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

Optimal. Leaf size=195 \[ -\frac {a^3 (b+c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {b} \sqrt {a+c x}}\right )}{4 b^{5/2} c^{5/2}}+\frac {a^2 (b+c) \sqrt {a+b x} \sqrt {a+c x}}{4 b^2 c^2 (b-c)}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 c (b-c)^2}+\frac {a x^2}{(b-c)^2}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b c (b-c)^2}+\frac {x^3 (b+c)}{3 (b-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6690, 80, 50, 63, 217, 206} \[ \frac {a^2 (b+c) \sqrt {a+b x} \sqrt {a+c x}}{4 b^2 c^2 (b-c)}-\frac {a^3 (b+c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {b} \sqrt {a+c x}}\right )}{4 b^{5/2} c^{5/2}}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 c (b-c)^2}+\frac {a x^2}{(b-c)^2}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b c (b-c)^2}+\frac {x^3 (b+c)}{3 (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 )^2} \, dx &=\frac {\int \left (2 a x+b \left (1+\frac {c}{b}\right ) x^2-2 x \sqrt {a+b x} \sqrt {a+c x}\right ) \, dx}{(b-c)^2}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}-\frac {2 \int x \sqrt {a+b x} \sqrt {a+c x} \, dx}{(b-c)^2}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b (b-c)^2 c}+\frac {(a (b+c)) \int \sqrt {a+b x} \sqrt {a+c x} \, dx}{b (b-c)^2 c}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 (b-c)^2 c}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b (b-c)^2 c}+\frac {\left (a^2 (b+c)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {a+c x}} \, dx}{4 b^2 (b-c) c}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}+\frac {a^2 (b+c) \sqrt {a+b x} \sqrt {a+c x}}{4 b^2 (b-c) c^2}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 (b-c)^2 c}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b (b-c)^2 c}-\frac {\left (a^3 (b+c)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {a+c x}} \, dx}{8 b^2 c^2}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}+\frac {a^2 (b+c) \sqrt {a+b x} \sqrt {a+c x}}{4 b^2 (b-c) c^2}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 (b-c)^2 c}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b (b-c)^2 c}-\frac {\left (a^3 (b+c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\frac {a c}{b}+\frac {c x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{4 b^3 c^2}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}+\frac {a^2 (b+c) \sqrt {a+b x} \sqrt {a+c x}}{4 b^2 (b-c) c^2}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 (b-c)^2 c}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b (b-c)^2 c}-\frac {\left (a^3 (b+c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {c x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {a+c x}}\right )}{4 b^3 c^2}\\ &=\frac {a x^2}{(b-c)^2}+\frac {(b+c) x^3}{3 (b-c)^2}+\frac {a^2 (b+c) \sqrt {a+b x} \sqrt {a+c x}}{4 b^2 (b-c) c^2}+\frac {a (b+c) (a+b x)^{3/2} \sqrt {a+c x}}{2 b^2 (b-c)^2 c}-\frac {2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 b (b-c)^2 c}-\frac {a^3 (b+c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {b} \sqrt {a+c x}}\right )}{4 b^{5/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 238, normalized size = 1.22 \[ \frac {\frac {3 a^4 (c-b)^3 (b+c) \sqrt {\frac {b (a+c x)}{a (b-c)}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a (b-c)}}\right )}{\sqrt {a (b-c)} \sqrt {a+c x}}+b \sqrt {c} \left (a^2 \left (3 b^2-2 b c+3 c^2\right ) \sqrt {a+b x} \sqrt {a+c x}+4 b^2 c^2 x^2 \left (-2 \sqrt {a+b x} \sqrt {a+c x}+b x+c x\right )-2 a b c x \left (b \sqrt {a+b x} \sqrt {a+c x}+c \sqrt {a+b x} \sqrt {a+c x}-6 b c x\right )\right )}{12 b^3 c^{5/2} (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[c]*(a^2*(3*b^2 - 2*b*c + 3*c^2)*Sqrt[a + b*x]*Sqrt[a + c*x] + 4*b^2*c^2*x^2*(b*x + c*x - 2*Sqrt[a + b*
