3.1960 \(\int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{d+e x} \, dx\)

Optimal. Leaf size=122 \[ -\frac{b x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^3 (a+b x)}+\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e} \]

[Out]

-((b*(b*d - a*e)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x))) + ((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(2*e) + ((b*d - a*e)^2*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 = 0.0683175, antiderivative size = 122, 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} \[ -\frac{b x \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^3 (a+b x)}+\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(b*d - a*e)*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x))) + ((a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(2*e) + ((b*d - a*e)^2*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} \, 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} \, 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} \, 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 (b d-a e)}{e^2}+\frac{b (a+b x)}{e}+\frac{(-b d+a e)^2}{e^2 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{b (b d-a e) x \sqrt{a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)}+\frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 e}+\frac{(b d-a e)^2 \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 = 0.0316657, size = 61, normalized size = 0.5 \[ \frac{\sqrt{(a+b x)^2} \left (b e x (4 a e-2 b d+b e x)+2 (b d-a e)^2 \log (d+e x)\right )}{2 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.01, size = 102, normalized size = 0.8 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ({x}^{2}{b}^{2}{e}^{2}+2\,\ln \left ( bex+bd \right ){a}^{2}{e}^{2}-4\,\ln \left ( bex+bd \right ) abde+2\,\ln \left ( bex+bd \right ){b}^{2}{d}^{2}+4\,xab{e}^{2}-2\,x{b}^{2}de+3\,{a}^{2}{e}^{2}-2\,abde \right ) }{2\,{e}^{3}}} \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),x)

[Out]

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

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69627, size = 135, normalized size = 1.11 \begin{align*} \frac{b^{2} e^{2} x^{2} - 2 \,{\left (b^{2} d e - 2 \, a b e^{2}\right )} x + 2 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (e x + d\right )}{2 \, 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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.379131, size = 44, normalized size = 0.36 \begin{align*} \frac{b^{2} x^{2}}{2 e} + \frac{x \left (2 a b e - b^{2} d\right )}{e^{2}} + \frac{\left (a e - b d\right )^{2} \log{\left (d + e x \right )}}{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),x)

[Out]

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

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Giac [A]  time = 1.14363, size = 131, normalized size = 1.07 \begin{align*}{\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 )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (b^{2} x^{2} e \mathrm{sgn}\left (b x + a\right ) - 2 \, b^{2} d x \mathrm{sgn}\left (b x + a\right ) + 4 \, a b x e \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-2\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),x, algorithm="giac")

[Out]

(b^2*d^2*sgn(b*x + a) - 2*a*b*d*e*sgn(b*x + a) + a^2*e^2*sgn(b*x + a))*e^(-3)*log(abs(x*e + d)) + 1/2*(b^2*x^2
*e*sgn(b*x + a) - 2*b^2*d*x*sgn(b*x + a) + 4*a*b*x*e*sgn(b*x + a))*e^(-2)