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]

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

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Rubi [A]  time = 0.23, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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]

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*}

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Mathematica [A]  time = 0.19, size = 142, normalized size = 0.90 \[ \frac {2 \left (-9 a \sqrt {b x+c}+9 c \sqrt {a+b x}-3 \sqrt {a} (a+3 c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+3 \sqrt {c} (3 a+c) \tanh ^{-1}\left (\frac {\sqrt {b x+c}}{\sqrt {c}}\right )+7 a \sqrt {a+b x}+4 b x \sqrt {a+b x}-7 c \sqrt {b x+c}-4 b x \sqrt {b x+c}\right )}{3 (a-c)^3} \]

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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fricas [A]  time = 0.50, size = 516, normalized size = 3.29 \[ \left [-\frac {3 \, {\left (a + 3 \, c\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 3 \, {\left (3 \, a + c\right )} \sqrt {c} \log \left (\frac {b x - 2 \, \sqrt {b x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt {b x + a} + 2 \, {\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt {b x + c}}{3 \, {\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}, -\frac {6 \, {\left (3 \, a + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {b x + c} \sqrt {-c}}{c}\right ) + 3 \, {\left (a + 3 \, c\right )} \sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt {b x + a} + 2 \, {\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt {b x + c}}{3 \, {\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}, \frac {6 \, \sqrt {-a} {\left (a + 3 \, c\right )} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - 3 \, {\left (3 \, a + c\right )} \sqrt {c} \log \left (\frac {b x - 2 \, \sqrt {b x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt {b x + a} - 2 \, {\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt {b x + c}}{3 \, {\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}, \frac {2 \, {\left (3 \, \sqrt {-a} {\left (a + 3 \, c\right )} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - 3 \, {\left (3 \, a + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {b x + c} \sqrt {-c}}{c}\right ) + {\left (4 \, b x + 7 \, a + 9 \, c\right )} \sqrt {b x + a} - {\left (4 \, b x + 9 \, a + 7 \, c\right )} \sqrt {b x + c}\right )}}{3 \, {\left (a^{3} - 3 \, a^{2} c + 3 \, a c^{2} - c^{3}\right )}}\right ] \]

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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giac [B]  time = 2.91, size = 2652, normalized size = 16.89 \[ \text {result too large to display} \]

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]