x]*Sqrt[a + c*x]) - 2*a*b*c*x*(-6*b*c*x + b*Sqrt[a + b*x]*Sqrt[a + c*x] + c*Sqrt[a + b*x]*Sqrt[a + c*x])) + (3
*a^4*(-b + c)^3*(b + c)*Sqrt[(b*(a + c*x))/(a*(b - c))]*ArcSinh[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[a*(b - c)]])/(Sqr
t[a*(b - c)]*Sqrt[a + c*x]))/(12*b^3*(b - c)^2*c^(5/2))

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fricas [A]  time = 0.52, size = 479, normalized size = 2.46 \[ \left [\frac {24 \, a b^{3} c^{3} x^{2} + 8 \, {\left (b^{4} c^{3} + b^{3} c^{4}\right )} x^{3} + 3 \, {\left (a^{3} b^{3} - a^{3} b^{2} c - a^{3} b c^{2} + a^{3} c^{3}\right )} \sqrt {b c} \log \left (a b^{2} + 2 \, a b c + a c^{2} + 2 \, {\left (2 \, b c - \sqrt {b c} {\left (b + c\right )}\right )} \sqrt {b x + a} \sqrt {c x + a} + 2 \, {\left (b^{2} c + b c^{2}\right )} x - 2 \, {\left (2 \, b c x + a b + a c\right )} \sqrt {b c}\right ) - 2 \, {\left (8 \, b^{3} c^{3} x^{2} - 3 \, a^{2} b^{3} c + 2 \, a^{2} b^{2} c^{2} - 3 \, a^{2} b c^{3} + 2 \, {\left (a b^{3} c^{2} + a b^{2} c^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {c x + a}}{24 \, {\left (b^{5} c^{3} - 2 \, b^{4} c^{4} + b^{3} c^{5}\right )}}, \frac {12 \, a b^{3} c^{3} x^{2} + 4 \, {\left (b^{4} c^{3} + b^{3} c^{4}\right )} x^{3} + 3 \, {\left (a^{3} b^{3} - a^{3} b^{2} c - a^{3} b c^{2} + a^{3} c^{3}\right )} \sqrt {-b c} \arctan \left (\frac {\sqrt {-b c} \sqrt {b x + a} \sqrt {c x + a} - \sqrt {-b c} a}{b c x}\right ) - {\left (8 \, b^{3} c^{3} x^{2} - 3 \, a^{2} b^{3} c + 2 \, a^{2} b^{2} c^{2} - 3 \, a^{2} b c^{3} + 2 \, {\left (a b^{3} c^{2} + a b^{2} c^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {c x + a}}{12 \, {\left (b^{5} c^{3} - 2 \, b^{4} c^{4} + b^{3} c^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(24*a*b^3*c^3*x^2 + 8*(b^4*c^3 + b^3*c^4)*x^3 + 3*(a^3*b^3 - a^3*b^2*c - a^3*b*c^2 + a^3*c^3)*sqrt(b*c)*
log(a*b^2 + 2*a*b*c + a*c^2 + 2*(2*b*c - sqrt(b*c)*(b + c))*sqrt(b*x + a)*sqrt(c*x + a) + 2*(b^2*c + b*c^2)*x
- 2*(2*b*c*x + a*b + a*c)*sqrt(b*c)) - 2*(8*b^3*c^3*x^2 - 3*a^2*b^3*c + 2*a^2*b^2*c^2 - 3*a^2*b*c^3 + 2*(a*b^3
*c^2 + a*b^2*c^3)*x)*sqrt(b*x + a)*sqrt(c*x + a))/(b^5*c^3 - 2*b^4*c^4 + b^3*c^5), 1/12*(12*a*b^3*c^3*x^2 + 4*
(b^4*c^3 + b^3*c^4)*x^3 + 3*(a^3*b^3 - a^3*b^2*c - a^3*b*c^2 + a^3*c^3)*sqrt(-b*c)*arctan((sqrt(-b*c)*sqrt(b*x
+ a)*sqrt(c*x + a) - sqrt(-b*c)*a)/(b*c*x)) - (8*b^3*c^3*x^2 - 3*a^2*b^3*c + 2*a^2*b^2*c^2 - 3*a^2*b*c^3 + 2*
(a*b^3*c^2 + a*b^2*c^3)*x)*sqrt(b*x + a)*sqrt(c*x + a))/(b^5*c^3 - 2*b^4*c^4 + b^3*c^5)]

