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

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

[Out]

1/3*(a+c)*x^3/(a-c)^2+1/2*b*x^4/(a-c)^2+5/12*(a+c)*(b*x+a)^(3/2)*(b*x+c)^(3/2)/b^3/(a-c)^2-1/2*x*(b*x+a)^(3/2)
*(b*x+c)^(3/2)/b^2/(a-c)^2-1/32*(4*a*c-5*(a+c)^2)*arctanh((b*x+a)^(1/2)/(b*x+c)^(1/2))/b^3+1/16*(4*a*c-5*(a+c)
^2)*(b*x+a)^(3/2)*(b*x+c)^(1/2)/b^3/(a-c)^2-1/32*(4*a*c-5*(a+c)^2)*(b*x+a)^(1/2)*(b*x+c)^(1/2)/b^3/(a-c)

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Rubi [A]  time = 0.37, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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 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 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 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 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])

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 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]

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*}

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Mathematica [A]  time = 1.29, size = 361, normalized size = 1.58 \[ \frac {-15 a^3 \sqrt {a+b x} \sqrt {b x+c}+\frac {3 \sqrt {b} (c-a)^3 \left (5 a^2+6 a c+5 c^2\right ) \sqrt {-\frac {b x+c}{a-c}} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {a+b x}}{\sqrt {b (c-a)}}\right )}{\sqrt {b (c-a)} \sqrt {b x+c}}+a^2 \sqrt {a+b x} \sqrt {b x+c} (10 b x+7 c)-16 b^3 x^3 \left (3 \sqrt {a+b x} \sqrt {b x+c}-2 c\right )-8 b^2 c x^2 \sqrt {a+b x} \sqrt {b x+c}-a \left (8 b^2 x^2 \sqrt {a+b x} \sqrt {b x+c}-7 c^2 \sqrt {a+b x} \sqrt {b x+c}+4 b c x \sqrt {a+b x} \sqrt {b x+c}-32 b^3 x^3\right )-15 c^3 \sqrt {a+b x} \sqrt {b x+c}+10 b c^2 x \sqrt {a+b x} \sqrt {b x+c}+48 b^4 x^4}{96 b^3 (a-c)^2} \]

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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fricas [A]  time = 0.46, size = 196, normalized size = 0.86 \[ \frac {96 \, b^{4} x^{4} + 64 \, {\left (a b^{3} + b^{3} c\right )} x^{3} - 2 \, {\left (48 \, b^{3} x^{3} + 15 \, a^{3} - 7 \, a^{2} c - 7 \, a c^{2} + 15 \, c^{3} + 8 \, {\left (a b^{2} + b^{2} c\right )} x^{2} - 2 \, {\left (5 \, a^{2} b - 2 \, a b c + 5 \, b c^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {b x + c} - 3 \, {\left (5 \, a^{4} - 4 \, a^{3} c - 2 \, a^{2} c^{2} - 4 \, a c^{3} + 5 \, c^{4}\right )} \log \left (-2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b x + c} - a - c\right )}{192 \, {\left (a^{2} b^{3} - 2 \, a b^{3} c + b^{3} c^{2}\right )}} \]

