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

Optimal. Leaf size=228 \[ -\frac{x (a+b x)^{3/2} (b x+c)^{3/2}}{2 b^2 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (b x+c)^{3/2}}{12 b^3 (a-c)^2}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{b x+c}}{16 b^3 (a-c)^2}-\frac{\left (4 a c-5 (a+c)^2\right ) \sqrt{a+b x} \sqrt{b x+c}}{32 b^3 (a-c)}-\frac{\left (4 a c-5 (a+c)^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{b x+c}}\right )}{32 b^3}+\frac{b x^4}{2 (a-c)^2}+\frac{x^3 (a+c)}{3 (a-c)^2} \]

[Out]

((a + c)*x^3)/(3*(a - c)^2) + (b*x^4)/(2*(a - c)^2) - ((4*a*c - 5*(a + c)^2)*Sqrt[a + b*x]*Sqrt[c + b*x])/(32*
b^3*(a - c)) + ((4*a*c - 5*(a + c)^2)*(a + b*x)^(3/2)*Sqrt[c + b*x])/(16*b^3*(a - c)^2) + (5*(a + c)*(a + b*x)
^(3/2)*(c + b*x)^(3/2))/(12*b^3*(a - c)^2) - (x*(a + b*x)^(3/2)*(c + b*x)^(3/2))/(2*b^2*(a - c)^2) - ((4*a*c -
5*(a + c)^2)*ArcTanh[Sqrt[a + b*x]/Sqrt[c + b*x]])/(32*b^3)

________________________________________________________________________________________

Rubi [A]  time = 0.373396, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {6689, 90, 80, 50, 63, 217, 206} \[ -\frac{x (a+b x)^{3/2} (b x+c)^{3/2}}{2 b^2 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (b x+c)^{3/2}}{12 b^3 (a-c)^2}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{b x+c}}{16 b^3 (a-c)^2}-\frac{\left (4 a c-5 (a+c)^2\right ) \sqrt{a+b x} \sqrt{b x+c}}{32 b^3 (a-c)}-\frac{\left (4 a c-5 (a+c)^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{b x+c}}\right )}{32 b^3}+\frac{b x^4}{2 (a-c)^2}+\frac{x^3 (a+c)}{3 (a-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + c)*x^3)/(3*(a - c)^2) + (b*x^4)/(2*(a - c)^2) - ((4*a*c - 5*(a + c)^2)*Sqrt[a + b*x]*Sqrt[c + b*x])/(32*
b^3*(a - c)) + ((4*a*c - 5*(a + c)^2)*(a + b*x)^(3/2)*Sqrt[c + b*x])/(16*b^3*(a - c)^2) + (5*(a + c)*(a + b*x)
^(3/2)*(c + b*x)^(3/2))/(12*b^3*(a - c)^2) - (x*(a + b*x)^(3/2)*(c + b*x)^(3/2))/(2*b^2*(a - c)^2) - ((4*a*c -
5*(a + c)^2)*ArcTanh[Sqrt[a + b*x]/Sqrt[c + b*x]])/(32*b^3)

