3.1840 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=160 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (-a B e-A b e+2 b B d)}{e^3 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{e^3 (a+b x) \sqrt{d+e x}}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.080696, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {770, 77} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (-a B e-A b e+2 b B d)}{e^3 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{e^3 (a+b x) \sqrt{d+e x}}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)*(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*(2*b*B*d - A*b*e
- a*B*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)) + (2*b*B*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/(3*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0600264, size = 85, normalized size = 0.53 \[ \frac{2 \sqrt{(a+b x)^2} \left (3 a e (-A e+2 B d+B e x)+3 A b e (2 d+e x)+b B \left (-8 d^2-4 d e x+e^2 x^2\right )\right )}{3 e^3 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(3*A*b*e*(2*d + e*x) + 3*a*e*(2*B*d - A*e + B*e*x) + b*B*(-8*d^2 - 4*d*e*x + e^2*x^2)))/(
3*e^3*(a + b*x)*Sqrt[d + e*x])

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Maple [A]  time = 0.004, size = 89, normalized size = 0.6 \begin{align*} -{\frac{-2\,B{x}^{2}b{e}^{2}-6\,Axb{e}^{2}-6\,aB{e}^{2}x+8\,Bxbde+6\,aA{e}^{2}-12\,Abde-12\,aBde+16\,Bb{d}^{2}}{3\, \left ( bx+a \right ){e}^{3}}\sqrt{ \left ( bx+a \right ) ^{2}}{\frac{1}{\sqrt{ex+d}}}} \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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.03474, size = 101, normalized size = 0.63 \begin{align*} \frac{2 \,{\left (b e x + 2 \, b d - a e\right )} A}{\sqrt{e x + d} e^{2}} + \frac{2 \,{\left (b e^{2} x^{2} - 8 \, b d^{2} + 6 \, a d e -{\left (4 \, b d e - 3 \, a e^{2}\right )} x\right )} B}{3 \, \sqrt{e x + d} e^{3}} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.33586, size = 174, normalized size = 1.09 \begin{align*} \frac{2 \,{\left (B b e^{2} x^{2} - 8 \, B b d^{2} - 3 \, A a e^{2} + 6 \,{\left (B a + A b\right )} d e -{\left (4 \, B b d e - 3 \,{\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{3 \,{\left (e^{4} x + d 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)^(3/2),x, algorithm="fricas")

[Out]

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

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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}}}{\left (d + e x\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral((A + B*x)*sqrt((a + b*x)**2)/(d + e*x)**(3/2), x)

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Giac [A]  time = 1.15766, size = 200, normalized size = 1.25 \begin{align*} \frac{2}{3} \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} B b e^{6} \mathrm{sgn}\left (b x + a\right ) - 6 \, \sqrt{x e + d} B b d e^{6} \mathrm{sgn}\left (b x + a\right ) + 3 \, \sqrt{x e + d} B a e^{7} \mathrm{sgn}\left (b x + a\right ) + 3 \, \sqrt{x e + d} A b e^{7} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-9\right )} - \frac{2 \,{\left (B b d^{2} \mathrm{sgn}\left (b x + a\right ) - B a d e \mathrm{sgn}\left (b x + a\right ) - A b d e \mathrm{sgn}\left (b x + a\right ) + A a e^{2} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{\sqrt{x e + 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)^(3/2),x, algorithm="giac")

[Out]

2/3*((x*e + d)^(3/2)*B*b*e^6*sgn(b*x + a) - 6*sqrt(x*e + d)*B*b*d*e^6*sgn(b*x + a) + 3*sqrt(x*e + d)*B*a*e^7*s
gn(b*x + a) + 3*sqrt(x*e + d)*A*b*e^7*sgn(b*x + a))*e^(-9) - 2*(B*b*d^2*sgn(b*x + a) - B*a*d*e*sgn(b*x + a) -
A*b*d*e*sgn(b*x + a) + A*a*e^2*sgn(b*x + a))*e^(-3)/sqrt(x*e + d)