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

Optimal. Leaf size=35 \[ \frac {(a+b x) \log (d+e x)}{e \sqrt {a^2+2 a b x+b^2 x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 31} \begin {gather*} \frac {(a+b x) \log (d+e x)}{e \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 31

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {a+b x}{(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 {a+b x}{\left (a b+b^2 x\right ) (d+e x)} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{d+e x} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(a+b x) \log (d+e x)}{e \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 26, normalized size = 0.74 \begin {gather*} \frac {(a+b x) \log (d+e x)}{e \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.54, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x}{(d+e x) \sqrt {a^2+2 a b x+b^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 10, normalized size = 0.29 \begin {gather*} \frac {\log \left (e x + d\right )}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e*x + d)/e

________________________________________________________________________________________

giac [A]  time = 0.16, size = 17, normalized size = 0.49 \begin {gather*} e^{\left (-1\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-1)*log(abs(x*e + d))*sgn(b*x + a)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 25, normalized size = 0.71 \begin {gather*} \frac {\left (b x +a \right ) \ln \left (e x +d \right )}{\sqrt {\left (b x +a \right )^{2}}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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((b*x+a)/(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(((2*a*b)/e>0)', see `assume?`
for more details)Is ((2*a*b)/e    -(2*b^2*d)/e^2)    ^2    -(4*b^2       *((-(2*a*b*d)/e)        +(b^2*d^2)/e^
2+a^2))     /e^2 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {a+b\,x}{\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((a + b*x)/(((a + b*x)^2)^(1/2)*(d + e*x)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 7, normalized size = 0.20 \begin {gather*} \frac {\log {\left (d + e x \right )}}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(d + e*x)/e

________________________________________________________________________________________