Rule 6689

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

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 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{c+b x}\right )^2} \, dx &=\frac{\int \left (a \left (1+\frac{c}{a}\right ) x^2+2 b x^3-2 x^2 \sqrt{a+b x} \sqrt{c+b x}\right ) \, dx}{(a-c)^2}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}-\frac{2 \int x^2 \sqrt{a+b x} \sqrt{c+b x} \, dx}{(a-c)^2}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}-\frac{\int \sqrt{a+b x} \sqrt{c+b x} \left (-a c-\frac{5}{2} b (a+c) x\right ) \, dx}{2 b^2 (a-c)^2}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (c+b x)^{3/2}}{12 b^3 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}+\frac{\left (4 a c-5 (a+c)^2\right ) \int \sqrt{a+b x} \sqrt{c+b x} \, dx}{8 b^2 (a-c)^2}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{c+b x}}{16 b^3 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (c+b x)^{3/2}}{12 b^3 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+b x}} \, dx}{32 b^2 (a-c)}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \sqrt{a+b x} \sqrt{c+b x}}{32 b^3 (a-c)}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{c+b x}}{16 b^3 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (c+b x)^{3/2}}{12 b^3 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+b x}} \, dx}{64 b^2}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \sqrt{a+b x} \sqrt{c+b x}}{32 b^3 (a-c)}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{c+b x}}{16 b^3 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (c+b x)^{3/2}}{12 b^3 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a+c+x^2}} \, dx,x,\sqrt{a+b x}\right )}{32 b^3}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \sqrt{a+b x} \sqrt{c+b x}}{32 b^3 (a-c)}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{c+b x}}{16 b^3 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (c+b x)^{3/2}}{12 b^3 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+b x}}\right )}{32 b^3}\\ &=\frac{(a+c) x^3}{3 (a-c)^2}+\frac{b x^4}{2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \sqrt{a+b x} \sqrt{c+b x}}{32 b^3 (a-c)}+\frac{\left (4 a c-5 (a+c)^2\right ) (a+b x)^{3/2} \sqrt{c+b x}}{16 b^3 (a-c)^2}+\frac{5 (a+c) (a+b x)^{3/2} (c+b x)^{3/2}}{12 b^3 (a-c)^2}-\frac{x (a+b x)^{3/2} (c+b x)^{3/2}}{2 b^2 (a-c)^2}+\frac{\left (5 a^2+6 a c+5 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{c+b x}}\right )}{32 b^3}\\ \end{align*}

Mathematica [A]  time = 1.36713, size = 361, normalized size = 1.58 \[ \frac{\frac{3 \sqrt{b} (c-a)^3 \left (5 a^2+6 a c+5 c^2\right ) \sqrt{-\frac{b x+c}{a-c}} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{a+b x}}{\sqrt{b (c-a)}}\right )}{\sqrt{b (c-a)} \sqrt{b x+c}}+a^2 \sqrt{a+b x} \sqrt{b x+c} (10 b x+7 c)-15 a^3 \sqrt{a+b x} \sqrt{b x+c}-a \left (8 b^2 x^2 \sqrt{a+b x} \sqrt{b x+c}-7 c^2 \sqrt{a+b x} \sqrt{b x+c}+4 b c x \sqrt{a+b x} \sqrt{b x+c}-32 b^3 x^3\right )-16 b^3 x^3 \left (3 \sqrt{a+b x} \sqrt{b x+c}-2 c\right )-8 b^2 c x^2 \sqrt{a+b x} \sqrt{b x+c}+10 b c^2 x \sqrt{a+b x} \sqrt{b x+c}-15 c^3 \sqrt{a+b x} \sqrt{b x+c}+48 b^4 x^4}{96 b^3 (a-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*b^4*x^4 - 15*a^3*Sqrt[a + b*x]*Sqrt[c + b*x] - 15*c^3*Sqrt[a + b*x]*Sqrt[c + b*x] + 10*b*c^2*x*Sqrt[a + b*
x]*Sqrt[c + b*x] - 8*b^2*c*x^2*Sqrt[a + b*x]*Sqrt[c + b*x] + a^2*Sqrt[a + b*x]*Sqrt[c + b*x]*(7*c + 10*b*x) -
16*b^3*x^3*(-2*c + 3*Sqrt[a + b*x]*Sqrt[c + b*x]) - a*(-32*b^3*x^3 - 7*c^2*Sqrt[a + b*x]*Sqrt[c + b*x] + 4*b*c
*x*Sqrt[a + b*x]*Sqrt[c + b*x] + 8*b^2*x^2*Sqrt[a + b*x]*Sqrt[c + b*x]) + (3*Sqrt[b]*(-a + c)^3*(5*a^2 + 6*a*c
+ 5*c^2)*Sqrt[-((c + b*x)/(a - c))]*ArcSinh[(Sqrt[b]*Sqrt[a + b*x])/Sqrt[b*(-a + c)]])/(Sqrt[b*(-a + c)]*Sqrt
[c + b*x]))/(96*b^3*(a - c)^2)

