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

Optimal. Leaf size=157 \[ -\frac {4 a \sqrt {a+b x}}{x (b-c)^3}+\frac {4 a \sqrt {a+c x}}{x (b-c)^3}+\frac {2 (b+3 c) \sqrt {a+b x}}{(b-c)^3}-\frac {2 (3 b+c) \sqrt {a+c x}}{(b-c)^3}-\frac {6 \sqrt {a} (b+c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{(b-c)^3}+\frac {6 \sqrt {a} (b+c) \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{(b-c)^3} \]

[Out]

-6*(b+c)*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2)/(b-c)^3+6*(b+c)*arctanh((c*x+a)^(1/2)/a^(1/2))*a^(1/2)/(b-c)^3
+2*(b+3*c)*(b*x+a)^(1/2)/(b-c)^3-4*a*(b*x+a)^(1/2)/(b-c)^3/x-2*(3*b+c)*(c*x+a)^(1/2)/(b-c)^3+4*a*(c*x+a)^(1/2)
/(b-c)^3/x

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 223, normalized size of antiderivative = 1.42, number of steps used = 14, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6690, 47, 63, 208, 50} \[ -\frac {4 a \sqrt {a+b x}}{x (b-c)^3}+\frac {4 a \sqrt {a+c x}}{x (b-c)^3}+\frac {2 (b+3 c) \sqrt {a+b x}}{(b-c)^3}-\frac {2 (3 b+c) \sqrt {a+c x}}{(b-c)^3}-\frac {2 \sqrt {a} (b+3 c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{(b-c)^3}-\frac {4 \sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{(b-c)^3}+\frac {4 \sqrt {a} c \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{(b-c)^3}+\frac {2 \sqrt {a} (3 b+c) \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{(b-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.88, size = 192, normalized size = 1.22 \[ \frac {2 \left (-(3 b+c) \sqrt {a+c x}+(b+3 c) \sqrt {a+b x}+\sqrt {a} (3 b+c) \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )-\sqrt {a} (b+3 c) \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {2 a \left (b x \sqrt {\frac {b x}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {b x}{a}+1}\right )+a+b x\right )}{x \sqrt {a+b x}}+\frac {2 a \left (c x \sqrt {\frac {c x}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x}{a}+1}\right )+a+c x\right )}{x \sqrt {a+c x}}\right )}{(b-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 260, normalized size = 1.66 \[ \left [-\frac {3 \, \sqrt {a} {\left (b + c\right )} x \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 3 \, \sqrt {a} {\left (b + c\right )} x \log \left (\frac {c x - 2 \, \sqrt {c x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, {\left ({\left (b + 3 \, c\right )} x - 2 \, a\right )} \sqrt {b x + a} + 2 \, {\left ({\left (3 \, b + c\right )} x - 2 \, a\right )} \sqrt {c x + a}}{{\left (b^{3} - 3 \, b^{2} c + 3 \, b c^{2} - c^{3}\right )} x}, \frac {2 \, {\left (3 \, \sqrt {-a} {\left (b + c\right )} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - 3 \, \sqrt {-a} {\left (b + c\right )} x \arctan \left (\frac {\sqrt {c x + a} \sqrt {-a}}{a}\right ) + {\left ({\left (b + 3 \, c\right )} x - 2 \, a\right )} \sqrt {b x + a} - {\left ({\left (3 \, b + c\right )} x - 2 \, a\right )} \sqrt {c x + a}\right )}}{{\left (b^{3} - 3 \, b^{2} c + 3 \, b c^{2} - c^{3}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 78.70, size = 2318, normalized size = 14.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*(3*b*abs(b) + c*abs(b))/(b^4 - 3*b^3*c + 3*b^2*c^2 - b*c^3) + 2*sqrt(b
*x + a)*a*b/((b^3 - 3*b^2*c + 3*b*c^2 - c^3)*x) - 3*(a*b^2 + a*b*c)*arctan(sqrt(b*x + a)/sqrt(-a))/((b^3 - 3*b
^2*c + 3*b*c^2 - c^3)*sqrt(-a)) - (sqrt(b*x + a)*b^2 + 3*sqrt(b*x + a)*b*c)/(b^3 - 3*b^2*c + 3*b*c^2 - c^3) +
