3.29.52 \(\int \frac {\sqrt {c+\sqrt {a x^2+x \sqrt {-b+a^2 x^2}}}}{\sqrt {-b+a^2 x^2}} \, dx\) [2852]

Optimal. Leaf size=294 \[ \frac {2 \sqrt {c+\sqrt {x \left (a x+\sqrt {-b+a^2 x^2}\right )}}}{a}-\frac {\sqrt {\sqrt {2} \sqrt {b}-2 \sqrt {a} c} \left (-\sqrt {b}+\sqrt {2} \sqrt {a} c\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c+\sqrt {x \left (a x+\sqrt {-b+a^2 x^2}\right )}}}{\sqrt {\sqrt {2} \sqrt {b}-2 \sqrt {a} c}}\right )}{a^{5/4} \left (-\sqrt {2} \sqrt {b}+2 \sqrt {a} c\right )}-\frac {\left (\sqrt {b}+\sqrt {2} \sqrt {a} c\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt {c+\sqrt {x \left (a x+\sqrt {-b+a^2 x^2}\right )}}}{\sqrt {\sqrt {2} \sqrt {b}+2 \sqrt {a} c}}\right )}{a^{5/4} \sqrt {\sqrt {2} \sqrt {b}+2 \sqrt {a} c}} \]

[Out]

2*(c+(x*(a*x+(a^2*x^2-b)^(1/2)))^(1/2))^(1/2)/a-(2^(1/2)*b^(1/2)-2*a^(1/2)*c)^(1/2)*(-b^(1/2)+2^(1/2)*a^(1/2)*
c)*arctan(2^(1/2)*a^(1/4)*(c+(x*(a*x+(a^2*x^2-b)^(1/2)))^(1/2))^(1/2)/(2^(1/2)*b^(1/2)-2*a^(1/2)*c)^(1/2))/a^(
5/4)/(-2^(1/2)*b^(1/2)+2*a^(1/2)*c)-(b^(1/2)+2^(1/2)*a^(1/2)*c)*arctanh(2^(1/2)*a^(1/4)*(c+(x*(a*x+(a^2*x^2-b)
^(1/2)))^(1/2))^(1/2)/(2^(1/2)*b^(1/2)+2*a^(1/2)*c)^(1/2))/a^(5/4)/(2^(1/2)*b^(1/2)+2*a^(1/2)*c)^(1/2)

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Rubi [F]
time = 0.29, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\sqrt {c+\sqrt {a x^2+x \sqrt {-b+a^2 x^2}}}}{\sqrt {-b+a^2 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[Sqrt[c + Sqrt[a*x^2 + x*Sqrt[-b + a^2*x^2]]]/Sqrt[-b + a^2*x^2],x]

[Out]

Defer[Int][Sqrt[c + Sqrt[a*x^2 + x*Sqrt[-b + a^2*x^2]]]/Sqrt[-b + a^2*x^2], x]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+\sqrt {a x^2+x \sqrt {-b+a^2 x^2}}}}{\sqrt {-b+a^2 x^2}} \, dx &=\int \frac {\sqrt {c+\sqrt {a x^2+x \sqrt {-b+a^2 x^2}}}}{\sqrt {-b+a^2 x^2}} \, dx\\ \end {align*}

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Mathematica [A]
time = 5.03, size = 275, normalized size = 0.94 \begin {gather*} \frac {4 \sqrt {a} \sqrt {c+\sqrt {x \left (a x+\sqrt {-b+a^2 x^2}\right )}}-\sqrt {2 \sqrt {2} \sqrt {a} \sqrt {b}-4 a c} \text {ArcTan}\left (\frac {\sqrt {2 \sqrt {2} \sqrt {a} \sqrt {b}-4 a c} \sqrt {c+\sqrt {x \left (a x+\sqrt {-b+a^2 x^2}\right )}}}{\sqrt {2} \sqrt {b}-2 \sqrt {a} c}\right )+\sqrt {2} \sqrt {-\sqrt {2} \sqrt {a} \sqrt {b}-2 a c} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {-\sqrt {2} \sqrt {a} \sqrt {b}-2 a c} \sqrt {c+\sqrt {x \left (a x+\sqrt {-b+a^2 x^2}\right )}}}{\sqrt {2} \sqrt {b}+2 \sqrt {a} c}\right )}{2 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + Sqrt[a*x^2 + x*Sqrt[-b + a^2*x^2]]]/Sqrt[-b + a^2*x^2],x]

[Out]

(4*Sqrt[a]*Sqrt[c + Sqrt[x*(a*x + Sqrt[-b + a^2*x^2])]] - Sqrt[2*Sqrt[2]*Sqrt[a]*Sqrt[b] - 4*a*c]*ArcTan[(Sqrt
[2*Sqrt[2]*Sqrt[a]*Sqrt[b] - 4*a*c]*Sqrt[c + Sqrt[x*(a*x + Sqrt[-b + a^2*x^2])]])/(Sqrt[2]*Sqrt[b] - 2*Sqrt[a]
*c)] + Sqrt[2]*Sqrt[-(Sqrt[2]*Sqrt[a]*Sqrt[b]) - 2*a*c]*ArcTan[(Sqrt[2]*Sqrt[-(Sqrt[2]*Sqrt[a]*Sqrt[b]) - 2*a*
c]*Sqrt[c + Sqrt[x*(a*x + Sqrt[-b + a^2*x^2])]])/(Sqrt[2]*Sqrt[b] + 2*Sqrt[a]*c)])/(2*a^(3/2))

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {c +\sqrt {a \,x^{2}+x \sqrt {a^{2} x^{2}-b}}}}{\sqrt {a^{2} x^{2}-b}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+(a*x^2+x*(a^2*x^2-b)^(1/2))^(1/2))^(1/2)/(a^2*x^2-b)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c + sqrt(a*x^2 + sqrt(a^2*x^2 - b)*x))/sqrt(a^2*x^2 - b), x)

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Fricas [F(-1)] Timed out
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((c+(a*x^2+x*(a^2*x^2-b)^(1/2))^(1/2))^(1/2)/(a^2*x^2-b)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c + \sqrt {a x^{2} + x \sqrt {a^{2} x^{2} - b}}}}{\sqrt {a^{2} x^{2} - b}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c + sqrt(a*x**2 + x*sqrt(a**2*x**2 - b)))/sqrt(a**2*x**2 - b), x)

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(4*

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c+\sqrt {x\,\sqrt {a^2\,x^2-b}+a\,x^2}}}{\sqrt {a^2\,x^2-b}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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