3.29.14 \(\int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\)

Optimal. Leaf size=276 \[ \frac {b^2 \log \left (\sqrt {a^2 x^2+b^2}+a x\right )}{2 a c^3}-\frac {b^2 \log \left (\sqrt {\sqrt {a^2 x^2+b^2}+a x}+c\right )}{a c^3}+\frac {\sqrt {a^2 x^2+b^2} \left (\left (3 b^2+2 c^4\right ) \sqrt {\sqrt {a^2 x^2+b^2}+a x}+4 a c^3 x+2 b^2 c\right )+\sqrt {\sqrt {a^2 x^2+b^2}+a x} \left (3 a b^2 x+2 a c^4 x-b^2 c^2\right )+4 a^2 c^3 x^2+2 a b^2 c x+2 b^2 c^3}{2 a c^3 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}-\frac {c \log \left (\sqrt {\sqrt {a^2 x^2+b^2}+a x}+c\right )}{a} \]

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Rubi [A]  time = 0.13, antiderivative size = 169, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2117, 1821, 1620} \begin {gather*} \frac {b^2 \log \left (\sqrt {a^2 x^2+b^2}+a x\right )}{2 a c^3}+\frac {b^2}{a c^2 \sqrt {\sqrt {a^2 x^2+b^2}+a x}}-\frac {\left (b^2+c^4\right ) \log \left (\sqrt {\sqrt {a^2 x^2+b^2}+a x}+c\right )}{a c^3}-\frac {b^2}{2 a c \left (\sqrt {a^2 x^2+b^2}+a x\right )}+\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*b^2/(a*c*(a*x + Sqrt[b^2 + a^2*x^2])) + b^2/(a*c^2*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]) + Sqrt[a*x + Sqrt[b^2
+ a^2*x^2]]/a + (b^2*Log[a*x + Sqrt[b^2 + a^2*x^2]])/(2*a*c^3) - ((b^2 + c^4)*Log[c + Sqrt[a*x + Sqrt[b^2 + a
^2*x^2]]])/(a*c^3)

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1821

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] -
1)*SubstFor[x^n, Pq, x]*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && PolyQ[Pq, x^n] && Intege
rQ[Simplify[(m + 1)/n]]

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 157, normalized size = 0.57 \begin {gather*} \frac {\frac {b^2 \log \left (\sqrt {a^2 x^2+b^2}+a x\right )}{2 c^3}+\frac {b^2}{c^2 \sqrt {\sqrt {a^2 x^2+b^2}+a x}}-\frac {\left (b^2+c^4\right ) \log \left (\sqrt {\sqrt {a^2 x^2+b^2}+a x}+c\right )}{c^3}-\frac {b^2}{2 c \left (\sqrt {a^2 x^2+b^2}+a x\right )}+\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])^(-1),x]

[Out]

(-1/2*b^2/(c*(a*x + Sqrt[b^2 + a^2*x^2])) + b^2/(c^2*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]) + Sqrt[a*x + Sqrt[b^2 +
a^2*x^2]] + (b^2*Log[a*x + Sqrt[b^2 + a^2*x^2]])/(2*c^3) - ((b^2 + c^4)*Log[c + Sqrt[a*x + Sqrt[b^2 + a^2*x^2]
]])/c^3)/a

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IntegrateAlgebraic [A]  time = 0.45, size = 276, normalized size = 1.00 \begin {gather*} \frac {2 b^2 c^3+2 a b^2 c x+4 a^2 c^3 x^2+\left (-b^2 c^2+3 a b^2 x+2 a c^4 x\right ) \sqrt {a x+\sqrt {b^2+a^2 x^2}}+\sqrt {b^2+a^2 x^2} \left (2 b^2 c+4 a c^3 x+\left (3 b^2+2 c^4\right ) \sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{2 a c^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {b^2 \log \left (a x+\sqrt {b^2+a^2 x^2}\right )}{2 a c^3}-\frac {b^2 \log \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{a c^3}-\frac {c \log \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + Sqrt[a*x + Sqrt[b^2 + a^2*x^2]])^(-1),x]

[Out]

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

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fricas [A]  time = 0.64, size = 134, normalized size = 0.49 \begin {gather*} \frac {a c^{2} x + 2 \, b^{2} \log \left (\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) - \sqrt {a^{2} x^{2} + b^{2}} c^{2} - 2 \, {\left (c^{4} + b^{2}\right )} \log \left (c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) + 2 \, {\left (c^{3} - a c x + \sqrt {a^{2} x^{2} + b^{2}} c\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{2 \, a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*c^2*x + 2*b^2*log(sqrt(a*x + sqrt(a^2*x^2 + b^2))) - sqrt(a^2*x^2 + b^2)*c^2 - 2*(c^4 + b^2)*log(c + sq
rt(a*x + sqrt(a^2*x^2 + b^2))) + 2*(c^3 - a*c*x + sqrt(a^2*x^2 + b^2)*c)*sqrt(a*x + sqrt(a^2*x^2 + b^2)))/(a*c
^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{c +\sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(c + sqrt(a*x + sqrt(a**2*x**2 + b**2))), x)

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