3.2947 \(\int x^3 \sqrt {a+b (c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=340 \[ -\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{55 b^{4/3} c^2 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}+\frac {6 a \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{55 b c^2}+\frac {2}{11} x^4 \sqrt {a+b \left (c x^2\right )^{3/2}} \]

[Out]

2/11*x^4*(a+b*(c*x^2)^(3/2))^(1/2)+6/55*a*(c*x^2)^(1/2)*(a+b*(c*x^2)^(3/2))^(1/2)/b/c^2-4/55*3^(3/4)*a^2*Ellip
ticF((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^(4/3)/c^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 = 0.25, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {368, 279, 321, 218} \[ -\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{55 b^{4/3} c^2 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}+\frac {6 a \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{55 b c^2}+\frac {2}{11} x^4 \sqrt {a+b \left (c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[a + b*(c*x^2)^(3/2)],x]

[Out]

(2*x^4*Sqrt[a + b*(c*x^2)^(3/2)])/11 + (6*a*Sqrt[c*x^2]*Sqrt[a + b*(c*x^2)^(3/2)])/(55*b*c^2) - (4*3^(3/4)*Sqr
t[2 + Sqrt[3]]*a^2*(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*Sqrt[3]])/(55*b^(4/3)*c^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 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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 368

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

Rubi steps

\begin {align*} \int x^3 \sqrt {a+b \left (c x^2\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int x^3 \sqrt {a+b x^3} \, dx,x,\sqrt {c x^2}\right )}{c^2}\\ &=\frac {2}{11} x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {a+b x^3}} \, dx,x,\sqrt {c x^2}\right )}{11 c^2}\\ &=\frac {2}{11} x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}+\frac {6 a \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{55 b c^2}-\frac {\left (6 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\sqrt {c x^2}\right )}{55 b c^2}\\ &=\frac {2}{11} x^4 \sqrt {a+b \left (c x^2\right )^{3/2}}+\frac {6 a \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{55 b c^2}-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \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 )}{55 b^{4/3} c^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 = 0.11, size = 109, normalized size = 0.32 \[ \frac {2 \sqrt {c x^2} \sqrt {a+b \left (c x^2\right )^{3/2}} \left (a \left (\frac {a+b \left (c x^2\right )^{3/2}}{a}\right )^{3/2}-a \, _2F_1\left (-\frac {1}{2},\frac {1}{3};\frac {4}{3};-\frac {b \left (c x^2\right )^{3/2}}{a}\right )\right )}{11 b c^2 \sqrt {\frac {a+b \left (c x^2\right )^{3/2}}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c*x^2]*Sqrt[a + b*(c*x^2)^(3/2)]*(a*((a + b*(c*x^2)^(3/2))/a)^(3/2) - a*Hypergeometric2F1[-1/2, 1/3, 4
/3, -((b*(c*x^2)^(3/2))/a)]))/(11*b*c^2*Sqrt[(a + b*(c*x^2)^(3/2))/a])

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\sqrt {c x^{2}} b c x^{2} + a} x^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*(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 \[ \int \sqrt {\left (c x^{2}\right )^{\frac {3}{2}} b + a} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*(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 \[ \int x^{3} \sqrt {a + b \left (c x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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