-2/3*sqrt(b*x + c)*(4*(a^3 - 3*a^2*c + 3*a*c^2 - c^3)*(b*x + a)/(a^6 - 6*a^5*c + 15*a^4*c^2 - 20*a^3*c^3 + 15*
a^2*c^4 - 6*a*c^5 + c^6) + (5*a^4 - 8*a^3*c - 6*a^2*c^2 + 16*a*c^3 - 7*c^4)/(a^6 - 6*a^5*c + 15*a^4*c^2 - 20*a
^3*c^3 + 15*a^2*c^4 - 6*a*c^5 + c^6)) + 2*(a^2 + 3*a*c)*arctan(sqrt(b*x + a)/sqrt(-a))/((a^3 - 3*a^2*c + 3*a*c
^2 - c^3)*sqrt(-a)) + 2/3*(4*(b*x + a)^(3/2)*a^6 + 3*sqrt(b*x + a)*a^7 - 24*(b*x + a)^(3/2)*a^5*c - 9*sqrt(b*x
+ a)*a^6*c + 60*(b*x + a)^(3/2)*a^4*c^2 - 9*sqrt(b*x + a)*a^5*c^2 - 80*(b*x + a)^(3/2)*a^3*c^3 + 75*sqrt(b*x
+ a)*a^4*c^3 + 60*(b*x + a)^(3/2)*a^2*c^4 - 135*sqrt(b*x + a)*a^3*c^4 - 24*(b*x + a)^(3/2)*a*c^5 + 117*sqrt(b*
x + a)*a^2*c^5 + 4*(b*x + a)^(3/2)*c^6 - 51*sqrt(b*x + a)*a*c^6 + 9*sqrt(b*x + a)*c^7)/(a^9 - 9*a^8*c + 36*a^7
*c^2 - 84*a^6*c^3 + 126*a^5*c^4 - 126*a^4*c^5 + 84*a^3*c^6 - 36*a^2*c^7 + 9*a*c^8 - c^9) - 2*(3*a^9*c - 14*a^8
*c^2 + 22*a^7*c^3 - 6*a^6*c^4 - 20*a^5*c^5 + 22*a^4*c^6 - 6*a^3*c^7 - 2*a^2*c^8 + a*c^9 - 2*(3*a^2*c^2 + a*c^3
+ (3*a*c^2 + c^3)*sqrt(a*c))*(a^3 - 3*a^2*c + 3*a*c^2 - c^3)^2*sgn(a^3 - 3*a^2*c + 3*a*c^2 - c^3) - 2*(3*a^2*
c^2 + a*c^3 - (3*a^2*c + a*c^2)*sqrt(a*c))*(a^3 - 3*a^2*c + 3*a*c^2 - c^3)^2 - (3*a^5*c^2 - 11*a^4*c^3 + 14*a^
3*c^4 - 6*a^2*c^5 - a*c^6 + c^7 - (3*a^5*c - 11*a^4*c^2 + 14*a^3*c^3 - 6*a^2*c^4 - a*c^5 + c^6)*sqrt(a*c))*abs
(-a^3 + 3*a^2*c - 3*a*c^2 + c^3)*sgn(a^3 - 3*a^2*c + 3*a*c^2 - c^3) - (3*a^6*c - 11*a^5*c^2 + 14*a^4*c^3 - 6*a
^3*c^4 - a^2*c^5 + a*c^6 + (3*a^5*c - 11*a^4*c^2 + 14*a^3*c^3 - 6*a^2*c^4 - a*c^5 + c^6)*sqrt(a*c))*abs(-a^3 +
3*a^2*c - 3*a*c^2 + c^3) + (3*a^9*c - 14*a^8*c^2 + 22*a^7*c^3 - 6*a^6*c^4 - 20*a^5*c^5 + 22*a^4*c^6 - 6*a^3*c
^7 - 2*a^2*c^8 + a*c^9 - (3*a^8*c - 14*a^7*c^2 + 22*a^6*c^3 - 6*a^5*c^4 - 20*a^4*c^5 + 22*a^3*c^6 - 6*a^2*c^7
- 2*a*c^8 + c^9)*sqrt(a*c))*sgn(a^3 - 3*a^2*c + 3*a*c^2 - c^3) + (3*a^9 - 14*a^8*c + 22*a^7*c^2 - 6*a^6*c^3 -
20*a^5*c^4 + 22*a^4*c^5 - 6*a^3*c^6 - 2*a^2*c^7 + a*c^8)*sqrt(a*c))*arctan(-(sqrt(b*x + a) - sqrt(b*x + c))/sq
rt(-(a^4 - 2*a^3*c + 2*a*c^3 - c^4 + sqrt((a^4 - 2*a^3*c + 2*a*c^3 - c^4)^2 - (a^5 - 5*a^4*c + 10*a^3*c^2 - 10
*a^2*c^3 + 5*a*c^4 - c^5)*(a^3 - 3*a^2*c + 3*a*c^2 - c^3)))/(a^3 - 3*a^2*c + 3*a*c^2 - c^3)))/((sqrt(-a)*a^8 -
