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

Optimal. Leaf size=112 \[ \frac {4 \sqrt {a^2 x^2+b^2} \left (a^2 x^3-b^2 x\right )}{3 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}}+\frac {2 \left (10 a^4 x^4-5 a^2 b^2 x^2-7 b^4\right )}{15 a \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 95, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2122, 270} \begin {gather*} -\frac {b^2}{a \sqrt {\sqrt {a^2 x^2+b^2}+a x}}+\frac {\left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}{6 a}-\frac {b^4}{10 a \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*b^4/(a*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2)) - b^2/(a*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]) + (a*x + Sqrt[b^2 +
a^2*x^2])^(3/2)/(6*a)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2122

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1*(i/c)^m)/(2^(2*m + 1)*e*f^(2*m)), Subst[Int[(x^n*(d^2 + a*f^2 - 2*d*x + x^2)^(2*m + 1))/(-d + x)^(2*(m +
1)), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, d, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && E
qQ[c*g - a*i, 0] && IntegerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {b^2+a^2 x^2}}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2+x^2\right )^2}{x^{7/2}} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{4 a}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {b^4}{x^{7/2}}+\frac {2 b^2}{x^{3/2}}+\sqrt {x}\right ) \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )}{4 a}\\ &=-\frac {b^4}{10 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}}-\frac {b^2}{a \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {\left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}{6 a}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 93, normalized size = 0.83 \begin {gather*} \frac {-\frac {4 b^2}{\sqrt {\sqrt {a^2 x^2+b^2}+a x}}+\frac {2}{3} \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}-\frac {2 b^4}{5 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}}}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b^4)/(5*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2)) - (4*b^2)/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] + (2*(a*x + Sqrt[b^2
+ a^2*x^2])^(3/2))/3)/(4*a)

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IntegrateAlgebraic [A]  time = 0.16, size = 112, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt {b^2+a^2 x^2} \left (-b^2 x+a^2 x^3\right )}{3 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}}+\frac {2 \left (-7 b^4-5 a^2 b^2 x^2+10 a^4 x^4\right )}{15 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[b^2 + a^2*x^2]*(-(b^2*x) + a^2*x^3))/(3*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2)) + (2*(-7*b^4 - 5*a^2*b^2*x^
2 + 10*a^4*x^4))/(15*a*(a*x + Sqrt[b^2 + a^2*x^2])^(5/2))

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fricas [A]  time = 0.44, size = 72, normalized size = 0.64 \begin {gather*} \frac {2 \, {\left (3 \, a^{3} x^{3} + 11 \, a b^{2} x - {\left (3 \, a^{2} x^{2} + 7 \, b^{2}\right )} \sqrt {a^{2} x^{2} + b^{2}}\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{15 \, a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^3*x^3 + 11*a*b^2*x - (3*a^2*x^2 + 7*b^2)*sqrt(a^2*x^2 + b^2))*sqrt(a*x + sqrt(a^2*x^2 + b^2))/(a*b^2
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a^{2} x^{2} + b^{2}}}{\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((a^2*x^2+b^2)^(1/2)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {a^{2} x^{2}+b^{2}}}{\sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a^2*x^2+b^2)^(1/2)/(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 {\sqrt {a^{2} x^{2} + b^{2}}}{\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((a^2*x^2+b^2)^(1/2)/(a*x+(a^2*x^2+b^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a^2\,x^2+b^2}}{\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((b^2 + a^2*x^2)^(1/2)/(a*x + (b^2 + a^2*x^2)^(1/2))^(1/2),x)

[Out]

int((b^2 + a^2*x^2)^(1/2)/(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 {\sqrt {a^{2} x^{2} + b^{2}}}{\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((a**2*x**2+b**2)**(1/2)/(a*x+(a**2*x**2+b**2)**(1/2))**(1/2),x)

[Out]

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

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