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

Optimal. Leaf size=163 \[ \frac {2 (b+3 c) (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {2 a (b+3 c) (a+b x)^{3/2}}{3 b^2 (b-c)^3}-\frac {2 (3 b+c) (a+c x)^{5/2}}{5 c^2 (b-c)^3}+\frac {2 a (3 b+c) (a+c x)^{3/2}}{3 c^2 (b-c)^3}+\frac {8 a (a+b x)^{3/2}}{3 b (b-c)^3}-\frac {8 a (a+c x)^{3/2}}{3 c (b-c)^3} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6690, 43} \[ \frac {2 (b+3 c) (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {2 a (b+3 c) (a+b x)^{3/2}}{3 b^2 (b-c)^3}-\frac {2 (3 b+c) (a+c x)^{5/2}}{5 c^2 (b-c)^3}+\frac {2 a (3 b+c) (a+c x)^{3/2}}{3 c^2 (b-c)^3}+\frac {8 a (a+b x)^{3/2}}{3 b (b-c)^3}-\frac {8 a (a+c x)^{3/2}}{3 c (b-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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

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Mathematica [A]  time = 0.44, size = 120, normalized size = 0.74 \[ \frac {2 \left (\frac {3 (b+3 c) (a+b x)^{5/2}}{b^2}-\frac {5 a (b+3 c) (a+b x)^{3/2}}{b^2}-\frac {3 (3 b+c) (a+c x)^{5/2}}{c^2}+\frac {5 a (3 b+c) (a+c x)^{3/2}}{c^2}+\frac {20 a (a+b x)^{3/2}}{b}-\frac {20 a (a+c x)^{3/2}}{c}\right )}{15 (b-c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((20*a*(a + b*x)^(3/2))/b - (5*a*(b + 3*c)*(a + b*x)^(3/2))/b^2 + (3*(b + 3*c)*(a + b*x)^(5/2))/b^2 - (20*a
*(a + c*x)^(3/2))/c + (5*a*(3*b + c)*(a + c*x)^(3/2))/c^2 - (3*(3*b + c)*(a + c*x)^(5/2))/c^2))/(15*(b - c)^3)

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fricas [A]  time = 0.50, size = 167, normalized size = 1.02 \[ \frac {2 \, {\left ({\left (6 \, a^{2} b c^{2} - 2 \, a^{2} c^{3} + {\left (b^{3} c^{2} + 3 \, b^{2} c^{3}\right )} x^{2} + {\left (7 \, a b^{2} c^{2} + a b c^{3}\right )} x\right )} \sqrt {b x + a} + {\left (2 \, a^{2} b^{3} - 6 \, a^{2} b^{2} c - {\left (3 \, b^{3} c^{2} + b^{2} c^{3}\right )} x^{2} - {\left (a b^{3} c + 7 \, a b^{2} c^{2}\right )} x\right )} \sqrt {c x + a}\right )}}{5 \, {\left (b^{5} c^{2} - 3 \, b^{4} c^{3} + 3 \, b^{3} c^{4} - b^{2} c^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*((6*a^2*b*c^2 - 2*a^2*c^3 + (b^3*c^2 + 3*b^2*c^3)*x^2 + (7*a*b^2*c^2 + a*b*c^3)*x)*sqrt(b*x + a) + (2*a^2*
b^3 - 6*a^2*b^2*c - (3*b^3*c^2 + b^2*c^3)*x^2 - (a*b^3*c + 7*a*b^2*c^2)*x)*sqrt(c*x + a))/(b^5*c^2 - 3*b^4*c^3
+ 3*b^3*c^4 - b^2*c^5)

