\(\int x \sqrt {a+b \sqrt {c x^2}} \, dx\) [2932]

Optimal result

Integrand size = 19, antiderivative size = 56 \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=-\frac {2 a \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b^2 c}+\frac {2 \left (a+b \sqrt {c x^2}\right )^{5/2}}{5 b^2 c} \]

[Out]

-2/3*a*(a+b*(c*x^2)^(1/2))^(3/2)/b^2/c+2/5*(a+b*(c*x^2)^(1/2))^(5/2)/b^2/c

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {375, 45} \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \left (a+b \sqrt {c x^2}\right )^{5/2}}{5 b^2 c}-\frac {2 a \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b^2 c} \]

[In]

Int[x*Sqrt[a + b*Sqrt[c*x^2]],x]

[Out]

(-2*a*(a + b*Sqrt[c*x^2])^(3/2))/(3*b^2*c) + (2*(a + b*Sqrt[c*x^2])^(5/2))/(5*b^2*c)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 375

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*} \text {integral}& = \frac {\text {Subst}\left (\int x \sqrt {a+b x} \, dx,x,\sqrt {c x^2}\right )}{c} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a \sqrt {a+b x}}{b}+\frac {(a+b x)^{3/2}}{b}\right ) \, dx,x,\sqrt {c x^2}\right )}{c} \\ & = -\frac {2 a \left (a+b \sqrt {c x^2}\right )^{3/2}}{3 b^2 c}+\frac {2 \left (a+b \sqrt {c x^2}\right )^{5/2}}{5 b^2 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.77 \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \left (a+b \sqrt {c x^2}\right )^{3/2} \left (-2 a+3 b \sqrt {c x^2}\right )}{15 b^2 c} \]

[In]

Integrate[x*Sqrt[a + b*Sqrt[c*x^2]],x]

[Out]

(2*(a + b*Sqrt[c*x^2])^(3/2)*(-2*a + 3*b*Sqrt[c*x^2]))/(15*b^2*c)

Maple [A] (verified)

Time = 4.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.64

method	result	size
default	\(\frac {2 \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {3}{2}} \left (-2 a +3 b \sqrt {c \,x^{2}}\right )}{15 b^{2} c}\)	\(36\)
derivativedivides	\(\frac {\frac {2 \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {3}{2}}}{3}}{b^{2} c}\)	\(41\)

[In]

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

[Out]

2/15*(a+b*(c*x^2)^(1/2))^(3/2)*(-2*a+3*b*(c*x^2)^(1/2))/b^2/c

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \, {\left (3 \, b^{2} c x^{2} + \sqrt {c x^{2}} a b - 2 \, a^{2}\right )} \sqrt {\sqrt {c x^{2}} b + a}}{15 \, b^{2} c} \]

[In]

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

[Out]

2/15*(3*b^2*c*x^2 + sqrt(c*x^2)*a*b - 2*a^2)*sqrt(sqrt(c*x^2)*b + a)/(b^2*c)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.77 \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\sqrt {c x^{2}} b + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (\sqrt {c x^{2}} b + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )}}{15 \, c} \]

[In]

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

[Out]

2/15*(3*(sqrt(c*x^2)*b + a)^(5/2)/b^2 - 5*(sqrt(c*x^2)*b + a)^(3/2)*a/b^2)/c

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (44) = 88\).

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.61 \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {2 \, {\left (\frac {5 \, {\left ({\left (b \sqrt {c} x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b \sqrt {c} x + a} a\right )} a}{b \sqrt {c}} + \frac {3 \, {\left (b \sqrt {c} x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b \sqrt {c} x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b \sqrt {c} x + a} a^{2}}{b \sqrt {c}}\right )}}{15 \, b \sqrt {c}} \]

[In]

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

[Out]

2/15*(5*((b*sqrt(c)*x + a)^(3/2) - 3*sqrt(b*sqrt(c)*x + a)*a)*a/(b*sqrt(c)) + (3*(b*sqrt(c)*x + a)^(5/2) - 10*
(b*sqrt(c)*x + a)^(3/2)*a + 15*sqrt(b*sqrt(c)*x + a)*a^2)/(b*sqrt(c)))/(b*sqrt(c))

Mupad [B] (verification not implemented)

Time = 6.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int x \sqrt {a+b \sqrt {c x^2}} \, dx=\frac {x^2\,\sqrt {a+b\,\sqrt {c}\,\sqrt {x^2}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},2;\ 3;\ -\frac {b\,\sqrt {c}\,\left |x\right |}{a}\right )}{2\,\sqrt {\frac {b\,\sqrt {c}\,\sqrt {x^2}}{a}+1}} \]

[In]

int(x*(a + b*(c*x^2)^(1/2))^(1/2),x)

[Out]

(x^2*(a + b*c^(1/2)*(x^2)^(1/2))^(1/2)*hypergeom([-1/2, 2], 3, -(b*c^(1/2)*abs(x))/a))/(2*((b*c^(1/2)*(x^2)^(1
/2))/a + 1)^(1/2))