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giac [B]  time = 3.47, size = 511, normalized size = 2.62 \[ -\frac {1}{12} \, \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b^{11} c^{4} {\left | b \right |} - 3 \, b^{10} c^{5} {\left | b \right |} + 3 \, b^{9} c^{6} {\left | b \right |} - b^{8} c^{7} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{17} c^{4} - 5 \, b^{16} c^{5} + 10 \, b^{15} c^{6} - 10 \, b^{14} c^{7} + 5 \, b^{13} c^{8} - b^{12} c^{9}} + \frac {a b^{12} c^{3} {\left | b \right |} - 10 \, a b^{11} c^{4} {\left | b \right |} + 24 \, a b^{10} c^{5} {\left | b \right |} - 22 \, a b^{9} c^{6} {\left | b \right |} + 7 \, a b^{8} c^{7} {\left | b \right |}}{b^{17} c^{4} - 5 \, b^{16} c^{5} + 10 \, b^{15} c^{6} - 10 \, b^{14} c^{7} + 5 \, b^{13} c^{8} - b^{12} c^{9}}\right )} - \frac {3 \, {\left (a^{2} b^{13} c^{2} {\left | b \right |} - 3 \, a^{2} b^{12} c^{3} {\left | b \right |} + 2 \, a^{2} b^{11} c^{4} {\left | b \right |} + 2 \, a^{2} b^{10} c^{5} {\left | b \right |} - 3 \, a^{2} b^{9} c^{6} {\left | b \right |} + a^{2} b^{8} c^{7} {\left | b \right |}\right )}}{b^{17} c^{4} - 5 \, b^{16} c^{5} + 10 \, b^{15} c^{6} - 10 \, b^{14} c^{7} + 5 \, b^{13} c^{8} - b^{12} c^{9}}\right )} \sqrt {b x + a} + \frac {{\left (b x + a\right )}^{3} b - 3 \, {\left (b x + a\right )} a^{2} b + {\left (b x + a\right )}^{3} c - 3 \, {\left (b x + a\right )}^{2} a c + 3 \, {\left (b x + a\right )} a^{2} c}{3 \, {\left (b^{5} - 2 \, b^{4} c + b^{3} c^{2}\right )}} + \frac {{\left (a^{3} b {\left | b \right |} + a^{3} c {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b c} \sqrt {b x + a} + \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c} \right |}\right )}{4 \, \sqrt {b c} b^{3} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*(2*(b*x + a)*(4*(b^11*c^4*abs(b) - 3*b^10*c^5*abs(b) + 3*b^9*c^6*abs
(b) - b^8*c^7*abs(b))*(b*x + a)/(b^17*c^4 - 5*b^16*c^5 + 10*b^15*c^6 - 10*b^14*c^7 + 5*b^13*c^8 - b^12*c^9) +
(a*b^12*c^3*abs(b) - 10*a*b^11*c^4*abs(b) + 24*a*b^10*c^5*abs(b) - 22*a*b^9*c^6*abs(b) + 7*a*b^8*c^7*abs(b))/(
b^17*c^4 - 5*b^16*c^5 + 10*b^15*c^6 - 10*b^14*c^7 + 5*b^13*c^8 - b^12*c^9)) - 3*(a^2*b^13*c^2*abs(b) - 3*a^2*b
^12*c^3*abs(b) + 2*a^2*b^11*c^4*abs(b) + 2*a^2*b^10*c^5*abs(b) - 3*a^2*b^9*c^6*abs(b) + a^2*b^8*c^7*abs(b))/(b
^17*c^4 - 5*b^16*c^5 + 10*b^15*c^6 - 10*b^14*c^7 + 5*b^13*c^8 - b^12*c^9))*sqrt(b*x + a) + 1/3*((b*x + a)^3*b
- 3*(b*x + a)*a^2*b + (b*x + a)^3*c - 3*(b*x + a)^2*a*c + 3*(b*x + a)*a^2*c)/(b^5 - 2*b^4*c + b^3*c^2) + 1/4*(
a^3*b*abs(b) + a^3*c*abs(b))*log(abs(-sqrt(b*c)*sqrt(b*x + a) + sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)))/(sqrt(b*
c)*b^3*c^2)

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maple [B]  time = 0.02, size = 517, normalized size = 2.65 \[ \frac {b \,x^{3}}{3 \left (b -c \right )^{2}}+\frac {c \,x^{3}}{3 \left (b -c \right )^{2}}+\frac {a \,x^{2}}{\left (b -c \right )^{2}}-\frac {\sqrt {b x +a}\, \sqrt {c x +a}\, \left (3 a^{3} b^{3} \ln \left (\frac {2 b c x +a b +a c +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}}{2 \sqrt {b c}}\right )-3 a^{3} b^{2} c \ln \left (\frac {2 b c x +a b +a c +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}}{2 \sqrt {b c}}\right )-3 a^{3} b \,c^{2} \ln \left (\frac {2 b c x +a b +a c +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}}{2 \sqrt {b c}}\right )+3 a^{3} c^{3} \ln \left (\frac {2 b c x +a b +a c +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}}{2 \sqrt {b c}}\right )+16 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, b^{2} c^{2} x^{2}+4 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, a \,b^{2} c x +4 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, a b \,c^{2} x -6 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, a^{2} b^{2}+4 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, a^{2} b c -6 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, a^{2} c^{2}\right )}{24 \left (b -c \right )^{2} \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \sqrt {b c}\, b^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3/(b-c)^2*b+1/3*x^3/(b-c)^2*c+a*x^2/(b-c)^2-1/24/(b-c)^2*(b*x+a)^(1/2)*(c*x+a)^(1/2)*(16*x^2*b^2*c^2*(b*
