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

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

[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))

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Rubi [A]  time = 0.3493, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, 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 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]

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 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 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 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])

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*}

Mathematica [A]  time = 0.676149, size = 238, normalized size = 1.22 \[ \frac{b \sqrt{c} \left (a^2 \left (3 b^2-2 b c+3 c^2\right ) \sqrt{a+b x} \sqrt{a+c x}+4 b^2 c^2 x^2 \left (-2 \sqrt{a+b x} \sqrt{a+c x}+b x+c x\right )-2 a b c x \left (b \sqrt{a+b x} \sqrt{a+c x}+c \sqrt{a+b x} \sqrt{a+c x}-6 b c x\right )\right )+\frac{3 a^4 (c-b)^3 (b+c) \sqrt{\frac{b (a+c x)}{a (b-c)}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a (b-c)}}\right )}{\sqrt{a (b-c)} \sqrt{a+c x}}}{12 b^3 c^{5/2} (b-c)^2} \]

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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Maple [B]  time = 0.014, size = 517, normalized size = 2.7 \begin{align*}{\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{1}{24\, \left ( b-c \right ) ^{2}{b}^{2}{c}^{2}}\sqrt{bx+a}\sqrt{cx+a} \left ( 16\,{x}^{2}{b}^{2}{c}^{2}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+3\,\ln \left ( 1/2\,{\frac{2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac}{\sqrt{bc}}} \right ){a}^{3}{b}^{3}-3\,\ln \left ( 1/2\,{\frac{2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac}{\sqrt{bc}}} \right ){a}^{3}{b}^{2}c-3\,\ln \left ( 1/2\,{\frac{2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac}{\sqrt{bc}}} \right ){a}^{3}b{c}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac}{\sqrt{bc}}} \right ){a}^{3}{c}^{3}+4\,\sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}xa{b}^{2}c+4\,\sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}xab{c}^{2}-6\,\sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{a}^{2}{b}^{2}+4\,\sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{a}^{2}bc-6\,\sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{a}^{2}{c}^{2} \right ){\frac{1}{\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}}}{\frac{1}{\sqrt{bc}}}} \end{align*}

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)^(1/2)*(b*c*
x^2+a*b*x+a*c*x+a^2)^(1/2)*x*a*b^2*c+4*(b*c)^(1/2)*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*x*a*b*c^2-6*(b*c)^(1/2)*(b*
c*x^2+a*b*x+a*c*x+a^2)^(1/2)*a^2*b^2+4*(b*c)^(1/2)*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*a^2*b*c-6*(b*c)^(1/2)*(b*c*
x^2+a*b*x+a*c*x+a^2)^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (\sqrt{b x + a} + \sqrt{c x + a}\right )}^{2}}\,{d x} \end{align*}

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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Fricas [A]  time = 1.34945, size = 1015, normalized size = 5.21 \begin{align*} \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 ] \end{align*}

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

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

[Out]

Timed out