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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giac [B]  time = 0.39, size = 797, normalized size = 3.50 \[ -\frac {1}{96} \, {\left (2 \, {\left (4 \, {\left (b x + a\right )} {\left (\frac {6 \, {\left (a^{5} b^{9} - 5 \, a^{4} b^{9} c + 10 \, a^{3} b^{9} c^{2} - 10 \, a^{2} b^{9} c^{3} + 5 \, a b^{9} c^{4} - b^{9} c^{5}\right )} {\left (b x + a\right )}}{a^{7} b^{12} - 7 \, a^{6} b^{12} c + 21 \, a^{5} b^{12} c^{2} - 35 \, a^{4} b^{12} c^{3} + 35 \, a^{3} b^{12} c^{4} - 21 \, a^{2} b^{12} c^{5} + 7 \, a b^{12} c^{6} - b^{12} c^{7}} - \frac {17 \, a^{6} b^{9} - 86 \, a^{5} b^{9} c + 175 \, a^{4} b^{9} c^{2} - 180 \, a^{3} b^{9} c^{3} + 95 \, a^{2} b^{9} c^{4} - 22 \, a b^{9} c^{5} + b^{9} c^{6}}{a^{7} b^{12} - 7 \, a^{6} b^{12} c + 21 \, a^{5} b^{12} c^{2} - 35 \, a^{4} b^{12} c^{3} + 35 \, a^{3} b^{12} c^{4} - 21 \, a^{2} b^{12} c^{5} + 7 \, a b^{12} c^{6} - b^{12} c^{7}}\right )} + \frac {59 \, a^{7} b^{9} - 301 \, a^{6} b^{9} c + 615 \, a^{5} b^{9} c^{2} - 625 \, a^{4} b^{9} c^{3} + 305 \, a^{3} b^{9} c^{4} - 39 \, a^{2} b^{9} c^{5} - 19 \, a b^{9} c^{6} + 5 \, b^{9} c^{7}}{a^{7} b^{12} - 7 \, a^{6} b^{12} c + 21 \, a^{5} b^{12} c^{2} - 35 \, a^{4} b^{12} c^{3} + 35 \, a^{3} b^{12} c^{4} - 21 \, a^{2} b^{12} c^{5} + 7 \, a b^{12} c^{6} - b^{12} c^{7}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (5 \, a^{8} b^{9} - 24 \, a^{7} b^{9} c + 44 \, a^{6} b^{9} c^{2} - 40 \, a^{5} b^{9} c^{3} + 30 \, a^{4} b^{9} c^{4} - 40 \, a^{3} b^{9} c^{5} + 44 \, a^{2} b^{9} c^{6} - 24 \, a b^{9} c^{7} + 5 \, b^{9} c^{8}\right )}}{a^{7} b^{12} - 7 \, a^{6} b^{12} c + 21 \, a^{5} b^{12} c^{2} - 35 \, a^{4} b^{12} c^{3} + 35 \, a^{3} b^{12} c^{4} - 21 \, a^{2} b^{12} c^{5} + 7 \, a b^{12} c^{6} - b^{12} c^{7}}\right )} \sqrt {b x + a} \sqrt {b x + c} + \frac {3 \, {\left (b x + a\right )}^{4} - 10 \, {\left (b x + a\right )}^{3} a + 12 \, {\left (b x + a\right )}^{2} a^{2} - 6 \, {\left (b x + a\right )} a^{3} + 2 \, {\left (b x + a\right )}^{3} c - 6 \, {\left (b x + a\right )}^{2} a c + 6 \, {\left (b x + a\right )} a^{2} c}{6 \, {\left (a^{2} b^{3} - 2 \, a b^{3} c + b^{3} c^{2}\right )}} - \frac {{\left (5 \, a^{2} + 6 \, a c + 5 \, c^{2}\right )} \log \left ({\left | -\sqrt {b x + a} + \sqrt {b x + c} \right |}\right )}{32 \, b^{3}} \]

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]

-1/96*(2*(4*(b*x + a)*(6*(a^5*b^9 - 5*a^4*b^9*c + 10*a^3*b^9*c^2 - 10*a^2*b^9*c^3 + 5*a*b^9*c^4 - b^9*c^5)*(b*
x + a)/(a^7*b^12 - 7*a^6*b^12*c + 21*a^5*b^12*c^2 - 35*a^4*b^12*c^3 + 35*a^3*b^12*c^4 - 21*a^2*b^12*c^5 + 7*a*
b^12*c^6 - b^12*c^7) - (17*a^6*b^9 - 86*a^5*b^9*c + 175*a^4*b^9*c^2 - 180*a^3*b^9*c^3 + 95*a^2*b^9*c^4 - 22*a*
b^9*c^5 + b^9*c^6)/(a^7*b^12 - 7*a^6*b^12*c + 21*a^5*b^12*c^2 - 35*a^4*b^12*c^3 + 35*a^3*b^12*c^4 - 21*a^2*b^1
2*c^5 + 7*a*b^12*c^6 - b^12*c^7)) + (59*a^7*b^9 - 301*a^6*b^9*c + 615*a^5*b^9*c^2 - 625*a^4*b^9*c^3 + 305*a^3*
b^9*c^4 - 39*a^2*b^9*c^5 - 19*a*b^9*c^6 + 5*b^9*c^7)/(a^7*b^12 - 7*a^6*b^12*c + 21*a^5*b^12*c^2 - 35*a^4*b^12*
c^3 + 35*a^3*b^12*c^4 - 21*a^2*b^12*c^5 + 7*a*b^12*c^6 - b^12*c^7))*(b*x + a) - 3*(5*a^8*b^9 - 24*a^7*b^9*c +
44*a^6*b^9*c^2 - 40*a^5*b^9*c^3 + 30*a^4*b^9*c^4 - 40*a^3*b^9*c^5 + 44*a^2*b^9*c^6 - 24*a*b^9*c^7 + 5*b^9*c^8)
/(a^7*b^12 - 7*a^6*b^12*c + 21*a^5*b^12*c^2 - 35*a^4*b^12*c^3 + 35*a^3*b^12*c^4 - 21*a^2*b^12*c^5 + 7*a*b^12*c
^6 - b^12*c^7))*sqrt(b*x + a)*sqrt(b*x + c) + 1/6*(3*(b*x + a)^4 - 10*(b*x + a)^3*a + 12*(b*x + a)^2*a^2 - 6*(
b*x + a)*a^3 + 2*(b*x + a)^3*c - 6*(b*x + a)^2*a*c + 6*(b*x + a)*a^2*c)/(a^2*b^3 - 2*a*b^3*c + b^3*c^2) - 1/32
*(5*a^2 + 6*a*c + 5*c^2)*log(abs(-sqrt(b*x + a) + sqrt(b*x + c)))/b^3