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Maple [C]  time = 0.019, size = 604, normalized size = 2.7 \begin{align*}{\frac{a{x}^{3}}{3\, \left ( a-c \right ) ^{2}}}+{\frac{c{x}^{3}}{3\, \left ( a-c \right ) ^{2}}}+{\frac{b{x}^{4}}{2\, \left ( a-c \right ) ^{2}}}-{\frac{{\it csgn} \left ( b \right ) }{192\, \left ( a-c \right ) ^{2}{b}^{3}}\sqrt{bx+a}\sqrt{bx+c} \left ( 96\,{\it csgn} \left ( b \right ){x}^{3}{b}^{3}\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+16\,{\it csgn} \left ( b \right ){x}^{2}a{b}^{2}\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+16\,{\it csgn} \left ( b \right ){x}^{2}{b}^{2}c\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}-20\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}x{a}^{2}b+8\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}xabc-20\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}xb{c}^{2}+30\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}{a}^{3}-14\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}{a}^{2}c-14\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}a{c}^{2}+30\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}{c}^{3}-15\,\ln \left ( 1/2\, \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ){\it csgn} \left ( b \right ) \right ){a}^{4}+12\,\ln \left ( 1/2\, \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ){\it csgn} \left ( b \right ) \right ){a}^{3}c+6\,\ln \left ( 1/2\, \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ){\it csgn} \left ( b \right ) \right ){a}^{2}{c}^{2}+12\,\ln \left ( 1/2\, \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ){\it csgn} \left ( b \right ) \right ) a{c}^{3}-15\,\ln \left ( 1/2\, \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ){\it csgn} \left ( b \right ) \right ){c}^{4} \right ){\frac{1}{\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3/(a-c)^2*a+1/3*x^3/(a-c)^2*c+1/2*b*x^4/(a-c)^2-1/192/(a-c)^2*(b*x+a)^(1/2)*(b*x+c)^(1/2)*(96*csgn(b)*x^
3*b^3*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+16*csgn(b)*x^2*a*b^2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+16*csgn(b)*x^2*b^2*
c*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)-20*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*x*a^2*b+8*csgn(b)*(b^2*x^2+a*b*x+
b*c*x+a*c)^(1/2)*x*a*b*c-20*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*x*b*c^2+30*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*
c)^(1/2)*a^3-14*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*a^2*c-14*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*a*c^2
+30*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*c^3-15*ln(1/2*(2*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+2*b*x+a+c
)*csgn(b))*a^4+12*ln(1/2*(2*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+2*b*x+a+c)*csgn(b))*a^3*c+6*ln(1/2*(2*csgn
(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+2*b*x+a+c)*csgn(b))*a^2*c^2+12*ln(1/2*(2*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)
^(1/2)+2*b*x+a+c)*csgn(b))*a*c^3-15*ln(1/2*(2*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+2*b*x+a+c)*csgn(b))*c^4)
*csgn(b)/b^3/(b^2*x^2+a*b*x+b*c*x+a*c)^(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{b x + c}\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)+(b*x+c)^(1/2))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.01601, size = 437, normalized size = 1.92 \begin{align*} \frac{96 \, b^{4} x^{4} + 64 \,{\left (a b^{3} + b^{3} c\right )} x^{3} - 2 \,{\left (48 \, b^{3} x^{3} + 15 \, a^{3} - 7 \, a^{2} c - 7 \, a c^{2} + 15 \, c^{3} + 8 \,{\left (a b^{2} + b^{2} c\right )} x^{2} - 2 \,{\left (5 \, a^{2} b - 2 \, a b c + 5 \, b c^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{b x + c} - 3 \,{\left (5 \, a^{4} - 4 \, a^{3} c - 2 \, a^{2} c^{2} - 4 \, a c^{3} + 5 \, c^{4}\right )} \log \left (-2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b x + c} - a - c\right )}{192 \,{\left (a^{2} b^{3} - 2 \, a b^{3} c + b^{3} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(96*b^4*x^4 + 64*(a*b^3 + b^3*c)*x^3 - 2*(48*b^3*x^3 + 15*a^3 - 7*a^2*c - 7*a*c^2 + 15*c^3 + 8*(a*b^2 +
b^2*c)*x^2 - 2*(5*a^2*b - 2*a*b*c + 5*b*c^2)*x)*sqrt(b*x + a)*sqrt(b*x + c) - 3*(5*a^4 - 4*a^3*c - 2*a^2*c^2 -
4*a*c^3 + 5*c^4)*log(-2*b*x + 2*sqrt(b*x + a)*sqrt(b*x + c) - a - c))/(a^2*b^3 - 2*a*b^3*c + b^3*c^2)

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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{b x + c}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out