + b*(c*x**2)**(3/2)), x)

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