3.2407 \(\int \frac {1}{(d+e x) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx\)

Optimal. Leaf size=96 \[ \frac {2 (b+2 c x) \log (b+2 c x)}{\sqrt {\frac {b^2}{c}+4 b x+4 c x^2} (2 c d-b e)}-\frac {2 (b+2 c x) \log (d+e x)}{\sqrt {\frac {b^2}{c}+4 b x+4 c x^2} (2 c d-b e)} \]

[Out]

2*(2*c*x+b)*ln(2*c*x+b)/(-b*e+2*c*d)/(1/c*b^2+4*b*x+4*c*x^2)^(1/2)-2*(2*c*x+b)*ln(e*x+d)/(-b*e+2*c*d)/(1/c*b^2
+4*b*x+4*c*x^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {646, 36, 31} \[ \frac {2 (b+2 c x) \log (b+2 c x)}{\sqrt {\frac {b^2}{c}+4 b x+4 c x^2} (2 c d-b e)}-\frac {2 (b+2 c x) \log (d+e x)}{\sqrt {\frac {b^2}{c}+4 b x+4 c x^2} (2 c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*Sqrt[b^2/(4*c) + b*x + c*x^2]),x]

[Out]

(2*(b + 2*c*x)*Log[b + 2*c*x])/((2*c*d - b*e)*Sqrt[b^2/c + 4*b*x + 4*c*x^2]) - (2*(b + 2*c*x)*Log[d + e*x])/((
2*c*d - b*e)*Sqrt[b^2/c + 4*b*x + 4*c*x^2])

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 {1}{(d+e x) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx &=\frac {\left (\frac {b}{2}+c x\right ) \int \frac {1}{\left (\frac {b}{2}+c x\right ) (d+e x)} \, dx}{\sqrt {\frac {b^2}{4 c}+b x+c x^2}}\\ &=\frac {\left (2 c \left (\frac {b}{2}+c x\right )\right ) \int \frac {1}{\frac {b}{2}+c x} \, dx}{(2 c d-b e) \sqrt {\frac {b^2}{4 c}+b x+c x^2}}-\frac {\left (2 e \left (\frac {b}{2}+c x\right )\right ) \int \frac {1}{d+e x} \, dx}{(2 c d-b e) \sqrt {\frac {b^2}{4 c}+b x+c x^2}}\\ &=\frac {2 (b+2 c x) \log (b+2 c x)}{(2 c d-b e) \sqrt {\frac {b^2}{c}+4 b x+4 c x^2}}-\frac {2 (b+2 c x) \log (d+e x)}{(2 c d-b e) \sqrt {\frac {b^2}{c}+4 b x+4 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.53 \[ \frac {2 (b+2 c x) (\log (b+2 c x)-\log (d+e x))}{\sqrt {\frac {(b+2 c x)^2}{c}} (2 c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*Sqrt[b^2/(4*c) + b*x + c*x^2]),x]

[Out]

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

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fricas [A]  time = 1.10, size = 294, normalized size = 3.06 \[ \left [-\frac {2 \, \sqrt {c} \log \left (\frac {16 \, c^{3} e^{2} x^{3} + 4 \, b c^{2} d^{2} + b^{3} e^{2} + 16 \, {\left (c^{3} d e + b c^{2} e^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} d^{2} + 4 \, b c^{2} d e + 3 \, b^{2} c e^{2}\right )} x + {\left (4 \, c^{2} d^{2} - b^{2} e^{2} + 4 \, {\left (2 \, c^{2} d e - b c e^{2}\right )} x\right )} \sqrt {c} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}}{4 \, c^{2} e x^{3} + b^{2} d + 4 \, {\left (c^{2} d + b c e\right )} x^{2} + {\left (4 \, b c d + b^{2} e\right )} x}\right )}{2 \, c d - b e}, -\frac {4 \, \sqrt {-c} \arctan \left (-\frac {{\left (4 \, c e x + 2 \, c d + b e\right )} \sqrt {-c} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}}{2 \, b c d - b^{2} e + 2 \, {\left (2 \, c^{2} d - b c e\right )} x}\right )}{2 \, c d - b e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(e*x+d)/(1/c*b^2+4*b*x+4*c*x^2)^(1/2),x, algorithm="fricas")

[Out]

[-2*sqrt(c)*log((16*c^3*e^2*x^3 + 4*b*c^2*d^2 + b^3*e^2 + 16*(c^3*d*e + b*c^2*e^2)*x^2 + 2*(4*c^3*d^2 + 4*b*c^
2*d*e + 3*b^2*c*e^2)*x + (4*c^2*d^2 - b^2*e^2 + 4*(2*c^2*d*e - b*c*e^2)*x)*sqrt(c)*sqrt((4*c^2*x^2 + 4*b*c*x +
b^2)/c))/(4*c^2*e*x^3 + b^2*d + 4*(c^2*d + b*c*e)*x^2 + (4*b*c*d + b^2*e)*x))/(2*c*d - b*e), -4*sqrt(-c)*arct
an(-(4*c*e*x + 2*c*d + b*e)*sqrt(-c)*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/c)/(2*b*c*d - b^2*e + 2*(2*c^2*d - b*c*e
)*x))/(2*c*d - b*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(e*x+d)/(1/c*b^2+4*b*x+4*c*x^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(2*c*t_nostep+b)]Discontinuities at zeroes of 2*c*t_nostep+b were not checkedsym2poly/r2sym(const gen & e,con
st index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.17, size = 58, normalized size = 0.60 \[ \frac {2 \left (2 c x +b \right ) \left (-\ln \left (2 c x +b \right )+\ln \left (e x +d \right )\right )}{\sqrt {\frac {4 c^{2} x^{2}+4 b c x +b^{2}}{c}}\, \left (b e -2 c d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(e*x+d)/(b^2/c+4*b*x+4*c*x^2)^(1/2),x)

[Out]

2*(2*c*x+b)*(ln(e*x+d)-ln(2*c*x+b))/((4*c^2*x^2+4*b*c*x+b^2)/c)^(1/2)/(b*e-2*c*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(2/(e*x+d)/(1/c*b^2+4*b*x+4*c*x^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(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {2}{\left (d+e\,x\right )\,\sqrt {4\,b\,x+4\,c\,x^2+\frac {b^2}{c}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/((d + e*x)*(4*b*x + 4*c*x^2 + b^2/c)^(1/2)),x)

[Out]

int(2/((d + e*x)*(4*b*x + 4*c*x^2 + b^2/c)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \int \frac {1}{d \sqrt {\frac {b^{2}}{c} + 4 b x + 4 c x^{2}} + e x \sqrt {\frac {b^{2}}{c} + 4 b x + 4 c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(e*x+d)/(1/c*b**2+4*b*x+4*c*x**2)**(1/2),x)

[Out]

2*Integral(1/(d*sqrt(b**2/c + 4*b*x + 4*c*x**2) + e*x*sqrt(b**2/c + 4*b*x + 4*c*x**2)), x)

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