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

Optimal. Leaf size=142 \[ \frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{2 b^{3/2} c^{3/2}}+\frac{2 a x}{(b-c)^2}-\frac{a \sqrt{a+b x} \sqrt{a+c x}}{2 b c (b-c)}-\frac{(a+b x)^{3/2} \sqrt{a+c x}}{b (b-c)^2}+\frac{x^2 (b+c)}{2 (b-c)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.227759, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6690, 50, 63, 217, 206} \[ \frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{2 b^{3/2} c^{3/2}}+\frac{2 a x}{(b-c)^2}-\frac{a \sqrt{a+b x} \sqrt{a+c x}}{2 b c (b-c)}-\frac{(a+b x)^{3/2} \sqrt{a+c x}}{b (b-c)^2}+\frac{x^2 (b+c)}{2 (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.579419, size = 177, normalized size = 1.25 \[ \frac{b \sqrt{c} \left (b c x \left (-2 \sqrt{a+b x} \sqrt{a+c x}+b x+c x\right )-a \left (b \sqrt{a+b x} \sqrt{a+c x}+c \sqrt{a+b x} \sqrt{a+c x}-4 b c x\right )\right )+\frac{(a (b-c))^{5/2} \sqrt{\frac{b (a+c x)}{a (b-c)}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a (b-c)}}\right )}{\sqrt{a+c x}}}{2 b^2 c^{3/2} (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.008, size = 385, normalized size = 2.7 \begin{align*}{\frac{b{x}^{2}}{2\, \left ( b-c \right ) ^{2}}}+{\frac{c{x}^{2}}{2\, \left ( b-c \right ) ^{2}}}+2\,{\frac{ax}{ \left ( b-c \right ) ^{2}}}-{\frac{1}{ \left ( b-c \right ) ^{2}c}\sqrt{bx+a} \left ( cx+a \right ) ^{{\frac{3}{2}}}}+{\frac{a}{2\, \left ( b-c \right ) ^{2}c}\sqrt{cx+a}\sqrt{bx+a}}-{\frac{a}{2\, \left ( b-c \right ) ^{2}b}\sqrt{cx+a}\sqrt{bx+a}}+{\frac{{a}^{2}b}{4\, \left ( b-c \right ) ^{2}c}\sqrt{ \left ( cx+a \right ) \left ( bx+a \right ) }\ln \left ({ \left ({\frac{ab}{2}}+{\frac{ac}{2}}+bcx \right ){\frac{1}{\sqrt{bc}}}}+\sqrt{bc{x}^{2}+ \left ( ab+ac \right ) x+{a}^{2}} \right ){\frac{1}{\sqrt{cx+a}}}{\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{bc}}}}-{\frac{{a}^{2}}{2\, \left ( b-c \right ) ^{2}}\sqrt{ \left ( cx+a \right ) \left ( bx+a \right ) }\ln \left ({ \left ({\frac{ab}{2}}+{\frac{ac}{2}}+bcx \right ){\frac{1}{\sqrt{bc}}}}+\sqrt{bc{x}^{2}+ \left ( ab+ac \right ) x+{a}^{2}} \right ){\frac{1}{\sqrt{cx+a}}}{\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{bc}}}}+{\frac{{a}^{2}c}{4\, \left ( b-c \right ) ^{2}b}\sqrt{ \left ( cx+a \right ) \left ( bx+a \right ) }\ln \left ({ \left ({\frac{ab}{2}}+{\frac{ac}{2}}+bcx \right ){\frac{1}{\sqrt{bc}}}}+\sqrt{bc{x}^{2}+ \left ( ab+ac \right ) x+{a}^{2}} \right ){\frac{1}{\sqrt{cx+a}}}{\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{bc}}}} \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))^2,x)

[Out]

