3.2098 \(\int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=148 \[ \frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x}}{e^3 (a+b x)}+\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) \sqrt{d+e x}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^3 (a+b x) (d+e x)^{3/2}} \]

[Out]

(-2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)*(d + e*x)^(3/2)) + (4*b*(b*d - a*e)*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)*Sqrt[d + e*x]) + (2*b^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^
3*(a + b*x))

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Rubi [A]  time = 0.0667385, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {770, 21, 43} \[ \frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x}}{e^3 (a+b x)}+\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) \sqrt{d+e x}}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{3 e^3 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)*(d + e*x)^(3/2)) + (4*b*(b*d - a*e)*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)*Sqrt[d + e*x]) + (2*b^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^
3*(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 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 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])

Rubi steps

\begin{align*} \int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{5/2}} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{(a+b x) \left (a b+b^2 x\right )}{(d+e x)^{5/2}} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{(a+b x)^2}{(d+e x)^{5/2}} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(-b d+a e)^2}{e^2 (d+e x)^{5/2}}-\frac{2 b (b d-a e)}{e^2 (d+e x)^{3/2}}+\frac{b^2}{e^2 \sqrt{d+e x}}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x) (d+e x)^{3/2}}+\frac{4 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt{d+e x}}+\frac{2 b^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0454454, size = 79, normalized size = 0.53 \[ -\frac{2 \sqrt{(a+b x)^2} \left (a^2 e^2+2 a b e (2 d+3 e x)+b^2 \left (-\left (8 d^2+12 d e x+3 e^2 x^2\right )\right )\right )}{3 e^3 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[(a + b*x)^2]*(a^2*e^2 + 2*a*b*e*(2*d + 3*e*x) - b^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2)))/(3*e^3*(a + b*x)
*(d + e*x)^(3/2))

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Maple [A]  time = 0.004, size = 78, normalized size = 0.5 \begin{align*} -{\frac{-6\,{x}^{2}{b}^{2}{e}^{2}+12\,xab{e}^{2}-24\,x{b}^{2}de+2\,{a}^{2}{e}^{2}+8\,abde-16\,{b}^{2}{d}^{2}}{3\, \left ( bx+a \right ){e}^{3}}\sqrt{ \left ( bx+a \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/(e*x+d)^(3/2)*(-3*b^2*e^2*x^2+6*a*b*e^2*x-12*b^2*d*e*x+a^2*e^2+4*a*b*d*e-8*b^2*d^2)*((b*x+a)^2)^(1/2)/e^3
/(b*x+a)

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Maxima [A]  time = 1.13715, size = 130, normalized size = 0.88 \begin{align*} -\frac{2 \,{\left (3 \, b e x + 2 \, b d + a e\right )} a}{3 \,{\left (e^{3} x + d e^{2}\right )} \sqrt{e x + d}} + \frac{2 \,{\left (3 \, b e^{2} x^{2} + 8 \, b d^{2} - 2 \, a d e + 3 \,{\left (4 \, b d e - a e^{2}\right )} x\right )} b}{3 \,{\left (e^{4} x + d e^{3}\right )} \sqrt{e x + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

-2/3*(3*b*e*x + 2*b*d + a*e)*a/((e^3*x + d*e^2)*sqrt(e*x + d)) + 2/3*(3*b*e^2*x^2 + 8*b*d^2 - 2*a*d*e + 3*(4*b
*d*e - a*e^2)*x)*b/((e^4*x + d*e^3)*sqrt(e*x + d))

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Fricas [A]  time = 0.961947, size = 174, normalized size = 1.18 \begin{align*} \frac{2 \,{\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 4 \, a b d e - a^{2} e^{2} + 6 \,{\left (2 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt{e x + d}}{3 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*b^2*e^2*x^2 + 8*b^2*d^2 - 4*a*b*d*e - a^2*e^2 + 6*(2*b^2*d*e - a*b*e^2)*x)*sqrt(e*x + d)/(e^5*x^2 + 2*d
*e^4*x + d^2*e^3)

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

[Out]

Timed out

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Giac [A]  time = 1.22577, size = 150, normalized size = 1.01 \begin{align*} 2 \, \sqrt{x e + d} b^{2} e^{\left (-3\right )} \mathrm{sgn}\left (b x + a\right ) + \frac{2 \,{\left (6 \,{\left (x e + d\right )} b^{2} d \mathrm{sgn}\left (b x + a\right ) - b^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) - 6 \,{\left (x e + d\right )} a b e \mathrm{sgn}\left (b x + a\right ) + 2 \, a b d e \mathrm{sgn}\left (b x + a\right ) - a^{2} e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{3 \,{\left (x e + d\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*((b*x+a)^2)^(1/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2*sqrt(x*e + d)*b^2*e^(-3)*sgn(b*x + a) + 2/3*(6*(x*e + d)*b^2*d*sgn(b*x + a) - b^2*d^2*sgn(b*x + a) - 6*(x*e
+ d)*a*b*e*sgn(b*x + a) + 2*a*b*d*e*sgn(b*x + a) - a^2*e^2*sgn(b*x + a))*e^(-3)/(x*e + d)^(3/2)