3.25.7 \(\int \frac {1}{(d+e x) \sqrt {\frac {b^2}{4 c}+b x+c x^2}} \, dx\) [2407]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {660, 36, 31} \begin {gather*} \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)} \end {gather*}

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 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*} \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 = 0.02, size = 51, normalized size = 0.53 \begin {gather*} \frac {2 (b+2 c x) (\log (b+2 c x)-\log (d+e x))}{(2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{c}}} \end {gather*}

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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Maple [A]
time = 0.75, size = 58, normalized size = 0.60

method	result	size
default	\(-\frac {2 \left (2 c x +b \right ) \left (\ln \left (2 c x +b \right )-\ln \left (e x +d \right )\right )}{\sqrt {\frac {4 c^{2} x^{2}+4 b c x +b^{2}}{c}}\, \left (b e -2 c d \right )}\)	\(58\)
risch	\(-\frac {2 \left (2 c x +b \right ) \ln \left (2 c x +b \right )}{\sqrt {\frac {\left (2 c x +b \right )^{2}}{c}}\, \left (b e -2 c d \right )}+\frac {2 \left (2 c x +b \right ) \ln \left (-e x -d \right )}{\sqrt {\frac {\left (2 c x +b \right )^{2}}{c}}\, \left (b e -2 c d \right )}\)	\(82\)

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,method=_RETURNVERBOSE)

[Out]

-2*(2*c*x+b)*(ln(2*c*x+b)-ln(e*x+d))/((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(e*x+d)/(b^2/c+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(2*c*d-%e*b>0)', see `assume?`
for more det

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Fricas [A]
time = 2.50, size = 288, normalized size = 3.00 \begin {gather*} \left [-\frac {2 \, \sqrt {c} \log \left (\frac {8 \, c^{3} d^{2} x + 4 \, b c^{2} d^{2} + {\left (8 \, c^{2} d x e + 4 \, c^{2} d^{2} - {\left (4 \, b c x + b^{2}\right )} e^{2}\right )} \sqrt {c} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}} + {\left (16 \, c^{3} x^{3} + 16 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} e^{2} + 8 \, {\left (2 \, c^{3} d x^{2} + b c^{2} d x\right )} e}{4 \, c^{2} d x^{2} + 4 \, b c d x + b^{2} d + {\left (4 \, c^{2} x^{3} + 4 \, b c x^{2} + b^{2} x\right )} e}\right )}{2 \, c d - b e}, -\frac {4 \, \sqrt {-c} \arctan \left (-\frac {{\left (2 \, c d + {\left (4 \, c x + b\right )} e\right )} \sqrt {-c} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}}{4 \, c^{2} d x + 2 \, b c d - {\left (2 \, b c x + b^{2}\right )} e}\right )}{2 \, c d - b e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-2*sqrt(c)*log((8*c^3*d^2*x + 4*b*c^2*d^2 + (8*c^2*d*x*e + 4*c^2*d^2 - (4*b*c*x + b^2)*e^2)*sqrt(c)*sqrt((4*c
^2*x^2 + 4*b*c*x + b^2)/c) + (16*c^3*x^3 + 16*b*c^2*x^2 + 6*b^2*c*x + b^3)*e^2 + 8*(2*c^3*d*x^2 + b*c^2*d*x)*e
)/(4*c^2*d*x^2 + 4*b*c*d*x + b^2*d + (4*c^2*x^3 + 4*b*c*x^2 + b^2*x)*e))/(2*c*d - b*e), -4*sqrt(-c)*arctan(-(2
*c*d + (4*c*x + b)*e)*sqrt(-c)*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/c)/(4*c^2*d*x + 2*b*c*d - (2*b*c*x + b^2)*e))/
(2*c*d - b*e)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(e*x+d)/(b**2/c+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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Giac [A]
time = 1.44, size = 94, normalized size = 0.98 \begin {gather*} \frac {2 \, c^{\frac {5}{2}} \log \left ({\left | 2 \, c x + b \right |}\right )}{2 \, c^{2} d {\left | c \right |} \mathrm {sgn}\left (2 \, c x + b\right ) - b c {\left | c \right |} e \mathrm {sgn}\left (2 \, c x + b\right )} - \frac {2 \, c^{\frac {3}{2}} e \log \left ({\left | x e + d \right |}\right )}{2 \, c d {\left | c \right |} e \mathrm {sgn}\left (2 \, c x + b\right ) - b {\left | c \right |} e^{2} \mathrm {sgn}\left (2 \, c x + b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c^(5/2)*log(abs(2*c*x + b))/(2*c^2*d*abs(c)*sgn(2*c*x + b) - b*c*abs(c)*e*sgn(2*c*x + b)) - 2*c^(3/2)*e*log(
abs(x*e + d))/(2*c*d*abs(c)*e*sgn(2*c*x + b) - b*abs(c)*e^2*sgn(2*c*x + b))

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

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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