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]

((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)

________________________________________________________________________________________

Rubi [A]  time = 0.21001, antiderivative size = 165, normalized size of antiderivative = 1., 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 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 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}{\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*}

Mathematica [A]  time = 0.871075, 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])

________________________________________________________________________________________

Maple [C]  time = 0.01, size = 431, normalized size = 2.6 \begin{align*}{\frac{a{x}^{2}}{2\, \left ( a-c \right ) ^{2}}}+{\frac{c{x}^{2}}{2\, \left ( a-c \right ) ^{2}}}+{\frac{2\,b{x}^{3}}{3\, \left ( a-c \right ) ^{2}}}-{\frac{{\it csgn} \left ( b \right ) }{24\, \left ( a-c \right ) ^{2}{b}^{2}}\sqrt{bx+a}\sqrt{bx+c} \left ( 16\,{\it csgn} \left ( b \right ){x}^{2}{b}^{2}\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+4\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}xab+4\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}xbc-6\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}{a}^{2}+4\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}ac-6\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}{c}^{2}+3\,\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}-3\,\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-3\,\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}^{2}+3\,\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}^{3} \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/((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*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*x*a*b+4*csgn(b)*(b^2*x^2+a*b*x+
b*c*x+a*c)^(1/2)*x*b*c-6*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*a^2+4*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)
*a*c-6*csgn(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)*c^2+3*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^3-3*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-3*ln(1/2*(2*csg
n(b)*(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)+2*b*x+a+c)*csgn(b))*a*c^2+3*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^3)*csgn(b)/b^2/(b^2*x^2+a*b*x+b*c*x+a*c)^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\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/((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)

________________________________________________________________________________________

Fricas [A]  time = 0.975902, size = 332, normalized size = 2.01 \begin{align*} \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 )}} \end{align*}

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)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\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/((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)

________________________________________________________________________________________

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

[Out]

Timed out