\(\int \sqrt {a+b \sqrt {c x^2}} \, dx\) [2938]

Optimal result

Integrand size = 17, antiderivative size = 34 \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 x \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {260, 32} \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 x \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b \sqrt {c x^2}} \]

[In]

Int[Sqrt[a + b*Sqrt[c*x^2]],x]

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 260

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*} \text {integral}& = \frac {x \text {Subst}\left (\int \sqrt {a+b x} \, dx,x,\sqrt {c x^2}\right )}{\sqrt {c x^2}} \\ & = \frac {2 x \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 x \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b \sqrt {c x^2}} \]

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^2]],x]

[Out]

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

Maple [A] (verified)

Time = 3.93 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {2 x \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {3}{2}}}{3 b \sqrt {c \,x^{2}}}\)	\(27\)

[In]

int((a+b*(c*x^2)^(1/2))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \, {\left (b c x^{2} + \sqrt {c x^{2}} a\right )} \sqrt {\sqrt {c x^{2}} b + a}}{3 \, b c x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\int \sqrt {a + b \sqrt {c x^{2}}}\, dx \]

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c*x**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {{\left ({\left (c^{\frac {3}{2}} + c\right )} b x + a {\left (c + \sqrt {c}\right )}\right )} \sqrt {b \sqrt {c} x + a}}{{\left (c^{2} + 2 \, c^{\frac {3}{2}} + c\right )} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53 \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \, {\left (b \sqrt {c} x + a\right )}^{\frac {3}{2}}}{3 \, b \sqrt {c}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \sqrt {c x^2}} \, dx=\int \sqrt {a+b\,\sqrt {c\,x^2}} \,d x \]

[In]

int((a + b*(c*x^2)^(1/2))^(1/2),x)

[Out]

int((a + b*(c*x^2)^(1/2))^(1/2), x)