3.1715 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=151 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{2 e^3 (a+b x) (d+e x)^2}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{e^3 (a+b x) (d+e x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.097895, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 77} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{2 e^3 (a+b x) (d+e x)^2}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{e^3 (a+b x) (d+e x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*d - a*e)*(B*d - A*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^3*(a + b*x)*(d + e*x)^2) + ((2*b*B*d - A*b*e - a
*B*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)*(d + e*x)) + (b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[d + e*
x])/(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} \, 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} \, 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}+\frac{b (-2 b B d+A b e+a B e)}{e^2 (d+e x)^2}+\frac{b^2 B}{e^2 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{(b d-a e) (B d-A e) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x) (d+e x)^2}+\frac{(2 b B d-A b e-a B e) \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) (d+e x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0507265, size = 89, normalized size = 0.59 \[ -\frac{\sqrt{(a+b x)^2} \left (a e (A e+B (d+2 e x))+b (A e (d+2 e x)-B d (3 d+4 e x))-2 b B (d+e x)^2 \log (d+e x)\right )}{2 e^3 (a+b x) (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.01, size = 117, normalized size = 0.8 \begin{align*} -{\frac{{\it csgn} \left ( bx+a \right ) \left ( -2\,B\ln \left ( bex+bd \right ){x}^{2}b{e}^{2}-4\,B\ln \left ( bex+bd \right ) xbde+2\,Axb{e}^{2}-2\,B\ln \left ( bex+bd \right ) b{d}^{2}+2\,aB{e}^{2}x-4\,Bxbde+aA{e}^{2}+Abde+aBde-3\,Bb{d}^{2} \right ) }{2\,{e}^{3} \left ( ex+d \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)/(e*x+d)^3,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60622, size = 227, normalized size = 1.5 \begin{align*} \frac{3 \, B b d^{2} - A a e^{2} -{\left (B a + A b\right )} d e + 2 \,{\left (2 \, B b d e -{\left (B a + A b\right )} e^{2}\right )} x + 2 \,{\left (B b e^{2} x^{2} + 2 \, B b d e x + B b d^{2}\right )} \log \left (e x + d\right )}{2 \,{\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)^3,x, algorithm="fricas")

[Out]

1/2*(3*B*b*d^2 - A*a*e^2 - (B*a + A*b)*d*e + 2*(2*B*b*d*e - (B*a + A*b)*e^2)*x + 2*(B*b*e^2*x^2 + 2*B*b*d*e*x
+ B*b*d^2)*log(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)

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Sympy [A]  time = 1.14237, size = 94, normalized size = 0.62 \begin{align*} \frac{B b \log{\left (d + e x \right )}}{e^{3}} - \frac{A a e^{2} + A b d e + B a d e - 3 B b d^{2} + x \left (2 A b e^{2} + 2 B a e^{2} - 4 B b d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{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)**3,x)

[Out]

B*b*log(d + e*x)/e**3 - (A*a*e**2 + A*b*d*e + B*a*d*e - 3*B*b*d**2 + x*(2*A*b*e**2 + 2*B*a*e**2 - 4*B*b*d*e))/
(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2)

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Giac [A]  time = 1.10246, size = 171, normalized size = 1.13 \begin{align*} B b e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm{sgn}\left (b x + a\right ) + \frac{{\left (2 \,{\left (2 \, B b d \mathrm{sgn}\left (b x + a\right ) - B a e \mathrm{sgn}\left (b x + a\right ) - A b e \mathrm{sgn}\left (b x + a\right )\right )} x +{\left (3 \, 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 (-1\right )}\right )} e^{\left (-2\right )}}{2 \,{\left (x e + d\right )}^{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)^3,x, algorithm="giac")

[Out]

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