3.2517 \(\int \frac {1}{\sqrt {a+b x} (e+f x) \sqrt {2 b e-a f+b f x}} \, dx\)

Optimal. Leaf size=59 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {f} \sqrt {a+b x} \sqrt {-a f+2 b e+b f x}}{b e-a f}\right )}{\sqrt {f} (b e-a f)} \]

[Out]

arctan(f^(1/2)*(b*x+a)^(1/2)*(b*f*x-a*f+2*b*e)^(1/2)/(-a*f+b*e))/(-a*f+b*e)/f^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {92, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {f} \sqrt {a+b x} \sqrt {-a f+2 b e+b f x}}{b e-a f}\right )}{\sqrt {f} (b e-a f)} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*x]*(e + f*x)*Sqrt[2*b*e - a*f + b*f*x]),x]

[Out]

ArcTan[(Sqrt[f]*Sqrt[a + b*x]*Sqrt[2*b*e - a*f + b*f*x])/(b*e - a*f)]/(Sqrt[f]*(b*e - a*f))

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 205

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 82, normalized size = 1.39 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x} \sqrt {f (a f-b e)}}{\sqrt {b e-a f} \sqrt {-a f+2 b e+b f x}}\right )}{\sqrt {b e-a f} \sqrt {f (a f-b e)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*x]*(e + f*x)*Sqrt[2*b*e - a*f + b*f*x]),x]

[Out]

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

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fricas [A]  time = 1.05, size = 208, normalized size = 3.53 \[ \left [\frac {\sqrt {-f} \log \left (-\frac {b^{2} f^{2} x^{2} + 2 \, b^{2} e f x - b^{2} e^{2} + 4 \, a b e f - 2 \, a^{2} f^{2} + 2 \, \sqrt {b f x + 2 \, b e - a f} {\left (b e - a f\right )} \sqrt {b x + a} \sqrt {-f}}{f^{2} x^{2} + 2 \, e f x + e^{2}}\right )}{2 \, {\left (b e f - a f^{2}\right )}}, \frac {\sqrt {f} \arctan \left (-\frac {\sqrt {b f x + 2 \, b e - a f} {\left (b e - a f\right )} \sqrt {b x + a} \sqrt {f}}{b^{2} f^{2} x^{2} + 2 \, b^{2} e f x + 2 \, a b e f - a^{2} f^{2}}\right )}{b e f - a f^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.89, size = 96, normalized size = 1.63 \[ -\frac {2 \, f^{\frac {3}{2}} \arctan \left (\frac {{\left (\sqrt {b f x - a f + 2 \, b e} \sqrt {f} - \sqrt {2 \, a f^{2} - 2 \, b f e + {\left (b f x - a f + 2 \, b e\right )} f}\right )}^{2}}{2 \, {\left (a f^{2} - b f e\right )}}\right )}{{\left (a f^{2} - b f e\right )} {\left | f \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*f^(3/2)*arctan(1/2*(sqrt(b*f*x - a*f + 2*b*e)*sqrt(f) - sqrt(2*a*f^2 - 2*b*f*e + (b*f*x - a*f + 2*b*e)*f))^
2/(a*f^2 - b*f*e))/((a*f^2 - b*f*e)*abs(f))

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maple [B]  time = 0.07, size = 154, normalized size = 2.61 \[ -\frac {\sqrt {b f x -a f +2 b e}\, \sqrt {b x +a}\, \ln \left (-\frac {2 \left (a^{2} f^{2}-2 a b e f +b^{2} e^{2}-\sqrt {-\frac {\left (a f -b e \right )^{2}}{f}}\, \sqrt {b^{2} f \,x^{2}+2 b^{2} e x -a^{2} f +2 a b e}\, f \right )}{f x +e}\right )}{\sqrt {-\frac {\left (a f -b e \right )^{2}}{f}}\, \sqrt {b^{2} f \,x^{2}+2 b^{2} e x -a^{2} f +2 a b e}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(-2*(a^2*f^2-2*a*b*e*f+b^2*e^2-(-(a*f-b*e)^2/f)^(1/2)*(b^2*f*x^2+2*b^2*e*x-a^2*f+2*a*b*e)^(1/2)*f)/(f*x+e))
*(b*f*x-a*f+2*b*e)^(1/2)*(b*x+a)^(1/2)/(-(a*f-b*e)^2/f)^(1/2)/(b^2*f*x^2+2*b^2*e*x-a^2*f+2*a*b*e)^(1/2)/f

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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(1/(f*x+e)/(b*x+a)^(1/2)/(b*f*x-a*f+2*b*e)^(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(a*f-b*e>0)', see `assume?` for
more details)Is a*f-b*e zero or nonzero?

