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

Optimal. Leaf size=157 \[ \frac{8 (a+b x)^{3/2}}{3 (a-c)^3}+\frac{2 (a+3 c) \sqrt{a+b x}}{(a-c)^3}-\frac{8 (b x+c)^{3/2}}{3 (a-c)^3}-\frac{2 (3 a+c) \sqrt{b x+c}}{(a-c)^3}-\frac{2 \sqrt{a} (a+3 c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{(a-c)^3}+\frac{2 \sqrt{c} (3 a+c) \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )}{(a-c)^3} \]

[Out]

(2*(a + 3*c)*Sqrt[a + b*x])/(a - c)^3 + (8*(a + b*x)^(3/2))/(3*(a - c)^3) - (2*(3*a + c)*Sqrt[c + b*x])/(a - c
)^3 - (8*(c + b*x)^(3/2))/(3*(a - c)^3) - (2*Sqrt[a]*(a + 3*c)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(a - c)^3 + (2*
Sqrt[c]*(3*a + c)*ArcTanh[Sqrt[c + b*x]/Sqrt[c]])/(a - c)^3

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Rubi [A]  time = 0.23279, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6689, 50, 63, 208} \[ \frac{8 (a+b x)^{3/2}}{3 (a-c)^3}+\frac{2 (a+3 c) \sqrt{a+b x}}{(a-c)^3}-\frac{8 (b x+c)^{3/2}}{3 (a-c)^3}-\frac{2 (3 a+c) \sqrt{b x+c}}{(a-c)^3}-\frac{2 \sqrt{a} (a+3 c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{(a-c)^3}+\frac{2 \sqrt{c} (3 a+c) \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )}{(a-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + 3*c)*Sqrt[a + b*x])/(a - c)^3 + (8*(a + b*x)^(3/2))/(3*(a - c)^3) - (2*(3*a + c)*Sqrt[c + b*x])/(a - c
)^3 - (8*(c + b*x)^(3/2))/(3*(a - c)^3) - (2*Sqrt[a]*(a + 3*c)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(a - c)^3 + (2*
Sqrt[c]*(3*a + c)*ArcTanh[Sqrt[c + b*x]/Sqrt[c]])/(a - c)^3

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

Mathematica [A]  time = 0.196875, size = 142, normalized size = 0.9 \[ \frac{2 \left (-9 a \sqrt{b x+c}+9 c \sqrt{a+b x}-3 \sqrt{a} (a+3 c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+3 \sqrt{c} (3 a+c) \tanh ^{-1}\left (\frac{\sqrt{b x+c}}{\sqrt{c}}\right )+7 a \sqrt{a+b x}+4 b x \sqrt{a+b x}-7 c \sqrt{b x+c}-4 b x \sqrt{b x+c}\right )}{3 (a-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a*Sqrt[a + b*x] + 9*c*Sqrt[a + b*x] + 4*b*x*Sqrt[a + b*x] - 9*a*Sqrt[c + b*x] - 7*c*Sqrt[c + b*x] - 4*b*
x*Sqrt[c + b*x] - 3*Sqrt[a]*(a + 3*c)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]] + 3*Sqrt[c]*(3*a + c)*ArcTanh[Sqrt[c + b*
x]/Sqrt[c]]))/(3*(a - c)^3)

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Maple [A]  time = 0.003, size = 181, normalized size = 1.2 \begin{align*}{\frac{a}{ \left ( a-c \right ) ^{3}} \left ( 2\,\sqrt{bx+a}-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) }+{\frac{8}{3\, \left ( a-c \right ) ^{3}} \left ( bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{8}{3\, \left ( a-c \right ) ^{3}} \left ( bx+c \right ) ^{{\frac{3}{2}}}}+3\,{\frac{c}{ \left ( a-c \right ) ^{3}} \left ( 2\,\sqrt{bx+a}-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) \right ) }-3\,{\frac{a}{ \left ( a-c \right ) ^{3}} \left ( 2\,\sqrt{bx+c}-2\,\sqrt{c}{\it Artanh} \left ({\frac{\sqrt{bx+c}}{\sqrt{c}}} \right ) \right ) }-{\frac{c}{ \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/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x)

