3.7.18 \(\int \frac {1}{(d+e x) (f+g x) \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=182 \[ \frac {e \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g) \sqrt {a e^2-b d e+c d^2}}-\frac {g \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g) \sqrt {a g^2-b f g+c f^2}} \]

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Rubi [A]  time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {960, 724, 206} \begin {gather*} \frac {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 f-d g) \sqrt {a e^2-b d e+c d^2}}-\frac {g \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{(e f-d g) \sqrt {a g^2-b f g+c f^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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])])/(Sqrt[c*d^2
- b*d*e + a*e^2]*(e*f - d*g)) - (g*ArcTanh[(b*f - 2*a*g + (2*c*f - b*g)*x)/(2*Sqrt[c*f^2 - b*f*g + a*g^2]*Sqr
t[a + b*x + c*x^2])])/((e*f - d*g)*Sqrt[c*f^2 - b*f*g + a*g^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 960

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 169, normalized size = 0.93 \begin {gather*} \frac {\frac {g \tanh ^{-1}\left (\frac {-2 a g+b (f-g x)+2 c f x}{2 \sqrt {a+x (b+c x)} \sqrt {g (a g-b f)+c f^2}}\right )}{\sqrt {g (a g-b f)+c f^2}}-\frac {e \tanh ^{-1}\left (\frac {-2 a e+b (d-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}}}{d g-e f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.87, size = 299, normalized size = 1.64 \begin {gather*} -\frac {2 e \sqrt {-a e^2+b d e-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+b x+c x^2}}{\sqrt {-a e^2+b d e-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2+b d e-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2+b d e-c d^2}}\right )}{(d g-e f) \left (a e^2-b d e+c d^2\right )}-\frac {2 g \sqrt {-a g^2+b f g-c f^2} \tan ^{-1}\left (-\frac {g \sqrt {a+b x+c x^2}}{\sqrt {-a g^2+b f g-c f^2}}+\frac {\sqrt {c} g x}{\sqrt {-a g^2+b f g-c f^2}}+\frac {\sqrt {c} f}{\sqrt {-a g^2+b f g-c f^2}}\right )}{(e f-d g) \left (a g^2-b f g+c f^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)*(f + g*x)*Sqrt[a + b*x + c*x^2]),x]

[Out]

(-2*e*Sqrt[-(c*d^2) + b*d*e - a*e^2]*ArcTan[(Sqrt[c]*d)/Sqrt[-(c*d^2) + b*d*e - a*e^2] + (Sqrt[c]*e*x)/Sqrt[-(
c*d^2) + b*d*e - a*e^2] - (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)*
(-(e*f) + d*g)) - (2*g*Sqrt[-(c*f^2) + b*f*g - a*g^2]*ArcTan[(Sqrt[c]*f)/Sqrt[-(c*f^2) + b*f*g - a*g^2] + (Sqr
t[c]*g*x)/Sqrt[-(c*f^2) + b*f*g - a*g^2] - (g*Sqrt[a + b*x + c*x^2])/Sqrt[-(c*f^2) + b*f*g - a*g^2]])/((e*f -
d*g)*(c*f^2 - b*f*g + a*g^2))

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fricas [B]  time = 123.75, size = 1952, normalized size = 10.73

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2

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maple [A]  time = 0.02, size = 327, normalized size = 1.80 \begin {gather*} \frac {\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 (d g -e f \right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\ln \left (\frac {\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {2 a \,g^{2}-2 b f g +2 c \,f^{2}}{g^{2}}+2 \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}\, \sqrt {\left (x +\frac {f}{g}\right )^{2} c +\frac {\left (b g -2 c f \right ) \left (x +\frac {f}{g}\right )}{g}+\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}}{x +\frac {f}{g}}\right )}{\left (d g -e f \right ) \sqrt {\frac {a \,g^{2}-b f g +c \,f^{2}}{g^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*g-e*f)/((a*g^2-b*f*g+c*f^2)/g^2)^(1/2)*ln(((b*g-2*c*f)*(x+f/g)/g+2*(a*g^2-b*f*g+c*f^2)/g^2+2*((a*g^2-b*f
*g+c*f^2)/g^2)^(1/2)*((x+f/g)^2*c+(b*g-2*c*f)*(x+f/g)/g+(a*g^2-b*f*g+c*f^2)/g^2)^(1/2))/(x+f/g))+1/(d*g-e*f)/(
(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]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {c x^{2} + b x + a} {\left (e x + d\right )} {\left (g x + f\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^2 + b*x + a)*(e*x + d)*(g*x + f)), x)

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

[Out]

int(1/((f + g*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 {1}{\left (d + e x\right ) \left (f + g x\right ) \sqrt {a + b x + c x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((d + e*x)*(f + g*x)*sqrt(a + b*x + c*x**2)), x)

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