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maple [C]  time = 0.02, size = 604, normalized size = 2.65 \[ \frac {b \,x^{4}}{2 \left (a -c \right )^{2}}+\frac {a \,x^{3}}{3 \left (a -c \right )^{2}}+\frac {c \,x^{3}}{3 \left (a -c \right )^{2}}-\frac {\sqrt {b x +a}\, \sqrt {b x +c}\, \left (96 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b^{3} x^{3} \mathrm {csgn}\relax (b )+16 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a \,b^{2} x^{2} \mathrm {csgn}\relax (b )+16 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b^{2} c \,x^{2} \mathrm {csgn}\relax (b )-15 a^{4} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )+12 a^{3} c \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )-20 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a^{2} b x \,\mathrm {csgn}\relax (b )+6 a^{2} c^{2} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )+8 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a b c x \,\mathrm {csgn}\relax (b )+12 a \,c^{3} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )-20 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b \,c^{2} x \,\mathrm {csgn}\relax (b )-15 c^{4} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )+30 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a^{3} \mathrm {csgn}\relax (b )-14 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a^{2} c \,\mathrm {csgn}\relax (b )-14 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a \,c^{2} \mathrm {csgn}\relax (b )+30 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, c^{3} \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{192 \left (a -c \right )^{2} \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b^{3}} \]

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*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*x*a^2*b+8*(b^2*x^2+a*b*x+b*c*x+a*
c)^(1/2)*csgn(b)*x*a*b*c-20*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*x*b*c^2+30*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)
*csgn(b)*a^3-14*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*a^2*c-14*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*a*c^2
+30*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*c^3-15*ln(1/2*(2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)+2*b*x+a+c
)*csgn(b))*a^4+12*ln(1/2*(2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)+2*b*x+a+c)*csgn(b))*a^3*c+6*ln(1/2*(2*(b^2
*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)+2*b*x+a+c)*csgn(b))*a^2*c^2+12*ln(1/2*(2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*c
sgn(b)+2*b*x+a+c)*csgn(b))*a*c^3-15*ln(1/2*(2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)+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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (\sqrt {b x + a} + \sqrt {b x + c}\right )}^{2}}\,{d x} \]

