3.415 \(\int \frac{1}{x^2 (\sqrt{a+b x}+\sqrt{c+b x})^3} \, dx\)

Optimal. Leaf size=162 \[ \frac{8 b \sqrt{a+b x}}{(a-c)^3}-\frac{8 b \sqrt{b x+c}}{(a-c)^3}-\frac{(a+3 c) \sqrt{a+b x}}{x (a-c)^3}+\frac{(3 a+c) \sqrt{b x+c}}{x (a-c)^3}-\frac{3 b (3 a+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{\sqrt{a} (a-c)^3}-\frac{3 b (a+3 c) \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )}{\sqrt{c} (c-a)^3} \]

[Out]

(8*b*Sqrt[a + b*x])/(a - c)^3 - ((a + 3*c)*Sqrt[a + b*x])/((a - c)^3*x) - (8*b*Sqrt[c + b*x])/(a - c)^3 + ((3*
a + c)*Sqrt[c + b*x])/((a - c)^3*x) - (3*b*(3*a + c)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(Sqrt[a]*(a - c)^3) - (3*
b*(a + 3*c)*ArcTanh[Sqrt[c + b*x]/Sqrt[c]])/(Sqrt[c]*(-a + c)^3)

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Rubi [A]  time = 0.273317, antiderivative size = 223, normalized size of antiderivative = 1.38, number of steps used = 14, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6689, 47, 63, 208, 50} \[ \frac{8 b \sqrt{a+b x}}{(a-c)^3}-\frac{8 b \sqrt{b x+c}}{(a-c)^3}-\frac{(a+3 c) \sqrt{a+b x}}{x (a-c)^3}+\frac{(3 a+c) \sqrt{b x+c}}{x (a-c)^3}-\frac{b (a+3 c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{\sqrt{a} (a-c)^3}-\frac{8 \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{(a-c)^3}+\frac{b (3 a+c) \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )}{\sqrt{c} (a-c)^3}+\frac{8 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )}{(a-c)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(Sqrt[a + b*x] + Sqrt[c + b*x])^3),x]

[Out]

(8*b*Sqrt[a + b*x])/(a - c)^3 - ((a + 3*c)*Sqrt[a + b*x])/((a - c)^3*x) - (8*b*Sqrt[c + b*x])/(a - c)^3 + ((3*
a + c)*Sqrt[c + b*x])/((a - c)^3*x) - (8*Sqrt[a]*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(a - c)^3 - (b*(a + 3*c)*Ar
cTanh[Sqrt[a + b*x]/Sqrt[a]])/(Sqrt[a]*(a - c)^3) + (8*b*Sqrt[c]*ArcTanh[Sqrt[c + b*x]/Sqrt[c]])/(a - c)^3 + (
b*(3*a + c)*ArcTanh[Sqrt[c + b*x]/Sqrt[c]])/((a - c)^3*Sqrt[c])

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.596263, size = 187, normalized size = 1.15 \[ \frac{b \left (-\frac{(a+3 c) \left (b x \sqrt{\frac{b x}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )+a+b x\right )}{b x \sqrt{a+b x}}+\frac{(3 a+c) \left (b x \sqrt{\frac{b x}{c}+1} \tanh ^{-1}\left (\sqrt{\frac{b x}{c}+1}\right )+b x+c\right )}{b x \sqrt{b x+c}}+8 \sqrt{a+b x}-8 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-8 \sqrt{b x+c}+8 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )\right )}{(a-c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(Sqrt[a + b*x] + Sqrt[c + b*x])^3),x]

[Out]

(b*(8*Sqrt[a + b*x] - 8*Sqrt[c + b*x] - 8*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]] + 8*Sqrt[c]*ArcTanh[Sqrt[c +
b*x]/Sqrt[c]] - ((a + 3*c)*(a + b*x + b*x*Sqrt[1 + (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]]))/(b*x*Sqrt[a + b*x]) +
((3*a + c)*(c + b*x + b*x*Sqrt[1 + (b*x)/c]*ArcTanh[Sqrt[1 + (b*x)/c]]))/(b*x*Sqrt[c + b*x])))/(a - c)^3

