3.2125 \(\int \frac{(a+b x) \sqrt{d+e x}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=41 \[ \frac{2 (a+b x) (d+e x)^{3/2}}{3 e \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(2*(a + b*x)*(d + e*x)^(3/2))/(3*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0289997, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {770, 21, 32} \[ \frac{2 (a+b x) (d+e x)^{3/2}}{3 e \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(d + e*x)^(3/2))/(3*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0135538, size = 32, normalized size = 0.78 \[ \frac{2 (a+b x) (d+e x)^{3/2}}{3 e \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(d + e*x)^(3/2))/(3*e*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.003, size = 27, normalized size = 0.7 \begin{align*}{\frac{2\,bx+2\,a}{3\,e} \left ( ex+d \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x+a)*(e*x+d)^(3/2)/e/((b*x+a)^2)^(1/2)

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Maxima [A]  time = 1.28418, size = 16, normalized size = 0.39 \begin{align*} \frac{2 \,{\left (e x + d\right )}^{\frac{3}{2}}}{3 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(e*x + d)^(3/2)/e

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Fricas [A]  time = 0.988282, size = 31, normalized size = 0.76 \begin{align*} \frac{2 \,{\left (e x + d\right )}^{\frac{3}{2}}}{3 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(e*x + d)^(3/2)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17807, size = 24, normalized size = 0.59 \begin{align*} \frac{2}{3} \,{\left (x e + d\right )}^{\frac{3}{2}} e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x*e + d)^(3/2)*e^(-1)*sgn(b*x + a)