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

Optimal. Leaf size=261 \[ \frac{8 a^2 (a+b x)^{3/2}}{3 b^2 (a-c)^3}-\frac{8 c^2 (b x+c)^{3/2}}{3 b^2 (a-c)^3}+\frac{8 (a+b x)^{7/2}}{7 b^2 (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{5/2}}{5 b^2 (a-c)^3}-\frac{16 a (a+b x)^{5/2}}{5 b^2 (a-c)^3}-\frac{2 a (a+3 c) (a+b x)^{3/2}}{3 b^2 (a-c)^3}-\frac{8 (b x+c)^{7/2}}{7 b^2 (a-c)^3}+\frac{16 c (b x+c)^{5/2}}{5 b^2 (a-c)^3}-\frac{2 (3 a+c) (b x+c)^{5/2}}{5 b^2 (a-c)^3}+\frac{2 c (3 a+c) (b x+c)^{3/2}}{3 b^2 (a-c)^3} \]

[Out]

(8*a^2*(a + b*x)^(3/2))/(3*b^2*(a - c)^3) - (2*a*(a + 3*c)*(a + b*x)^(3/2))/(3*b^2*(a - c)^3) - (16*a*(a + b*x
)^(5/2))/(5*b^2*(a - c)^3) + (2*(a + 3*c)*(a + b*x)^(5/2))/(5*b^2*(a - c)^3) + (8*(a + b*x)^(7/2))/(7*b^2*(a -
c)^3) - (8*c^2*(c + b*x)^(3/2))/(3*b^2*(a - c)^3) + (2*c*(3*a + c)*(c + b*x)^(3/2))/(3*b^2*(a - c)^3) + (16*c
*(c + b*x)^(5/2))/(5*b^2*(a - c)^3) - (2*(3*a + c)*(c + b*x)^(5/2))/(5*b^2*(a - c)^3) - (8*(c + b*x)^(7/2))/(7
*b^2*(a - c)^3)

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Rubi [A]  time = 0.236346, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {6689, 43} \[ \frac{8 a^2 (a+b x)^{3/2}}{3 b^2 (a-c)^3}-\frac{8 c^2 (b x+c)^{3/2}}{3 b^2 (a-c)^3}+\frac{8 (a+b x)^{7/2}}{7 b^2 (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{5/2}}{5 b^2 (a-c)^3}-\frac{16 a (a+b x)^{5/2}}{5 b^2 (a-c)^3}-\frac{2 a (a+3 c) (a+b x)^{3/2}}{3 b^2 (a-c)^3}-\frac{8 (b x+c)^{7/2}}{7 b^2 (a-c)^3}+\frac{16 c (b x+c)^{5/2}}{5 b^2 (a-c)^3}-\frac{2 (3 a+c) (b x+c)^{5/2}}{5 b^2 (a-c)^3}+\frac{2 c (3 a+c) (b x+c)^{3/2}}{3 b^2 (a-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(a + b*x)^(3/2))/(3*b^2*(a - c)^3) - (2*a*(a + 3*c)*(a + b*x)^(3/2))/(3*b^2*(a - c)^3) - (16*a*(a + b*x
)^(5/2))/(5*b^2*(a - c)^3) + (2*(a + 3*c)*(a + b*x)^(5/2))/(5*b^2*(a - c)^3) + (8*(a + b*x)^(7/2))/(7*b^2*(a -
c)^3) - (8*c^2*(c + b*x)^(3/2))/(3*b^2*(a - c)^3) + (2*c*(3*a + c)*(c + b*x)^(3/2))/(3*b^2*(a - c)^3) + (16*c
*(c + b*x)^(5/2))/(5*b^2*(a - c)^3) - (2*(3*a + c)*(c + b*x)^(5/2))/(5*b^2*(a - c)^3) - (8*(c + b*x)^(7/2))/(7
*b^2*(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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x}{\left (\sqrt{a+b x}+\sqrt{c+b x}\right )^3} \, dx &=\frac{\int \left (a \left (1+\frac{3 c}{a}\right ) x \sqrt{a+b x}+4 b x^2 \sqrt{a+b x}-3 a \left (1+\frac{c}{3 a}\right ) x \sqrt{c+b x}-4 b x^2 \sqrt{c+b x}\right ) \, dx}{(a-c)^3}\\ &=\frac{(4 b) \int x^2 \sqrt{a+b x} \, dx}{(a-c)^3}-\frac{(4 b) \int x^2 \sqrt{c+b x} \, dx}{(a-c)^3}-\frac{(3 a+c) \int x \sqrt{c+b x} \, dx}{(a-c)^3}+\frac{(a+3 c) \int x \sqrt{a+b x} \, dx}{(a-c)^3}\\ &=\frac{(4 b) \int \left (\frac{a^2 \sqrt{a+b x}}{b^2}-\frac{2 a (a+b x)^{3/2}}{b^2}+\frac{(a+b x)^{5/2}}{b^2}\right ) \, dx}{(a-c)^3}-\frac{(4 b) \int \left (\frac{c^2 \sqrt{c+b x}}{b^2}-\frac{2 c (c+b x)^{3/2}}{b^2}+\frac{(c+b x)^{5/2}}{b^2}\right ) \, dx}{(a-c)^3}-\frac{(3 a+c) \int \left (-\frac{c \sqrt{c+b x}}{b}+\frac{(c+b x)^{3/2}}{b}\right ) \, dx}{(a-c)^3}+\frac{(a+3 c) \int \left (-\frac{a \sqrt{a+b x}}{b}+\frac{(a+b x)^{3/2}}{b}\right ) \, dx}{(a-c)^3}\\ &=\frac{8 a^2 (a+b x)^{3/2}}{3 b^2 (a-c)^3}-\frac{2 a (a+3 c) (a+b x)^{3/2}}{3 b^2 (a-c)^3}-\frac{16 a (a+b x)^{5/2}}{5 b^2 (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{5/2}}{5 b^2 (a-c)^3}+\frac{8 (a+b x)^{7/2}}{7 b^2 (a-c)^3}-\frac{8 c^2 (c+b x)^{3/2}}{3 b^2 (a-c)^3}+\frac{2 c (3 a+c) (c+b x)^{3/2}}{3 b^2 (a-c)^3}+\frac{16 c (c+b x)^{5/2}}{5 b^2 (a-c)^3}-\frac{2 (3 a+c) (c+b x)^{5/2}}{5 b^2 (a-c)^3}-\frac{8 (c+b x)^{7/2}}{7 b^2 (a-c)^3}\\ \end{align*}

