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

Optimal result

Integrand size = 28, antiderivative size = 131 \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a+b x}{(b d-a e) (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) \log (d+e x)}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 46} \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a+b x}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)}+\frac {b (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {b (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.89 \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {-\frac {\sqrt {a^2} e (-b d+a e) x}{a d (d+e x)}-\frac {e \sqrt {(a+b x)^2}}{d+e x}+\frac {\left (a-\sqrt {a^2}\right ) b \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )}{a}-\frac {\left (a+\sqrt {a^2}\right ) b \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{a}+\frac {\left (a+\sqrt {a^2}\right ) b \log \left (2 a e x+d \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )\right )}{a}+\frac {\left (-a+\sqrt {a^2}\right ) b \log \left (-2 a e x+d \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{a}}{2 (b d-a e)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 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	\(-\frac {\sqrt {\left (b x +a \right )^{2}}}{\left (b x +a \right ) \left (a e -b d \right ) \left (e x +d \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b \ln \left (e x +d \right )}{\left (b x +a \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\)	\(135\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {b d - a e + {\left (b e x + b d\right )} \log \left (b x + a\right ) - {\left (b e x + b d\right )} \log \left (e x + d\right )}{b^{2} d^{3} - 2 \, a b d^{2} e + a^{2} d e^{2} + {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Exception raised: ValueError} \]

[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(a*e-b*d>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx={\left (\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} - \frac {b e \log \left ({\left | e x + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} + \frac {1}{{\left (b d - a e\right )} {\left (e x + d\right )}}\right )} \mathrm {sgn}\left (b x + a\right ) \]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^2} \,d x \]

[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)