3.1961 \(\int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=132 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) (d+e x)}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^3 (a+b x)}+\frac{b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \]

[Out]

(b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x)) - ((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a
+ b*x)*(d + e*x)) - (2*b*(b*d - a*e)*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 = 0.0791044, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) (d+e x)}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^3 (a+b x)}+\frac{b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x)) - ((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a
+ b*x)*(d + e*x)) - (2*b*(b*d - a*e)*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 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)^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)^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)^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^2}{e^2}+\frac{(-b d+a e)^2}{e^2 (d+e x)^2}-\frac{2 b (b d-a e)}{e^2 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=\frac{b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)}-\frac{(b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) (d+e x)}-\frac{2 b (b d-a e) \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 = 0.0536121, size = 87, normalized size = 0.66 \[ \frac{\sqrt{(a+b x)^2} \left (-a^2 e^2-2 b (d+e x) (b d-a e) \log (d+e x)+2 a b d e+b^2 \left (-d^2+d e x+e^2 x^2\right )\right )}{e^3 (a+b x) (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.013, size = 131, normalized size = 1. \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ( 2\,\ln \left ( bex+bd \right ) xab{e}^{2}-2\,\ln \left ( bex+bd \right ) x{b}^{2}de+{x}^{2}{b}^{2}{e}^{2}+2\,\ln \left ( bex+bd \right ) abde-2\,\ln \left ( bex+bd \right ){b}^{2}{d}^{2}+xab{e}^{2}+x{b}^{2}de-{a}^{2}{e}^{2}+3\,abde-{b}^{2}{d}^{2} \right ) }{{e}^{3} \left ( ex+d \right ) }} \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)^2,x)

[Out]

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

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61301, size = 184, normalized size = 1.39 \begin{align*} \frac{b^{2} e^{2} x^{2} + b^{2} d e x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} - 2 \,{\left (b^{2} d^{2} - a b d e +{\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.522427, size = 60, normalized size = 0.45 \begin{align*} \frac{b^{2} x}{e^{2}} + \frac{2 b \left (a e - b d\right ) \log{\left (d + e x \right )}}{e^{3}} - \frac{a^{2} e^{2} - 2 a b d e + b^{2} d^{2}}{d e^{3} + e^{4} x} \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)**2,x)

[Out]

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

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Giac [A]  time = 1.10182, size = 136, normalized size = 1.03 \begin{align*} b^{2} x e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) - 2 \,{\left (b^{2} d \mathrm{sgn}\left (b x + a\right ) - a b e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (b^{2} d^{2} \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 )}}{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)^2,x, algorithm="giac")

[Out]

b^2*x*e^(-2)*sgn(b*x + a) - 2*(b^2*d*sgn(b*x + a) - a*b*e*sgn(b*x + a))*e^(-3)*log(abs(x*e + d)) - (b^2*d^2*sg
n(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)