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

Optimal. Leaf size=131 \[ \frac{4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^4 d^2}-\frac{4 a \left (a^2-b^2 c\right ) \sqrt{a+b \sqrt{c+d x}}}{b^4 d^2}+\frac{4 \left (a+b \sqrt{c+d x}\right )^{7/2}}{7 b^4 d^2}-\frac{12 a \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^4 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0939594, antiderivative size = 131, 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} \[ \frac{4 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^4 d^2}-\frac{4 a \left (a^2-b^2 c\right ) \sqrt{a+b \sqrt{c+d x}}}{b^4 d^2}+\frac{4 \left (a+b \sqrt{c+d x}\right )^{7/2}}{7 b^4 d^2}-\frac{12 a \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

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

Mathematica [A]  time = 0.100955, size = 84, normalized size = 0.64 \[ \frac{4 \sqrt{a+b \sqrt{c+d x}} \left (24 a^2 b \sqrt{c+d x}-48 a^3+2 a b^2 (26 c-9 d x)+5 b^3 \sqrt{c+d x} (3 d x-4 c)\right )}{105 b^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[a + b*Sqrt[c + d*x]]*(-48*a^3 + 2*a*b^2*(26*c - 9*d*x) + 24*a^2*b*Sqrt[c + d*x] + 5*b^3*Sqrt[c + d*x]*
(-4*c + 3*d*x)))/(105*b^4*d^2)

________________________________________________________________________________________

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

________________________________________________________________________________________

Maxima [A]  time = 1.10432, size = 126, normalized size = 0.96 \begin{align*} \frac{4 \,{\left (15 \,{\left (\sqrt{d x + c} b + a\right )}^{\frac{7}{2}} - 63 \,{\left (\sqrt{d x + c} b + a\right )}^{\frac{5}{2}} a - 35 \,{\left (b^{2} c - 3 \, a^{2}\right )}{\left (\sqrt{d x + c} b + a\right )}^{\frac{3}{2}} + 105 \,{\left (a b^{2} c - a^{3}\right )} \sqrt{\sqrt{d x + c} b + a}\right )}}{105 \, 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]

4/105*(15*(sqrt(d*x + c)*b + a)^(7/2) - 63*(sqrt(d*x + c)*b + a)^(5/2)*a - 35*(b^2*c - 3*a^2)*(sqrt(d*x + c)*b
+ a)^(3/2) + 105*(a*b^2*c - a^3)*sqrt(sqrt(d*x + c)*b + a))/(b^4*d^2)

________________________________________________________________________________________

Fricas [A]  time = 2.34615, size = 178, normalized size = 1.36 \begin{align*} -\frac{4 \,{\left (18 \, a b^{2} d x - 52 \, a b^{2} c + 48 \, a^{3} -{\left (15 \, b^{3} d x - 20 \, b^{3} c + 24 \, a^{2} b\right )} \sqrt{d x + c}\right )} \sqrt{\sqrt{d x + c} b + a}}{105 \, 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]

-4/105*(18*a*b^2*d*x - 52*a*b^2*c + 48*a^3 - (15*b^3*d*x - 20*b^3*c + 24*a^2*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 \frac{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 = 1.22184, size = 412, normalized size = 3.15 \begin{align*} -\frac{4 \,{\left (35 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )} b^{2} 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}} a b^{2} c \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - 15 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{3} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) + 63 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{2} a \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^{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}} a^{3} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right )\right )}}{105 \, b^{4} d^{2}{\left | b \right |}} \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]

-4/105*(35*sqrt((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)*b^2*c*sgn((sqrt(d*x + c)*b + a)*b - a*b) - 10
5*sqrt((sqrt(d*x + c)*b + a)*b^2)*a*b^2*c*sgn((sqrt(d*x + c)*b + a)*b - a*b) - 15*sqrt((sqrt(d*x + c)*b + a)*b
^2)*(sqrt(d*x + c)*b + a)^3*sgn((sqrt(d*x + c)*b + a)*b - a*b) + 63*sqrt((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x
+ c)*b + a)^2*a*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^2*sgn((sqrt(d*x + c)*b + a)*b - a*b) + 105*sqrt((sqrt(d*x + c)*b + a)*b^2)*a^3*sgn((sqrt(d*x + c)*b + a)*b
- a*b))/(b^4*d^2*abs(b))