a^8*sqrt(-c) - 8*sqrt(-a)*a^7*c + 8*a^7*sqrt(-c)*c + 28*sqrt(-a)*a^6*c^2 - 28*a^6*sqrt(-c)*c^2 - 56*sqrt(-a)*
a^5*c^3 + 56*a^5*sqrt(-c)*c^3 + 70*sqrt(-a)*a^4*c^4 - 70*a^4*sqrt(-c)*c^4 - 56*sqrt(-a)*a^3*c^5 + 56*a^3*sqrt(
-c)*c^5 + 28*sqrt(-a)*a^2*c^6 - 28*a^2*sqrt(-c)*c^6 - 8*sqrt(-a)*a*c^7 + 8*a*sqrt(-c)*c^7 + sqrt(-a)*c^8 - sqr
t(-c)*c^8)*abs(-a^3 + 3*a^2*c - 3*a*c^2 + c^3)) + 2*(3*a^9*c - 14*a^8*c^2 + 22*a^7*c^3 - 6*a^6*c^4 - 20*a^5*c^
5 + 22*a^4*c^6 - 6*a^3*c^7 - 2*a^2*c^8 + a*c^9 + 2*(3*a^2*c^2 + a*c^3 + (3*a*c^2 + c^3)*sqrt(a*c))*(a^3 - 3*a^
2*c + 3*a*c^2 - c^3)^2*sgn(a^3 - 3*a^2*c + 3*a*c^2 - c^3) - 2*(3*a^2*c^2 + a*c^3 + (3*a^2*c + a*c^2)*sqrt(a*c)
)*(a^3 - 3*a^2*c + 3*a*c^2 - c^3)^2 - (3*a^5*c^2 - 11*a^4*c^3 + 14*a^3*c^4 - 6*a^2*c^5 - a*c^6 + c^7 - (3*a^5*
c - 11*a^4*c^2 + 14*a^3*c^3 - 6*a^2*c^4 - a*c^5 + c^6)*sqrt(a*c))*abs(-a^3 + 3*a^2*c - 3*a*c^2 + c^3)*sgn(a^3
- 3*a^2*c + 3*a*c^2 - c^3) + (3*a^6*c - 11*a^5*c^2 + 14*a^4*c^3 - 6*a^3*c^4 - a^2*c^5 + a*c^6 + (3*a^5*c - 11*
a^4*c^2 + 14*a^3*c^3 - 6*a^2*c^4 - a*c^5 + c^6)*sqrt(a*c))*abs(-a^3 + 3*a^2*c - 3*a*c^2 + c^3) - (3*a^9*c - 14
*a^8*c^2 + 22*a^7*c^3 - 6*a^6*c^4 - 20*a^5*c^5 + 22*a^4*c^6 - 6*a^3*c^7 - 2*a^2*c^8 + a*c^9 + (3*a^8*c - 14*a^
7*c^2 + 22*a^6*c^3 - 6*a^5*c^4 - 20*a^4*c^5 + 22*a^3*c^6 - 6*a^2*c^7 - 2*a*c^8 + c^9)*sqrt(a*c))*sgn(a^3 - 3*a
^2*c + 3*a*c^2 - c^3) + (3*a^9 - 14*a^8*c + 22*a^7*c^2 - 6*a^6*c^3 - 20*a^5*c^4 + 22*a^4*c^5 - 6*a^3*c^6 - 2*a
^2*c^7 + a*c^8)*sqrt(a*c))*arctan(-(sqrt(b*x + a) - sqrt(b*x + c))/sqrt(-(a^4 - 2*a^3*c + 2*a*c^3 - c^4 - sqrt
((a^4 - 2*a^3*c + 2*a*c^3 - c^4)^2 - (a^5 - 5*a^4*c + 10*a^3*c^2 - 10*a^2*c^3 + 5*a*c^4 - c^5)*(a^3 - 3*a^2*c
+ 3*a*c^2 - c^3)))/(a^3 - 3*a^2*c + 3*a*c^2 - c^3)))/((sqrt(-a)*a^8 - a^8*sqrt(-c) - 8*sqrt(-a)*a^7*c + 8*a^7*
sqrt(-c)*c + 28*sqrt(-a)*a^6*c^2 - 28*a^6*sqrt(-c)*c^2 - 56*sqrt(-a)*a^5*c^3 + 56*a^5*sqrt(-c)*c^3 + 70*sqrt(-
a)*a^4*c^4 - 70*a^4*sqrt(-c)*c^4 - 56*sqrt(-a)*a^3*c^5 + 56*a^3*sqrt(-c)*c^5 + 28*sqrt(-a)*a^2*c^6 - 28*a^2*sq
rt(-c)*c^6 - 8*sqrt(-a)*a*c^7 + 8*a*sqrt(-c)*c^7 + sqrt(-a)*c^8 - sqrt(-c)*c^8)*abs(-a^3 + 3*a^2*c - 3*a*c^2 +
c^3))

