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

Optimal. Leaf size=132 \[ \frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} e}-\frac {(B d-A e) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e \sqrt {a e^2-b d e+c d^2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {843, 621, 206, 724} \begin {gather*} \frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} e}-\frac {(B d-A 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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(Sqrt[c]*e) - ((B*d - A*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 {A+B x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx &=\frac {B \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{e}+\frac {(-B d+A e) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e}\\ &=\frac {(2 B) \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 (-B d+A 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 {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} e}-\frac {(B d-A 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 = 0.15, size = 126, normalized size = 0.95 \begin {gather*} \frac {\frac {(B d-A 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}}+\frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((B*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c] + ((B*d - A*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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IntegrateAlgebraic [A]  time = 0.63, size = 152, normalized size = 1.15 \begin {gather*} -\frac {2 (B d-A e) \sqrt {-a e^2+b d e-c d^2} \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 )}{e \left (a e^2-b d e+c d^2\right )}-\frac {B \log \left (-2 \sqrt {c} e \sqrt {a+b x+c x^2}+b e+2 c e x\right )}{\sqrt {c} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 42.74, size = 1079, normalized size = 8.17 \begin {gather*} \left [\frac {{\left (B c d^{2} - B b d e + B 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 ) - {\left (B c d - A c 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 )}{2 \, {\left (c^{2} d^{2} e - b c d e^{2} + a c e^{3}\right )}}, -\frac {2 \, {\left (B c d^{2} - B b d e + B 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 ) + {\left (B c d - A c 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 )}{2 \, {\left (c^{2} d^{2} e - b c d e^{2} + a c e^{3}\right )}}, -\frac {2 \, {\left (B c d - A c 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 ) - {\left (B c d^{2} - B b d e + B 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 )}{2 \, {\left (c^{2} d^{2} e - b c d e^{2} + a c e^{3}\right )}}, -\frac {{\left (B c d - A c 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 ) + {\left (B c d^{2} - B b d e + B 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^{2} d^{2} e - b c d e^{2} + a c e^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((B*c*d^2 - B*b*d*e + B*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) - (B*c*d - A*c*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)))/
(c^2*d^2*e - b*c*d*e^2 + a*c*e^3), -1/2*(2*(B*c*d^2 - B*b*d*e + B*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)) + (B*c*d - A*c*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)))/(c^2*d^2*e - b*c*d*e^2 + a*c*e^3), -1/2*(2*(B*c*d - A*c*e)*sqrt(-c*d^2 + b*d*e
- a*e^2)*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)) - (B*c*d^2 - B*
b*d*e + B*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^2*d^2*e - b*c*d*e^2 + a*c*e^3), -((B*c*d - A*c*e)*sqrt(-c*d^2 + b*d*e - a*e^2)*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)) + (B*c*d^2 - B*b*d*e + B*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^2*d^2*e - b*c*d*e^2 + a*c*e^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(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 = 349, normalized size = 2.64 \begin {gather*} -\frac {A \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 {B 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 {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B/e*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(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))*A+1/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))*B*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((B*x+A)/(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 \begin {gather*} \int \frac {A+B\,x}{\left (d+e\,x\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*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 \begin {gather*} \int \frac {A + B x}{\left (d + e x\right ) \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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