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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mupad [B]  time = 81.17, size = 1358, normalized size = 5.96 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*(a + c))/(3*(a - c)^2) - ((((a + b*x)^(1/2) - a^(1/2))^15*((3*a*c)/8 + (5*a^2)/16 + (5*c^2)/16))/(b^3*((c
+ b*x)^(1/2) - c^(1/2))^15) + (((a + b*x)^(1/2) - a^(1/2))^3*((23*a*c^3)/12 + (23*a^3*c)/12 - (115*a^4)/48 -
(115*c^4)/48 + (349*a^2*c^2)/8))/(((c + b*x)^(1/2) - c^(1/2))^3*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) + (((a + b*x)
^(1/2) - a^(1/2))^13*((23*a*c^3)/12 + (23*a^3*c)/12 - (115*a^4)/48 - (115*c^4)/48 + (349*a^2*c^2)/8))/(((c + b
*x)^(1/2) - c^(1/2))^13*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) + (((a + b*x)^(1/2) - a^(1/2))^5*((3917*a*c^3)/12 + (
3917*a^3*c)/12 + (383*a^4)/48 + (383*c^4)/48 + (7279*a^2*c^2)/8))/(((c + b*x)^(1/2) - c^(1/2))^5*(a^2*b^3 + b^
3*c^2 - 2*a*b^3*c)) + (((a + b*x)^(1/2) - a^(1/2))^11*((3917*a*c^3)/12 + (3917*a^3*c)/12 + (383*a^4)/48 + (383
*c^4)/48 + (7279*a^2*c^2)/8))/(((c + b*x)^(1/2) - c^(1/2))^11*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) + (((a + b*x)^(
1/2) - a^(1/2))^7*((17567*a*c^3)/12 + (17567*a^3*c)/12 + (2789*a^4)/48 + (2789*c^4)/48 + (28213*a^2*c^2)/8))/(
((c + b*x)^(1/2) - c^(1/2))^7*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) + (((a + b*x)^(1/2) - a^(1/2))^9*((17567*a*c^3)
/12 + (17567*a^3*c)/12 + (2789*a^4)/48 + (2789*c^4)/48 + (28213*a^2*c^2)/8))/(((c + b*x)^(1/2) - c^(1/2))^9*(a
^2*b^3 + b^3*c^2 - 2*a*b^3*c)) + (((a + b*x)^(1/2) - a^(1/2))*((3*a*c)/8 + (5*a^2)/16 + (5*c^2)/16))/(b^3*((c
+ b*x)^(1/2) - c^(1/2))) - (a^(1/2)*c^(1/2)*(192*a*c^2 + 192*a^2*c)*((a + b*x)^(1/2) - a^(1/2))^4)/(((c + b*x)
^(1/2) - c^(1/2))^4*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) - (a^(1/2)*c^(1/2)*(192*a*c^2 + 192*a^2*c)*((a + b*x)^(1/
2) - a^(1/2))^12)/(((c + b*x)^(1/2) - c^(1/2))^12*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) - (a^(1/2)*c^(1/2)*((a + b*
x)^(1/2) - a^(1/2))^6*((5120*a*c^2)/3 + (5120*a^2*c)/3 + 256*a^3 + 256*c^3))/(((c + b*x)^(1/2) - c^(1/2))^6*(a
^2*b^3 + b^3*c^2 - 2*a*b^3*c)) - (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^10*((5120*a*c^2)/3 + (5120*a^2*c
)/3 + 256*a^3 + 256*c^3))/(((c + b*x)^(1/2) - c^(1/2))^10*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)) - (a^(1/2)*c^(1/2)*
((a + b*x)^(1/2) - a^(1/2))^8*((10112*a*c^2)/3 + (10112*a^2*c)/3 + 512*a^3 + 512*c^3))/(((c + b*x)^(1/2) - c^(
1/2))^8*(a^2*b^3 + b^3*c^2 - 2*a*b^3*c)))/((28*((a + b*x)^(1/2) - a^(1/2))^4)/((c + b*x)^(1/2) - c^(1/2))^4 -
(8*((a + b*x)^(1/2) - a^(1/2))^2)/((c + b*x)^(1/2) - c^(1/2))^2 - (56*((a + b*x)^(1/2) - a^(1/2))^6)/((c + b*x
)^(1/2) - c^(1/2))^6 + (70*((a + b*x)^(1/2) - a^(1/2))^8)/((c + b*x)^(1/2) - c^(1/2))^8 - (56*((a + b*x)^(1/2)
- a^(1/2))^10)/((c + b*x)^(1/2) - c^(1/2))^10 + (28*((a + b*x)^(1/2) - a^(1/2))^12)/((c + b*x)^(1/2) - c^(1/2
))^12 - (8*((a + b*x)^(1/2) - a^(1/2))^14)/((c + b*x)^(1/2) - c^(1/2))^14 + ((a + b*x)^(1/2) - a^(1/2))^16/((c
+ b*x)^(1/2) - c^(1/2))^16 + 1) + (b*x^4)/(2*(a - c)^2) + (atanh(((a + b*x)^(1/2) - a^(1/2))/((c + b*x)^(1/2)
- c^(1/2)))*(6*a*c + 5*a^2 + 5*c^2))/(16*b^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (\sqrt {a + b x} + \sqrt {b x + c}\right )^{2}}\, dx \]

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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