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

Optimal. Leaf size=165 \[ -\frac {2 (a+b x)^{3/2} (b x+c)^{3/2}}{3 b^2 (a-c)^2}+\frac {(a+c) (a+b x)^{3/2} \sqrt {b x+c}}{2 b^2 (a-c)^2}-\frac {(a+c) \sqrt {a+b x} \sqrt {b x+c}}{4 b^2 (a-c)}-\frac {(a+c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {b x+c}}\right )}{4 b^2}+\frac {2 b x^3}{3 (a-c)^2}+\frac {x^2 (a+c)}{2 (a-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6689, 80, 50, 63, 217, 206} \[ -\frac {2 (a+b x)^{3/2} (b x+c)^{3/2}}{3 b^2 (a-c)^2}+\frac {(a+c) (a+b x)^{3/2} \sqrt {b x+c}}{2 b^2 (a-c)^2}-\frac {(a+c) \sqrt {a+b x} \sqrt {b x+c}}{4 b^2 (a-c)}-\frac {(a+c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {b x+c}}\right )}{4 b^2}+\frac {2 b x^3}{3 (a-c)^2}+\frac {x^2 (a+c)}{2 (a-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.88, size = 229, normalized size = 1.39 \[ \frac {3 a^2 \sqrt {a+b x} \sqrt {b x+c}-2 a \left (b x \sqrt {a+b x} \sqrt {b x+c}+c \sqrt {a+b x} \sqrt {b x+c}-3 b^2 x^2\right )+(4 b x+3 c) \left (-2 b x \sqrt {a+b x} \sqrt {b x+c}+c \sqrt {a+b x} \sqrt {b x+c}+2 b^2 x^2\right )}{12 b^2 (a-c)^2}-\frac {(a+c) \sqrt {b (c-a)} \sqrt {-\frac {b x+c}{a-c}} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {a+b x}}{\sqrt {b (c-a)}}\right )}{4 b^{5/2} \sqrt {b x+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 149, normalized size = 0.90 \[ \frac {16 \, b^{3} x^{3} + 12 \, {\left (a b^{2} + b^{2} c\right )} x^{2} - 2 \, {\left (8 \, b^{2} x^{2} - 3 \, a^{2} + 2 \, a c - 3 \, c^{2} + 2 \, {\left (a b + b c\right )} x\right )} \sqrt {b x + a} \sqrt {b x + c} + 3 \, {\left (a^{3} - a^{2} c - a c^{2} + c^{3}\right )} \log \left (-2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b x + c} - a - c\right )}{24 \, {\left (a^{2} b^{2} - 2 \, a b^{2} c + b^{2} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(16*b^3*x^3 + 12*(a*b^2 + b^2*c)*x^2 - 2*(8*b^2*x^2 - 3*a^2 + 2*a*c - 3*c^2 + 2*(a*b + b*c)*x)*sqrt(b*x +
a)*sqrt(b*x + c) + 3*(a^3 - a^2*c - a*c^2 + c^3)*log(-2*b*x + 2*sqrt(b*x + a)*sqrt(b*x + c) - a - c))/(a^2*b^
2 - 2*a*b^2*c + b^2*c^2)

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giac [B]  time = 0.27, size = 445, normalized size = 2.70 \[ -\frac {{\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (a^{3} b^{2} - 3 \, a^{2} b^{2} c + 3 \, a b^{2} c^{2} - b^{2} c^{3}\right )} {\left (b x + a\right )}}{a^{5} b^{3} - 5 \, a^{4} b^{3} c + 10 \, a^{3} b^{3} c^{2} - 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{3} c^{4} - b^{3} c^{5}} - \frac {7 \, a^{4} b^{2} - 22 \, a^{3} b^{2} c + 24 \, a^{2} b^{2} c^{2} - 10 \, a b^{2} c^{3} + b^{2} c^{4}}{a^{5} b^{3} - 5 \, a^{4} b^{3} c + 10 \, a^{3} b^{3} c^{2} - 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{3} c^{4} - b^{3} c^{5}}\right )} + \frac {3 \, {\left (a^{5} b^{2} - 3 \, a^{4} b^{2} c + 2 \, a^{3} b^{2} c^{2} + 2 \, a^{2} b^{2} c^{3} - 3 \, a b^{2} c^{4} + b^{2} c^{5}\right )}}{a^{5} b^{3} - 5 \, a^{4} b^{3} c + 10 \, a^{3} b^{3} c^{2} - 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{3} c^{4} - b^{3} c^{5}}\right )} \sqrt {b x + a} \sqrt {b x + c} - \frac {3 \, {\left (a + c\right )} \log \left ({\left | -\sqrt {b x + a} + \sqrt {b x + c} \right |}\right )}{b} - \frac {2 \, {\left (4 \, {\left (b x + a\right )}^{3} - 9 \, {\left (b x + a\right )}^{2} a + 6 \, {\left (b x + a\right )} a^{2} + 3 \, {\left (b x + a\right )}^{2} c - 6 \, {\left (b x + a\right )} a c\right )}}{a^{2} b - 2 \, a b c + b c^{2}}}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((2*(b*x + a)*(4*(a^3*b^2 - 3*a^2*b^2*c + 3*a*b^2*c^2 - b^2*c^3)*(b*x + a)/(a^5*b^3 - 5*a^4*b^3*c + 10*a
^3*b^3*c^2 - 10*a^2*b^3*c^3 + 5*a*b^3*c^4 - b^3*c^5) - (7*a^4*b^2 - 22*a^3*b^2*c + 24*a^2*b^2*c^2 - 10*a*b^2*c
