3.442 \(\int \frac{1}{(\sqrt{a+b x}+\sqrt{a+c x})^3} \, dx\)

Optimal. Leaf size=164 \[ -\frac{2 a \sqrt{a+b x}}{x^2 (b-c)^3}+\frac{2 a \sqrt{a+c x}}{x^2 (b-c)^3}-\frac{(2 b+3 c) \sqrt{a+b x}}{x (b-c)^3}+\frac{(3 b+2 c) \sqrt{a+c x}}{x (b-c)^3}-\frac{3 b c \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{\sqrt{a} (b-c)^3}+\frac{3 b c \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{\sqrt{a} (b-c)^3} \]

[Out]

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

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Rubi [A]  time = 0.1781, antiderivative size = 275, normalized size of antiderivative = 1.68, number of steps used = 16, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6690, 47, 51, 63, 208} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{\sqrt{a} (b-c)^3}-\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{\sqrt{a} (b-c)^3}-\frac{2 a \sqrt{a+b x}}{x^2 (b-c)^3}+\frac{2 a \sqrt{a+c x}}{x^2 (b-c)^3}-\frac{b \sqrt{a+b x}}{x (b-c)^3}-\frac{(b+3 c) \sqrt{a+b x}}{x (b-c)^3}+\frac{c \sqrt{a+c x}}{x (b-c)^3}+\frac{(3 b+c) \sqrt{a+c x}}{x (b-c)^3}-\frac{b (b+3 c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{\sqrt{a} (b-c)^3}+\frac{c (3 b+c) \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{\sqrt{a} (b-c)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x] + Sqrt[a + c*x])^(-3),x]

[Out]

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

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 208

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

Rubi steps

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

Mathematica [C]  time = 0.260628, size = 182, normalized size = 1.11 \[ \frac{-\frac{8 b^2 (a+b x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b x}{a}+1\right )}{a^2}+\frac{8 c^2 (a+c x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x}{a}+1\right )}{a^2}-\frac{3 (b+3 c) \left (b x \sqrt{\frac{b x}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )+a+b x\right )}{x \sqrt{a+b x}}+\frac{3 (3 b+c) \left (c x \sqrt{\frac{c x}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{c x}{a}+1}\right )+a+c x\right )}{x \sqrt{a+c x}}}{3 (b-c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x] + Sqrt[a + c*x])^(-3),x]

[Out]

((-3*(b + 3*c)*(a + b*x + b*x*Sqrt[1 + (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]]))/(x*Sqrt[a + b*x]) + (3*(3*b + c)*
(a + c*x + c*x*Sqrt[1 + (c*x)/a]*ArcTanh[Sqrt[1 + (c*x)/a]]))/(x*Sqrt[a + c*x]) - (8*b^2*(a + b*x)^(3/2)*Hyper
geometric2F1[3/2, 3, 5/2, 1 + (b*x)/a])/a^2 + (8*c^2*(a + c*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x)/
a])/a^2)/(3*(b - c)^3)

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Maple [B]  time = 0.003, size = 300, normalized size = 1.8 \begin{align*} 2\,{\frac{{b}^{2}}{ \left ( b-c \right ) ^{3}} \left ( -1/2\,{\frac{\sqrt{bx+a}}{bx}}-1/2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }+8\,{\frac{a{b}^{2}}{ \left ( b-c \right ) ^{3}} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( -1/8\,{\frac{ \left ( bx+a \right ) ^{3/2}}{a}}-1/8\,\sqrt{bx+a} \right ) }+1/8\,{\frac{1}{{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-8\,{\frac{{c}^{2}a}{ \left ( b-c \right ) ^{3}} \left ({\frac{1}{{c}^{2}{x}^{2}} \left ( -1/8\,{\frac{ \left ( cx+a \right ) ^{3/2}}{a}}-1/8\,\sqrt{cx+a} \right ) }+1/8\,{\frac{1}{{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{cx+a}}{\sqrt{a}}} \right ) } \right ) }+6\,{\frac{bc}{ \left ( b-c \right ) ^{3}} \left ( -1/2\,{\frac{\sqrt{bx+a}}{bx}}-1/2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-6\,{\frac{bc}{ \left ( b-c \right ) ^{3}} \left ( -1/2\,{\frac{\sqrt{cx+a}}{cx}}-1/2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{cx+a}}{\sqrt{a}}} \right ) } \right ) }-2\,{\frac{{c}^{2}}{ \left ( b-c \right ) ^{3}} \left ( -1/2\,{\frac{\sqrt{cx+a}}{cx}}-1/2\,{\frac{1}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{cx+a}}{\sqrt{a}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x)

[Out]

2/(b-c)^3*b^2*(-1/2*(b*x+a)^(1/2)/b/x-1/2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2))+8/(b-c)^3*a*b^2*((-1/8/a*(b*
x+a)^(3/2)-1/8*(b*x+a)^(1/2))/b^2/x^2+1/8/a^(3/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-8/(b-c)^3*a*c^2*((-1/8/a*(c*
x+a)^(3/2)-1/8*(c*x+a)^(1/2))/c^2/x^2+1/8/a^(3/2)*arctanh((c*x+a)^(1/2)/a^(1/2)))+6/(b-c)^3*c*b*(-1/2*(b*x+a)^
(1/2)/b/x-1/2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(1/2))-6/(b-c)^3*b*c*(-1/2*(c*x+a)^(1/2)/c/x-1/2/a^(1/2)*arctan
h((c*x+a)^(1/2)/a^(1/2)))-2/(b-c)^3*c^2*(-1/2*(c*x+a)^(1/2)/c/x-1/2/a^(1/2)*arctanh((c*x+a)^(1/2)/a^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (\sqrt{b x + a} + \sqrt{c x + a}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x, algorithm="maxima")

[Out]

integrate((sqrt(b*x + a) + sqrt(c*x + a))^(-3), x)

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Fricas [A]  time = 1.32698, size = 703, normalized size = 4.29 \begin{align*} \left [-\frac{3 \, \sqrt{a} b c x^{2} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 3 \, \sqrt{a} b c x^{2} \log \left (\frac{c x - 2 \, \sqrt{c x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (2 \, a^{2} +{\left (2 \, a b + 3 \, a c\right )} x\right )} \sqrt{b x + a} - 2 \,{\left (2 \, a^{2} +{\left (3 \, a b + 2 \, a c\right )} x\right )} \sqrt{c x + a}}{2 \,{\left (a b^{3} - 3 \, a b^{2} c + 3 \, a b c^{2} - a c^{3}\right )} x^{2}}, \frac{3 \, \sqrt{-a} b c x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - 3 \, \sqrt{-a} b c x^{2} \arctan \left (\frac{\sqrt{c x + a} \sqrt{-a}}{a}\right ) -{\left (2 \, a^{2} +{\left (2 \, a b + 3 \, a c\right )} x\right )} \sqrt{b x + a} +{\left (2 \, a^{2} +{\left (3 \, a b + 2 \, a c\right )} x\right )} \sqrt{c x + a}}{{\left (a b^{3} - 3 \, a b^{2} c + 3 \, a b c^{2} - a c^{3}\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\sqrt{a + b x} + \sqrt{a + c x}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)**(1/2)+(c*x+a)**(1/2))**3,x)

[Out]

Integral((sqrt(a + b*x) + sqrt(a + c*x))**(-3), 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(1/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x, algorithm="giac")

[Out]

Timed out