Mathematica [A]  time = 0.269916, size = 214, normalized size = 0.82 \[ \frac{2 \left (-a^2 \sqrt{a+b x} (3 b x+14 c)+6 a^3 \sqrt{a+b x}+a \left (b^2 x^2 \left (11 \sqrt{a+b x}-21 \sqrt{b x+c}\right )+7 b c x \left (\sqrt{a+b x}-\sqrt{b x+c}\right )+14 c^2 \sqrt{b x+c}\right )+b^2 c x^2 \left (21 \sqrt{a+b x}-11 \sqrt{b x+c}\right )+20 b^3 x^3 \left (\sqrt{a+b x}-\sqrt{b x+c}\right )-6 c^3 \sqrt{b x+c}+3 b c^2 x \sqrt{b x+c}\right )}{35 b^2 (a-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*a^3*Sqrt[a + b*x] - 6*c^3*Sqrt[c + b*x] + 3*b*c^2*x*Sqrt[c + b*x] - a^2*Sqrt[a + b*x]*(14*c + 3*b*x) + b
^2*c*x^2*(21*Sqrt[a + b*x] - 11*Sqrt[c + b*x]) + 20*b^3*x^3*(Sqrt[a + b*x] - Sqrt[c + b*x]) + a*(14*c^2*Sqrt[c
+ b*x] + b^2*x^2*(11*Sqrt[a + b*x] - 21*Sqrt[c + b*x]) + 7*b*c*x*(Sqrt[a + b*x] - Sqrt[c + b*x]))))/(35*b^2*(
a - c)^3)

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Maple [A]  time = 0.003, size = 222, normalized size = 0.9 \begin{align*} 2\,{\frac{a \left ( 1/5\, \left ( bx+a \right ) ^{5/2}-1/3\,a \left ( bx+a \right ) ^{3/2} \right ) }{ \left ( a-c \right ) ^{3}{b}^{2}}}+6\,{\frac{c \left ( 1/5\, \left ( bx+a \right ) ^{5/2}-1/3\,a \left ( bx+a \right ) ^{3/2} \right ) }{ \left ( a-c \right ) ^{3}{b}^{2}}}-6\,{\frac{a \left ( 1/5\, \left ( bx+c \right ) ^{5/2}-1/3\,c \left ( bx+c \right ) ^{3/2} \right ) }{ \left ( a-c \right ) ^{3}{b}^{2}}}-2\,{\frac{c \left ( 1/5\, \left ( bx+c \right ) ^{5/2}-1/3\,c \left ( bx+c \right ) ^{3/2} \right ) }{ \left ( a-c \right ) ^{3}{b}^{2}}}+8\,{\frac{1/7\, \left ( bx+a \right ) ^{7/2}-2/5\,a \left ( bx+a \right ) ^{5/2}+1/3\,{a}^{2} \left ( bx+a \right ) ^{3/2}}{ \left ( a-c \right ) ^{3}{b}^{2}}}-8\,{\frac{1/7\, \left ( bx+c \right ) ^{7/2}-2/5\,c \left ( bx+c \right ) ^{5/2}+1/3\,{c}^{2} \left ( bx+c \right ) ^{3/2}}{ \left ( a-c \right ) ^{3}{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a-c)^3*a/b^2*(1/5*(b*x+a)^(5/2)-1/3*a*(b*x+a)^(3/2))+6/(a-c)^3*c/b^2*(1/5*(b*x+a)^(5/2)-1/3*a*(b*x+a)^(3/2)
)-6/(a-c)^3*a/b^2*(1/5*(b*x+c)^(5/2)-1/3*c*(b*x+c)^(3/2))-2/(a-c)^3*c/b^2*(1/5*(b*x+c)^(5/2)-1/3*c*(b*x+c)^(3/
2))+8/(a-c)^3/b^2*(1/7*(b*x+a)^(7/2)-2/5*a*(b*x+a)^(5/2)+1/3*a^2*(b*x+a)^(3/2))-8/(a-c)^3/b^2*(1/7*(b*x+c)^(7/
2)-2/5*c*(b*x+c)^(5/2)+1/3*c^2*(b*x+c)^(3/2))

