3.16.92 \(\int \frac {1}{(d+e x) \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [1592]

Optimal. Leaf size=86 \[ \frac {(a+b x) \log (a+b x)}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (d+e x)}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {660, 36, 31} \begin {gather*} \frac {(a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {(a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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) \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)} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b \left (a b+b^2 x\right )\right ) \int \frac {1}{a b+b^2 x} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (e \left (a b+b^2 x\right )\right ) \int \frac {1}{d+e x} \, dx}{b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(a+b x) \log (a+b x)}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (d+e x)}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 42, normalized size = 0.49 \begin {gather*} \frac {(a+b x) (\log (a+b x)-\log (d+e x))}{(b d-a e) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.54, size = 42, normalized size = 0.49

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (\ln \left (b x +a \right )-\ln \left (e x +d \right )\right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )}\)	\(42\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (a e -b d \right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) \left (a e -b d \right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)/((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

________________________________________________________________________________________

Fricas [A]
time = 3.78, size = 28, normalized size = 0.33 \begin {gather*} \frac {\log \left (b x + a\right ) - \log \left (x e + d\right )}{b d - a e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (56) = 112\).
time = 0.17, size = 128, normalized size = 1.49 \begin {gather*} \frac {\log {\left (x + \frac {- \frac {a^{2} e^{2}}{a e - b d} + \frac {2 a b d e}{a e - b d} + a e - \frac {b^{2} d^{2}}{a e - b d} + b d}{2 b e} \right )}}{a e - b d} - \frac {\log {\left (x + \frac {\frac {a^{2} e^{2}}{a e - b d} - \frac {2 a b d e}{a e - b d} + a e + \frac {b^{2} d^{2}}{a e - b d} + b d}{2 b e} \right )}}{a e - b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.62, size = 56, normalized size = 0.65 \begin {gather*} {\left (\frac {b \log \left ({\left | b x + a \right |}\right )}{b^{2} d - a b e} - \frac {e \log \left ({\left | x e + d \right |}\right )}{b d e - a e^{2}}\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)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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 )} \,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)),x)

[Out]

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

________________________________________________________________________________________