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

Optimal. Leaf size=176 \[ \frac {g \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (-b e g-2 c d g+4 c e f)}{2 c^{3/2} e^2}+\frac {(e f-d g)^2 \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^2 \sqrt {a e^2-b d e+c d^2}}+\frac {g^2 \sqrt {a+b x+c x^2}}{c e} \]

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Rubi [A]  time = 0.30, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1653, 843, 621, 206, 724} \begin {gather*} \frac {g \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (-b e g-2 c d g+4 c e f)}{2 c^{3/2} e^2}+\frac {(e f-d g)^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^2 \sqrt {a e^2-b d e+c d^2}}+\frac {g^2 \sqrt {a+b x+c x^2}}{c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 170, normalized size = 0.97 \begin {gather*} \frac {-\frac {g \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) (b e g+2 c d g-4 c e f)}{c^{3/2}}+\frac {2 (e f-d g)^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 )}{\sqrt {e (a e-b d)+c d^2}}+\frac {2 e g^2 \sqrt {a+x (b+c x)}}{c}}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.77, size = 210, normalized size = 1.19 \begin {gather*} \frac {\log \left (-2 c^{3/2} \sqrt {a+b x+c x^2}+b c+2 c^2 x\right ) \left (b e g^2+2 c d g^2-4 c e f g\right )}{2 c^{3/2} e^2}+\frac {2 \left (d^2 g^2-2 d e f g+e^2 f^2\right ) \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^2 \left (a e^2-b d e+c d^2\right )}+\frac {g^2 \sqrt {a+b x+c x^2}}{c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(g^2*Sqrt[a + b*x + c*x^2])/(c*e) + (2*Sqrt[-(c*d^2) + b*d*e - a*e^2]*(e^2*f^2 - 2*d*e*f*g + d^2*g^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^2*(c*d^2 - b*d*e + a*e^
2)) + ((-4*c*e*f*g + 2*c*d*g^2 + b*e*g^2)*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[a + b*x + c*x^2]])/(2*c^(3/2)*e^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((g*x+f)^2/(e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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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((g*x+f)^2/(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 = 0.02, size = 613, normalized size = 3.48 \begin {gather*} -\frac {d^{2} g^{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 )}{\sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e^{3}}+\frac {2 d f g \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 {f^{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 )}{\sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e}-\frac {b \,g^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}} e}-\frac {d \,g^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}\, e^{2}}+\frac {2 f g \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}\, e}+\frac {\sqrt {c \,x^{2}+b x +a}\, g^{2}}{c e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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