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

Optimal. Leaf size=102 \[ \frac {\left (8 a f-b \left (\frac {4 d f}{e}+e\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{2 \sqrt {f} \sqrt {d+e x+f x^2}}\right )}{8 f^{3/2}}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}+\frac {b \sqrt {d+e x+f x^2}}{4 f} \]

[Out]

1/8*(8*a*f-b*(e+4*d*f/e))*arctanh(1/2*(2*f*x+e)/f^(1/2)/(f*x^2+e*x+d)^(1/2))/f^(3/2)+1/4*b*(f*x^2+e*x+d)^(1/2)
/f+1/2*b*x*(f*x^2+e*x+d)^(1/2)/e

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Rubi [A]  time = 0.09, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1661, 640, 621, 206} \[ \frac {\left (8 a f-b \left (\frac {4 d f}{e}+e\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{2 \sqrt {f} \sqrt {d+e x+f x^2}}\right )}{8 f^{3/2}}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}+\frac {b \sqrt {d+e x+f x^2}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d + e*x + f*x^2])/(4*f) + (b*x*Sqrt[d + e*x + f*x^2])/(2*e) + ((8*a*f - b*(e + (4*d*f)/e))*ArcTanh[(e
+ 2*f*x)/(2*Sqrt[f]*Sqrt[d + e*x + f*x^2])])/(8*f^(3/2))

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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {a+b x+\frac {b f x^2}{e}}{\sqrt {d+e x+f x^2}} \, dx &=\frac {b x \sqrt {d+e x+f x^2}}{2 e}+\frac {\int \frac {\left (2 a-\frac {b d}{e}\right ) f+\frac {b f x}{2}}{\sqrt {d+e x+f x^2}} \, dx}{2 f}\\ &=\frac {b \sqrt {d+e x+f x^2}}{4 f}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}+\frac {\left (-b e+8 a f-\frac {4 b d f}{e}\right ) \int \frac {1}{\sqrt {d+e x+f x^2}} \, dx}{8 f}\\ &=\frac {b \sqrt {d+e x+f x^2}}{4 f}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}+\frac {\left (-b e+8 a f-\frac {4 b d f}{e}\right ) \operatorname {Subst}\left (\int \frac {1}{4 f-x^2} \, dx,x,\frac {e+2 f x}{\sqrt {d+e x+f x^2}}\right )}{4 f}\\ &=\frac {b \sqrt {d+e x+f x^2}}{4 f}+\frac {b x \sqrt {d+e x+f x^2}}{2 e}-\frac {\left (b e-8 a f+\frac {4 b d f}{e}\right ) \tanh ^{-1}\left (\frac {e+2 f x}{2 \sqrt {f} \sqrt {d+e x+f x^2}}\right )}{8 f^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 87, normalized size = 0.85 \[ \frac {2 b \sqrt {f} (e+2 f x) \sqrt {d+x (e+f x)}-\left (b \left (4 d f+e^2\right )-8 a e f\right ) \tanh ^{-1}\left (\frac {e+2 f x}{2 \sqrt {f} \sqrt {d+x (e+f x)}}\right )}{8 e f^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sqrt[f]*(e + 2*f*x)*Sqrt[d + x*(e + f*x)] - (-8*a*e*f + b*(e^2 + 4*d*f))*ArcTanh[(e + 2*f*x)/(2*Sqrt[f]*S
qrt[d + x*(e + f*x)])])/(8*e*f^(3/2))

