3.438 \(\int \frac {x^4}{(\sqrt {a+b x}+\sqrt {a+c x})^3} \, dx\)

Optimal. Leaf size=277 \[ \frac {2 a^2 (b+3 c) (a+b x)^{3/2}}{3 b^3 (b-c)^3}-\frac {8 a^2 (a+b x)^{3/2}}{3 b^2 (b-c)^3}-\frac {2 a^2 (3 b+c) (a+c x)^{3/2}}{3 c^3 (b-c)^3}+\frac {8 a^2 (a+c x)^{3/2}}{3 c^2 (b-c)^3}+\frac {2 (b+3 c) (a+b x)^{7/2}}{7 b^3 (b-c)^3}-\frac {4 a (b+3 c) (a+b x)^{5/2}}{5 b^3 (b-c)^3}+\frac {8 a (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {2 (3 b+c) (a+c x)^{7/2}}{7 c^3 (b-c)^3}+\frac {4 a (3 b+c) (a+c x)^{5/2}}{5 c^3 (b-c)^3}-\frac {8 a (a+c x)^{5/2}}{5 c^2 (b-c)^3} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6690, 43} \[ \frac {2 a^2 (b+3 c) (a+b x)^{3/2}}{3 b^3 (b-c)^3}-\frac {8 a^2 (a+b x)^{3/2}}{3 b^2 (b-c)^3}-\frac {2 a^2 (3 b+c) (a+c x)^{3/2}}{3 c^3 (b-c)^3}+\frac {8 a^2 (a+c x)^{3/2}}{3 c^2 (b-c)^3}+\frac {2 (b+3 c) (a+b x)^{7/2}}{7 b^3 (b-c)^3}-\frac {4 a (b+3 c) (a+b x)^{5/2}}{5 b^3 (b-c)^3}+\frac {8 a (a+b x)^{5/2}}{5 b^2 (b-c)^3}-\frac {2 (3 b+c) (a+c x)^{7/2}}{7 c^3 (b-c)^3}+\frac {4 a (3 b+c) (a+c x)^{5/2}}{5 c^3 (b-c)^3}-\frac {8 a (a+c x)^{5/2}}{5 c^2 (b-c)^3} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[a + b*x] + Sqrt[a + c*x])^3,x]

[Out]

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

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

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Mathematica [A]  time = 0.39, size = 271, normalized size = 0.98 \[ \frac {2 \left (8 a^3 \left (b^4 \left (-\sqrt {a+c x}\right )+2 b^3 c \sqrt {a+c x}+c^4 \sqrt {a+b x}-2 b c^3 \sqrt {a+b x}\right )+4 a^2 b c x \left (b^3 \sqrt {a+c x}-2 b^2 c \sqrt {a+c x}-c^3 \sqrt {a+b x}+2 b c^2 \sqrt {a+b x}\right )+5 b^3 c^3 x^3 \left (-3 b \sqrt {a+c x}+3 c \sqrt {a+b x}+b \sqrt {a+b x}-c \sqrt {a+c x}\right )+a b^2 c^2 x^2 \left (-3 b^2 \sqrt {a+c x}+3 c^2 \sqrt {a+b x}+29 b c \left (\sqrt {a+b x}-\sqrt {a+c x}\right )\right )\right )}{35 b^3 c^3 (b-c)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[a + b*x] + Sqrt[a + c*x])^3,x]

[Out]

(2*(5*b^3*c^3*x^3*(b*Sqrt[a + b*x] + 3*c*Sqrt[a + b*x] - 3*b*Sqrt[a + c*x] - c*Sqrt[a + c*x]) + 4*a^2*b*c*x*(2
*b*c^2*Sqrt[a + b*x] - c^3*Sqrt[a + b*x] + b^3*Sqrt[a + c*x] - 2*b^2*c*Sqrt[a + c*x]) + 8*a^3*(-2*b*c^3*Sqrt[a
+ b*x] + c^4*Sqrt[a + b*x] - b^4*Sqrt[a + c*x] + 2*b^3*c*Sqrt[a + c*x]) + a*b^2*c^2*x^2*(3*c^2*Sqrt[a + b*x]
- 3*b^2*Sqrt[a + c*x] + 29*b*c*(Sqrt[a + b*x] - Sqrt[a + c*x]))))/(35*b^3*(b - c)^3*c^3)