4*((sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))*a^2*b^3*c*abs(b) - (sqrt(b*c)*sqrt(b*x + a)
- sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))*a^2*b^2*c^2*abs(b) + (sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)
*b*c - a*b*c))^3*a*b*c*abs(b))/((a^2*b^4 - 2*a^2*b^3*c + a^2*b^2*c^2 - 2*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2
+ (b*x + a)*b*c - a*b*c))^2*a*b^2 - 2*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^2*a*b*c
+ (sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^4)*(b^3 - 3*b^2*c + 3*b*c^2 - c^3)) + 3*(2*
(a*b^4*c - a*b^2*c^3)*(a*b^4 - 3*a*b^3*c + 3*a*b^2*c^2 - a*b*c^3)^2*sqrt(-a)*abs(b)*sgn(b^3 - 3*b^2*c + 3*b*c^
2 - c^3) + 2*(a*b^4 - 3*a*b^3*c + 3*a*b^2*c^2 - a*b*c^3)^2*(a*b^4 - a*b^2*c^2)*sqrt(-a*b*c)*abs(b) + (a^2*b^8
- 4*a^2*b^7*c + 5*a^2*b^6*c^2 - 5*a^2*b^4*c^4 + 4*a^2*b^3*c^5 - a^2*b^2*c^6)*sqrt(-a*b*c)*abs(-a*b^4 + 3*a*b^3
*c - 3*a*b^2*c^2 + a*b*c^3)*abs(b)*sgn(b^3 - 3*b^2*c + 3*b*c^2 - c^3) + (a^2*b^9 - 4*a^2*b^8*c + 5*a^2*b^7*c^2
- 5*a^2*b^5*c^4 + 4*a^2*b^4*c^5 - a^2*b^3*c^6)*sqrt(-a)*abs(-a*b^4 + 3*a*b^3*c - 3*a*b^2*c^2 + a*b*c^3)*abs(b
) + (a^3*b^12*c - 5*a^3*b^11*c^2 + 8*a^3*b^10*c^3 - 14*a^3*b^8*c^5 + 14*a^3*b^7*c^6 - 8*a^3*b^5*c^8 + 5*a^3*b^
4*c^9 - a^3*b^3*c^10)*sqrt(-a)*abs(b)*sgn(b^3 - 3*b^2*c + 3*b*c^2 - c^3) + (a^3*b^12 - 5*a^3*b^11*c + 8*a^3*b^
10*c^2 - 14*a^3*b^8*c^4 + 14*a^3*b^7*c^5 - 8*a^3*b^5*c^7 + 5*a^3*b^4*c^8 - a^3*b^3*c^9)*sqrt(-a*b*c)*abs(b))*a
rctan(-(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))/sqrt(-(a*b^5 - 2*a*b^4*c + 2*a*b^2*c^3
- a*b*c^4 + sqrt((a*b^5 - 2*a*b^4*c + 2*a*b^2*c^3 - a*b*c^4)^2 - (a^2*b^7 - 5*a^2*b^6*c + 10*a^2*b^5*c^2 - 10*
a^2*b^4*c^3 + 5*a^2*b^3*c^4 - a^2*b^2*c^5)*(b^3 - 3*b^2*c + 3*b*c^2 - c^3)))/(b^3 - 3*b^2*c + 3*b*c^2 - c^3)))
/((b^11 - 9*b^10*c + 36*b^9*c^2 - 84*b^8*c^3 + 126*b^7*c^4 - 126*b^6*c^5 + 84*b^5*c^6 - 36*b^4*c^7 + 9*b^3*c^8
- b^2*c^9)*a^2*abs(-a*b^4 + 3*a*b^3*c - 3*a*b^2*c^2 + a*b*c^3)) - 3*(2*(a*b^4*c - a*b^2*c^3)*(a*b^4 - 3*a*b^3
*c + 3*a*b^2*c^2 - a*b*c^3)^2*sqrt(-a)*abs(b)*sgn(b^3 - 3*b^2*c + 3*b*c^2 - c^3) + 2*(a*b^4 - 3*a*b^3*c + 3*a*
b^2*c^2 - a*b*c^3)^2*(a*b^4 - a*b^2*c^2)*sqrt(-a*b*c)*abs(b) + (a^2*b^8 - 4*a^2*b^7*c + 5*a^2*b^6*c^2 - 5*a^2*
b^4*c^4 + 4*a^2*b^3*c^5 - a^2*b^2*c^6)*sqrt(-a*b*c)*abs(-a*b^4 + 3*a*b^3*c - 3*a*b^2*c^2 + a*b*c^3)*abs(b)*sgn
(b^3 - 3*b^2*c + 3*b*c^2 - c^3) + (a^2*b^9 - 4*a^2*b^8*c + 5*a^2*b^7*c^2 - 5*a^2*b^5*c^4 + 4*a^2*b^4*c^5 - a^2
*b^3*c^6)*sqrt(-a)*abs(-a*b^4 + 3*a*b^3*c - 3*a*b^2*c^2 + a*b*c^3)*abs(b) + (a^3*b^12*c - 5*a^3*b^11*c^2 + 8*a
^3*b^10*c^3 - 14*a^3*b^8*c^5 + 14*a^3*b^7*c^6 - 8*a^3*b^5*c^8 + 5*a^3*b^4*c^9 - a^3*b^3*c^10)*sqrt(-a)*abs(b)*
sgn(b^3 - 3*b^2*c + 3*b*c^2 - c^3) + (a^3*b^12 - 5*a^3*b^11*c + 8*a^3*b^10*c^2 - 14*a^3*b^8*c^4 + 14*a^3*b^7*c
^5 - 8*a^3*b^5*c^7 + 5*a^3*b^4*c^8 - a^3*b^3*c^9)*sqrt(-a*b*c)*abs(b))*arctan(-(sqrt(b*c)*sqrt(b*x + a) - sqrt
(a*b^2 + (b*x + a)*b*c - a*b*c))/sqrt(-(a*b^5 - 2*a*b^4*c + 2*a*b^2*c^3 - a*b*c^4 - sqrt((a*b^5 - 2*a*b^4*c +
2*a*b^2*c^3 - a*b*c^4)^2 - (a^2*b^7 - 5*a^2*b^6*c + 10*a^2*b^5*c^2 - 10*a^2*b^4*c^3 + 5*a^2*b^3*c^4 - a^2*b^2*
c^5)*(b^3 - 3*b^2*c + 3*b*c^2 - c^3)))/(b^3 - 3*b^2*c + 3*b*c^2 - c^3)))/((b^11 - 9*b^10*c + 36*b^9*c^2 - 84*b
^8*c^3 + 126*b^7*c^4 - 126*b^6*c^5 + 84*b^5*c^6 - 36*b^4*c^7 + 9*b^3*c^8 - b^2*c^9)*a^2*abs(-a*b^4 + 3*a*b^3*c
- 3*a*b^2*c^2 + a*b*c^3)))/b