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maple [A]  time = 0.01, size = 181, normalized size = 1.15 \[ \frac {\left (-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\right ) a}{\left (a -c \right )^{3}}-\frac {3 \left (-2 \sqrt {c}\, \arctanh \left (\frac {\sqrt {b x +c}}{\sqrt {c}}\right )+2 \sqrt {b x +c}\right ) a}{\left (a -c \right )^{3}}+\frac {3 \left (-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\right ) c}{\left (a -c \right )^{3}}-\frac {\left (-2 \sqrt {c}\, \arctanh \left (\frac {\sqrt {b x +c}}{\sqrt {c}}\right )+2 \sqrt {b x +c}\right ) c}{\left (a -c \right )^{3}}+\frac {8 \left (b x +a \right )^{\frac {3}{2}}}{3 \left (a -c \right )^{3}}-\frac {8 \left (b x +c \right )^{\frac {3}{2}}}{3 \left (a -c \right )^{3}} \]

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

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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mupad [B]  time = 27.72, size = 4060, normalized size = 25.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a^(1/2)*(16*a + 16*c))/(3*a*c^2 - 3*a^2*c + a^3 - c^3) + (c^(1/2)*(16*a + 16*c))/(3*a*c^2 - 3*a^2*c + a^3
- c^3))*((a + b*x)^(1/2) - a^(1/2)))/((c + b*x)^(1/2) - c^(1/2)) + (((a^(1/2)*(12*a + 20*c))/(3*a*c^2 - 3*a^2*
c + a^3 - c^3) + (c^(1/2)*(20*a + 12*c))/(3*a*c^2 - 3*a^2*c + a^3 - c^3))*((a + b*x)^(1/2) - a^(1/2))^2)/((c +
b*x)^(1/2) - c^(1/2))^2 + (a^(1/2)*((28*a)/3 + 12*c))/(3*a*c^2 - 3*a^2*c + a^3 - c^3) + (c^(1/2)*(12*a + (28*
c)/3))/(3*a*c^2 - 3*a^2*c + a^3 - c^3))/((3*((a + b*x)^(1/2) - a^(1/2)))/((c + b*x)^(1/2) - c^(1/2)) + (3*((a
+ b*x)^(1/2) - a^(1/2))^2)/((c + b*x)^(1/2) - c^(1/2))^2 + ((a + b*x)^(1/2) - a^(1/2))^3/((c + b*x)^(1/2) - c^
(1/2))^3 + 1) + (log(((a + b*x)^(1/2) - a^(1/2))/((c + b*x)^(1/2) - c^(1/2)))*(a*(a^(1/2) + 3*c^(1/2)) + c*(3*
a^(1/2) + c^(1/2))))/(3*a*c^2 - 3*a^2*c + a^3 - c^3) + (atan(((((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2
*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))^(1/2)*((6*a*c^(11/2) - 6*a^(11/2)*c + 2*a^(3/2)*c^5 - 2*a^5*c^(3/2)
+ 12*a^3*c^(7/2) - 12*a^(7/2)*c^3 - 16*a^2*c^(9/2) + 16*a^(9/2)*c^2)/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5
- 20*a^4*c^4 + 15*a^5*c^3 - 6*a^6*c^2) + (((a^(1/2)*c^(15/2) - 5*a^(3/2)*c^(13/2) + 9*a^(5/2)*c^(11/2) - 5*a^(
7/2)*c^(9/2) - 5*a^(9/2)*c^(7/2) + 9*a^(11/2)*c^(5/2) - 5*a^(13/2)*c^(3/2) + a^(15/2)*c^(1/2))/(a*c^7 + a^7*c
- 6*a^2*c^6 + 15*a^3*c^5 - 20*a^4*c^4 + 15*a^5*c^3 - 6*a^6*c^2) - (2*((a + b*x)^(1/2) - a^(1/2))*(a*c^9 + a^9*
c - 7*a^2*c^8 + 22*a^3*c^7 - 41*a^4*c^6 + 50*a^5*c^5 - 41*a^6*c^4 + 22*a^7*c^3 - 7*a^8*c^2))/(((c + b*x)^(1/2)
- c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6*a^7*c^3 + a^8*c^2)))*((a^(1/2)*c^(
3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))^(1/2)*(a^(1/2)*c^3 - 3*a^(5/2)*c
- 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3/2) + 2*a^(3/2)*c^2))/(a*c^6 - a^6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4
