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*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))

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Rubi [A]  time = 0.042182, antiderivative size = 118, normalized size of antiderivative = 1., 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 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]

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])

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 = 0.0297499, 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))

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Maple [A]  time = 0.003, size = 44, normalized size = 0.4 \begin{align*}{\frac{6\,Bb{x}^{2}+10\,Abx+10\,aBx+30\,aA}{15\,bx+15\,a}\sqrt{x}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

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)

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Maxima [A]  time = 1.0606, size = 46, normalized size = 0.39 \begin{align*} \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}} \end{align*}

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)

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Fricas [A]  time = 1.52014, size = 72, normalized size = 0.61 \begin{align*} \frac{2}{15} \,{\left (3 \, B b x^{2} + 15 \, A a + 5 \,{\left (B a + A b\right )} x\right )} \sqrt{x} \end{align*}

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)

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

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)

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Giac [A]  time = 1.15125, size = 72, normalized size = 0.61 \begin{align*} \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 ) \end{align*}

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)