3.411 \(\int \frac{x^2}{(\sqrt{a+b x}+\sqrt{c+b x})^3} \, dx\)

Optimal. Leaf size=375 \[ \frac{24 a^2 (a+b x)^{5/2}}{5 b^3 (a-c)^3}+\frac{2 a^2 (a+3 c) (a+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac{8 a^3 (a+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac{24 c^2 (b x+c)^{5/2}}{5 b^3 (a-c)^3}+\frac{8 c^3 (b x+c)^{3/2}}{3 b^3 (a-c)^3}-\frac{2 c^2 (3 a+c) (b x+c)^{3/2}}{3 b^3 (a-c)^3}+\frac{8 (a+b x)^{9/2}}{9 b^3 (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac{24 a (a+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac{4 a (a+3 c) (a+b x)^{5/2}}{5 b^3 (a-c)^3}-\frac{8 (b x+c)^{9/2}}{9 b^3 (a-c)^3}+\frac{24 c (b x+c)^{7/2}}{7 b^3 (a-c)^3}-\frac{2 (3 a+c) (b x+c)^{7/2}}{7 b^3 (a-c)^3}+\frac{4 c (3 a+c) (b x+c)^{5/2}}{5 b^3 (a-c)^3} \]

[Out]

(-8*a^3*(a + b*x)^(3/2))/(3*b^3*(a - c)^3) + (2*a^2*(a + 3*c)*(a + b*x)^(3/2))/(3*b^3*(a - c)^3) + (24*a^2*(a
+ b*x)^(5/2))/(5*b^3*(a - c)^3) - (4*a*(a + 3*c)*(a + b*x)^(5/2))/(5*b^3*(a - c)^3) - (24*a*(a + b*x)^(7/2))/(
7*b^3*(a - c)^3) + (2*(a + 3*c)*(a + b*x)^(7/2))/(7*b^3*(a - c)^3) + (8*(a + b*x)^(9/2))/(9*b^3*(a - c)^3) + (
8*c^3*(c + b*x)^(3/2))/(3*b^3*(a - c)^3) - (2*c^2*(3*a + c)*(c + b*x)^(3/2))/(3*b^3*(a - c)^3) - (24*c^2*(c +
b*x)^(5/2))/(5*b^3*(a - c)^3) + (4*c*(3*a + c)*(c + b*x)^(5/2))/(5*b^3*(a - c)^3) + (24*c*(c + b*x)^(7/2))/(7*
b^3*(a - c)^3) - (2*(3*a + c)*(c + b*x)^(7/2))/(7*b^3*(a - c)^3) - (8*(c + b*x)^(9/2))/(9*b^3*(a - c)^3)

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Rubi [A]  time = 0.37164, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {6689, 43} \[ \frac{24 a^2 (a+b x)^{5/2}}{5 b^3 (a-c)^3}+\frac{2 a^2 (a+3 c) (a+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac{8 a^3 (a+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac{24 c^2 (b x+c)^{5/2}}{5 b^3 (a-c)^3}+\frac{8 c^3 (b x+c)^{3/2}}{3 b^3 (a-c)^3}-\frac{2 c^2 (3 a+c) (b x+c)^{3/2}}{3 b^3 (a-c)^3}+\frac{8 (a+b x)^{9/2}}{9 b^3 (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac{24 a (a+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac{4 a (a+3 c) (a+b x)^{5/2}}{5 b^3 (a-c)^3}-\frac{8 (b x+c)^{9/2}}{9 b^3 (a-c)^3}+\frac{24 c (b x+c)^{7/2}}{7 b^3 (a-c)^3}-\frac{2 (3 a+c) (b x+c)^{7/2}}{7 b^3 (a-c)^3}+\frac{4 c (3 a+c) (b x+c)^{5/2}}{5 b^3 (a-c)^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(Sqrt[a + b*x] + Sqrt[c + b*x])^3,x]

[Out]