*c^3 + 5*a^5*c^2) - (2*((a + b*x)^(1/2) - a^(1/2))*(3*a^(3/2)*c^7 - 3*a^7*c^(3/2) + 8*a^6*c^(5/2) - 8*a^(5/2)*
c^6 - 6*a^5*c^(7/2) + 6*a^(7/2)*c^5 + a^3*c^(11/2) - a^(11/2)*c^3))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*
a^3*c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6*a^7*c^3 + a^8*c^2)))*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/
2) + a^3*c^(1/2) + 2*a^2*c^(3/2) + 2*a^(3/2)*c^2)*1i)/(a*c^6 - a^6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5
*a^5*c^2) - (((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))^(1/2)*(
(((a^(1/2)*c^(15/2) - 5*a^(3/2)*c^(13/2) + 9*a^(5/2)*c^(11/2) - 5*a^(7/2)*c^(9/2) - 5*a^(9/2)*c^(7/2) + 9*a^(1
1/2)*c^(5/2) - 5*a^(13/2)*c^(3/2) + a^(15/2)*c^(1/2))/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5 - 20*a^4*c^4 + 1
5*a^5*c^3 - 6*a^6*c^2) - (2*((a + b*x)^(1/2) - a^(1/2))*(a*c^9 + a^9*c - 7*a^2*c^8 + 22*a^3*c^7 - 41*a^4*c^6 +
50*a^5*c^5 - 41*a^6*c^4 + 22*a^7*c^3 - 7*a^8*c^2))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4
*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6*a^7*c^3 + a^8*c^2)))*((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2*a*c +
a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))^(1/2)*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3/
2) + 2*a^(3/2)*c^2))/(a*c^6 - a^6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2) - (6*a*c^(11/2) - 6*a^(
11/2)*c + 2*a^(3/2)*c^5 - 2*a^5*c^(3/2) + 12*a^3*c^(7/2) - 12*a^(7/2)*c^3 - 16*a^2*c^(9/2) + 16*a^(9/2)*c^2)/(
a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5 - 20*a^4*c^4 + 15*a^5*c^3 - 6*a^6*c^2) + (2*((a + b*x)^(1/2) - a^(1/2))
*(3*a^(3/2)*c^7 - 3*a^7*c^(3/2) + 8*a^6*c^(5/2) - 8*a^(5/2)*c^6 - 6*a^5*c^(7/2) + 6*a^(7/2)*c^5 + a^3*c^(11/2)
- a^(11/2)*c^3))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6
*a^7*c^3 + a^8*c^2)))*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3/2) + 2*a^(3/2)*c^2)*
1i)/(a*c^6 - a^6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2))/((2*(a^(1/2)*c^(9/2) - 4*a^(3/2)*c^(7/2
) + 6*a^(5/2)*c^(5/2) - 4*a^(7/2)*c^(3/2) + a^(9/2)*c^(1/2)))/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5 - 20*a^4
*c^4 + 15*a^5*c^3 - 6*a^6*c^2) - (((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2*a*c + a^(1/2)*c^(3/2) + a^(3
/2)*c^(1/2)))^(1/2)*((6*a*c^(11/2) - 6*a^(11/2)*c + 2*a^(3/2)*c^5 - 2*a^5*c^(3/2) + 12*a^3*c^(7/2) - 12*a^(7/2
)*c^3 - 16*a^2*c^(9/2) + 16*a^(9/2)*c^2)/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5 - 20*a^4*c^4 + 15*a^5*c^3 - 6
*a^6*c^2) + (((a^(1/2)*c^(15/2) - 5*a^(3/2)*c^(13/2) + 9*a^(5/2)*c^(11/2) - 5*a^(7/2)*c^(9/2) - 5*a^(9/2)*c^(7
/2) + 9*a^(11/2)*c^(5/2) - 5*a^(13/2)*c^(3/2) + a^(15/2)*c^(1/2))/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5 - 20
*a^4*c^4 + 15*a^5*c^3 - 6*a^6*c^2) - (2*((a + b*x)^(1/2) - a^(1/2))*(a*c^9 + a^9*c - 7*a^2*c^8 + 22*a^3*c^7 -
41*a^4*c^6 + 50*a^5*c^5 - 41*a^6*c^4 + 22*a^7*c^3 - 7*a^8*c^2))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*a^3*
