3.2976 \(\int \sqrt{a+b (c x^3)^{3/2}} \, dx\)

Optimal. Leaf size=91 \[ \frac{9 a x \sqrt{\frac{b \left (c x^3\right )^{3/2}}{a}+1} \, _2F_1\left (\frac{2}{9},\frac{1}{2};\frac{11}{9};-\frac{b \left (c x^3\right )^{3/2}}{a}\right )}{13 \sqrt{a+b \left (c x^3\right )^{3/2}}}+\frac{4}{13} x \sqrt{a+b \left (c x^3\right )^{3/2}} \]

[Out]

(4*x*Sqrt[a + b*(c*x^3)^(3/2)])/13 + (9*a*x*Sqrt[1 + (b*(c*x^3)^(3/2))/a]*Hypergeometric2F1[2/9, 1/2, 11/9, -(
(b*(c*x^3)^(3/2))/a)])/(13*Sqrt[a + b*(c*x^3)^(3/2)])

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Rubi [A]  time = 0.0522511, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {255, 243, 279, 365, 364} \[ \frac{9 a x \sqrt{\frac{b \left (c x^3\right )^{3/2}}{a}+1} \, _2F_1\left (\frac{2}{9},\frac{1}{2};\frac{11}{9};-\frac{b \left (c x^3\right )^{3/2}}{a}\right )}{13 \sqrt{a+b \left (c x^3\right )^{3/2}}}+\frac{4}{13} x \sqrt{a+b \left (c x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c*x^3)^(3/2)],x]

[Out]

(4*x*Sqrt[a + b*(c*x^3)^(3/2)])/13 + (9*a*x*Sqrt[1 + (b*(c*x^3)^(3/2))/a]*Hypergeometric2F1[2/9, 1/2, 11/9, -(
(b*(c*x^3)^(3/2))/a)])/(13*Sqrt[a + b*(c*x^3)^(3/2)])

Rule 255

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Subst[Int[(a + b*c^n
*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b, c, p, q}, x] && Fraction
Q[n]

Rule 243

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k - 1)*(a + b*
x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, p}, x] && FractionQ[n]

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 365

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sqrt{a+b \left (c x^3\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int \sqrt{a+b c^{3/2} x^{9/2}} \, dx,\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\operatorname{Subst}\left (2 \operatorname{Subst}\left (\int x \sqrt{a+b c^{3/2} x^9} \, dx,x,\sqrt{x}\right ),\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\frac{4}{13} x \sqrt{a+b \left (c x^3\right )^{3/2}}+\operatorname{Subst}\left (\frac{1}{13} (18 a) \operatorname{Subst}\left (\int \frac{x}{\sqrt{a+b c^{3/2} x^9}} \, dx,x,\sqrt{x}\right ),\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\frac{4}{13} x \sqrt{a+b \left (c x^3\right )^{3/2}}+\operatorname{Subst}\left (\frac{\left (18 a \sqrt{1+\frac{b c^{3/2} x^{9/2}}{a}}\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+\frac{b c^{3/2} x^9}{a}}} \, dx,x,\sqrt{x}\right )}{13 \sqrt{a+b c^{3/2} x^{9/2}}},\sqrt{x},\frac{\sqrt{c x^3}}{\sqrt{c} x}\right )\\ &=\frac{4}{13} x \sqrt{a+b \left (c x^3\right )^{3/2}}+\frac{9 a x \sqrt{1+\frac{b \left (c x^3\right )^{3/2}}{a}} \, _2F_1\left (\frac{2}{9},\frac{1}{2};\frac{11}{9};-\frac{b \left (c x^3\right )^{3/2}}{a}\right )}{13 \sqrt{a+b \left (c x^3\right )^{3/2}}}\\ \end{align*}

Mathematica [F]  time = 0.0160173, size = 0, normalized size = 0. \[ \int \sqrt{a+b \left (c x^3\right )^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \sqrt{a+b \left ( c{x}^{3} \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (c x^{3}\right )^{\frac{3}{2}} b + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \left (c x^{3}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\left (c x^{3}\right )^{\frac{3}{2}} b + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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