3.786 \(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x}} \, dx\)

Optimal. Leaf size=118 \[ \frac {2 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 (a+b x)}+\frac {2 a A \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b B x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

[Out]

2/3*(A*b+B*a)*x^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2/5*b*B*x^(5/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2*a*A*x^(1/2)*((b*x+
a)^2)^(1/2)/(b*x+a)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 76} \[ \frac {2 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 (a+b x)}+\frac {2 a A \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b B x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (2*(A*b + a*B)*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(3*(a + b*x)) + (2*b*B*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 49, normalized size = 0.42 \[ \frac {2 \sqrt {x} \sqrt {(a+b x)^2} (5 a (3 A+B x)+b x (5 A+3 B x))}{15 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Sqrt[(a + b*x)^2]*(5*a*(3*A + B*x) + b*x*(5*A + 3*B*x)))/(15*(a + b*x))

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 27, normalized size = 0.23 \[ \frac {2}{15} \, {\left (3 \, B b x^{2} + 15 \, A a + 5 \, {\left (B a + A b\right )} x\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*B*b*x^2 + 15*A*a + 5*(B*a + A*b)*x)*sqrt(x)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 53, normalized size = 0.45 \[ \frac {2}{5} \, B b x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, B a x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{3} \, A b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a \sqrt {x} \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*B*b*x^(5/2)*sgn(b*x + a) + 2/3*B*a*x^(3/2)*sgn(b*x + a) + 2/3*A*b*x^(3/2)*sgn(b*x + a) + 2*A*a*sqrt(x)*sgn
(b*x + a)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 44, normalized size = 0.37 \[ \frac {2 \left (3 B b \,x^{2}+5 A b x +5 B a x +15 A a \right ) \sqrt {\left (b x +a \right )^{2}}\, \sqrt {x}}{15 \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*x^(1/2)*(3*B*b*x^2+5*A*b*x+5*B*a*x+15*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)

________________________________________________________________________________________

maxima [A]  time = 0.50, size = 34, normalized size = 0.29 \[ \frac {2}{15} \, {\left (3 \, b x^{2} + 5 \, a x\right )} B \sqrt {x} + \frac {2 \, {\left (b x^{2} + 3 \, a x\right )} A}{3 \, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b*x^2 + 5*a*x)*B*sqrt(x) + 2/3*(b*x^2 + 3*a*x)*A/sqrt(x)

________________________________________________________________________________________

mupad [B]  time = 1.31, size = 56, normalized size = 0.47 \[ \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^3}{5}+\frac {x^2\,\left (10\,A\,b+10\,B\,a\right )}{15\,b}+\frac {2\,A\,a\,x}{b}\right )}{x^{3/2}+\frac {a\,\sqrt {x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a + b*x)^2)^(1/2)*((2*B*x^3)/5 + (x^2*(10*A*b + 10*B*a))/(15*b) + (2*A*a*x)/b))/(x^(3/2) + (a*x^(1/2))/b)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________