________________________________________________________________________________________

maple [A]  time = 0.01, size = 237, normalized size = 1.51 \[ \frac {8 \left (-\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {b x +a}}{2 b x}\right ) a b}{\left (b -c \right )^{3}}-\frac {8 \left (-\frac {\arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\sqrt {c x +a}}{2 c x}\right ) a c}{\left (b -c \right )^{3}}+\frac {\left (-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\right ) b}{\left (b -c \right )^{3}}-\frac {3 \left (-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )+2 \sqrt {c x +a}\right ) b}{\left (b -c \right )^{3}}+\frac {3 \left (-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}\right ) c}{\left (b -c \right )^{3}}-\frac {\left (-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )+2 \sqrt {c x +a}\right ) c}{\left (b -c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 7.49, size = 559, normalized size = 3.56 \[ \frac {2\,\sqrt {a}\,b^2\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )\,\left (\frac {8\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}-\frac {2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}+\frac {3\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}+1\right )-2\,\sqrt {a}\,c^2\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )\,\left (\frac {2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^4}+\frac {3\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^3}\right )+2\,\sqrt {a}\,b\,c\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )\,\left (\frac {8\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}-\frac {14\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}+\frac {3\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}-\frac {3\,\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^3}\right )}{{\left (b-c\right )}^3\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (b-\frac {c\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^(1/2)*b^2*((a + c*x)^(1/2) - a^(1/2))*((8*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2) - a^(1/2)) - (2*(
(a + b*x)^(1/2) - a^(1/2))^2)/((a + c*x)^(1/2) - a^(1/2))^2 + (3*log(((a + b*x)^(1/2) - a^(1/2))/((a + c*x)^(1
/2) - a^(1/2)))*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2) - a^(1/2)) + 1) - 2*a^(1/2)*c^2*((a + c*x)^(1/2)
- a^(1/2))*((2*((a + b*x)^(1/2) - a^(1/2))^2)/((a + c*x)^(1/2) - a^(1/2))^2 - ((a + b*x)^(1/2) - a^(1/2))^4/(
(a + c*x)^(1/2) - a^(1/2))^4 + (3*log(((a + b*x)^(1/2) - a^(1/2))/((a + c*x)^(1/2) - a^(1/2)))*((a + b*x)^(1/2
) - a^(1/2))^3)/((a + c*x)^(1/2) - a^(1/2))^3) + 2*a^(1/2)*b*c*((a + c*x)^(1/2) - a^(1/2))*((8*((a + b*x)^(1/2
) - a^(1/2)))/((a + c*x)^(1/2) - a^(1/2)) - (14*((a + b*x)^(1/2) - a^(1/2))^2)/((a + c*x)^(1/2) - a^(1/2))^2 +
(3*log(((a + b*x)^(1/2) - a^(1/2))/((a + c*x)^(1/2) - a^(1/2)))*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2)
- a^(1/2)) - (3*log(((a + b*x)^(1/2) - a^(1/2))/((a + c*x)^(1/2) - a^(1/2)))*((a + b*x)^(1/2) - a^(1/2))^3)/(
(a + c*x)^(1/2) - a^(1/2))^3))/((b - c)^3*((a + b*x)^(1/2) - a^(1/2))*(b - (c*((a + b*x)^(1/2) - a^(1/2))^2)/(
(a + c*x)^(1/2) - a^(1/2))^2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________