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

Optimal. Leaf size=141 \[ \frac {\sqrt {a+b x+c x^2} (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )}-\frac {e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {806, 724, 206} \begin {gather*} \frac {\sqrt {a+b x+c x^2} (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )}-\frac {e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c*d - b*e)*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) - ((b^2 - 4*a*c)*e*ArcTanh[(b*d - 2*
a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(2*(c*d^2 - b*d*e + a*e^2)^(3/2
))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rubi steps

\begin {align*} \int \frac {b+2 c x}{(d+e x)^2 \sqrt {a+b x+c x^2}} \, dx &=\frac {(2 c d-b e) \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac {\left (\left (b^2-4 a c\right ) e\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 \left (c d^2-b d e+a e^2\right )}\\ &=\frac {(2 c d-b e) \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {\left (\left (b^2-4 a c\right ) e\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{c d^2-b d e+a e^2}\\ &=\frac {(2 c d-b e) \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac {\left (b^2-4 a c\right ) e \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 138, normalized size = 0.98 \begin {gather*} \frac {e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{2 \left (e (a e-b d)+c d^2\right )^{3/2}}+\frac {\sqrt {a+x (b+c x)} (2 c d-b e)}{(d+e x) \left (e (a e-b d)+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c*d - b*e)*Sqrt[a + x*(b + c*x)])/((c*d^2 + e*(-(b*d) + a*e))*(d + e*x)) + ((b^2 - 4*a*c)*e*ArcTanh[(-(b*d
) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(2*(c*d^2 + e*(-(b*d)
+ a*e))^(3/2))

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IntegrateAlgebraic [A]  time = 0.83, size = 140, normalized size = 0.99 \begin {gather*} \frac {\left (4 a c e-b^2 e\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+b x+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2+b d e-c d^2}}\right )}{\left (-a e^2+b d e-c d^2\right )^{3/2}}+\frac {\sqrt {a+b x+c x^2} (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*c*d - b*e)*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) + ((-(b^2*e) + 4*a*c*e)*ArcTan[(Sqrt
[c]*d + Sqrt[c]*e*x - e*Sqrt[a + b*x + c*x^2])/Sqrt[-(c*d^2) + b*d*e - a*e^2]])/(-(c*d^2) + b*d*e - a*e^2)^(3/
2)

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fricas [B]  time = 0.64, size = 686, normalized size = 4.87 \begin {gather*} \left [-\frac {{\left ({\left (b^{2} - 4 \, a c\right )} e^{2} x + {\left (b^{2} - 4 \, a c\right )} d e\right )} \sqrt {c d^{2} - b d e + a e^{2}} \log \left (\frac {8 \, a b d e - 8 \, a^{2} e^{2} - {\left (b^{2} + 4 \, a c\right )} d^{2} - {\left (8 \, c^{2} d^{2} - 8 \, b c d e + {\left (b^{2} + 4 \, a c\right )} e^{2}\right )} x^{2} - 4 \, \sqrt {c d^{2} - b d e + a e^{2}} \sqrt {c x^{2} + b x + a} {\left (b d - 2 \, a e + {\left (2 \, c d - b e\right )} x\right )} - 2 \, {\left (4 \, b c d^{2} + 4 \, a b e^{2} - {\left (3 \, b^{2} + 4 \, a c\right )} d e\right )} x}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}, -\frac {{\left ({\left (b^{2} - 4 \, a c\right )} e^{2} x + {\left (b^{2} - 4 \, a c\right )} d e\right )} \sqrt {-c d^{2} + b d e - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e - a e^{2}} \sqrt {c x^{2} + b x + a} {\left (b d - 2 \, a e + {\left (2 \, c d - b e\right )} x\right )}}{2 \, {\left (a c d^{2} - a b d e + a^{2} e^{2} + {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} x^{2} + {\left (b c d^{2} - b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e - 2 \, a b d^{2} e^{3} + a^{2} d e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} - 2 \, a b d e^{4} + a^{2} e^{5} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(((b^2 - 4*a*c)*e^2*x + (b^2 - 4*a*c)*d*e)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2
+ 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b
*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*
x + d^2)) - 4*(2*c^2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)*sqrt(c*x^2 + b*x + a))/(c^2*d^5 - 2*b*
c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4 + (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e
^5 + (b^2 + 2*a*c)*d^2*e^3)*x), -1/2*(((b^2 - 4*a*c)*e^2*x + (b^2 - 4*a*c)*d*e)*sqrt(-c*d^2 + b*d*e - a*e^2)*a
rctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d
*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) - 2*(2*c^2*d^3 - 3*b*c*d^
2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)*sqrt(c*x^2 + b*x + a))/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4
+ (b^2 + 2*a*c)*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4 + a^2*e^5 + (b^2 + 2*a*c)*d^2*e^3)*x)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 860, normalized size = 6.10 \begin {gather*} \frac {b^{2} \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {2 b c d \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e}+\frac {2 c^{2} d^{2} \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e^{2}}-\frac {\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, b}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}+\frac {2 \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, c d}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) e}-\frac {2 c \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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