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Maple [A]  time = 0.003, size = 252, normalized size = 1.6 \begin{align*} 2\,{\frac{ab}{ \left ( a-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 ( a-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{ab}{ \left ( a-c \right ) ^{3}} \left ( -1/2\,{\frac{\sqrt{bx+c}}{bx}}-1/2\,{\frac{1}{\sqrt{c}}{\it Artanh} \left ({\frac{\sqrt{bx+c}}{\sqrt{c}}} \right ) } \right ) }-2\,{\frac{bc}{ \left ( a-c \right ) ^{3}} \left ( -1/2\,{\frac{\sqrt{bx+c}}{bx}}-1/2\,{\frac{1}{\sqrt{c}}{\it Artanh} \left ({\frac{\sqrt{bx+c}}{\sqrt{c}}} \right ) } \right ) }+4\,{\frac{b}{ \left ( a-c \right ) ^{3}} \left ( 2\,\sqrt{bx+a}-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) }-4\,{\frac{b}{ \left ( a-c \right ) ^{3}} \left ( 2\,\sqrt{bx+c}-2\,\sqrt{c}{\it Artanh} \left ({\frac{\sqrt{bx+c}}{\sqrt{c}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2}{\left (\sqrt{b x + a} + \sqrt{b x + c}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.11474, size = 1558, normalized size = 9.62 \begin{align*} \left [-\frac{3 \,{\left (3 \, a b c + b c^{2}\right )} \sqrt{a} x \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 3 \,{\left (a^{2} b + 3 \, a b c\right )} \sqrt{c} x \log \left (\frac{b x - 2 \, \sqrt{b x + c} \sqrt{c} + 2 \, c}{x}\right ) - 2 \,{\left (8 \, a b c x - a^{2} c - 3 \, a c^{2}\right )} \sqrt{b x + a} + 2 \,{\left (8 \, a b c x - 3 \, a^{2} c - a c^{2}\right )} \sqrt{b x + c}}{2 \,{\left (a^{4} c - 3 \, a^{3} c^{2} + 3 \, a^{2} c^{3} - a c^{4}\right )} x}, -\frac{6 \,{\left (a^{2} b + 3 \, a b c\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{b x + c} \sqrt{-c}}{c}\right ) + 3 \,{\left (3 \, a b c + b c^{2}\right )} \sqrt{a} x \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (8 \, a b c x - a^{2} c - 3 \, a c^{2}\right )} \sqrt{b x + a} + 2 \,{\left (8 \, a b c x - 3 \, a^{2} c - a c^{2}\right )} \sqrt{b x + c}}{2 \,{\left (a^{4} c - 3 \, a^{3} c^{2} + 3 \, a^{2} c^{3} - a c^{4}\right )} x}, \frac{6 \,{\left (3 \, a b c + b c^{2}\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - 3 \,{\left (a^{2} b + 3 \, a b c\right )} \sqrt{c} x \log \left (\frac{b x - 2 \, \sqrt{b x + c} \sqrt{c} + 2 \, c}{x}\right ) + 2 \,{\left (8 \, a b c x - a^{2} c - 3 \, a c^{2}\right )} \sqrt{b x + a} - 2 \,{\left (8 \, a b c x - 3 \, a^{2} c - a c^{2}\right )} \sqrt{b x + c}}{2 \,{\left (a^{4} c - 3 \, a^{3} c^{2} + 3 \, a^{2} c^{3} - a c^{4}\right )} x}, \frac{3 \,{\left (3 \, a b c + b c^{2}\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - 3 \,{\left (a^{2} b + 3 \, a b c\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{b x + c} \sqrt{-c}}{c}\right ) +{\left (8 \, a b c x - a^{2} c - 3 \, a c^{2}\right )} \sqrt{b x + a} -{\left (8 \, a b c x - 3 \, a^{2} c - a c^{2}\right )} \sqrt{b x + c}}{{\left (a^{4} c - 3 \, a^{3} c^{2} + 3 \, a^{2} c^{3} - a c^{4}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(3*(3*a*b*c + b*c^2)*sqrt(a)*x*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 3*(a^2*b + 3*a*b*c)*sqrt(c
)*x*log((b*x - 2*sqrt(b*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a*b*c*x - a^2*c - 3*a*c^2)*sqrt(b*x + a) + 2*(8*a*b*c*
x - 3*a^2*c - a*c^2)*sqrt(b*x + c))/((a^4*c - 3*a^3*c^2 + 3*a^2*c^3 - a*c^4)*x), -1/2*(6*(a^2*b + 3*a*b*c)*sqr
t(-c)*x*arctan(sqrt(b*x + c)*sqrt(-c)/c) + 3*(3*a*b*c + b*c^2)*sqrt(a)*x*log((b*x + 2*sqrt(b*x + a)*sqrt(a) +
2*a)/x) - 2*(8*a*b*c*x - a^2*c - 3*a*c^2)*sqrt(b*x + a) + 2*(8*a*b*c*x - 3*a^2*c - a*c^2)*sqrt(b*x + c))/((a^4
*c - 3*a^3*c^2 + 3*a^2*c^3 - a*c^4)*x), 1/2*(6*(3*a*b*c + b*c^2)*sqrt(-a)*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) -
3*(a^2*b + 3*a*b*c)*sqrt(c)*x*log((b*x - 2*sqrt(b*x + c)*sqrt(c) + 2*c)/x) + 2*(8*a*b*c*x - a^2*c - 3*a*c^2)*
sqrt(b*x + a) - 2*(8*a*b*c*x - 3*a^2*c - a*c^2)*sqrt(b*x + c))/((a^4*c - 3*a^3*c^2 + 3*a^2*c^3 - a*c^4)*x), (3
*(3*a*b*c + b*c^2)*sqrt(-a)*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) - 3*(a^2*b + 3*a*b*c)*sqrt(-c)*x*arctan(sqrt(b*
x + c)*sqrt(-c)/c) + (8*a*b*c*x - a^2*c - 3*a*c^2)*sqrt(b*x + a) - (8*a*b*c*x - 3*a^2*c - a*c^2)*sqrt(b*x + c)
)/((a^4*c - 3*a^3*c^2 + 3*a^2*c^3 - a*c^4)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (\sqrt{a + b x} + \sqrt{b x + c}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out