^3 + b^2*c^4)/(a^5*b^3 - 5*a^4*b^3*c + 10*a^3*b^3*c^2 - 10*a^2*b^3*c^3 + 5*a*b^3*c^4 - b^3*c^5)) + 3*(a^5*b^2
- 3*a^4*b^2*c + 2*a^3*b^2*c^2 + 2*a^2*b^2*c^3 - 3*a*b^2*c^4 + b^2*c^5)/(a^5*b^3 - 5*a^4*b^3*c + 10*a^3*b^3*c^2
- 10*a^2*b^3*c^3 + 5*a*b^3*c^4 - b^3*c^5))*sqrt(b*x + a)*sqrt(b*x + c) - 3*(a + c)*log(abs(-sqrt(b*x + a) + s
qrt(b*x + c)))/b - 2*(4*(b*x + a)^3 - 9*(b*x + a)^2*a + 6*(b*x + a)*a^2 + 3*(b*x + a)^2*c - 6*(b*x + a)*a*c)/(
a^2*b - 2*a*b*c + b*c^2))/b

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maple [C]  time = 0.02, size = 431, normalized size = 2.61 \[ \frac {2 b \,x^{3}}{3 \left (a -c \right )^{2}}+\frac {a \,x^{2}}{2 \left (a -c \right )^{2}}+\frac {c \,x^{2}}{2 \left (a -c \right )^{2}}-\frac {\sqrt {b x +a}\, \sqrt {b x +c}\, \left (16 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b^{2} x^{2} \mathrm {csgn}\relax (b )+3 a^{3} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )-3 a^{2} c \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )+4 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a b x \,\mathrm {csgn}\relax (b )-3 a \,c^{2} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )+4 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b c x \,\mathrm {csgn}\relax (b )+3 c^{3} \ln \left (\frac {\left (2 b x +a +c +2 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{2}\right )-6 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a^{2} \mathrm {csgn}\relax (b )+4 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, a c \,\mathrm {csgn}\relax (b )-6 \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, c^{2} \mathrm {csgn}\relax (b )\right ) \mathrm {csgn}\relax (b )}{24 \left (a -c \right )^{2} \sqrt {b^{2} x^{2}+a b x +b c x +a c}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2/(a-c)^2*a+1/2*x^2/(a-c)^2*c+2/3*b*x^3/(a-c)^2-1/24/(a-c)^2*(b*x+a)^(1/2)*(b*x+c)^(1/2)*(16*csgn(b)*x^2
*b^2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+4*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*x*a*b+4*(b^2*x^2+a*b*x+b*c*x+a*
c)^(1/2)*csgn(b)*x*b*c-6*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*a^2+4*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)
*a*c-6*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b)*c^2+3*ln(1/2*(2*b*x+a+c+2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(
b))*csgn(b))*a^3-3*ln(1/2*(2*b*x+a+c+2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b))*csgn(b))*a^2*c-3*ln(1/2*(2*b*x
+a+c+2*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*csgn(b))*csgn(b))*a*c^2+3*ln(1/2*(2*b*x+a+c+2*(b^2*x^2+a*b*x+b*c*x+a*c)
^(1/2)*csgn(b))*csgn(b))*c^3)*csgn(b)/b^2/(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 \[ \int \frac {x}{{\left (\sqrt {b x + a} + \sqrt {b x + c}\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 37.52, size = 1012, normalized size = 6.13 \[ \frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {a}{2}+\frac {c}{2}\right )}{b^2\,\left (\sqrt {c+b\,x}-\sqrt {c}\right )}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (\frac {a}{2}+\frac {c}{2}\right )}{b^2\,{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^{11}}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {17\,a^3}{6}+\frac {101\,a^2\,c}{2}+\frac {101\,a\,c^2}{2}+\frac {17\,c^3}{6}\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^3\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (\frac {17\,a^3}{6}+\frac {101\,a^2\,c}{2}+\frac {101\,a\,c^2}{2}+\frac {17\,c^3}{6}\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^9\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (19\,a^3+269\,a^2\,c+269\,a\,c^2+19\,c^3\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^5\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (19\,a^3+269\,a^2\,c+269\,a\,c^2+19\,c^3\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^7\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}+\frac {16\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^2\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}+\frac {16\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^{10}\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (64\,a^2+192\,a\,c+64\,c^2\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^4\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8\,\left (64\,a^2+192\,a\,c+64\,c^2\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^8\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (128\,a^2+\frac {1312\,a\,c}{3}+128\,c^2\right )}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^6\,\left (a^2\,b^2-2\,a\,b^2\,c+b^2\,c^2\right )}}{\frac {15\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^4}-\frac {6\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^2}-\frac {20\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^6}+\frac {15\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^8}-\frac {6\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^{10}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}}{{\left (\sqrt {c+b\,x}-\sqrt {c}\right )}^{12}}+1}-\frac {\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {c+b\,x}-\sqrt {c}}\right )\,\left (a+c\right )}{2\,b^2}+\frac {x^2\,\left (a+c\right )}{2\,{\left (a-c\right )}^2}+\frac {2\,b\,x^3}{3\,{\left (a-c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x)^(1/2) - a^(1/2))*(a/2 + c/2))/(b^2*((c + b*x)^(1/2) - c^(1/2))) + (((a + b*x)^(1/2) - a^(1/2))^11
*(a/2 + c/2))/(b^2*((c + b*x)^(1/2) - c^(1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^3*((101*a*c^2)/2 + (101*a^2*
c)/2 + (17*a^3)/6 + (17*c^3)/6))/(((c + b*x)^(1/2) - c^(1/2))^3*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) - (((a + b*x)
^(1/2) - a^(1/2))^9*((101*a*c^2)/2 + (101*a^2*c)/2 + (17*a^3)/6 + (17*c^3)/6))/(((c + b*x)^(1/2) - c^(1/2))^9*
(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) - (((a + b*x)^(1/2) - a^(1/2))^5*(269*a*c^2 + 269*a^2*c + 19*a^3 + 19*c^3))/(
((c + b*x)^(1/2) - c^(1/2))^5*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) - (((a + b*x)^(1/2) - a^(1/2))^7*(269*a*c^2 + 2
69*a^2*c + 19*a^3 + 19*c^3))/(((c + b*x)^(1/2) - c^(1/2))^7*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) + (16*a^(3/2)*c^(
3/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(((c + b*x)^(1/2) - c^(1/2))^2*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) + (16*a^(3
/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^10)/(((c + b*x)^(1/2) - c^(1/2))^10*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) +
(a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^4*(192*a*c + 64*a^2 + 64*c^2))/(((c + b*x)^(1/2) - c^(1/2))^4*(a
^2*b^2 + b^2*c^2 - 2*a*b^2*c)) + (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^8*(192*a*c + 64*a^2 + 64*c^2))/(
((c + b*x)^(1/2) - c^(1/2))^8*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)) + (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^
6*((1312*a*c)/3 + 128*a^2 + 128*c^2))/(((c + b*x)^(1/2) - c^(1/2))^6*(a^2*b^2 + b^2*c^2 - 2*a*b^2*c)))/((15*((
a + b*x)^(1/2) - a^(1/2))^4)/((c + b*x)^(1/2) - c^(1/2))^4 - (6*((a + b*x)^(1/2) - a^(1/2))^2)/((c + b*x)^(1/2
) - c^(1/2))^2 - (20*((a + b*x)^(1/2) - a^(1/2))^6)/((c + b*x)^(1/2) - c^(1/2))^6 + (15*((a + b*x)^(1/2) - a^(
1/2))^8)/((c + b*x)^(1/2) - c^(1/2))^8 - (6*((a + b*x)^(1/2) - a^(1/2))^10)/((c + b*x)^(1/2) - c^(1/2))^10 + (
(a + b*x)^(1/2) - a^(1/2))^12/((c + b*x)^(1/2) - c^(1/2))^12 + 1) - (atanh(((a + b*x)^(1/2) - a^(1/2))/((c + b
*x)^(1/2) - c^(1/2)))*(a + c))/(2*b^2) + (x^2*(a + c))/(2*(a - c)^2) + (2*b*x^3)/(3*(a - c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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