3.1548 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{d+e x} \, dx\)

Optimal. Leaf size=80 \[ \frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{e (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^2 (a+b x)} \]

[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)

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Rubi [A]  time = 0.04, antiderivative size = 80, 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} \[ \frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{e (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^2 (a+b x)} \]

Antiderivative was successfully verified.

[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 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} \, dx &=\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*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.52 \[ \frac {\sqrt {(a+b x)^2} ((a e-b d) \log (d+e x)+b e x)}{e^2 (a+b x)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.79, size = 25, normalized size = 0.31 \[ \frac {b e x - {\left (b d - a e\right )} \log \left (e x + d\right )}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.15, size = 45, normalized size = 0.56 \[ b x e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) - {\left (b d \mathrm {sgn}\left (b x + a\right ) - a e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-2\right )} \log \left ({\left | x e + d \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.06, size = 44, normalized size = 0.55 \[ \frac {\left (a e \ln \left (b e x +b d \right )-b d \ln \left (b e x +b d \right )+b e x +a e \right ) \mathrm {csgn}\left (b x +a \right )}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)/(d + e*x),x)

[Out]

int(((a + b*x)^2)^(1/2)/(d + e*x), x)

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sympy [A]  time = 0.17, size = 20, normalized size = 0.25 \[ \frac {b x}{e} + \frac {\left (a e - b d\right ) \log {\left (d + e x \right )}}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/e + (a*e - b*d)*log(d + e*x)/e**2

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