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]

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

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Rubi [A]  time = 0.18, 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 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]

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]

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*}

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Mathematica [C]  time = 0.27, 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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fricas [A]  time = 0.47, size = 297, normalized size = 1.81 \[ \left [-\frac {3 \, \sqrt {a} b c x^{2} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 3 \, \sqrt {a} b c x^{2} \log \left (\frac {c x - 2 \, \sqrt {c x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, a^{2} + {\left (2 \, a b + 3 \, a c\right )} x\right )} \sqrt {b x + a} - 2 \, {\left (2 \, a^{2} + {\left (3 \, a b + 2 \, a c\right )} x\right )} \sqrt {c x + a}}{2 \, {\left (a b^{3} - 3 \, a b^{2} c + 3 \, a b c^{2} - a c^{3}\right )} x^{2}}, \frac {3 \, \sqrt {-a} b c x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - 3 \, \sqrt {-a} b c x^{2} \arctan \left (\frac {\sqrt {c x + a} \sqrt {-a}}{a}\right ) - {\left (2 \, a^{2} + {\left (2 \, a b + 3 \, a c\right )} x\right )} \sqrt {b x + a} + {\left (2 \, a^{2} + {\left (3 \, a b + 2 \, a c\right )} x\right )} \sqrt {c x + a}}{{\left (a b^{3} - 3 \, a b^{2} c + 3 \, a b c^{2} - a c^{3}\right )} x^{2}}\right ] \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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maple [B]  time = 0.02, size = 300, normalized size = 1.83 \[ \frac {8 \left (\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}+\frac {-\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x +a}}{8}}{b^{2} x^{2}}\right ) a \,b^{2}}{\left (b -c \right )^{3}}-\frac {8 \left (\frac {\arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}+\frac {-\frac {\left (c x +a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {c x +a}}{8}}{c^{2} x^{2}}\right ) a \,c^{2}}{\left (b -c \right )^{3}}+\frac {2 \left (-\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {b x +a}}{2 b x}\right ) b^{2}}{\left (b -c \right )^{3}}+\frac {6 \left (-\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {b x +a}}{2 b x}\right ) b c}{\left (b -c \right )^{3}}-\frac {6 \left (-\frac {\arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {c x +a}}{2 c x}\right ) b c}{\left (b -c \right )^{3}}-\frac {2 \left (-\frac {\arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {c x +a}}{2 c x}\right ) c^{2}}{\left (b -c \right )^{3}} \]

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*(b*x+
a)^(3/2)/a-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*(c*x+
a)^(3/2)/a-1/8*(c*x+a)^(1/2))/x^2/c^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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (\sqrt {b x + a} + \sqrt {c x + a}\right )}^{3}}\,{d x} \]

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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mupad [B]  time = 5.74, size = 287, normalized size = 1.75 \[ \frac {c^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{4\,\sqrt {a}\,{\left (b-c\right )}^3\,{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}-\frac {\left (\frac {\sqrt {a}\,b^2}{4\,\left (a\,b^3-3\,a\,b^2\,c+3\,a\,b\,c^2-a\,c^3\right )}-\frac {\sqrt {a}\,\left (b^2+c\,b\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\left (\sqrt {a+c\,x}-\sqrt {a}\right )\,\left (a\,b^3-3\,a\,b^2\,c+3\,a\,b\,c^2-a\,c^3\right )}\right )\,{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+\frac {3\,b\,c\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )}{\sqrt {a}\,\left (b^3-3\,b^2\,c+3\,b\,c^2-c^3\right )}-\frac {c\,\left (b+c\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a}\,{\left (b-c\right )}^3\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\sqrt {a + b x} + \sqrt {a + c x}\right )^{3}}\, dx \]

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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