1/2*x^2/(b-c)^2*b+1/2*x^2/(b-c)^2*c+2*a*x/(b-c)^2-1/(b-c)^2/c*(b*x+a)^(1/2)*(c*x+a)^(3/2)+1/2/(b-c)^2/c*(c*x+a
)^(1/2)*(b*x+a)^(1/2)*a-1/2/(b-c)^2/b*(c*x+a)^(1/2)*(b*x+a)^(1/2)*a+1/4/(b-c)^2/c*((c*x+a)*(b*x+a))^(1/2)/(c*x
+a)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*b+1/2*a*c+b*c*x)/(b*c)^(1/2)+(b*c*x^2+(a*b+a*c)*x+a^2)^(1/2))/(b*c)^(1/2)*a^
2*b-1/2/(b-c)^2*((c*x+a)*(b*x+a))^(1/2)/(c*x+a)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*b+1/2*a*c+b*c*x)/(b*c)^(1/2)+(b*
c*x^2+(a*b+a*c)*x+a^2)^(1/2))/(b*c)^(1/2)*a^2+1/4/(b-c)^2*c/b*((c*x+a)*(b*x+a))^(1/2)/(c*x+a)^(1/2)/(b*x+a)^(1
/2)*ln((1/2*a*b+1/2*a*c+b*c*x)/(b*c)^(1/2)+(b*c*x^2+(a*b+a*c)*x+a^2)^(1/2))/(b*c)^(1/2)*a^2

________________________________________________________________________________________

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 )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.36518, size = 813, normalized size = 5.73 \begin{align*} \left [\frac{8 \, a b^{2} c^{2} x + 2 \,{\left (b^{3} c^{2} + b^{2} c^{3}\right )} x^{2} +{\left (a^{2} b^{2} - 2 \, a^{2} b c + a^{2} c^{2}\right )} \sqrt{b c} \log \left (a b^{2} + 2 \, a b c + a c^{2} + 2 \,{\left (2 \, b c + \sqrt{b c}{\left (b + c\right )}\right )} \sqrt{b x + a} \sqrt{c x + a} + 2 \,{\left (b^{2} c + b c^{2}\right )} x + 2 \,{\left (2 \, b c x + a b + a c\right )} \sqrt{b c}\right ) - 2 \,{\left (2 \, b^{2} c^{2} x + a b^{2} c + a b c^{2}\right )} \sqrt{b x + a} \sqrt{c x + a}}{4 \,{\left (b^{4} c^{2} - 2 \, b^{3} c^{3} + b^{2} c^{4}\right )}}, \frac{4 \, a b^{2} c^{2} x +{\left (b^{3} c^{2} + b^{2} c^{3}\right )} x^{2} -{\left (a^{2} b^{2} - 2 \, a^{2} b c + a^{2} c^{2}\right )} \sqrt{-b c} \arctan \left (\frac{\sqrt{-b c} \sqrt{b x + a} \sqrt{c x + a} - \sqrt{-b c} a}{b c x}\right ) -{\left (2 \, b^{2} c^{2} x + a b^{2} c + a b c^{2}\right )} \sqrt{b x + a} \sqrt{c x + a}}{2 \,{\left (b^{4} c^{2} - 2 \, b^{3} c^{3} + b^{2} 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))^2,x, algorithm="fricas")

[Out]

[1/4*(8*a*b^2*c^2*x + 2*(b^3*c^2 + b^2*c^3)*x^2 + (a^2*b^2 - 2*a^2*b*c + a^2*c^2)*sqrt(b*c)*log(a*b^2 + 2*a*b*
c + a*c^2 + 2*(2*b*c + sqrt(b*c)*(b + c))*sqrt(b*x + a)*sqrt(c*x + a) + 2*(b^2*c + b*c^2)*x + 2*(2*b*c*x + a*b
+ a*c)*sqrt(b*c)) - 2*(2*b^2*c^2*x + a*b^2*c + a*b*c^2)*sqrt(b*x + a)*sqrt(c*x + a))/(b^4*c^2 - 2*b^3*c^3 + b
^2*c^4), 1/2*(4*a*b^2*c^2*x + (b^3*c^2 + b^2*c^3)*x^2 - (a^2*b^2 - 2*a^2*b*c + a^2*c^2)*sqrt(-b*c)*arctan((sqr
t(-b*c)*sqrt(b*x + a)*sqrt(c*x + a) - sqrt(-b*c)*a)/(b*c*x)) - (2*b^2*c^2*x + a*b^2*c + a*b*c^2)*sqrt(b*x + a)
*sqrt(c*x + a))/(b^4*c^2 - 2*b^3*c^3 + b^2*c^4)]

________________________________________________________________________________________

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 )^{2}}\, 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))**2,x)

[Out]

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

________________________________________________________________________________________

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))^2,x, algorithm="giac")

[Out]

Timed out