3.1601 \(\int \frac{1}{(d+e x) (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac{e^2 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{e^2 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{e}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{1}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

[Out]

e/((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(e^2*(a + b*x)*Log[a + b*x])/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^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 = 0.0969946, antiderivative size = 165, 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} \[ \frac{e^2 (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{e^2 (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{e}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{1}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

e/((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(e^2*(a + b*x)*Log[a + b*x])/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^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) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^3 (d+e x)} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac{1}{b^2 (b d-a e) (a+b x)^3}-\frac{e}{b^2 (b d-a e)^2 (a+b x)^2}+\frac{e^2}{b^2 (b d-a e)^3 (a+b x)}-\frac{e^3}{b^3 (b d-a e)^3 (d+e x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{e}{(b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{1}{2 (b d-a e) (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^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{e^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 = 0.0550638, size = 92, normalized size = 0.56 \[ \frac{-2 e^2 (a+b x)^2 \log (d+e x)-(b d-a e) (b (d-2 e x)-3 a e)+2 e^2 (a+b x)^2 \log (a+b x)}{2 (a+b x) \sqrt{(a+b x)^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.203, size = 155, normalized size = 0.9 \begin{align*}{\frac{ \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 ) xab{e}^{2}-4\,\ln \left ( bx+a \right ) xab{e}^{2}+2\,\ln \left ( ex+d \right ){a}^{2}{e}^{2}-2\,\ln \left ( bx+a \right ){a}^{2}{e}^{2}+2\,xab{e}^{2}-2\,x{b}^{2}de+3\,{a}^{2}{e}^{2}-4\,abde+{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) }{2\, \left ( ae-bd \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.49254, size = 491, normalized size = 2.98 \begin{align*} -\frac{b^{2} d^{2} - 4 \, a b d e + 3 \, 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 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (b x + a\right ) + 2 \,{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (e x + d\right )}{2 \,{\left (a^{2} b^{3} d^{3} - 3 \, a^{3} b^{2} d^{2} e + 3 \, a^{4} b d e^{2} - a^{5} e^{3} +{\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} x^{2} + 2 \,{\left (a b^{4} d^{3} - 3 \, a^{2} b^{3} d^{2} e + 3 \, a^{3} b^{2} d e^{2} - a^{4} b e^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x