(-8*a^3*(a + b*x)^(3/2))/(3*b^3*(a - c)^3) + (2*a^2*(a + 3*c)*(a + b*x)^(3/2))/(3*b^3*(a - c)^3) + (24*a^2*(a
+ b*x)^(5/2))/(5*b^3*(a - c)^3) - (4*a*(a + 3*c)*(a + b*x)^(5/2))/(5*b^3*(a - c)^3) - (24*a*(a + b*x)^(7/2))/(
7*b^3*(a - c)^3) + (2*(a + 3*c)*(a + b*x)^(7/2))/(7*b^3*(a - c)^3) + (8*(a + b*x)^(9/2))/(9*b^3*(a - c)^3) + (
8*c^3*(c + b*x)^(3/2))/(3*b^3*(a - c)^3) - (2*c^2*(3*a + c)*(c + b*x)^(3/2))/(3*b^3*(a - c)^3) - (24*c^2*(c +
b*x)^(5/2))/(5*b^3*(a - c)^3) + (4*c*(3*a + c)*(c + b*x)^(5/2))/(5*b^3*(a - c)^3) + (24*c*(c + b*x)^(7/2))/(7*
b^3*(a - c)^3) - (2*(3*a + c)*(c + b*x)^(7/2))/(7*b^3*(a - c)^3) - (8*(c + b*x)^(9/2))/(9*b^3*(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^2}{\left (\sqrt{a+b x}+\sqrt{c+b x}\right )^3} \, dx &=\frac{\int \left (a \left (1+\frac{3 c}{a}\right ) x^2 \sqrt{a+b x}+4 b x^3 \sqrt{a+b x}-3 a \left (1+\frac{c}{3 a}\right ) x^2 \sqrt{c+b x}-4 b x^3 \sqrt{c+b x}\right ) \, dx}{(a-c)^3}\\ &=\frac{(4 b) \int x^3 \sqrt{a+b x} \, dx}{(a-c)^3}-\frac{(4 b) \int x^3 \sqrt{c+b x} \, dx}{(a-c)^3}-\frac{(3 a+c) \int x^2 \sqrt{c+b x} \, dx}{(a-c)^3}+\frac{(a+3 c) \int x^2 \sqrt{a+b x} \, dx}{(a-c)^3}\\ &=\frac{(4 b) \int \left (-\frac{a^3 \sqrt{a+b x}}{b^3}+\frac{3 a^2 (a+b x)^{3/2}}{b^3}-\frac{3 a (a+b x)^{5/2}}{b^3}+\frac{(a+b x)^{7/2}}{b^3}\right ) \, dx}{(a-c)^3}-\frac{(4 b) \int \left (-\frac{c^3 \sqrt{c+b x}}{b^3}+\frac{3 c^2 (c+b x)^{3/2}}{b^3}-\frac{3 c (c+b x)^{5/2}}{b^3}+\frac{(c+b x)^{7/2}}{b^3}\right ) \, dx}{(a-c)^3}-\frac{(3 a+c) \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{(a+3 c) \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{8 a^3 (a+b x)^{3/2}}{3 b^3 (a-c)^3}+\frac{2 a^2 (a+3 c) (a+b x)^{3/2}}{3 b^3 (a-c)^3}+\frac{24 a^2 (a+b x)^{5/2}}{5 b^3 (a-c)^3}-\frac{4 a (a+3 c) (a+b x)^{5/2}}{5 b^3 (a-c)^3}-\frac{24 a (a+b x)^{7/2}}{7 b^3 (a-c)^3}+\frac{2 (a+3 c) (a+b x)^{7/2}}{7 b^3 (a-c)^3}+\frac{8 (a+b x)^{9/2}}{9 b^3 (a-c)^3}+\frac{8 c^3 (c+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac{2 c^2 (3 a+c) (c+b x)^{3/2}}{3 b^3 (a-c)^3}-\frac{24 c^2 (c+b x)^{5/2}}{5 b^3 (a-c)^3}+\frac{4 c (3 a+c) (c+b x)^{5/2}}{5 b^3 (a-c)^3}+\frac{24 c (c+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac{2 (3 a+c) (c+b x)^{7/2}}{7 b^3 (a-c)^3}-\frac{8 (c+b x)^{9/2}}{9 b^3 (a-c)^3}\\ \end{align*}

Mathematica [A]  time = 0.412515, size = 282, normalized size = 0.75 \[ \frac{2 \left (4 a^3 \sqrt{a+b x} (5 b x+18 c)-3 a^2 b x \sqrt{a+b x} (5 b x+12 c)-40 a^4 \sqrt{a+b x}+a \left (27 b^2 c x^2 \left (\sqrt{a+b x}-\sqrt{b x+c}\right )+5 b^3 x^3 \left (13 \sqrt{a+b x}-27 \sqrt{b x+c}\right )-72 c^3 \sqrt{b x+c}+36 b c^2 x \sqrt{b x+c}\right )+5 \left (b^3 c x^3 \left (27 \sqrt{a+b x}-13 \sqrt{b x+c}\right )+28 b^4 x^4 \left (\sqrt{a+b x}-\sqrt{b x+c}\right )+3 b^2 c^2 x^2 \sqrt{b x+c}+8 c^4 \sqrt{b x+c}-4 b c^3 x \sqrt{b x+c}\right )\right )}{315 b^3 (a-c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(Sqrt[a + b*x] + Sqrt[c + b*x])^3,x]

[Out]

(2*(-40*a^4*Sqrt[a + b*x] - 3*a^2*b*x*Sqrt[a + b*x]*(12*c + 5*b*x) + 4*a^3*Sqrt[a + b*x]*(18*c + 5*b*x) + a*(-
72*c^3*Sqrt[c + b*x] + 36*b*c^2*x*Sqrt[c + b*x] + 5*b^3*x^3*(13*Sqrt[a + b*x] - 27*Sqrt[c + b*x]) + 27*b^2*c*x
^2*(Sqrt[a + b*x] - Sqrt[c + b*x])) + 5*(8*c^4*Sqrt[c + b*x] - 4*b*c^3*x*Sqrt[c + b*x] + 3*b^2*c^2*x^2*Sqrt[c
+ b*x] + b^3*c*x^3*(27*Sqrt[a + b*x] - 13*Sqrt[c + b*x]) + 28*b^4*x^4*(Sqrt[a + b*x] - Sqrt[c + b*x]))))/(315*
b^3*(a - c)^3)

