\(\int \sqrt {a+b (c x^3)^{3/2}} \, dx\) [2976]

Optimal result

Integrand size = 17, antiderivative size = 91 \[ \int \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\frac {4}{13} x \sqrt {a+b \left (c x^3\right )^{3/2}}+\frac {9 a x \sqrt {1+\frac {b \left (c x^3\right )^{3/2}}{a}} \operatorname {Hypergeometric2F1}\left (\frac {2}{9},\frac {1}{2},\frac {11}{9},-\frac {b \left (c x^3\right )^{3/2}}{a}\right )}{13 \sqrt {a+b \left (c x^3\right )^{3/2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {261, 249, 285, 372, 371} \[ \int \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\frac {9 a x \sqrt {\frac {b \left (c x^3\right )^{3/2}}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {2}{9},\frac {1}{2},\frac {11}{9},-\frac {b \left (c x^3\right )^{3/2}}{a}\right )}{13 \sqrt {a+b \left (c x^3\right )^{3/2}}}+\frac {4}{13} x \sqrt {a+b \left (c x^3\right )^{3/2}} \]

[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 249

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 261

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 285

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \sqrt {a+b c^{3/2} x^{9/2}} \, dx,\sqrt {x},\frac {\sqrt {c x^3}}{\sqrt {c} x}\right ) \\ & = \text {Subst}\left (2 \text {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}}+\text {Subst}\left (\frac {1}{13} (18 a) \text {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}}+\text {Subst}\left (\frac {\left (18 a \sqrt {1+\frac {b c^{3/2} x^{9/2}}{a}}\right ) \text {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]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(sqrt(c*x^3)*b*c*x^3 + a), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \left (c x^3\right )^{3/2}} \, dx=\int \sqrt {a+b\,{\left (c\,x^3\right )}^{3/2}} \,d x \]

[In]

int((a + b*(c*x^3)^(3/2))^(1/2),x)

[Out]

int((a + b*(c*x^3)^(3/2))^(1/2), x)