3.1962 \(\int \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=140 \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) (d+e x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^3 (a+b x) (d+e x)^2}+\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0813509, antiderivative size = 140, 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} \[ \frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) (d+e x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^3 (a+b x) (d+e x)^2}+\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

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)^2*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*e)*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/(e^3*(a + b*x)*(d + e*x)) + (b^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)^3} \, 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)^3} \, 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)^3} \, 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)^3}-\frac{2 b (b d-a e)}{e^2 (d+e x)^2}+\frac{b^2}{e^2 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{(b d-a e)^2 \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 e) \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) (d+e x)}+\frac{b^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 = 0.0378832, size = 73, normalized size = 0.52 \[ \frac{\sqrt{(a+b x)^2} \left ((b d-a e) (a e+3 b d+4 b e x)+2 b^2 (d+e x)^2 \log (d+e x)\right )}{2 e^3 (a+b x) (d+e x)^2} \]

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

________________________________________________________________________________________

Maple [C]  time = 0.011, size = 112, normalized size = 0.8 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ( 2\,\ln \left ( bex+bd \right ){x}^{2}{b}^{2}{e}^{2}+4\,\ln \left ( bex+bd \right ) x{b}^{2}de+2\,\ln \left ( bex+bd \right ){b}^{2}{d}^{2}-4\,xab{e}^{2}+4\,x{b}^{2}de-{a}^{2}{e}^{2}-2\,abde+3\,{b}^{2}{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*ln(b*e*x+b*d)*x^2*b^2*e^2+4*ln(b*e*x+b*d)*x*b^2*d*e+2*ln(b*e*x+b*d)*b^2*d^2-4*x*a*b*e^2+4*x
*b^2*d*e-a^2*e^2-2*a*b*d*e+3*b^2*d^2)/e^3/(e*x+d)^2

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [A]  time = 1.49011, size = 205, normalized size = 1.46 \begin{align*} \frac{3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \,{\left (b^{2} d e - a b e^{2}\right )} x + 2 \,{\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} 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^2*d^2 - 2*a*b*d*e - a^2*e^2 + 4*(b^2*d*e - a*b*e^2)*x + 2*(b^2*e^2*x^2 + 2*b^2*d*e*x + b^2*d^2)*log(e
*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)

________________________________________________________________________________________

Sympy [A]  time = 0.648105, size = 80, normalized size = 0.57 \begin{align*} \frac{b^{2} \log{\left (d + e x \right )}}{e^{3}} - \frac{a^{2} e^{2} + 2 a b d e - 3 b^{2} d^{2} + x \left (4 a b e^{2} - 4 b^{2} 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**2*log(d + e*x)/e**3 - (a**2*e**2 + 2*a*b*d*e - 3*b**2*d**2 + x*(4*a*b*e**2 - 4*b**2*d*e))/(2*d**2*e**3 + 4*
d*e**4*x + 2*e**5*x**2)

________________________________________________________________________________________

Giac [A]  time = 1.11509, size = 142, normalized size = 1.01 \begin{align*} b^{2} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm{sgn}\left (b x + a\right ) + \frac{{\left (4 \,{\left (b^{2} d \mathrm{sgn}\left (b x + a\right ) - a b e \mathrm{sgn}\left (b x + a\right )\right )} x +{\left (3 \, 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 (-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^2*e^(-3)*log(abs(x*e + d))*sgn(b*x + a) + 1/2*(4*(b^2*d*sgn(b*x + a) - a*b*e*sgn(b*x + a))*x + (3*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^(-1))*e^(-2)/(x*e + d)^2