3.16.93 \(\int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [1593]

Optimal. Leaf size=131 \[ \frac {a+b x}{(b d-a e) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) \log (d+e x)}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 = {660, 46} \begin {gather*} \frac {a+b x}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)}+\frac {b (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {b (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 660

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 69, normalized size = 0.53 \begin {gather*} \frac {(a+b x) (b d-a e+b (d+e x) \log (a+b x)-b (d+e x) \log (d+e x))}{(b d-a e)^2 \sqrt {(a+b x)^2} (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.56, size = 81, normalized size = 0.62

method	result	size
default	\(\frac {\left (b x +a \right ) \left (\ln \left (b x +a \right ) b e x -\ln \left (e x +d \right ) b e x +\ln \left (b x +a \right ) b d -\ln \left (e x +d \right ) b d -a e +b d \right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{2} \left (e x +d \right )}\)	\(81\)
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}}{\left (b x +a \right ) \left (a e -b d \right ) \left (e x +d \right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b \ln \left (e x +d \right )}{\left (b x +a \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 3.52, size = 97, normalized size = 0.74 \begin {gather*} \frac {b d - a e + {\left (b x e + b d\right )} \log \left (b x + a\right ) - {\left (b x e + b d\right )} \log \left (x e + d\right )}{b^{2} d^{3} + a^{2} x e^{3} - {\left (2 \, a b d x - a^{2} d\right )} e^{2} + {\left (b^{2} d^{2} x - 2 \, a b d^{2}\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (90) = 180\).
time = 0.37, size = 233, normalized size = 1.78 \begin {gather*} - \frac {b \log {\left (x + \frac {- \frac {a^{3} b e^{3}}{\left (a e - b d\right )^{2}} + \frac {3 a^{2} b^{2} d e^{2}}{\left (a e - b d\right )^{2}} - \frac {3 a b^{3} d^{2} e}{\left (a e - b d\right )^{2}} + a b e + \frac {b^{4} d^{3}}{\left (a e - b d\right )^{2}} + b^{2} d}{2 b^{2} e} \right )}}{\left (a e - b d\right )^{2}} + \frac {b \log {\left (x + \frac {\frac {a^{3} b e^{3}}{\left (a e - b d\right )^{2}} - \frac {3 a^{2} b^{2} d e^{2}}{\left (a e - b d\right )^{2}} + \frac {3 a b^{3} d^{2} e}{\left (a e - b d\right )^{2}} + a b e - \frac {b^{4} d^{3}}{\left (a e - b d\right )^{2}} + b^{2} d}{2 b^{2} e} \right )}}{\left (a e - b d\right )^{2}} - \frac {1}{a d e - b d^{2} + x \left (a e^{2} - b d e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.16, size = 103, normalized size = 0.79 \begin {gather*} {\left (\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} - \frac {b e \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} + \frac {1}{{\left (b d - a e\right )} {\left (x e + d\right )}}\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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