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

Optimal. Leaf size=147 \[ \frac{2 a^2 (a+b x)^{3/2}}{3 b^3 (b-c)}-\frac{2 a^2 (a+c x)^{3/2}}{3 c^3 (b-c)}+\frac{2 (a+b x)^{7/2}}{7 b^3 (b-c)}-\frac{4 a (a+b x)^{5/2}}{5 b^3 (b-c)}-\frac{2 (a+c x)^{7/2}}{7 c^3 (b-c)}+\frac{4 a (a+c x)^{5/2}}{5 c^3 (b-c)} \]

[Out]

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

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Rubi [A]  time = 0.121838, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2103, 43} \[ \frac{2 a^2 (a+b x)^{3/2}}{3 b^3 (b-c)}-\frac{2 a^2 (a+c x)^{3/2}}{3 c^3 (b-c)}+\frac{2 (a+b x)^{7/2}}{7 b^3 (b-c)}-\frac{4 a (a+b x)^{5/2}}{5 b^3 (b-c)}-\frac{2 (a+c x)^{7/2}}{7 c^3 (b-c)}+\frac{4 a (a+c x)^{5/2}}{5 c^3 (b-c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2103

Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[c/(e*(b*c - a*d)
), Int[(u*Sqrt[a + b*x])/x, x], x] - Dist[a/(f*(b*c - a*d)), Int[(u*Sqrt[c + d*x])/x, x], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a*e^2 - c*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^3}{\sqrt{a+b x}+\sqrt{a+c x}} \, dx &=\frac{\int x^2 \sqrt{a+b x} \, dx}{b-c}-\frac{\int x^2 \sqrt{a+c x} \, dx}{b-c}\\ &=\frac{\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}-\frac{\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}\\ &=\frac{2 a^2 (a+b x)^{3/2}}{3 b^3 (b-c)}-\frac{4 a (a+b x)^{5/2}}{5 b^3 (b-c)}+\frac{2 (a+b x)^{7/2}}{7 b^3 (b-c)}-\frac{2 a^2 (a+c x)^{3/2}}{3 (b-c) c^3}+\frac{4 a (a+c x)^{5/2}}{5 (b-c) c^3}-\frac{2 (a+c x)^{7/2}}{7 (b-c) c^3}\\ \end{align*}

Mathematica [A]  time = 0.218664, size = 147, normalized size = 1. \[ \frac{2 a^2 (a+b x)^{3/2}}{3 b^3 (b-c)}-\frac{2 a^2 (a+c x)^{3/2}}{3 c^3 (b-c)}+\frac{2 (a+b x)^{7/2}}{7 b^3 (b-c)}-\frac{4 a (a+b x)^{5/2}}{5 b^3 (b-c)}-\frac{2 (a+c x)^{7/2}}{7 c^3 (b-c)}+\frac{4 a (a+c x)^{5/2}}{5 c^3 (b-c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 90, normalized size = 0.6 \begin{align*} 2\,{\frac{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}}{ \left ( b-c \right ){b}^{3}}}-2\,{\frac{1/7\, \left ( cx+a \right ) ^{7/2}-2/5\,a \left ( cx+a \right ) ^{5/2}+1/3\,{a}^{2} \left ( cx+a \right ) ^{3/2}}{ \left ( b-c \right ){c}^{3}}} \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)),x)

[Out]

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

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

[Out]

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

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Fricas [A]  time = 1.20086, size = 250, normalized size = 1.7 \begin{align*} \frac{2 \,{\left ({\left (15 \, b^{3} c^{3} x^{3} + 3 \, a b^{2} c^{3} x^{2} - 4 \, a^{2} b c^{3} x + 8 \, a^{3} c^{3}\right )} \sqrt{b x + a} -{\left (15 \, b^{3} c^{3} x^{3} + 3 \, a b^{3} c^{2} x^{2} - 4 \, a^{2} b^{3} c x + 8 \, a^{3} b^{3}\right )} \sqrt{c x + a}\right )}}{105 \,{\left (b^{4} c^{3} - b^{3} c^{4}\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)),x, algorithm="fricas")

[Out]

2/105*((15*b^3*c^3*x^3 + 3*a*b^2*c^3*x^2 - 4*a^2*b*c^3*x + 8*a^3*c^3)*sqrt(b*x + a) - (15*b^3*c^3*x^3 + 3*a*b^
3*c^2*x^2 - 4*a^2*b^3*c*x + 8*a^3*b^3)*sqrt(c*x + a))/(b^4*c^3 - b^3*c^4)

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

[Out]

Integral(x**3/(sqrt(a + b*x) + sqrt(a + c*x)), 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)),x, algorithm="giac")

[Out]

Timed out