3.31.61 \(\int \frac {b^2+a x}{(-b^2+a x) \sqrt {b+\sqrt {b^2+a x^2}}} \, dx\)

Optimal. Leaf size=471 \[ \frac {2 \left (\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {\sqrt {a+b^2}+b}+\sqrt {2} \sqrt {b} \sqrt {a+b^2} \sqrt {\sqrt {a+b^2}+b}-\sqrt {2} b^{3/2} \sqrt {\sqrt {a+b^2}+b}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {\sqrt {a x^2+b^2}-\sqrt {a} x}}{\sqrt {b} \sqrt {\sqrt {a+b^2}+b}}\right )}{a^{5/4}}-\frac {2 \left (-\sqrt {2} \sqrt {a} \sqrt {b} \sqrt {\sqrt {a+b^2}-b}+\sqrt {2} \sqrt {b} \sqrt {a+b^2} \sqrt {\sqrt {a+b^2}-b}+\sqrt {2} b^{3/2} \sqrt {\sqrt {a+b^2}-b}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {\sqrt {a x^2+b^2}-\sqrt {a} x}}{\sqrt {b} \sqrt {\sqrt {a+b^2}-b}}\right )}{a^{5/4}}+\frac {\sqrt {2} \sqrt {\sqrt {a x^2+b^2}-\sqrt {a} x} \left (\sqrt {a} x-b\right )}{\sqrt {a} b}+\frac {\sqrt {2} \sqrt {a x^2+b^2} \sqrt {\sqrt {a x^2+b^2}-\sqrt {a} x}}{\sqrt {a} b}-\frac {2 \sqrt {2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {\sqrt {a x^2+b^2}-\sqrt {a} x}}{\sqrt {b}}\right )}{\sqrt {a}} \]

________________________________________________________________________________________

Rubi [F]  time = 0.67, 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 {b^2+a x}{\left (-b^2+a x\right ) \sqrt {b+\sqrt {b^2+a x^2}}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b^2+a x}{\left (-b^2+a x\right ) \sqrt {b+\sqrt {b^2+a x^2}}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 4.76, size = 1138, normalized size = 2.42 \begin {gather*} \frac {2 x}{\sqrt {b+\sqrt {b^2+a x^2}}}+\frac {2 \sqrt {2} \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {2} \sqrt {b} \sqrt {b+\sqrt {b^2+a x^2}}}-\frac {\sqrt {b+\sqrt {b^2+a x^2}}}{\sqrt {2} \sqrt {b}}\right )}{\sqrt {a}}-2 i \sqrt {a} b^3 \text {RootSum}\left [16 a^4 b^4-8 a^2 b^2 \text {$\#$1}^4-16 a b^4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}}{-4 a^2 b^2-8 a b^4+\text {$\#$1}^4}\&\right ]-\frac {i b^2 \text {RootSum}\left [16 a^4 b^4-8 a^2 b^2 \text {$\#$1}^4-16 a b^4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}^3}{-4 a^2 b^2-8 a b^4+\text {$\#$1}^4}\&\right ]}{\sqrt {a}}+4 i a^{3/2} b^4 \text {RootSum}\left [16 a^4 b^4-8 a^2 b^2 \text {$\#$1}^4-16 a b^4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right )}{4 a^2 b^2 \text {$\#$1}+8 a b^4 \text {$\#$1}-\text {$\#$1}^5}\&\right ]+8 i a^{5/2} b^5 \text {RootSum}\left [16 a^4 b^4-8 a^2 b^2 \text {$\#$1}^4-16 a b^4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right )}{4 a^2 b^2 \text {$\#$1}^3+8 a b^4 \text {$\#$1}^3-\text {$\#$1}^7}\&\right ]-i b \text {RootSum}\left [16 a^4 b^4-8 a^2 b^2 \text {$\#$1}^4-16 a b^4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {8 a^3 b^3 \log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right )+4 a^2 b^2 \log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}^2+8 a b^4 \log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}^2-2 a b \log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}^4-4 b^3 \log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}^4-\log \left (\frac {i a x}{\sqrt {b+\sqrt {b^2+a x^2}}}-i \sqrt {a} \sqrt {b+\sqrt {b^2+a x^2}}-\text {$\#$1}\right ) \text {$\#$1}^6}{4 a^2 b^2 \text {$\#$1}^3+8 a b^4 \text {$\#$1}^3-\text {$\#$1}^7}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b^2 + a*x)/((-b^2 + a*x)*Sqrt[b + Sqrt[b^2 + a*x^2]]),x]

[Out]

(2*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] + (2*Sqrt[2]*Sqrt[b]*ArcTan[(Sqrt[a]*x)/(Sqrt[2]*Sqrt[b]*Sqrt[b + Sqrt[b^2 +
 a*x^2]]) - Sqrt[b + Sqrt[b^2 + a*x^2]]/(Sqrt[2]*Sqrt[b])])/Sqrt[a] - (2*I)*Sqrt[a]*b^3*RootSum[16*a^4*b^4 - 8
*a^2*b^2*#1^4 - 16*a*b^4*#1^4 + #1^8 & , (Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^
2 + a*x^2]] - #1]*#1)/(-4*a^2*b^2 - 8*a*b^4 + #1^4) & ] - (I*b^2*RootSum[16*a^4*b^4 - 8*a^2*b^2*#1^4 - 16*a*b^
4*#1^4 + #1^8 & , (Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]*#1^3)
/(-4*a^2*b^2 - 8*a*b^4 + #1^4) & ])/Sqrt[a] + (4*I)*a^(3/2)*b^4*RootSum[16*a^4*b^4 - 8*a^2*b^2*#1^4 - 16*a*b^4
*#1^4 + #1^8 & , Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]/(4*a^2*
b^2*#1 + 8*a*b^4*#1 - #1^5) & ] + (8*I)*a^(5/2)*b^5*RootSum[16*a^4*b^4 - 8*a^2*b^2*#1^4 - 16*a*b^4*#1^4 + #1^8
 & , Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]/(4*a^2*b^2*#1^3 + 8
*a*b^4*#1^3 - #1^7) & ] - I*b*RootSum[16*a^4*b^4 - 8*a^2*b^2*#1^4 - 16*a*b^4*#1^4 + #1^8 & , (8*a^3*b^3*Log[(I
*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1] + 4*a^2*b^2*Log[(I*a*x)/Sqrt[b
 + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]*#1^2 + 8*a*b^4*Log[(I*a*x)/Sqrt[b + Sqrt[b
^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]*#1^2 - 2*a*b*Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]
] - I*Sqrt[a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]*#1^4 - 4*b^3*Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[
a]*Sqrt[b + Sqrt[b^2 + a*x^2]] - #1]*#1^4 - Log[(I*a*x)/Sqrt[b + Sqrt[b^2 + a*x^2]] - I*Sqrt[a]*Sqrt[b + Sqrt[
b^2 + a*x^2]] - #1]*#1^6)/(4*a^2*b^2*#1^3 + 8*a*b^4*#1^3 - #1^7) & ]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {b^{2} + a x}{{\left (b^{2} - a x\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {a x +b^{2}}{\left (a x -b^{2}\right ) \sqrt {b +\sqrt {a \,x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________