c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)+3*ln(1/2*(2*b*c*x+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)+a*b+a
*c)/(b*c)^(1/2))*a^3*b^3-3*ln(1/2*(2*b*c*x+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)+a*b+a*c)/(b*c)^(1/2))
*a^3*b^2*c-3*ln(1/2*(2*b*c*x+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)+a*b+a*c)/(b*c)^(1/2))*a^3*b*c^2+3*l
n(1/2*(2*b*c*x+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)+a*b+a*c)/(b*c)^(1/2))*a^3*c^3+4*(b*c*x^2+a*b*x+a*
c*x+a^2)^(1/2)*(b*c)^(1/2)*x*a*b^2*c+4*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)*x*a*b*c^2-6*(b*c*x^2+a*b*x+
a*c*x+a^2)^(1/2)*(b*c)^(1/2)*a^2*b^2+4*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*(b*c)^(1/2)*a^2*b*c-6*(b*c*x^2+a*b*x+a*
c*x+a^2)^(1/2)*(b*c)^(1/2)*a^2*c^2)/(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)/b^2/c^2/(b*c)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{{\left (\sqrt {b x + a} + \sqrt {c x + a}\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 18.15, size = 1107, normalized size = 5.68 \[ \frac {\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (128\,a^3\,b^3\,c+\frac {1312\,a^3\,b^2\,c^2}{3}+128\,a^3\,b\,c^3\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^6}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (19\,a^3\,b^3\,c+269\,a^3\,b^2\,c^2+269\,a^3\,b\,c^3+19\,a^3\,c^4\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^7}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (19\,a^3\,b^4+269\,a^3\,b^3\,c+269\,a^3\,b^2\,c^2+19\,a^3\,b\,c^3\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (64\,a^3\,b^4+192\,a^3\,b^3\,c+64\,a^3\,b^2\,c^2\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^4}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8\,\left (64\,a^3\,b^2\,c^2+192\,a^3\,b\,c^3+64\,a^3\,c^4\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^8}+\frac {16\,a^3\,b^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}+\frac {16\,a^3\,c^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^{10}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (a^3\,b^3\,c^3-a^3\,b^2\,c^4-a^3\,b\,c^5+a^3\,c^6\right )}{2\,b^2\,{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^{11}}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (17\,a^3\,b^5+303\,a^3\,b^4\,c+303\,a^3\,b^3\,c^2+17\,a^3\,b^2\,c^3\right )}{6\,c\,{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^3}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (17\,a^3\,b^3\,c^2+303\,a^3\,b^2\,c^3+303\,a^3\,b\,c^4+17\,a^3\,c^5\right )}{6\,b\,{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^9}+\frac {\left (a^3\,b+a^3\,c\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (b^5-2\,b^4\,c+b^3\,c^2\right )}{2\,c^2\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )}}{b^8-2\,b^7\,c+b^6\,c^2+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}\,\left (b^2\,c^6-2\,b\,c^7+c^8\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^{12}}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2\,\left (6\,b^7\,c-12\,b^6\,c^2+6\,b^5\,c^3\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}\,\left (6\,b^3\,c^5-12\,b^2\,c^6+6\,b\,c^7\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^{10}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (15\,b^6\,c^2-30\,b^5\,c^3+15\,b^4\,c^4\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^4}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8\,\left (15\,b^4\,c^4-30\,b^3\,c^5+15\,b^2\,c^6\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^8}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (20\,b^5\,c^3-40\,b^4\,c^4+20\,b^3\,c^5\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^6}}+\frac {x^3\,\left (b+c\right )}{3\,{\left (b-c\right )}^2}+\frac {a\,x^2}{{\left (b-c\right )}^2}-\frac {a^3\,\mathrm {atanh}\left (\frac {\sqrt {c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )}\right )\,\left (b+c\right )}{2\,b^{5/2}\,c^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((a + b*x)^(1/2) - a^(1/2))^6*(128*a^3*b*c^3 + 128*a^3*b^3*c + (1312*a^3*b^2*c^2)/3))/((a + c*x)^(1/2) - a^(
1/2))^6 - (((a + b*x)^(1/2) - a^(1/2))^7*(19*a^3*c^4 + 269*a^3*b*c^3 + 19*a^3*b^3*c + 269*a^3*b^2*c^2))/((a +
c*x)^(1/2) - a^(1/2))^7 - (((a + b*x)^(1/2) - a^(1/2))^5*(19*a^3*b^4 + 19*a^3*b*c^3 + 269*a^3*b^3*c + 269*a^3*
b^2*c^2))/((a + c*x)^(1/2) - a^(1/2))^5 + (((a + b*x)^(1/2) - a^(1/2))^4*(64*a^3*b^4 + 192*a^3*b^3*c + 64*a^3*
b^2*c^2))/((a + c*x)^(1/2) - a^(1/2))^4 + (((a + b*x)^(1/2) - a^(1/2))^8*(64*a^3*c^4 + 192*a^3*b*c^3 + 64*a^3*
b^2*c^2))/((a + c*x)^(1/2) - a^(1/2))^8 + (16*a^3*b^4*((a + b*x)^(1/2) - a^(1/2))^2)/((a + c*x)^(1/2) - a^(1/2
))^2 + (16*a^3*c^4*((a + b*x)^(1/2) - a^(1/2))^10)/((a + c*x)^(1/2) - a^(1/2))^10 + (((a + b*x)^(1/2) - a^(1/2
))^11*(a^3*c^6 - a^3*b*c^5 - a^3*b^2*c^4 + a^3*b^3*c^3))/(2*b^2*((a + c*x)^(1/2) - a^(1/2))^11) - (((a + b*x)^
(1/2) - a^(1/2))^3*(17*a^3*b^5 + 303*a^3*b^4*c + 17*a^3*b^2*c^3 + 303*a^3*b^3*c^2))/(6*c*((a + c*x)^(1/2) - a^
(1/2))^3) - (((a + b*x)^(1/2) - a^(1/2))^9*(17*a^3*c^5 + 303*a^3*b*c^4 + 303*a^3*b^2*c^3 + 17*a^3*b^3*c^2))/(6
*b*((a + c*x)^(1/2) - a^(1/2))^9) + ((a^3*b + a^3*c)*((a + b*x)^(1/2) - a^(1/2))*(b^5 - 2*b^4*c + b^3*c^2))/(2
*c^2*((a + c*x)^(1/2) - a^(1/2))))/(b^8 - 2*b^7*c + b^6*c^2 + (((a + b*x)^(1/2) - a^(1/2))^12*(c^8 - 2*b*c^7 +
b^2*c^6))/((a + c*x)^(1/2) - a^(1/2))^12 - (((a + b*x)^(1/2) - a^(1/2))^2*(6*b^7*c + 6*b^5*c^3 - 12*b^6*c^2))
/((a + c*x)^(1/2) - a^(1/2))^2 - (((a + b*x)^(1/2) - a^(1/2))^10*(6*b*c^7 - 12*b^2*c^6 + 6*b^3*c^5))/((a + c*x
)^(1/2) - a^(1/2))^10 + (((a + b*x)^(1/2) - a^(1/2))^4*(15*b^4*c^4 - 30*b^5*c^3 + 15*b^6*c^2))/((a + c*x)^(1/2
) - a^(1/2))^4 + (((a + b*x)^(1/2) - a^(1/2))^8*(15*b^2*c^6 - 30*b^3*c^5 + 15*b^4*c^4))/((a + c*x)^(1/2) - a^(
1/2))^8 - (((a + b*x)^(1/2) - a^(1/2))^6*(20*b^3*c^5 - 40*b^4*c^4 + 20*b^5*c^3))/((a + c*x)^(1/2) - a^(1/2))^6
) + (x^3*(b + c))/(3*(b - c)^2) + (a*x^2)/(b - c)^2 - (a^3*atanh((c^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2
)*((a + c*x)^(1/2) - a^(1/2))))*(b + c))/(2*b^(5/2)*c^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\sqrt {a + b x} + \sqrt {a + c x}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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