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

Optimal. Leaf size=64 \[ \frac{(a-c)^2}{10 b \left (\sqrt{a+b x}+\sqrt{b x+c}\right )^5}-\frac{1}{2 b \left (\sqrt{a+b x}+\sqrt{b x+c}\right )} \]

[Out]

(a - c)^2/(10*b*(Sqrt[a + b*x] + Sqrt[c + b*x])^5) - 1/(2*b*(Sqrt[a + b*x] + Sqrt[c + b*x]))

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Rubi [B]  time = 0.0946979, antiderivative size = 151, normalized size of antiderivative = 2.36, number of steps used = 6, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {6689, 43} \[ \frac{8 (a+b x)^{5/2}}{5 b (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{3/2}}{3 b (a-c)^3}-\frac{8 a (a+b x)^{3/2}}{3 b (a-c)^3}-\frac{8 (b x+c)^{5/2}}{5 b (a-c)^3}+\frac{8 c (b x+c)^{3/2}}{3 b (a-c)^3}-\frac{2 (3 a+c) (b x+c)^{3/2}}{3 b (a-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.151964, size = 151, normalized size = 2.36 \[ \frac{8 (a+b x)^{5/2}}{5 b (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{3/2}}{3 b (a-c)^3}-\frac{8 a (a+b x)^{3/2}}{3 b (a-c)^3}-\frac{8 (b x+c)^{5/2}}{5 b (a-c)^3}+\frac{8 c (b x+c)^{3/2}}{3 b (a-c)^3}-\frac{2 (3 a+c) (b x+c)^{3/2}}{3 b (a-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.003, size = 146, normalized size = 2.3 \begin{align*}{\frac{2\,a}{3\,b \left ( a-c \right ) ^{3}} \left ( bx+a \right ) ^{{\frac{3}{2}}}}+2\,{\frac{c \left ( bx+a \right ) ^{3/2}}{b \left ( a-c \right ) ^{3}}}-2\,{\frac{a \left ( bx+c \right ) ^{3/2}}{b \left ( a-c \right ) ^{3}}}-{\frac{2\,c}{3\,b \left ( a-c \right ) ^{3}} \left ( bx+c \right ) ^{{\frac{3}{2}}}}+8\,{\frac{1/5\, \left ( bx+a \right ) ^{5/2}-1/3\,a \left ( bx+a \right ) ^{3/2}}{b \left ( a-c \right ) ^{3}}}-8\,{\frac{1/5\, \left ( bx+c \right ) ^{5/2}-1/3\,c \left ( bx+c \right ) ^{3/2}}{b \left ( a-c \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [B]  time = 0.988146, size = 228, normalized size = 3.56 \begin{align*} \frac{2 \,{\left ({\left (4 \, b^{2} x^{2} - a^{2} + 5 \, a c +{\left (3 \, a b + 5 \, b c\right )} x\right )} \sqrt{b x + a} -{\left (4 \, b^{2} x^{2} + 5 \, a c - c^{2} +{\left (5 \, a b + 3 \, b c\right )} x\right )} \sqrt{b x + c}\right )}}{5 \,{\left (a^{3} b - 3 \, a^{2} b c + 3 \, a b c^{2} - b c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*((4*b^2*x^2 - a^2 + 5*a*c + (3*a*b + 5*b*c)*x)*sqrt(b*x + a) - (4*b^2*x^2 + 5*a*c - c^2 + (5*a*b + 3*b*c)*
x)*sqrt(b*x + c))/(a^3*b - 3*a^2*b*c + 3*a*b*c^2 - b*c^3)

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Sympy [A]  time = 1.91095, size = 384, normalized size = 6. \begin{align*} \begin{cases} - \frac{2 a}{5 a b \sqrt{a + b x} + 15 a b \sqrt{b x + c} + 20 b^{2} x \sqrt{a + b x} + 20 b^{2} x \sqrt{b x + c} + 15 b c \sqrt{a + b x} + 5 b c \sqrt{b x + c}} - \frac{4 b x}{5 a b \sqrt{a + b x} + 15 a b \sqrt{b x + c} + 20 b^{2} x \sqrt{a + b x} + 20 b^{2} x \sqrt{b x + c} + 15 b c \sqrt{a + b x} + 5 b c \sqrt{b x + c}} - \frac{2 c}{5 a b \sqrt{a + b x} + 15 a b \sqrt{b x + c} + 20 b^{2} x \sqrt{a + b x} + 20 b^{2} x \sqrt{b x + c} + 15 b c \sqrt{a + b x} + 5 b c \sqrt{b x + c}} - \frac{6 \sqrt{a + b x} \sqrt{b x + c}}{5 a b \sqrt{a + b x} + 15 a b \sqrt{b x + c} + 20 b^{2} x \sqrt{a + b x} + 20 b^{2} x \sqrt{b x + c} + 15 b c \sqrt{a + b x} + 5 b c \sqrt{b x + c}} & \text{for}\: b \neq 0 \\\frac{x}{\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(1/((b*x+a)**(1/2)+(b*x+c)**(1/2))**3,x)

[Out]

Piecewise((-2*a/(5*a*b*sqrt(a + b*x) + 15*a*b*sqrt(b*x + c) + 20*b**2*x*sqrt(a + b*x) + 20*b**2*x*sqrt(b*x + c
) + 15*b*c*sqrt(a + b*x) + 5*b*c*sqrt(b*x + c)) - 4*b*x/(5*a*b*sqrt(a + b*x) + 15*a*b*sqrt(b*x + c) + 20*b**2*
x*sqrt(a + b*x) + 20*b**2*x*sqrt(b*x + c) + 15*b*c*sqrt(a + b*x) + 5*b*c*sqrt(b*x + c)) - 2*c/(5*a*b*sqrt(a +
b*x) + 15*a*b*sqrt(b*x + c) + 20*b**2*x*sqrt(a + b*x) + 20*b**2*x*sqrt(b*x + c) + 15*b*c*sqrt(a + b*x) + 5*b*c
*sqrt(b*x + c)) - 6*sqrt(a + b*x)*sqrt(b*x + c)/(5*a*b*sqrt(a + b*x) + 15*a*b*sqrt(b*x + c) + 20*b**2*x*sqrt(a
+ b*x) + 20*b**2*x*sqrt(b*x + c) + 15*b*c*sqrt(a + b*x) + 5*b*c*sqrt(b*x + c)), Ne(b, 0)), (x/(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(1/((b*x+a)^(1/2)+(b*x+c)^(1/2))^3,x, algorithm="giac")

[Out]

Timed out