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Maple [A]  time = 0.003, size = 294, normalized size = 0.8 \begin{align*} 2\,{\frac{a \left ( 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} \right ) }{ \left ( a-c \right ) ^{3}{b}^{3}}}+6\,{\frac{c \left ( 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} \right ) }{ \left ( a-c \right ) ^{3}{b}^{3}}}-6\,{\frac{a \left ( 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} \right ) }{ \left ( a-c \right ) ^{3}{b}^{3}}}-2\,{\frac{c \left ( 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} \right ) }{ \left ( a-c \right ) ^{3}{b}^{3}}}+8\,{\frac{1/9\, \left ( bx+a \right ) ^{9/2}-3/7\, \left ( bx+a \right ) ^{7/2}a+3/5\,{a}^{2} \left ( bx+a \right ) ^{5/2}-1/3\,{a}^{3} \left ( bx+a \right ) ^{3/2}}{ \left ( a-c \right ) ^{3}{b}^{3}}}-8\,{\frac{1/9\, \left ( bx+c \right ) ^{9/2}-3/7\,c \left ( bx+c \right ) ^{7/2}+3/5\,{c}^{2} \left ( bx+c \right ) ^{5/2}-1/3\,{c}^{3} \left ( bx+c \right ) ^{3/2}}{ \left ( a-c \right ) ^{3}{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a-c)^3*a/b^3*(1/7*(b*x+a)^(7/2)-2/5*a*(b*x+a)^(5/2)+1/3*a^2*(b*x+a)^(3/2))+6/(a-c)^3*c/b^3*(1/7*(b*x+a)^(7/
2)-2/5*a*(b*x+a)^(5/2)+1/3*a^2*(b*x+a)^(3/2))-6/(a-c)^3*a/b^3*(1/7*(b*x+c)^(7/2)-2/5*c*(b*x+c)^(5/2)+1/3*c^2*(
b*x+c)^(3/2))-2/(a-c)^3*c/b^3*(1/7*(b*x+c)^(7/2)-2/5*c*(b*x+c)^(5/2)+1/3*c^2*(b*x+c)^(3/2))+8/(a-c)^3/b^3*(1/9
*(b*x+a)^(9/2)-3/7*(b*x+a)^(7/2)*a+3/5*a^2*(b*x+a)^(5/2)-1/3*a^3*(b*x+a)^(3/2))-8/(a-c)^3/b^3*(1/9*(b*x+c)^(9/
2)-3/7*c*(b*x+c)^(7/2)+3/5*c^2*(b*x+c)^(5/2)-1/3*c^3*(b*x+c)^(3/2))

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

[Out]

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

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Fricas [A]  time = 0.945516, size = 452, normalized size = 1.21 \begin{align*} \frac{2 \,{\left ({\left (140 \, b^{4} x^{4} - 40 \, a^{4} + 72 \, a^{3} c + 5 \,{\left (13 \, a b^{3} + 27 \, b^{3} c\right )} x^{3} - 3 \,{\left (5 \, a^{2} b^{2} - 9 \, a b^{2} c\right )} x^{2} + 4 \,{\left (5 \, a^{3} b - 9 \, a^{2} b c\right )} x\right )} \sqrt{b x + a} -{\left (140 \, b^{4} x^{4} + 72 \, a c^{3} - 40 \, c^{4} + 5 \,{\left (27 \, a b^{3} + 13 \, b^{3} c\right )} x^{3} + 3 \,{\left (9 \, a b^{2} c - 5 \, b^{2} c^{2}\right )} x^{2} - 4 \,{\left (9 \, a b c^{2} - 5 \, b c^{3}\right )} x\right )} \sqrt{b x + c}\right )}}{315 \,{\left (a^{3} b^{3} - 3 \, a^{2} b^{3} c + 3 \, a b^{3} c^{2} - b^{3} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*((140*b^4*x^4 - 40*a^4 + 72*a^3*c + 5*(13*a*b^3 + 27*b^3*c)*x^3 - 3*(5*a^2*b^2 - 9*a*b^2*c)*x^2 + 4*(5*a
^3*b - 9*a^2*b*c)*x)*sqrt(b*x + a) - (140*b^4*x^4 + 72*a*c^3 - 40*c^4 + 5*(27*a*b^3 + 13*b^3*c)*x^3 + 3*(9*a*b
^2*c - 5*b^2*c^2)*x^2 - 4*(9*a*b*c^2 - 5*b*c^3)*x)*sqrt(b*x + c))/(a^3*b^3 - 3*a^2*b^3*c + 3*a*b^3*c^2 - b^3*c
^3)

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

[Out]

Integral(x**2/(sqrt(a + b*x) + sqrt(b*x + c))**3, x)

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

[Out]

Timed out