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

Optimal. Leaf size=490 \[ \frac {\sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^2}-\frac {e \sqrt {a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{g (e f-d g)^2}+\frac {(2 c f-b g) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 g (e f-d g) \sqrt {a g^2-b f g+c f^2}}+\frac {\sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^2}+\frac {e (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^2} \]

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Rubi [A]  time = 0.70, antiderivative size = 490, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {960, 734, 843, 621, 206, 724, 732} \begin {gather*} \frac {\sqrt {a e^2-b d e+c d^2} \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)^2}-\frac {e \sqrt {a g^2-b f g+c f^2} \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 )}{g (e f-d g)^2}+\frac {(2 c f-b 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 )}{2 g (e f-d g) \sqrt {a g^2-b f g+c f^2}}+\frac {\sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)}-\frac {(2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^2}+\frac {e (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a + b*x + c*x^2]/((e*f - d*g)*(f + g*x)) - ((2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c
*x^2])])/(2*Sqrt[c]*(e*f - d*g)^2) + (e*(2*c*f - b*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/
(2*Sqrt[c]*g*(e*f - d*g)^2) - (Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(g*(e*f - d*g))
+ (Sqrt[c*d^2 - b*d*e + a*e^2]*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*f - d*g)^2 + ((2*c*f - b*g)*ArcTanh[(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])])/(2*g*(e*f - d*g)*Sqrt[c*f^2 - b*f*g + a*g^2]) - (e*Sqrt[c*f^2 - b*f*g + a*g
^2]*ArcTanh[(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])])/(g*(e*f -
d*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 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 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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]

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

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Mathematica [A]  time = 0.52, size = 222, normalized size = 0.45 \begin {gather*} \frac {2 \sqrt {e (a e-b d)+c d^2} \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 )-\frac {(2 a e g-b (d g+e f)+2 c d f) \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 {2 \sqrt {a+x (b+c x)} (e f-d g)}{f+g x}}{2 (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.26, size = 351, normalized size = 0.72 \begin {gather*} \frac {2 \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 )}{(e f-d g)^2}+\frac {\left (-2 c d f \sqrt {-a g^2+b f g-c f^2}+b d g \sqrt {-a g^2+b f g-c f^2}+b e f \sqrt {-a g^2+b f g-c f^2}-2 a e g \sqrt {-a g^2+b f g-c f^2}\right ) \tan ^{-1}\left (\frac {-g \sqrt {a+b x+c x^2}+\sqrt {c} f+\sqrt {c} g x}{\sqrt {-a g^2+b f g-c f^2}}\right )}{(e f-d g)^2 \left (a g^2-b f g+c f^2\right )}+\frac {\sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out

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maple [B]  time = 0.02, size = 3162, normalized size = 6.45 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-g/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)/(x+f/g)*((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)^(3/2)+g/(
d*g-e*f)/(a*g^2-b*f*g+c*f^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)*b-1/(d*g-e*f)/(
a*g^2-b*f*g+c*f^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)*c*f-1/(d*g-e*f)/(a*g^2-b*
f*g+c*f^2)*ln(((x+f/g)*c+1/2*(b*g-2*c*f)/g)/c^(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))*c^(1/2)*f*b+1/g/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)*ln(((x+f/g)*c+1/2*(b*g-2*c*f)/g)/c^(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))*c^(3/2)*f^2-1/2*g/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)/((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))*a*b+1/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)
/((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))*a*c*f+1/2/(d*g-e*f)/(a*g^
2-b*f*g+c*f^2)/((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))*b^2*f-3/2/g
/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)/((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))*b*f^2*c+1/g^2/(d*g-e*f)/(a*g^2-b*f*g+c*f^2)/((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))*c^2*f^3+g/(d*g-e*f)*c/(a*g^2-b*f*g+c*f^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+g/(d*g-e*f)*c^(1/2)/(a*g^2-b*f*g+c*f^2)*ln(((x+f/g)*c+1/2*(b*g-2*c*f)/g)/c^(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))*a-1/(d*g-e*f)^2*e*((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)-1/2/(d*g-e*f)^2*e*ln(((x+f/g)*c+1/2*(b*g-2*c*f)/g)/c^(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))/c^(1/2)*b+1/(d*g-e*f)^2*e/g*ln(((x+f/g)*c+1/2*(b*g-2*c*f)/g
)/c^(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))*c^(1/2)*f+1/(d*g-e*f)^2*e/((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))*a-1/(d*g-e*f)^2*e/g/((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))*b*f+1/(d*g-e*f)^2*e/g^2/((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))*c*f^2+1/(d*g-e*f)^2*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)+1/2/(d*g-e*f)^2*e*ln(((x+d/e)*c+1/2*(b*e-2*c*d)/e)/c^(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))/c^(1/2)*b-1/(d*g-e*f)^2*ln(((x+d/e)*c+1/2*(b*e-
2*c*d)/e)/c^(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))*c^(1/2)*d-1/(d*g-e*f)^2*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/(d*g-e*f)^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-1/(d*g-e*f)^2/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))*c*d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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