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fricas [A]  time = 0.48, size = 225, normalized size = 0.81 \[ -\frac {2 \, {\left ({\left (16 \, a^{3} b c^{3} - 8 \, a^{3} c^{4} - 5 \, {\left (b^{4} c^{3} + 3 \, b^{3} c^{4}\right )} x^{3} - {\left (29 \, a b^{3} c^{3} + 3 \, a b^{2} c^{4}\right )} x^{2} - 4 \, {\left (2 \, a^{2} b^{2} c^{3} - a^{2} b c^{4}\right )} x\right )} \sqrt {b x + a} + {\left (8 \, a^{3} b^{4} - 16 \, a^{3} b^{3} c + 5 \, {\left (3 \, b^{4} c^{3} + b^{3} c^{4}\right )} x^{3} + {\left (3 \, a b^{4} c^{2} + 29 \, a b^{3} c^{3}\right )} x^{2} - 4 \, {\left (a^{2} b^{4} c - 2 \, a^{2} b^{3} c^{2}\right )} x\right )} \sqrt {c x + a}\right )}}{35 \, {\left (b^{6} c^{3} - 3 \, b^{5} c^{4} + 3 \, b^{4} c^{5} - b^{3} c^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x, algorithm="fricas")

[Out]

-2/35*((16*a^3*b*c^3 - 8*a^3*c^4 - 5*(b^4*c^3 + 3*b^3*c^4)*x^3 - (29*a*b^3*c^3 + 3*a*b^2*c^4)*x^2 - 4*(2*a^2*b
^2*c^3 - a^2*b*c^4)*x)*sqrt(b*x + a) + (8*a^3*b^4 - 16*a^3*b^3*c + 5*(3*b^4*c^3 + b^3*c^4)*x^3 + (3*a*b^4*c^2
+ 29*a*b^3*c^3)*x^2 - 4*(a^2*b^4*c - 2*a^2*b^3*c^2)*x)*sqrt(c*x + a))/(b^6*c^3 - 3*b^5*c^4 + 3*b^4*c^5 - b^3*c
^6)

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giac [B]  time = 5.58, size = 932, normalized size = 3.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x, algorithm="giac")

[Out]

