3.5.27 \(\int \frac {x^2}{\sqrt {a+b x}+\sqrt {a+c x}} \, dx\) [427]

Optimal. Leaf size=95 \[ -\frac {2 a (a+b x)^{3/2}}{3 b^2 (b-c)}+\frac {2 (a+b x)^{5/2}}{5 b^2 (b-c)}+\frac {2 a (a+c x)^{3/2}}{3 (b-c) c^2}-\frac {2 (a+c x)^{5/2}}{5 (b-c) c^2} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2128, 45} \begin {gather*} \frac {2 (a+b x)^{5/2}}{5 b^2 (b-c)}-\frac {2 a (a+b x)^{3/2}}{3 b^2 (b-c)}-\frac {2 (a+c x)^{5/2}}{5 c^2 (b-c)}+\frac {2 a (a+c x)^{3/2}}{3 c^2 (b-c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 2128

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]

Rubi steps

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

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Mathematica [A]
time = 0.74, size = 113, normalized size = 1.19 \begin {gather*} \frac {6 b^2 c^2 x^2 \left (\sqrt {a+b x}-\sqrt {a+c x}\right )+2 a b c x \left (c \sqrt {a+b x}-b \sqrt {a+c x}\right )+a^2 \left (-4 c^2 \sqrt {a+b x}+4 b^2 \sqrt {a+c x}\right )}{15 b^2 (b-c) c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b^2*c^2*x^2*(Sqrt[a + b*x] - Sqrt[a + c*x]) + 2*a*b*c*x*(c*Sqrt[a + b*x] - b*Sqrt[a + c*x]) + a^2*(-4*c^2*S
qrt[a + b*x] + 4*b^2*Sqrt[a + c*x]))/(15*b^2*(b - c)*c^2)

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Maple [A]
time = 0.03, size = 66, normalized size = 0.69

method	result	size
default	\(\frac {\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {2 \left (b x +a \right )^{\frac {3}{2}} a}{3}}{\left (b -c \right ) b^{2}}-\frac {2 \left (\frac {\left (c x +a \right )^{\frac {5}{2}}}{5}-\frac {\left (c x +a \right )^{\frac {3}{2}} a}{3}\right )}{\left (b -c \right ) c^{2}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((b*x+a)^(1/2)+(c*x+a)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 92, normalized size = 0.97 \begin {gather*} \frac {2 \, {\left ({\left (3 \, b^{2} c^{2} x^{2} + a b c^{2} x - 2 \, a^{2} c^{2}\right )} \sqrt {b x + a} - {\left (3 \, b^{2} c^{2} x^{2} + a b^{2} c x - 2 \, a^{2} b^{2}\right )} \sqrt {c x + a}\right )}}{15 \, {\left (b^{3} c^{2} - b^{2} c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {a + b x} + \sqrt {a + c x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (79) = 158\).
time = 4.90, size = 255, normalized size = 2.68 \begin {gather*} -\frac {2}{15} \, \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c} {\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (b^{9} c^{3} {\left | b \right |} - b^{8} c^{4} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{14} c^{3} - 2 \, b^{13} c^{4} + b^{12} c^{5}} + \frac {a b^{10} c^{2} {\left | b \right |} - 7 \, a b^{9} c^{3} {\left | b \right |} + 6 \, a b^{8} c^{4} {\left | b \right |}}{b^{14} c^{3} - 2 \, b^{13} c^{4} + b^{12} c^{5}}\right )} - \frac {2 \, a^{2} b^{11} c {\left | b \right |} - a^{2} b^{10} c^{2} {\left | b \right |} - 4 \, a^{2} b^{9} c^{3} {\left | b \right |} + 3 \, a^{2} b^{8} c^{4} {\left | b \right |}}{b^{14} c^{3} - 2 \, b^{13} c^{4} + b^{12} c^{5}}\right )} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a\right )}}{15 \, {\left (b^{3} - b^{2} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.86, size = 129, normalized size = 1.36 \begin {gather*} \frac {2\,x^2\,\sqrt {a+b\,x}}{5\,\left (b-c\right )}-\frac {2\,x^2\,\sqrt {a+c\,x}}{5\,\left (b-c\right )}-\frac {4\,a^2\,\sqrt {a+b\,x}}{15\,b^2\,\left (b-c\right )}+\frac {4\,a^2\,\sqrt {a+c\,x}}{15\,c^2\,\left (b-c\right )}+\frac {2\,a\,x\,\sqrt {a+b\,x}}{15\,b\,\left (b-c\right )}-\frac {2\,a\,x\,\sqrt {a+c\,x}}{15\,c\,\left (b-c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x^2*(a + b*x)^(1/2))/(5*(b - c)) - (2*x^2*(a + c*x)^(1/2))/(5*(b - c)) - (4*a^2*(a + b*x)^(1/2))/(15*b^2*(b
- c)) + (4*a^2*(a + c*x)^(1/2))/(15*c^2*(b - c)) + (2*a*x*(a + b*x)^(1/2))/(15*b*(b - c)) - (2*a*x*(a + c*x)^
(1/2))/(15*c*(b - c))

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