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

[Out]

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

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Fricas [A]  time = 0.998509, size = 343, normalized size = 1.31 \begin{align*} \frac{2 \,{\left ({\left (20 \, b^{3} x^{3} + 6 \, a^{3} - 14 \, a^{2} c +{\left (11 \, a b^{2} + 21 \, b^{2} c\right )} x^{2} -{\left (3 \, a^{2} b - 7 \, a b c\right )} x\right )} \sqrt{b x + a} -{\left (20 \, b^{3} x^{3} - 14 \, a c^{2} + 6 \, c^{3} +{\left (21 \, a b^{2} + 11 \, b^{2} c\right )} x^{2} +{\left (7 \, a b c - 3 \, b c^{2}\right )} x\right )} \sqrt{b x + c}\right )}}{35 \,{\left (a^{3} b^{2} - 3 \, a^{2} b^{2} c + 3 \, a b^{2} c^{2} - b^{2} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*((20*b^3*x^3 + 6*a^3 - 14*a^2*c + (11*a*b^2 + 21*b^2*c)*x^2 - (3*a^2*b - 7*a*b*c)*x)*sqrt(b*x + a) - (20*
b^3*x^3 - 14*a*c^2 + 6*c^3 + (21*a*b^2 + 11*b^2*c)*x^2 + (7*a*b*c - 3*b*c^2)*x)*sqrt(b*x + c))/(a^3*b^2 - 3*a^
2*b^2*c + 3*a*b^2*c^2 - b^2*c^3)

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Sympy [A]  time = 2.07677, size = 942, normalized size = 3.61 \begin{align*} \begin{cases} \frac{12 a^{2}}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{54 a b x}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{44 a c}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{36 a \sqrt{a + b x} \sqrt{b x + c}}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{40 b^{2} x^{2}}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{54 b c x}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{30 b x \sqrt{a + b x} \sqrt{b x + c}}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{12 c^{2}}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} + \frac{36 c \sqrt{a + b x} \sqrt{b x + c}}{35 a b^{2} \sqrt{a + b x} + 105 a b^{2} \sqrt{b x + c} + 140 b^{3} x \sqrt{a + b x} + 140 b^{3} x \sqrt{b x + c} + 105 b^{2} c \sqrt{a + b x} + 35 b^{2} c \sqrt{b x + c}} & \text{for}\: b \neq 0 \\\frac{x^{2}}{2 \left (\sqrt{a} + \sqrt{c}\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((12*a**2/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x) + 140*b**3*x
*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)) + 54*a*b*x/(35*a*b**2*sqrt(a + b*x) + 105
*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x) + 140*b**3*x*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b*
*2*c*sqrt(b*x + c)) + 44*a*c/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x) +
140*b**3*x*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)) + 36*a*sqrt(a + b*x)*sqrt(b*x +
c)/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x) + 140*b**3*x*sqrt(b*x + c)
+ 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)) + 40*b**2*x**2/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqr
t(b*x + c) + 140*b**3*x*sqrt(a + b*x) + 140*b**3*x*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b
*x + c)) + 54*b*c*x/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x) + 140*b**3*
x*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)) + 30*b*x*sqrt(a + b*x)*sqrt(b*x + c)/(35
*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x) + 140*b**3*x*sqrt(b*x + c) + 105*b
**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)) + 12*c**2/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) +
140*b**3*x*sqrt(a + b*x) + 140*b**3*x*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)) + 3
6*c*sqrt(a + b*x)*sqrt(b*x + c)/(35*a*b**2*sqrt(a + b*x) + 105*a*b**2*sqrt(b*x + c) + 140*b**3*x*sqrt(a + b*x)
+ 140*b**3*x*sqrt(b*x + c) + 105*b**2*c*sqrt(a + b*x) + 35*b**2*c*sqrt(b*x + c)), Ne(b, 0)), (x**2/(2*(sqrt(a
) + sqrt(c))**3), True))

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

[Out]

Timed out