-2/35*sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*(((b*x + a)*(5*(3*b^22*c^5*abs(b) - 17*b^21*c^6*abs(b) + 39*b^20*c^7
*abs(b) - 45*b^19*c^8*abs(b) + 25*b^18*c^9*abs(b) - 3*b^17*c^10*abs(b) - 3*b^16*c^11*abs(b) + b^15*c^12*abs(b)
)*(b*x + a)/(b^29*c^5 - 9*b^28*c^6 + 36*b^27*c^7 - 84*b^26*c^8 + 126*b^25*c^9 - 126*b^24*c^10 + 84*b^23*c^11 -
36*b^22*c^12 + 9*b^21*c^13 - b^20*c^14) + (3*a*b^23*c^4*abs(b) - 34*a*b^22*c^5*abs(b) + 126*a*b^21*c^6*abs(b)
- 210*a*b^20*c^7*abs(b) + 140*a*b^19*c^8*abs(b) + 42*a*b^18*c^9*abs(b) - 126*a*b^17*c^10*abs(b) + 74*a*b^16*c
^11*abs(b) - 15*a*b^15*c^12*abs(b))/(b^29*c^5 - 9*b^28*c^6 + 36*b^27*c^7 - 84*b^26*c^8 + 126*b^25*c^9 - 126*b^
24*c^10 + 84*b^23*c^11 - 36*b^22*c^12 + 9*b^21*c^13 - b^20*c^14)) - (4*a^2*b^24*c^3*abs(b) - 26*a^2*b^23*c^4*a
bs(b) + 85*a^2*b^22*c^5*abs(b) - 203*a^2*b^21*c^6*abs(b) + 385*a^2*b^20*c^7*abs(b) - 539*a^2*b^19*c^8*abs(b) +
511*a^2*b^18*c^9*abs(b) - 305*a^2*b^17*c^10*abs(b) + 103*a^2*b^16*c^11*abs(b) - 15*a^2*b^15*c^12*abs(b))/(b^2
9*c^5 - 9*b^28*c^6 + 36*b^27*c^7 - 84*b^26*c^8 + 126*b^25*c^9 - 126*b^24*c^10 + 84*b^23*c^11 - 36*b^22*c^12 +
9*b^21*c^13 - b^20*c^14))*(b*x + a) + (8*a^3*b^25*c^2*abs(b) - 60*a^3*b^24*c^3*abs(b) + 187*a^3*b^23*c^4*abs(b
) - 296*a^3*b^22*c^5*abs(b) + 196*a^3*b^21*c^6*abs(b) + 112*a^3*b^20*c^7*abs(b) - 350*a^3*b^19*c^8*abs(b) + 32
8*a^3*b^18*c^9*abs(b) - 164*a^3*b^17*c^10*abs(b) + 44*a^3*b^16*c^11*abs(b) - 5*a^3*b^15*c^12*abs(b))/(b^29*c^5
- 9*b^28*c^6 + 36*b^27*c^7 - 84*b^26*c^8 + 126*b^25*c^9 - 126*b^24*c^10 + 84*b^23*c^11 - 36*b^22*c^12 + 9*b^2
1*c^13 - b^20*c^14)) + 2/35*(5*(b*x + a)^(7/2)*b + 14*(b*x + a)^(5/2)*a*b - 35*(b*x + a)^(3/2)*a^2*b + 15*(b*x
+ a)^(7/2)*c - 42*(b*x + a)^(5/2)*a*c + 35*(b*x + a)^(3/2)*a^2*c)/(b^6 - 3*b^5*c + 3*b^4*c^2 - b^3*c^3)

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maple [A]  time = 0.01, size = 246, normalized size = 0.89 \[ \frac {8 \left (-\frac {\left (b x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (b x +a \right )^{\frac {5}{2}}}{5}\right ) a}{\left (b -c \right )^{3} b^{2}}-\frac {8 \left (-\frac {\left (c x +a \right )^{\frac {3}{2}} a}{3}+\frac {\left (c x +a \right )^{\frac {5}{2}}}{5}\right ) a}{\left (b -c \right )^{3} c^{2}}-\frac {6 \left (\frac {\left (c x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {2 \left (c x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (c x +a \right )^{\frac {7}{2}}}{7}\right ) b}{\left (b -c \right )^{3} c^{3}}+\frac {\frac {2 \left (b x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {4 \left (b x +a \right )^{\frac {5}{2}} a}{5}+\frac {2 \left (b x +a \right )^{\frac {7}{2}}}{7}}{\left (b -c \right )^{3} b^{2}}+\frac {6 \left (\frac {\left (b x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {2 \left (b x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (b x +a \right )^{\frac {7}{2}}}{7}\right ) c}{\left (b -c \right )^{3} b^{3}}-\frac {2 \left (\frac {\left (c x +a \right )^{\frac {3}{2}} a^{2}}{3}-\frac {2 \left (c x +a \right )^{\frac {5}{2}} a}{5}+\frac {\left (c x +a \right )^{\frac {7}{2}}}{7}\right )}{\left (b -c \right )^{3} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (\sqrt {b x + a} + \sqrt {c x + a}\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x+a)^(1/2)+(c*x+a)^(1/2))^3,x, algorithm="maxima")

[Out]

integrate(x^4/(sqrt(b*x + a) + sqrt(c*x + a))^3, x)

