3.627 \(\int x \sqrt{a+b \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=133 \[ \frac{4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^4 d^2}-\frac{4 a \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^4 d^2}+\frac{4 \left (a+b \sqrt{c+d x}\right )^{9/2}}{9 b^4 d^2}-\frac{12 a \left (a+b \sqrt{c+d x}\right )^{7/2}}{7 b^4 d^2} \]

[Out]

(-4*a*(a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(3/2))/(3*b^4*d^2) + (4*(3*a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(5/2))
/(5*b^4*d^2) - (12*a*(a + b*Sqrt[c + d*x])^(7/2))/(7*b^4*d^2) + (4*(a + b*Sqrt[c + d*x])^(9/2))/(9*b^4*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0958616, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {371, 1398, 772} \[ \frac{4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^4 d^2}-\frac{4 a \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^4 d^2}+\frac{4 \left (a+b \sqrt{c+d x}\right )^{9/2}}{9 b^4 d^2}-\frac{12 a \left (a+b \sqrt{c+d x}\right )^{7/2}}{7 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

(-4*a*(a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(3/2))/(3*b^4*d^2) + (4*(3*a^2 - b^2*c)*(a + b*Sqrt[c + d*x])^(5/2))
/(5*b^4*d^2) - (12*a*(a + b*Sqrt[c + d*x])^(7/2))/(7*b^4*d^2) + (4*(a + b*Sqrt[c + d*x])^(9/2))/(9*b^4*d^2)

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int x \sqrt{a+b \sqrt{c+d x}} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{a+b \sqrt{x}} (-c+x) \, dx,x,c+d x\right )}{d^2}\\ &=\frac{2 \operatorname{Subst}\left (\int x \sqrt{a+b x} \left (-c+x^2\right ) \, dx,x,\sqrt{c+d x}\right )}{d^2}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{\left (-a^3+a b^2 c\right ) \sqrt{a+b x}}{b^3}+\frac{\left (3 a^2-b^2 c\right ) (a+b x)^{3/2}}{b^3}-\frac{3 a (a+b x)^{5/2}}{b^3}+\frac{(a+b x)^{7/2}}{b^3}\right ) \, dx,x,\sqrt{c+d x}\right )}{d^2}\\ &=-\frac{4 a \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^4 d^2}+\frac{4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^4 d^2}-\frac{12 a \left (a+b \sqrt{c+d x}\right )^{7/2}}{7 b^4 d^2}+\frac{4 \left (a+b \sqrt{c+d x}\right )^{9/2}}{9 b^4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0871912, size = 84, normalized size = 0.63 \[ \frac{4 \left (a+b \sqrt{c+d x}\right )^{3/2} \left (24 a^2 b \sqrt{c+d x}-16 a^3+6 a b^2 (2 c-5 d x)+7 b^3 \sqrt{c+d x} (5 d x-4 c)\right )}{315 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

(4*(a + b*Sqrt[c + d*x])^(3/2)*(-16*a^3 + 6*a*b^2*(2*c - 5*d*x) + 24*a^2*b*Sqrt[c + d*x] + 7*b^3*Sqrt[c + d*x]
*(-4*c + 5*d*x)))/(315*b^4*d^2)

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 94, normalized size = 0.7 \begin{align*} 4\,{\frac{1/9\, \left ( a+b\sqrt{dx+c} \right ) ^{9/2}-3/7\, \left ( a+b\sqrt{dx+c} \right ) ^{7/2}a+1/5\, \left ( -{b}^{2}c+3\,{a}^{2} \right ) \left ( a+b\sqrt{dx+c} \right ) ^{5/2}-1/3\, \left ( -{b}^{2}c+{a}^{2} \right ) a \left ( a+b\sqrt{dx+c} \right ) ^{3/2}}{{b}^{4}{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d^2/b^4*(1/9*(a+b*(d*x+c)^(1/2))^(9/2)-3/7*(a+b*(d*x+c)^(1/2))^(7/2)*a+1/5*(-b^2*c+3*a^2)*(a+b*(d*x+c)^(1/2)
)^(5/2)-1/3*(-b^2*c+a^2)*a*(a+b*(d*x+c)^(1/2))^(3/2))

________________________________________________________________________________________

Maxima [A]  time = 1.10106, size = 126, normalized size = 0.95 \begin{align*} \frac{4 \,{\left (35 \,{\left (\sqrt{d x + c} b + a\right )}^{\frac{9}{2}} - 135 \,{\left (\sqrt{d x + c} b + a\right )}^{\frac{7}{2}} a - 63 \,{\left (b^{2} c - 3 \, a^{2}\right )}{\left (\sqrt{d x + c} b + a\right )}^{\frac{5}{2}} + 105 \,{\left (a b^{2} c - a^{3}\right )}{\left (\sqrt{d x + c} b + a\right )}^{\frac{3}{2}}\right )}}{315 \, b^{4} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(35*(sqrt(d*x + c)*b + a)^(9/2) - 135*(sqrt(d*x + c)*b + a)^(7/2)*a - 63*(b^2*c - 3*a^2)*(sqrt(d*x + c)*
b + a)^(5/2) + 105*(a*b^2*c - a^3)*(sqrt(d*x + c)*b + a)^(3/2))/(b^4*d^2)

________________________________________________________________________________________

Fricas [A]  time = 2.39281, size = 240, normalized size = 1.8 \begin{align*} \frac{4 \,{\left (35 \, b^{4} d^{2} x^{2} - 28 \, b^{4} c^{2} + 36 \, a^{2} b^{2} c - 16 \, a^{4} +{\left (7 \, b^{4} c - 6 \, a^{2} b^{2}\right )} d x +{\left (5 \, a b^{3} d x - 16 \, a b^{3} c + 8 \, a^{3} b\right )} \sqrt{d x + c}\right )} \sqrt{\sqrt{d x + c} b + a}}{315 \, b^{4} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(35*b^4*d^2*x^2 - 28*b^4*c^2 + 36*a^2*b^2*c - 16*a^4 + (7*b^4*c - 6*a^2*b^2)*d*x + (5*a*b^3*d*x - 16*a*b
^3*c + 8*a^3*b)*sqrt(d*x + c))*sqrt(sqrt(d*x + c)*b + a)/(b^4*d^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{a + b \sqrt{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.16013, size = 460, normalized size = 3.46 \begin{align*} -\frac{4 \,{\left (63 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{2} b^{4} c \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - 105 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )} a b^{4} c \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - 35 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{4} b^{2} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) + 135 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{3} a b^{2} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - 189 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{2} a^{2} b^{2} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) + 105 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )} a^{3} b^{2} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right )\right )}{\left | b \right |}}{315 \, b^{8} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*(63*sqrt((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)^2*b^4*c*sgn((sqrt(d*x + c)*b + a)*b - a*b) -
105*sqrt((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)*a*b^4*c*sgn((sqrt(d*x + c)*b + a)*b - a*b) - 35*sqrt
((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)^4*b^2*sgn((sqrt(d*x + c)*b + a)*b - a*b) + 135*sqrt((sqrt(d*
x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)^3*a*b^2*sgn((sqrt(d*x + c)*b + a)*b - a*b) - 189*sqrt((sqrt(d*x + c)*
b + a)*b^2)*(sqrt(d*x + c)*b + a)^2*a^2*b^2*sgn((sqrt(d*x + c)*b + a)*b - a*b) + 105*sqrt((sqrt(d*x + c)*b + a
)*b^2)*(sqrt(d*x + c)*b + a)*a^3*b^2*sgn((sqrt(d*x + c)*b + a)*b - a*b))*abs(b)/(b^8*d^2)