3.402 \(\int \frac {x}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2104, 43} \[ \frac {2 (a+b x)^{5/2}}{5 b^2 (a-c)}-\frac {2 a (a+b x)^{3/2}}{3 b^2 (a-c)}-\frac {2 (b x+c)^{5/2}}{5 b^2 (a-c)}+\frac {2 c (b x+c)^{3/2}}{3 b^2 (a-c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2104

Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> -Dist[d/(e*(b*c - a*d
)), Int[u*Sqrt[a + b*x], x], x] + Dist[b/(f*(b*c - a*d)), Int[u*Sqrt[c + d*x], x], x] /; FreeQ[{a, b, c, d, e,
 f}, x] && NeQ[b*c - a*d, 0] && EqQ[b*e^2 - d*f^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 95, normalized size = 1.00 \[ \frac {2 (a+b x)^{5/2}}{5 b^2 (a-c)}-\frac {2 a (a+b x)^{3/2}}{3 b^2 (a-c)}-\frac {2 (b x+c)^{5/2}}{5 b^2 (a-c)}+\frac {2 c (b x+c)^{3/2}}{3 b^2 (a-c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 70, normalized size = 0.74 \[ \frac {2 \, {\left ({\left (3 \, b^{2} x^{2} + a b x - 2 \, a^{2}\right )} \sqrt {b x + a} - {\left (3 \, b^{2} x^{2} + b c x - 2 \, c^{2}\right )} \sqrt {b x + c}\right )}}{15 \, {\left (a b^{2} - b^{2} c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 206, normalized size = 2.17 \[ -\frac {2 \, {\left ({\left ({\left (b x + a\right )} {\left (\frac {3 \, {\left (a b^{2} - b^{2} c\right )} {\left (b x + a\right )}}{a^{2} b^{3} - 2 \, a b^{3} c + b^{3} c^{2}} - \frac {6 \, a^{2} b^{2} - 7 \, a b^{2} c + b^{2} c^{2}}{a^{2} b^{3} - 2 \, a b^{3} c + b^{3} c^{2}}\right )} + \frac {3 \, a^{3} b^{2} - 4 \, a^{2} b^{2} c - a b^{2} c^{2} + 2 \, b^{2} c^{3}}{a^{2} b^{3} - 2 \, a b^{3} c + b^{3} c^{2}}\right )} \sqrt {b x + c} - \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a}{a b - b c}\right )}}{15 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(((b*x + a)*(3*(a*b^2 - b^2*c)*(b*x + a)/(a^2*b^3 - 2*a*b^3*c + b^3*c^2) - (6*a^2*b^2 - 7*a*b^2*c + b^2*
c^2)/(a^2*b^3 - 2*a*b^3*c + b^3*c^2)) + (3*a^3*b^2 - 4*a^2*b^2*c - a*b^2*c^2 + 2*b^2*c^3)/(a^2*b^3 - 2*a*b^3*c
 + b^3*c^2))*sqrt(b*x + c) - (3*(b*x + a)^(5/2) - 5*(b*x + a)^(3/2)*a)/(a*b - b*c))/b

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maple [A]  time = 0.00, size = 66, normalized size = 0.69 \[ \frac {-\frac {2 \left (b x +a \right )^{\frac {3}{2}} a}{3}+\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5}}{\left (a -c \right ) b^{2}}-\frac {2 \left (-\frac {\left (b x +c \right )^{\frac {3}{2}} c}{3}+\frac {\left (b x +c \right )^{\frac {5}{2}}}{5}\right )}{\left (a -c \right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {b x + a} + \sqrt {b x + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.70, size = 129, normalized size = 1.36 \[ \frac {2\,x^2\,\sqrt {a+b\,x}}{5\,\left (a-c\right )}-\frac {2\,x^2\,\sqrt {c+b\,x}}{5\,\left (a-c\right )}-\frac {4\,a^2\,\sqrt {a+b\,x}}{15\,b^2\,\left (a-c\right )}+\frac {4\,c^2\,\sqrt {c+b\,x}}{15\,b^2\,\left (a-c\right )}+\frac {2\,a\,x\,\sqrt {a+b\,x}}{15\,b\,\left (a-c\right )}-\frac {2\,c\,x\,\sqrt {c+b\,x}}{15\,b\,\left (a-c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a + b x} + \sqrt {b x + c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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