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

Optimal. Leaf size=287 \[ \frac {4 c \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {4 c \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \]

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Rubi [A]  time = 0.41, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {911, 93, 208} \begin {gather*} \frac {4 c \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-e \left (\sqrt {b^2-4 a c}+b\right )} \sqrt {2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}-\frac {4 c \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}}{\sqrt {f+g x} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}}\right )}{\sqrt {b^2-4 a c} \sqrt {2 c d-e \left (b-\sqrt {b^2-4 a c}\right )} \sqrt {2 c f-g \left (b-\sqrt {b^2-4 a c}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 911

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x
] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.75, size = 267, normalized size = 0.93 \begin {gather*} \frac {4 c \left (\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {g \left (b-\sqrt {b^2-4 a c}\right )-2 c f}}{\sqrt {f+g x} \sqrt {e \left (b-\sqrt {b^2-4 a c}\right )-2 c d}}\right )}{\sqrt {e \left (b-\sqrt {b^2-4 a c}\right )-2 c d} \sqrt {g \left (b-\sqrt {b^2-4 a c}\right )-2 c f}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}}{\sqrt {f+g x} \sqrt {e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}}\right )}{\sqrt {e \left (\sqrt {b^2-4 a c}+b\right )-2 c d} \sqrt {g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}}\right )}{\sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 5.45, size = 566, normalized size = 1.97 \begin {gather*} \frac {\left (\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {a e^2-b d e+c d^2}-2 \sqrt {2} c d \sqrt {a e^2-b d e+c d^2}+\sqrt {2} b e \sqrt {a e^2-b d e+c d^2}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {f+g x} \sqrt {a e^2-b d e+c d^2}}{\sqrt {d+e x} \sqrt {-d g \sqrt {b^2-4 a c}+e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}\right )}{\sqrt {b^2-4 a c} \left (-a e^2+b d e-c d^2\right ) \sqrt {-d g \sqrt {b^2-4 a c}+e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}+\frac {\left (\sqrt {2} e \sqrt {b^2-4 a c} \sqrt {a e^2-b d e+c d^2}+2 \sqrt {2} c d \sqrt {a e^2-b d e+c d^2}-\sqrt {2} b e \sqrt {a e^2-b d e+c d^2}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {f+g x} \sqrt {a e^2-b d e+c d^2}}{\sqrt {d+e x} \sqrt {d g \sqrt {b^2-4 a c}-e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}}\right )}{\sqrt {b^2-4 a c} \left (-a e^2+b d e-c d^2\right ) \sqrt {d g \sqrt {b^2-4 a c}-e f \sqrt {b^2-4 a c}-2 a e g+b d g+b e f-2 c d f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 5507, normalized size = 19.19 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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