3.1549 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=85 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x) (d+e x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 85, 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 = {646, 43} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x) (d+e x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.59 \[ \frac {\sqrt {(a+b x)^2} (-a e+b (d+e x) \log (d+e x)+b d)}{e^2 (a+b x) (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 37, normalized size = 0.44 \[ \frac {b d - a e + {\left (b e x + b d\right )} \log \left (e x + d\right )}{e^{3} x + d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 51, normalized size = 0.60 \[ b e^{\left (-2\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (b d \mathrm {sgn}\left (b x + a\right ) - a e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-2\right )}}{x e + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.06, size = 51, normalized size = 0.60 \[ \frac {\left (b e x \ln \left (b e x +b d \right )+b d \ln \left (b e x +b d \right )-a e +b d \right ) \mathrm {csgn}\left (b x +a \right )}{\left (e x +d \right ) e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x+a)^2)^(1/2)/(e*x+d)^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 details)Is a*e-b*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.21, size = 27, normalized size = 0.32 \[ \frac {b \log {\left (d + e x \right )}}{e^{2}} + \frac {- a e + b d}{d e^{2} + e^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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