[Out]

1/(a-c)^3*a*(2*(b*x+a)^(1/2)-2*a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))+8/3*(b*x+a)^(3/2)/(a-c)^3-8/3*(b*x+c)^(
3/2)/(a-c)^3+3/(a-c)^3*c*(2*(b*x+a)^(1/2)-2*a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-3/(a-c)^3*a*(2*(b*x+c)^(1/
2)-2*c^(1/2)*arctanh((b*x+c)^(1/2)/c^(1/2)))-1/(a-c)^3*c*(2*(b*x+c)^(1/2)-2*c^(1/2)*arctanh((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{\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/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.041, size = 1283, normalized size = 8.17 \begin{align*} \left [-\frac{3 \,{\left (a + 3 \, c\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 3 \,{\left (3 \, a + c\right )} \sqrt{c} \log \left (\frac{b x - 2 \, \sqrt{b x + c} \sqrt{c} + 2 \, c}{x}\right ) - 2 \,{\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt{b x + a} + 2 \,{\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt{b x + c}}{3 \,{\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}, -\frac{6 \,{\left (3 \, a + c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{b x + c} \sqrt{-c}}{c}\right ) + 3 \,{\left (a + 3 \, c\right )} \sqrt{a} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) - 2 \,{\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt{b x + a} + 2 \,{\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt{b x + c}}{3 \,{\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}, \frac{6 \, \sqrt{-a}{\left (a + 3 \, c\right )} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - 3 \,{\left (3 \, a + c\right )} \sqrt{c} \log \left (\frac{b x - 2 \, \sqrt{b x + c} \sqrt{c} + 2 \, c}{x}\right ) + 2 \,{\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt{b x + a} - 2 \,{\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt{b x + c}}{3 \,{\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}, \frac{2 \,{\left (3 \, \sqrt{-a}{\left (a + 3 \, c\right )} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - 3 \,{\left (3 \, a + c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{b x + c} \sqrt{-c}}{c}\right ) +{\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt{b x + a} -{\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt{b x + c}\right )}}{3 \,{\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*(a + 3*c)*sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 3*(3*a + c)*sqrt(c)*log((b*x - 2*sqr
t(b*x + c)*sqrt(c) + 2*c)/x) - 2*(4*b*x + 7*a + 9*c)*sqrt(b*x + a) + 2*(4*b*x + 9*a + 7*c)*sqrt(b*x + c))/(a^3
- 3*a^2*c + 3*a*c^2 - c^3), -1/3*(6*(3*a + c)*sqrt(-c)*arctan(sqrt(b*x + c)*sqrt(-c)/c) + 3*(a + 3*c)*sqrt(a)
*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) - 2*(4*b*x + 7*a + 9*c)*sqrt(b*x + a) + 2*(4*b*x + 9*a + 7*c)*sq
rt(b*x + c))/(a^3 - 3*a^2*c + 3*a*c^2 - c^3), 1/3*(6*sqrt(-a)*(a + 3*c)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - 3*(
3*a + c)*sqrt(c)*log((b*x - 2*sqrt(b*x + c)*sqrt(c) + 2*c)/x) + 2*(4*b*x + 7*a + 9*c)*sqrt(b*x + a) - 2*(4*b*x
+ 9*a + 7*c)*sqrt(b*x + c))/(a^3 - 3*a^2*c + 3*a*c^2 - c^3), 2/3*(3*sqrt(-a)*(a + 3*c)*arctan(sqrt(b*x + a)*s
qrt(-a)/a) - 3*(3*a + c)*sqrt(-c)*arctan(sqrt(b*x + c)*sqrt(-c)/c) + (4*b*x + 7*a + 9*c)*sqrt(b*x + a) - (4*b*
x + 9*a + 7*c)*sqrt(b*x + c))/(a^3 - 3*a^2*c + 3*a*c^2 - c^3)]

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

[Out]

Integral(1/(x*(sqrt(a + b*x) + sqrt(b*x + c))**3), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError