3.2124 \(\int \frac{(a+b x) (d+e x)^{3/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=41 \[ \frac{2 (a+b x) (d+e x)^{5/2}}{5 e \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(2*(a + b*x)*(d + e*x)^(5/2))/(5*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0294393, 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} \[ \frac{2 (a+b x) (d+e x)^{5/2}}{5 e \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(d + e*x)^(5/2))/(5*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) (d+e x)^{3/2}}{\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) (d+e x)^{3/2}}{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 (d+e x)^{3/2} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (a+b x) (d+e x)^{5/2}}{5 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0179308, size = 32, normalized size = 0.78 \[ \frac{2 (a+b x) (d+e x)^{5/2}}{5 e \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(d + e*x)^(5/2))/(5*e*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.001, size = 27, normalized size = 0.7 \begin{align*}{\frac{2\,bx+2\,a}{5\,e} \left ( ex+d \right ) ^{{\frac{5}{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)^(3/2)/((b*x+a)^2)^(1/2),x)

[Out]

2/5*(b*x+a)*(e*x+d)^(5/2)/e/((b*x+a)^2)^(1/2)

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Maxima [A]  time = 1.20687, size = 38, normalized size = 0.93 \begin{align*} \frac{2 \,{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt{e x + d}}{5 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

2/5*(e^2*x^2 + 2*d*e*x + d^2)*sqrt(e*x + d)/e

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Fricas [A]  time = 0.995322, size = 63, normalized size = 1.54 \begin{align*} \frac{2 \,{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt{e x + d}}{5 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

2/5*(e^2*x^2 + 2*d*e*x + d^2)*sqrt(e*x + d)/e

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.12794, size = 69, normalized size = 1.68 \begin{align*} \frac{2}{15} \,{\left (5 \,{\left (x e + d\right )}^{\frac{3}{2}} d \mathrm{sgn}\left (b x + a\right ) +{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^(3/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/15*(5*(x*e + d)^(3/2)*d*sgn(b*x + a) + (3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*sgn(b*x + a))*e^(-1)