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mupad [B]  time = 8.48, size = 998, normalized size = 16.92 \[ \frac {2\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {a}\,\sqrt {2\,b\,e-a\,f}\,{\left (f\,\left (b^2\,e^2+a\,f\,\left (a\,f-2\,b\,e\right )\right )\right )}^{5/2}+\frac {b\,e\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,{\left (f\,\left (b^2\,e^2+a\,f\,\left (a\,f-2\,b\,e\right )\right )\right )}^{5/2}}{\sqrt {2\,b\,e-a\,f}-\sqrt {2\,b\,e-a\,f+b\,f\,x}}}{2\,b^6\,e^6\,f^2+2\,a^3\,f^5\,{\left (a\,f-2\,b\,e\right )}^3+6\,a^2\,b^2\,e^2\,f^4\,{\left (a\,f-2\,b\,e\right )}^2+6\,a\,b^4\,e^4\,f^3\,\left (a\,f-2\,b\,e\right )}\right )-\mathrm {atan}\left (\frac {b^2\,e^2\,f^3\,\left (a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f\right )\,\left (\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {64}{b\,e\,f^2\,\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}\,\left (2\,a\,f^2\,\left (a\,f-2\,b\,e\right )+2\,b^2\,e^2\,f\right )}+\frac {8\,{\left (b^2\,e^2+a\,f\,\left (a\,f-2\,b\,e\right )\right )}^2}{b^3\,e^3\,f\,{\left (f\,\left (b^2\,e^2+a\,f\,\left (a\,f-2\,b\,e\right )\right )\right )}^{5/2}}-\frac {32\,a\,\left (a\,f-2\,b\,e\right )}{b^2\,e^2\,f^2\,\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}\,\left (4\,b^3\,e^3+4\,a\,b\,e\,f\,\left (a\,f-2\,b\,e\right )\right )}\right )}{\sqrt {2\,b\,e-a\,f}-\sqrt {2\,b\,e-a\,f+b\,f\,x}}-\frac {\left (\frac {f^2\,{\left (b^2\,e^2+a\,f\,\left (a\,f-2\,b\,e\right )\right )}^2\,\left (\frac {4}{b^2\,e^2\,f^2}-\frac {12\,a\,f^2\,\left (a\,f-2\,b\,e\right )+12\,b^2\,e^2\,f}{b^2\,e^2\,f^2\,\left (a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f\right )}\right )}{b\,e\,{\left (f\,\left (b^2\,e^2+a\,f\,\left (a\,f-2\,b\,e\right )\right )\right )}^{5/2}}+\frac {32\,a\,\left (a\,f-2\,b\,e\right )}{b^2\,e^2\,f\,\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}\,\left (4\,b^3\,e^3+4\,a\,b\,e\,f\,\left (a\,f-2\,b\,e\right )\right )}\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{{\left (\sqrt {2\,b\,e-a\,f}-\sqrt {2\,b\,e-a\,f+b\,f\,x}\right )}^3}+\frac {\left (\frac {32\,\sqrt {a}\,\sqrt {2\,b\,e-a\,f}}{b^2\,e^2\,f\,\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}\,\left (2\,a\,f^2\,\left (a\,f-2\,b\,e\right )+2\,b^2\,e^2\,f\right )}+\frac {64\,\sqrt {a}\,\sqrt {2\,b\,e-a\,f}}{b\,e\,f^2\,\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}\,\left (4\,b^3\,e^3+4\,a\,b\,e\,f\,\left (a\,f-2\,b\,e\right )\right )}\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {2\,b\,e-a\,f}-\sqrt {2\,b\,e-a\,f+b\,f\,x}\right )}^2}+\frac {32\,\sqrt {a}\,\sqrt {2\,b\,e-a\,f}}{b^2\,e^2\,f^2\,\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}\,\left (2\,a\,f^2\,\left (a\,f-2\,b\,e\right )+2\,b^2\,e^2\,f\right )}\right )}{16}\right )\right )}{\sqrt {a\,f^2\,\left (a\,f-2\,b\,e\right )+b^2\,e^2\,f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(atan((2*a^(1/2)*(2*b*e - a*f)^(1/2)*(f*(b^2*e^2 + a*f*(a*f - 2*b*e)))^(5/2) + (b*e*((a + b*x)^(1/2) - a^(1
/2))*(f*(b^2*e^2 + a*f*(a*f - 2*b*e)))^(5/2))/((2*b*e - a*f)^(1/2) - (2*b*e - a*f + b*f*x)^(1/2)))/(2*b^6*e^6*
f^2 + 2*a^3*f^5*(a*f - 2*b*e)^3 + 6*a^2*b^2*e^2*f^4*(a*f - 2*b*e)^2 + 6*a*b^4*e^4*f^3*(a*f - 2*b*e))) - atan((
b^2*e^2*f^3*(a*f^2*(a*f - 2*b*e) + b^2*e^2*f)*((((a + b*x)^(1/2) - a^(1/2))*(64/(b*e*f^2*(a*f^2*(a*f - 2*b*e)
+ b^2*e^2*f)^(1/2)*(2*a*f^2*(a*f - 2*b*e) + 2*b^2*e^2*f)) + (8*(b^2*e^2 + a*f*(a*f - 2*b*e))^2)/(b^3*e^3*f*(f*
(b^2*e^2 + a*f*(a*f - 2*b*e)))^(5/2)) - (32*a*(a*f - 2*b*e))/(b^2*e^2*f^2*(a*f^2*(a*f - 2*b*e) + b^2*e^2*f)^(1
/2)*(4*b^3*e^3 + 4*a*b*e*f*(a*f - 2*b*e)))))/((2*b*e - a*f)^(1/2) - (2*b*e - a*f + b*f*x)^(1/2)) - (((f^2*(b^2
*e^2 + a*f*(a*f - 2*b*e))^2*(4/(b^2*e^2*f^2) - (12*a*f^2*(a*f - 2*b*e) + 12*b^2*e^2*f)/(b^2*e^2*f^2*(a*f^2*(a*
f - 2*b*e) + b^2*e^2*f))))/(b*e*(f*(b^2*e^2 + a*f*(a*f - 2*b*e)))^(5/2)) + (32*a*(a*f - 2*b*e))/(b^2*e^2*f*(a*
f^2*(a*f - 2*b*e) + b^2*e^2*f)^(1/2)*(4*b^3*e^3 + 4*a*b*e*f*(a*f - 2*b*e))))*((a + b*x)^(1/2) - a^(1/2))^3)/((
2*b*e - a*f)^(1/2) - (2*b*e - a*f + b*f*x)^(1/2))^3 + (((32*a^(1/2)*(2*b*e - a*f)^(1/2))/(b^2*e^2*f*(a*f^2*(a*
f - 2*b*e) + b^2*e^2*f)^(1/2)*(2*a*f^2*(a*f - 2*b*e) + 2*b^2*e^2*f)) + (64*a^(1/2)*(2*b*e - a*f)^(1/2))/(b*e*f
^2*(a*f^2*(a*f - 2*b*e) + b^2*e^2*f)^(1/2)*(4*b^3*e^3 + 4*a*b*e*f*(a*f - 2*b*e))))*((a + b*x)^(1/2) - a^(1/2))
^2)/((2*b*e - a*f)^(1/2) - (2*b*e - a*f + b*f*x)^(1/2))^2 + (32*a^(1/2)*(2*b*e - a*f)^(1/2))/(b^2*e^2*f^2*(a*f
^2*(a*f - 2*b*e) + b^2*e^2*f)^(1/2)*(2*a*f^2*(a*f - 2*b*e) + 2*b^2*e^2*f))))/16)))/(a*f^2*(a*f - 2*b*e) + b^2*
e^2*f)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b x} \left (e + f x\right ) \sqrt {- a f + 2 b e + b f x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + b*x)*(e + f*x)*sqrt(-a*f + 2*b*e + b*f*x)), x)

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