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

Optimal. Leaf size=118 \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac {b f^2+2 e^2 x}{2 e f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{8 e^3}+\frac {f \left (b f^2+2 e^2 x\right ) \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}{4 e^2}+d x+\frac {e x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {612, 621, 206} \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac {b f^2+2 e^2 x}{2 e f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{8 e^3}+\frac {f \left (b f^2+2 e^2 x\right ) \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}{4 e^2}+d x+\frac {e x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 120, normalized size = 1.02 \[ \frac {1}{8} \left (\frac {\left (4 a e^2 f^2-b^2 f^4\right ) \log \left (2 e \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )+b f^2\right )}{e^3}+4 f x \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+\frac {2 b f^3 \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}{e^2}+8 d x+4 e x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 123, normalized size = 1.04 \[ \frac {4 \, e^{4} x^{2} + 8 \, d e^{3} x + {\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \log \left (-b f^{2} - 2 \, e^{2} x + 2 \, e f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \, {\left (b e f^{3} + 2 \, e^{3} f x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}}{8 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 111, normalized size = 0.94 \[ \frac {1}{2} \, x^{2} e + d x + \frac {{\left ({\left (b^{2} f^{4} - 4 \, a f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | -b f^{2} - 2 \, {\left (x e - \sqrt {b f^{2} x + a f^{2} + x^{2} e^{2}}\right )} e \right |}\right ) + 2 \, \sqrt {b f^{2} x + a f^{2} + x^{2} e^{2}} {\left (b f^{2} e^{\left (-2\right )} + 2 \, x\right )}\right )} {\left | f \right |}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 173, normalized size = 1.47 \[ -\frac {b^{2} f^{3} \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{8 \sqrt {\frac {e^{2}}{f^{2}}}\, e^{2}}+\frac {a f \ln \left (\frac {\frac {b}{2}+\frac {e^{2} x}{f^{2}}}{\sqrt {\frac {e^{2}}{f^{2}}}}+\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\right )}{2 \sqrt {\frac {e^{2}}{f^{2}}}}+\frac {e \,x^{2}}{2}+\frac {\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, b \,f^{3}}{4 e^{2}}+d x +\frac {\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, f x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*x+1/2*e*x^2+1/2*f*(b*x+e^2/f^2*x^2+a)^(1/2)*x+1/4/e^2*f^3*(b*x+e^2/f^2*x^2+a)^(1/2)*b+1/2*f*ln((1/2*b+e^2/f^
2*x)/(e^2/f^2)^(1/2)+(b*x+e^2/f^2*x^2+a)^(1/2))/(e^2/f^2)^(1/2)*a-1/8/e^2*f^3*ln((1/2*b+e^2/f^2*x)/(e^2/f^2)^(
1/2)+(b*x+e^2/f^2*x^2+a)^(1/2))/(e^2/f^2)^(1/2)*b^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(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(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^2*f^2-4*a*e^2>0)', see `assu
me?` for more details)Is b^2*f^2-4*a*e^2    positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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