c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6*a^7*c^3 + a^8*c^2)))*((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2
))*(2*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))^(1/2)*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) +
2*a^2*c^(3/2) + 2*a^(3/2)*c^2))/(a*c^6 - a^6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2) - (2*((a +
b*x)^(1/2) - a^(1/2))*(3*a^(3/2)*c^7 - 3*a^7*c^(3/2) + 8*a^6*c^(5/2) - 8*a^(5/2)*c^6 - 6*a^5*c^(7/2) + 6*a^(7/
2)*c^5 + a^3*c^(11/2) - a^(11/2)*c^3))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4*c^6 - 20*a^5
*c^5 + 15*a^6*c^4 - 6*a^7*c^3 + a^8*c^2)))*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3
/2) + 2*a^(3/2)*c^2))/(a*c^6 - a^6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2) - (((a^(1/2)*c^(3/2) -
2*a*c + a^(3/2)*c^(1/2))*(2*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))^(1/2)*((((a^(1/2)*c^(15/2) - 5*a^(3/2)*
c^(13/2) + 9*a^(5/2)*c^(11/2) - 5*a^(7/2)*c^(9/2) - 5*a^(9/2)*c^(7/2) + 9*a^(11/2)*c^(5/2) - 5*a^(13/2)*c^(3/2
) + a^(15/2)*c^(1/2))/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a^3*c^5 - 20*a^4*c^4 + 15*a^5*c^3 - 6*a^6*c^2) - (2*((a
+ b*x)^(1/2) - a^(1/2))*(a*c^9 + a^9*c - 7*a^2*c^8 + 22*a^3*c^7 - 41*a^4*c^6 + 50*a^5*c^5 - 41*a^6*c^4 + 22*a^
7*c^3 - 7*a^8*c^2))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 -
6*a^7*c^3 + a^8*c^2)))*((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2
)))^(1/2)*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3/2) + 2*a^(3/2)*c^2))/(a*c^6 - a^
6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2) - (6*a*c^(11/2) - 6*a^(11/2)*c + 2*a^(3/2)*c^5 - 2*a^5*
c^(3/2) + 12*a^3*c^(7/2) - 12*a^(7/2)*c^3 - 16*a^2*c^(9/2) + 16*a^(9/2)*c^2)/(a*c^7 + a^7*c - 6*a^2*c^6 + 15*a
^3*c^5 - 20*a^4*c^4 + 15*a^5*c^3 - 6*a^6*c^2) + (2*((a + b*x)^(1/2) - a^(1/2))*(3*a^(3/2)*c^7 - 3*a^7*c^(3/2)
+ 8*a^6*c^(5/2) - 8*a^(5/2)*c^6 - 6*a^5*c^(7/2) + 6*a^(7/2)*c^5 + a^3*c^(11/2) - a^(11/2)*c^3))/(((c + b*x)^(1
/2) - c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6*a^7*c^3 + a^8*c^2)))*(a^(1/2)*c
^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3/2) + 2*a^(3/2)*c^2))/(a*c^6 - a^6*c - 5*a^2*c^5 + 10
*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2) + (4*((a + b*x)^(1/2) - a^(1/2))*(6*a^3*c^4 - a^6*c - 5*a^2*c^5 - a*c^6 + 6
*a^4*c^3 - 5*a^5*c^2 + 3*a^(3/2)*c^(11/2) + 4*a^(5/2)*c^(9/2) - 14*a^(7/2)*c^(7/2) + 4*a^(9/2)*c^(5/2) + 3*a^(
11/2)*c^(3/2)))/(((c + b*x)^(1/2) - c^(1/2))*(a^2*c^8 - 6*a^3*c^7 + 15*a^4*c^6 - 20*a^5*c^5 + 15*a^6*c^4 - 6*a
^7*c^3 + a^8*c^2))))*((a^(1/2)*c^(3/2) - 2*a*c + a^(3/2)*c^(1/2))*(2*a*c + a^(1/2)*c^(3/2) + a^(3/2)*c^(1/2)))
^(1/2)*(a^(1/2)*c^3 - 3*a^(5/2)*c - 3*a*c^(5/2) + a^3*c^(1/2) + 2*a^2*c^(3/2) + 2*a^(3/2)*c^2)*2i)/(a*c^6 - a^
6*c - 5*a^2*c^5 + 10*a^3*c^4 - 10*a^4*c^3 + 5*a^5*c^2)

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

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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