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fricas [A]  time = 1.21, size = 205, normalized size = 2.01 \[ \left [-\frac {{\left (b e^{2} + 4 \, {\left (b d - 2 \, a e\right )} f\right )} \sqrt {f} \log \left (-8 \, f^{2} x^{2} - 8 \, e f x - e^{2} - 4 \, \sqrt {f x^{2} + e x + d} {\left (2 \, f x + e\right )} \sqrt {f} - 4 \, d f\right ) - 4 \, {\left (2 \, b f^{2} x + b e f\right )} \sqrt {f x^{2} + e x + d}}{16 \, e f^{2}}, \frac {{\left (b e^{2} + 4 \, {\left (b d - 2 \, a e\right )} f\right )} \sqrt {-f} \arctan \left (\frac {\sqrt {f x^{2} + e x + d} {\left (2 \, f x + e\right )} \sqrt {-f}}{2 \, {\left (f^{2} x^{2} + e f x + d f\right )}}\right ) + 2 \, {\left (2 \, b f^{2} x + b e f\right )} \sqrt {f x^{2} + e x + d}}{8 \, e f^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((b*e^2 + 4*(b*d - 2*a*e)*f)*sqrt(f)*log(-8*f^2*x^2 - 8*e*f*x - e^2 - 4*sqrt(f*x^2 + e*x + d)*(2*f*x +
e)*sqrt(f) - 4*d*f) - 4*(2*b*f^2*x + b*e*f)*sqrt(f*x^2 + e*x + d))/(e*f^2), 1/8*((b*e^2 + 4*(b*d - 2*a*e)*f)*s
qrt(-f)*arctan(1/2*sqrt(f*x^2 + e*x + d)*(2*f*x + e)*sqrt(-f)/(f^2*x^2 + e*f*x + d*f)) + 2*(2*b*f^2*x + b*e*f)
*sqrt(f*x^2 + e*x + d))/(e*f^2)]

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giac [A]  time = 0.26, size = 84, normalized size = 0.82 \[ \frac {1}{4} \, \sqrt {f x^{2} + x e + d} {\left (2 \, b x e^{\left (-1\right )} + \frac {b}{f}\right )} + \frac {{\left (4 \, b d f - 8 \, a f e + b e^{2}\right )} e^{\left (-1\right )} \log \left ({\left | -2 \, {\left (\sqrt {f} x - \sqrt {f x^{2} + x e + d}\right )} \sqrt {f} - e \right |}\right )}{8 \, f^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(f*x^2 + x*e + d)*(2*b*x*e^(-1) + b/f) + 1/8*(4*b*d*f - 8*a*f*e + b*e^2)*e^(-1)*log(abs(-2*(sqrt(f)*x
- sqrt(f*x^2 + x*e + d))*sqrt(f) - e))/f^(3/2)

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maple [A]  time = 0.03, size = 136, normalized size = 1.33 \[ \frac {a \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{\sqrt {f}}-\frac {b d \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{2 e \sqrt {f}}-\frac {b e \ln \left (\frac {f x +\frac {e}{2}}{\sqrt {f}}+\sqrt {f \,x^{2}+e x +d}\right )}{8 f^{\frac {3}{2}}}+\frac {\sqrt {f \,x^{2}+e x +d}\, b x}{2 e}+\frac {\sqrt {f \,x^{2}+e x +d}\, b}{4 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x*(f*x^2+e*x+d)^(1/2)/e+1/4*b*(f*x^2+e*x+d)^(1/2)/f-1/8*e*b/f^(3/2)*ln((1/2*e+f*x)/f^(1/2)+(f*x^2+e*x+d)
^(1/2))-1/2/e*b/f^(1/2)*d*ln((1/2*e+f*x)/f^(1/2)+(f*x^2+e*x+d)^(1/2))+a*ln((1/2*e+f*x)/f^(1/2)+(f*x^2+e*x+d)^(
1/2))/f^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x+b*f*x^2/e)/(f*x^2+e*x+d)^(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(4*d*f-e^2>0)', see `assume?` f
or more details)Is 4*d*f-e^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,x+\frac {b\,f\,x^2}{e}}{\sqrt {f\,x^2+e\,x+d}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a e}{\sqrt {d + e x + f x^{2}}}\, dx + \int \frac {b e x}{\sqrt {d + e x + f x^{2}}}\, dx + \int \frac {b f x^{2}}{\sqrt {d + e x + f x^{2}}}\, dx}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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