3.440 \(\int \frac{x^2}{(\sqrt{a+b x}+\sqrt{a+c x})^3} \, dx\)

Optimal. Leaf size=155 \[ -\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{(b-c)^3}+\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{(b-c)^3}+\frac{8 a \sqrt{a+b x}}{(b-c)^3}-\frac{8 a \sqrt{a+c x}}{(b-c)^3}+\frac{2 (b+3 c) (a+b x)^{3/2}}{3 b (b-c)^3}-\frac{2 (3 b+c) (a+c x)^{3/2}}{3 c (b-c)^3} \]

[Out]

(8*a*Sqrt[a + b*x])/(b - c)^3 + (2*(b + 3*c)*(a + b*x)^(3/2))/(3*b*(b - c)^3) - (8*a*Sqrt[a + c*x])/(b - c)^3
- (2*(3*b + c)*(a + c*x)^(3/2))/(3*(b - c)^3*c) - (8*a^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(b - c)^3 + (8*a^
(3/2)*ArcTanh[Sqrt[a + c*x]/Sqrt[a]])/(b - c)^3

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Rubi [A]  time = 0.202475, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6690, 50, 63, 208} \[ -\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{(b-c)^3}+\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{(b-c)^3}+\frac{8 a \sqrt{a+b x}}{(b-c)^3}-\frac{8 a \sqrt{a+c x}}{(b-c)^3}+\frac{2 (b+3 c) (a+b x)^{3/2}}{3 b (b-c)^3}-\frac{2 (3 b+c) (a+c x)^{3/2}}{3 c (b-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*Sqrt[a + b*x])/(b - c)^3 + (2*(b + 3*c)*(a + b*x)^(3/2))/(3*b*(b - c)^3) - (8*a*Sqrt[a + c*x])/(b - c)^3
- (2*(3*b + c)*(a + c*x)^(3/2))/(3*(b - c)^3*c) - (8*a^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(b - c)^3 + (8*a^
(3/2)*ArcTanh[Sqrt[a + c*x]/Sqrt[a]])/(b - c)^3

Rule 6690

Int[(u_.)*((e_.)*Sqrt[(a_.) + (b_.)*(x_)^(n_.)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)^(n_.)])^(m_), x_Symbol] :> Dis
t[(b*e^2 - d*f^2)^m, Int[ExpandIntegrand[(u*x^(m*n))/(e*Sqrt[a + b*x^n] - f*Sqrt[c + d*x^n])^m, x], x], x] /;
FreeQ[{a, b, c, d, e, f, n}, x] && ILtQ[m, 0] && EqQ[a*e^2 - c*f^2, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^2}{\left (\sqrt{a+b x}+\sqrt{a+c x}\right )^3} \, dx &=\frac{\int \left (b \left (1+\frac{3 c}{b}\right ) \sqrt{a+b x}+\frac{4 a \sqrt{a+b x}}{x}-3 b \left (1+\frac{c}{3 b}\right ) \sqrt{a+c x}-\frac{4 a \sqrt{a+c x}}{x}\right ) \, dx}{(b-c)^3}\\ &=\frac{2 (b+3 c) (a+b x)^{3/2}}{3 b (b-c)^3}-\frac{2 (3 b+c) (a+c x)^{3/2}}{3 (b-c)^3 c}+\frac{(4 a) \int \frac{\sqrt{a+b x}}{x} \, dx}{(b-c)^3}-\frac{(4 a) \int \frac{\sqrt{a+c x}}{x} \, dx}{(b-c)^3}\\ &=\frac{8 a \sqrt{a+b x}}{(b-c)^3}+\frac{2 (b+3 c) (a+b x)^{3/2}}{3 b (b-c)^3}-\frac{8 a \sqrt{a+c x}}{(b-c)^3}-\frac{2 (3 b+c) (a+c x)^{3/2}}{3 (b-c)^3 c}+\frac{\left (4 a^2\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{(b-c)^3}-\frac{\left (4 a^2\right ) \int \frac{1}{x \sqrt{a+c x}} \, dx}{(b-c)^3}\\ &=\frac{8 a \sqrt{a+b x}}{(b-c)^3}+\frac{2 (b+3 c) (a+b x)^{3/2}}{3 b (b-c)^3}-\frac{8 a \sqrt{a+c x}}{(b-c)^3}-\frac{2 (3 b+c) (a+c x)^{3/2}}{3 (b-c)^3 c}+\frac{\left (8 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{b (b-c)^3}-\frac{\left (8 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x}\right )}{(b-c)^3 c}\\ &=\frac{8 a \sqrt{a+b x}}{(b-c)^3}+\frac{2 (b+3 c) (a+b x)^{3/2}}{3 b (b-c)^3}-\frac{8 a \sqrt{a+c x}}{(b-c)^3}-\frac{2 (3 b+c) (a+c x)^{3/2}}{3 (b-c)^3 c}-\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{(b-c)^3}+\frac{8 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{(b-c)^3}\\ \end{align*}

