\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx\) [1548]

Optimal result

Integrand size = 28, antiderivative size = 80 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx=\frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{e (a+b x)}-\frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^2 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, 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, 45} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx=\frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{e (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^2 (a+b x)} \]

[In]

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

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {a b+b^2 x}{d+e x} \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^2}{e}-\frac {b (b d-a e)}{e (d+e x)}\right ) \, dx}{a b+b^2 x} \\ & = \frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{e (a+b x)}-\frac {(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^2 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx=\frac {\sqrt {(a+b x)^2} (b e x+(-b d+a e) \log (d+e x))}{e^2 (a+b x)} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (\ln \left (-b e x -b d \right ) a e -\ln \left (-b e x -b d \right ) b d +b e x +a e \right )}{e^{2}}\)	\(48\)
risch	\(\frac {b x \sqrt {\left (b x +a \right )^{2}}}{e \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{2}}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.31 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx=\frac {b e x - {\left (b d - a e\right )} \log \left (e x + d\right )}{e^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d),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 = 45, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx=\frac {b x \mathrm {sgn}\left (b x + a\right )}{e} - \frac {{\left (b d \mathrm {sgn}\left (b x + a\right ) - a e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x),x)

[Out]

int(((a + b*x)^2)^(1/2)/(d + e*x), x)