3.1594 \(\int \frac{1}{(d+e x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=182 \[ \frac{b (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}+\frac{a+b x}{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac{b^2 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{b^2 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

[Out]

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

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Rubi [A]  time = 0.0813618, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 44} \[ \frac{b (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}+\frac{a+b x}{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac{b^2 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{b^2 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (\frac{b^2}{(b d-a e)^3 (a+b x)}-\frac{e}{b (b d-a e) (d+e x)^3}-\frac{e}{(b d-a e)^2 (d+e x)^2}-\frac{b e}{(b d-a e)^3 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{a+b x}{2 (b d-a e) (d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b (a+b x)}{(b d-a e)^2 (d+e x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b^2 (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b^2 (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0650985, size = 97, normalized size = 0.53 \[ \frac{(a+b x) \left (2 b^2 (d+e x)^2 \log (a+b x)+(b d-a e) (-a e+3 b d+2 b e x)-2 b^2 (d+e x)^2 \log (d+e x)\right )}{2 \sqrt{(a+b x)^2} (d+e x)^2 (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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Maple [A]  time = 0.165, size = 163, normalized size = 0.9 \begin{align*}{\frac{ \left ( bx+a \right ) \left ( 2\,\ln \left ( ex+d \right ){x}^{2}{b}^{2}{e}^{2}-2\,\ln \left ( bx+a \right ){x}^{2}{b}^{2}{e}^{2}+4\,\ln \left ( ex+d \right ) x{b}^{2}de-4\,\ln \left ( bx+a \right ) x{b}^{2}de+2\,\ln \left ( ex+d \right ){b}^{2}{d}^{2}-2\,\ln \left ( bx+a \right ){b}^{2}{d}^{2}+2\,xab{e}^{2}-2\,x{b}^{2}de-{a}^{2}{e}^{2}+4\,abde-3\,{b}^{2}{d}^{2} \right ) }{2\, \left ( ae-bd \right ) ^{3} \left ( ex+d \right ) ^{2}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x)

[Out]

1/2*(b*x+a)*(2*ln(e*x+d)*x^2*b^2*e^2-2*ln(b*x+a)*x^2*b^2*e^2+4*ln(e*x+d)*x*b^2*d*e-4*ln(b*x+a)*x*b^2*d*e+2*ln(
e*x+d)*b^2*d^2-2*ln(b*x+a)*b^2*d^2+2*x*a*b*e^2-2*x*b^2*d*e-a^2*e^2+4*a*b*d*e-3*b^2*d^2)/((b*x+a)^2)^(1/2)/(a*e
-b*d)^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(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61354, size = 490, normalized size = 2.69 \begin{align*} \frac{3 \, b^{2} d^{2} - 4 \, a b d e + a^{2} e^{2} + 2 \,{\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 (b x + a\right ) - 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 (b^{3} d^{5} - 3 \, a b^{2} d^{4} e + 3 \, a^{2} b d^{3} e^{2} - a^{3} d^{2} e^{3} +{\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} x^{2} + 2 \,{\left (b^{3} d^{4} e - 3 \, a b^{2} d^{3} e^{2} + 3 \, a^{2} b d^{2} e^{3} - a^{3} d e^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.47254, size = 381, normalized size = 2.09 \begin{align*} \frac{b^{2} \log{\left (x + \frac{- \frac{a^{4} b^{2} e^{4}}{\left (a e - b d\right )^{3}} + \frac{4 a^{3} b^{3} d e^{3}}{\left (a e - b d\right )^{3}} - \frac{6 a^{2} b^{4} d^{2} e^{2}}{\left (a e - b d\right )^{3}} + \frac{4 a b^{5} d^{3} e}{\left (a e - b d\right )^{3}} + a b^{2} e - \frac{b^{6} d^{4}}{\left (a e - b d\right )^{3}} + b^{3} d}{2 b^{3} e} \right )}}{\left (a e - b d\right )^{3}} - \frac{b^{2} \log{\left (x + \frac{\frac{a^{4} b^{2} e^{4}}{\left (a e - b d\right )^{3}} - \frac{4 a^{3} b^{3} d e^{3}}{\left (a e - b d\right )^{3}} + \frac{6 a^{2} b^{4} d^{2} e^{2}}{\left (a e - b d\right )^{3}} - \frac{4 a b^{5} d^{3} e}{\left (a e - b d\right )^{3}} + a b^{2} e + \frac{b^{6} d^{4}}{\left (a e - b d\right )^{3}} + b^{3} d}{2 b^{3} e} \right )}}{\left (a e - b d\right )^{3}} + \frac{- a e + 3 b d + 2 b e x}{2 a^{2} d^{2} e^{2} - 4 a b d^{3} e + 2 b^{2} d^{4} + x^{2} \left (2 a^{2} e^{4} - 4 a b d e^{3} + 2 b^{2} d^{2} e^{2}\right ) + x \left (4 a^{2} d e^{3} - 8 a b d^{2} e^{2} + 4 b^{2} d^{3} e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

b**2*log(x + (-a**4*b**2*e**4/(a*e - b*d)**3 + 4*a**3*b**3*d*e**3/(a*e - b*d)**3 - 6*a**2*b**4*d**2*e**2/(a*e
- b*d)**3 + 4*a*b**5*d**3*e/(a*e - b*d)**3 + a*b**2*e - b**6*d**4/(a*e - b*d)**3 + b**3*d)/(2*b**3*e))/(a*e -
b*d)**3 - b**2*log(x + (a**4*b**2*e**4/(a*e - b*d)**3 - 4*a**3*b**3*d*e**3/(a*e - b*d)**3 + 6*a**2*b**4*d**2*e
**2/(a*e - b*d)**3 - 4*a*b**5*d**3*e/(a*e - b*d)**3 + a*b**2*e + b**6*d**4/(a*e - b*d)**3 + b**3*d)/(2*b**3*e)
)/(a*e - b*d)**3 + (-a*e + 3*b*d + 2*b*e*x)/(2*a**2*d**2*e**2 - 4*a*b*d**3*e + 2*b**2*d**4 + x**2*(2*a**2*e**4
- 4*a*b*d*e**3 + 2*b**2*d**2*e**2) + x*(4*a**2*d*e**3 - 8*a*b*d**2*e**2 + 4*b**2*d**3*e))

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Giac [A]  time = 1.16502, size = 235, normalized size = 1.29 \begin{align*} \frac{1}{2} \,{\left (\frac{2 \, b^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}} - \frac{2 \, b^{2} e \log \left ({\left | x e + d \right |}\right )}{b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}} + \frac{3 \, b^{2} d^{2} - 4 \, a b d e + a^{2} e^{2} + 2 \,{\left (b^{2} d e - a b e^{2}\right )} x}{{\left (b d - a e\right )}^{3}{\left (x e + d\right )}^{2}}\right )} \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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