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mupad [B]  time = 3.34, size = 429, normalized size = 1.55 \[ \frac {x^2\,\left (\frac {12\,a\,\left (3\,b+c\right )}{7\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )\,\sqrt {a+c\,x}}{5\,c}-\frac {2\,a\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}-\frac {4\,a\,\left (\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}-\frac {12\,a\,\left (b^2+3\,c\,b\right )}{7\,b\,{\left (b-c\right )}^3}\right )}{5\,b}\right )\,\sqrt {a+b\,x}}{3\,b^2}+\frac {x\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}-\frac {4\,a\,\left (\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}-\frac {12\,a\,\left (b^2+3\,c\,b\right )}{7\,b\,{\left (b-c\right )}^3}\right )}{5\,b}\right )\,\sqrt {a+b\,x}}{3\,b}+\frac {2\,a\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}+\frac {4\,a\,\left (\frac {12\,a\,\left (3\,b+c\right )}{7\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )}{5\,c}\right )\,\sqrt {a+c\,x}}{3\,c^2}+\frac {x^2\,\left (\frac {2\,a\,\left (5\,b+3\,c\right )}{{\left (b-c\right )}^3}-\frac {12\,a\,\left (b^2+3\,c\,b\right )}{7\,b\,{\left (b-c\right )}^3}\right )\,\sqrt {a+b\,x}}{5\,b}-\frac {x\,\left (\frac {8\,a^2}{{\left (b-c\right )}^3}+\frac {4\,a\,\left (\frac {12\,a\,\left (3\,b+c\right )}{7\,{\left (b-c\right )}^3}-\frac {2\,a\,\left (3\,b+5\,c\right )}{{\left (b-c\right )}^3}\right )}{5\,c}\right )\,\sqrt {a+c\,x}}{3\,c}-\frac {2\,x^3\,\left (3\,b+c\right )\,\sqrt {a+c\,x}}{7\,{\left (b-c\right )}^3}+\frac {2\,x^3\,\left (b^2+3\,c\,b\right )\,\sqrt {a+b\,x}}{7\,b\,{\left (b-c\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + b*x)^(1/2) + (a + c*x)^(1/2))^3,x)

[Out]

(x^2*((12*a*(3*b + c))/(7*(b - c)^3) - (2*a*(3*b + 5*c))/(b - c)^3)*(a + c*x)^(1/2))/(5*c) - (2*a*((8*a^2)/(b
- c)^3 - (4*a*((2*a*(5*b + 3*c))/(b - c)^3 - (12*a*(3*b*c + b^2))/(7*b*(b - c)^3)))/(5*b))*(a + b*x)^(1/2))/(3
*b^2) + (x*((8*a^2)/(b - c)^3 - (4*a*((2*a*(5*b + 3*c))/(b - c)^3 - (12*a*(3*b*c + b^2))/(7*b*(b - c)^3)))/(5*
b))*(a + b*x)^(1/2))/(3*b) + (2*a*((8*a^2)/(b - c)^3 + (4*a*((12*a*(3*b + c))/(7*(b - c)^3) - (2*a*(3*b + 5*c)
)/(b - c)^3))/(5*c))*(a + c*x)^(1/2))/(3*c^2) + (x^2*((2*a*(5*b + 3*c))/(b - c)^3 - (12*a*(3*b*c + b^2))/(7*b*
(b - c)^3))*(a + b*x)^(1/2))/(5*b) - (x*((8*a^2)/(b - c)^3 + (4*a*((12*a*(3*b + c))/(7*(b - c)^3) - (2*a*(3*b
+ 5*c))/(b - c)^3))/(5*c))*(a + c*x)^(1/2))/(3*c) - (2*x^3*(3*b + c)*(a + c*x)^(1/2))/(7*(b - c)^3) + (2*x^3*(
3*b*c + b^2)*(a + b*x)^(1/2))/(7*b*(b - c)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/((b*x+a)**(1/2)+(c*x+a)**(1/2))**3,x)

[Out]

Timed out

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