Mathematica [A]  time = 0.270552, size = 119, normalized size = 0.77 \[ \frac{2 \left (-12 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+12 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )+\frac{(b+3 c) (a+b x)^{3/2}}{b}-\frac{(3 b+c) (a+c x)^{3/2}}{c}+12 a \sqrt{a+b x}-12 a \sqrt{a+c x}\right )}{3 (b-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(12*a*Sqrt[a + b*x] + ((b + 3*c)*(a + b*x)^(3/2))/b - 12*a*Sqrt[a + c*x] - ((3*b + c)*(a + c*x)^(3/2))/c -
12*a^(3/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]] + 12*a^(3/2)*ArcTanh[Sqrt[a + c*x]/Sqrt[a]]))/(3*(b - c)^3)

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Maple [A]  time = 0.004, size = 148, normalized size = 1. \begin{align*}{\frac{2}{3\, \left ( b-c \right ) ^{3}} \left ( bx+a \right ) ^{{\frac{3}{2}}}}+2\,{\frac{c \left ( bx+a \right ) ^{3/2}}{ \left ( b-c \right ) ^{3}b}}-2\,{\frac{b \left ( cx+a \right ) ^{3/2}}{ \left ( b-c \right ) ^{3}c}}-{\frac{2}{3\, \left ( b-c \right ) ^{3}} \left ( cx+a \right ) ^{{\frac{3}{2}}}}+4\,{\frac{a}{ \left ( b-c \right ) ^{3}} \left ( 2\,\sqrt{bx+a}-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) }-4\,{\frac{a}{ \left ( b-c \right ) ^{3}} \left ( 2\,\sqrt{cx+a}-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{cx+a}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/(b-c)^3*(b*x+a)^(3/2)+2/(b-c)^3*c*(b*x+a)^(3/2)/b-2/(b-c)^3*b*(c*x+a)^(3/2)/c-2/3/(b-c)^3*(c*x+a)^(3/2)+4/
(b-c)^3*a*(2*(b*x+a)^(1/2)-2*a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-4/(b-c)^3*a*(2*(c*x+a)^(1/2)-2*a^(1/2)*ar
ctanh((c*x+a)^(1/2)/a^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (\sqrt{b x + a} + \sqrt{c x + a}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.31923, size = 755, normalized size = 4.87 \begin{align*} \left [-\frac{2 \,{\left (6 \, a^{\frac{3}{2}} b c \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 6 \, a^{\frac{3}{2}} b c \log \left (\frac{c x - 2 \, \sqrt{c x + a} \sqrt{a} + 2 \, a}{x}\right ) -{\left (13 \, a b c + 3 \, a c^{2} +{\left (b^{2} c + 3 \, b c^{2}\right )} x\right )} \sqrt{b x + a} +{\left (3 \, a b^{2} + 13 \, a b c +{\left (3 \, b^{2} c + b c^{2}\right )} x\right )} \sqrt{c x + a}\right )}}{3 \,{\left (b^{4} c - 3 \, b^{3} c^{2} + 3 \, b^{2} c^{3} - b c^{4}\right )}}, \frac{2 \,{\left (12 \, \sqrt{-a} a b c \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - 12 \, \sqrt{-a} a b c \arctan \left (\frac{\sqrt{c x + a} \sqrt{-a}}{a}\right ) +{\left (13 \, a b c + 3 \, a c^{2} +{\left (b^{2} c + 3 \, b c^{2}\right )} x\right )} \sqrt{b x + a} -{\left (3 \, a b^{2} + 13 \, a b c +{\left (3 \, b^{2} c + b c^{2}\right )} x\right )} \sqrt{c x + a}\right )}}{3 \,{\left (b^{4} c - 3 \, b^{3} c^{2} + 3 \, b^{2} c^{3} - b c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-2/3*(6*a^(3/2)*b*c*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 6*a^(3/2)*b*c*log((c*x - 2*sqrt(c*x + a)*s
qrt(a) + 2*a)/x) - (13*a*b*c + 3*a*c^2 + (b^2*c + 3*b*c^2)*x)*sqrt(b*x + a) + (3*a*b^2 + 13*a*b*c + (3*b^2*c +
b*c^2)*x)*sqrt(c*x + a))/(b^4*c - 3*b^3*c^2 + 3*b^2*c^3 - b*c^4), 2/3*(12*sqrt(-a)*a*b*c*arctan(sqrt(b*x + a)
*sqrt(-a)/a) - 12*sqrt(-a)*a*b*c*arctan(sqrt(c*x + a)*sqrt(-a)/a) + (13*a*b*c + 3*a*c^2 + (b^2*c + 3*b*c^2)*x)
*sqrt(b*x + a) - (3*a*b^2 + 13*a*b*c + (3*b^2*c + b*c^2)*x)*sqrt(c*x + a))/(b^4*c - 3*b^3*c^2 + 3*b^2*c^3 - b*
c^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\left (\sqrt{a + b x} + \sqrt{a + c x}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out