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giac [B]  time = 5.23, size = 480, normalized size = 2.94 \[ -\frac {2}{5} \, \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c} {\left ({\left (b x + a\right )} {\left (\frac {{\left (3 \, b^{12} c^{3} {\left | b \right |} - 8 \, b^{11} c^{4} {\left | b \right |} + 6 \, b^{10} c^{5} {\left | b \right |} - b^{8} c^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{18} c^{3} - 6 \, b^{17} c^{4} + 15 \, b^{16} c^{5} - 20 \, b^{15} c^{6} + 15 \, b^{14} c^{7} - 6 \, b^{13} c^{8} + b^{12} c^{9}} + \frac {a b^{13} c^{2} {\left | b \right |} - 2 \, a b^{12} c^{3} {\left | b \right |} - 2 \, a b^{11} c^{4} {\left | b \right |} + 8 \, a b^{10} c^{5} {\left | b \right |} - 7 \, a b^{9} c^{6} {\left | b \right |} + 2 \, a b^{8} c^{7} {\left | b \right |}}{b^{18} c^{3} - 6 \, b^{17} c^{4} + 15 \, b^{16} c^{5} - 20 \, b^{15} c^{6} + 15 \, b^{14} c^{7} - 6 \, b^{13} c^{8} + b^{12} c^{9}}\right )} - \frac {2 \, a^{2} b^{14} c {\left | b \right |} - 11 \, a^{2} b^{13} c^{2} {\left | b \right |} + 25 \, a^{2} b^{12} c^{3} {\left | b \right |} - 30 \, a^{2} b^{11} c^{4} {\left | b \right |} + 20 \, a^{2} b^{10} c^{5} {\left | b \right |} - 7 \, a^{2} b^{9} c^{6} {\left | b \right |} + a^{2} b^{8} c^{7} {\left | b \right |}}{b^{18} c^{3} - 6 \, b^{17} c^{4} + 15 \, b^{16} c^{5} - 20 \, b^{15} c^{6} + 15 \, b^{14} c^{7} - 6 \, b^{13} c^{8} + b^{12} c^{9}}\right )} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} b + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} c - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a c\right )}}{5 \, {\left (b^{5} - 3 \, b^{4} c + 3 \, b^{3} c^{2} - b^{2} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*((b*x + a)*((3*b^12*c^3*abs(b) - 8*b^11*c^4*abs(b) + 6*b^10*c^5*abs(b
) - b^8*c^7*abs(b))*(b*x + a)/(b^18*c^3 - 6*b^17*c^4 + 15*b^16*c^5 - 20*b^15*c^6 + 15*b^14*c^7 - 6*b^13*c^8 +
b^12*c^9) + (a*b^13*c^2*abs(b) - 2*a*b^12*c^3*abs(b) - 2*a*b^11*c^4*abs(b) + 8*a*b^10*c^5*abs(b) - 7*a*b^9*c^6
*abs(b) + 2*a*b^8*c^7*abs(b))/(b^18*c^3 - 6*b^17*c^4 + 15*b^16*c^5 - 20*b^15*c^6 + 15*b^14*c^7 - 6*b^13*c^8 +
b^12*c^9)) - (2*a^2*b^14*c*abs(b) - 11*a^2*b^13*c^2*abs(b) + 25*a^2*b^12*c^3*abs(b) - 30*a^2*b^11*c^4*abs(b) +
20*a^2*b^10*c^5*abs(b) - 7*a^2*b^9*c^6*abs(b) + a^2*b^8*c^7*abs(b))/(b^18*c^3 - 6*b^17*c^4 + 15*b^16*c^5 - 20
*b^15*c^6 + 15*b^14*c^7 - 6*b^13*c^8 + b^12*c^9)) + 2/5*((b*x + a)^(5/2)*b + 5*(b*x + a)^(3/2)*a*b + 3*(b*x +
a)^(5/2)*c - 5*(b*x + a)^(3/2)*a*c)/(b^5 - 3*b^4*c + 3*b^3*c^2 - b^2*c^3)

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maple [A]  time = 0.00, size = 172, normalized size = 1.06 \[ \frac {8 \left (b x +a \right )^{\frac {3}{2}} a}{3 \left (b -c \right )^{3} b}-\frac {8 \left (c x +a \right )^{\frac {3}{2}} a}{3 \left (b -c \right )^{3} c}-\frac {6 \left (-\frac {\left (c x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (c x +a \right )^{\frac {5}{2}}}{5}\right ) b}{\left (b -c \right )^{3} c^{2}}+\frac {-\frac {2 \left (b x +a \right )^{\frac {3}{2}} a}{3}+\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}}{\left (b -c \right )^{3} b}+\frac {6 \left (-\frac {\left (b x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (b x +a \right )^{\frac {5}{2}}}{5}\right ) c}{\left (b -c \right )^{3} b^{2}}-\frac {2 \left (-\frac {\left (c x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (c x +a \right )^{\frac {5}{2}}}{5}\right )}{\left (b -c \right )^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.36, size = 268, normalized size = 1.64 \[ \frac {\left (\frac {8\,a^2}{{\left (b-c\right )}^3}+\frac {2\,a\,\left (\frac {8\,a\,\left (b+3\,c\right )}{5\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}\right )}{3\,b}\right )\,\sqrt {a+b\,x}}{b}-\frac {\left (\frac {8\,a^2}{{\left (b-c\right )}^3}+\frac {2\,a\,\left (\frac {8\,a\,\left (3\,b+c\right )}{5\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )}{3\,c}\right )\,\sqrt {a+c\,x}}{c}-\frac {x\,\left (\frac {8\,a\,\left (b+3\,c\right )}{5\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}\right )\,\sqrt {a+b\,x}}{3\,b}+\frac {x\,\left (\frac {8\,a\,\left (3\,b+c\right )}{5\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )\,\sqrt {a+c\,x}}{3\,c}+\frac {2\,x^2\,\left (b+3\,c\right )\,\sqrt {a+b\,x}}{5\,{\left (b-c\right )}^3}-\frac {2\,x^2\,\left (3\,b+c\right )\,\sqrt {a+c\,x}}{5\,{\left (b-c\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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