3.1577 \(\int \frac {b+2 c x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=133 \[ \frac {2 \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e \sqrt {a e^2-b d e+c d^2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {843, 621, 206, 724} \[ \frac {2 \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e \sqrt {a e^2-b d e+c d^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/e - ((2*c*d - b*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])])/(e*Sqrt[c*d^2 - b*d*e + a*e^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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 127, normalized size = 0.95 \[ \frac {\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{\sqrt {e (a e-b d)+c d^2}}+2 \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 3.27, size = 1048, normalized size = 7.88 \[ \left [\frac {2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - \sqrt {c d^{2} - b d e + a e^{2}} {\left (2 \, c d - b e\right )} \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 )}{2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}}, -\frac {\sqrt {-c d^{2} + b d e - a e^{2}} {\left (2 \, c d - b e\right )} \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 ) - {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right )}{c d^{2} e - b d e^{2} + a e^{3}}, -\frac {4 \, {\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + \sqrt {c d^{2} - b d e + a e^{2}} {\left (2 \, c d - b e\right )} \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 )}{2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}}, -\frac {\sqrt {-c d^{2} + b d e - a e^{2}} {\left (2 \, c d - b e\right )} \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 (c d^{2} - b d e + a e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right )}{c d^{2} e - b d e^{2} + a e^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.06, size = 350, normalized size = 2.63 \[ -\frac {b \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e}+\frac {2 c d \ln \left (\frac {\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e^{2}}+\frac {2 \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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