3.18.58 \(\int \frac {\sqrt {b+a x}}{\sqrt {a b x+\sqrt {b+a x}}} \, dx\)

Optimal. Leaf size=118 \[ \frac {\sqrt {b (a x+b)+\sqrt {a x+b}-b^2} \left (2 b \sqrt {a x+b}-3\right )}{2 a b^2}+\frac {\left (-4 b^3-3\right ) \log \left (2 \sqrt {b} \sqrt {b (a x+b)+\sqrt {a x+b}-b^2}-2 b \sqrt {a x+b}-1\right )}{4 a b^{5/2}} \]

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Rubi [A]  time = 0.25, antiderivative size = 148, normalized size of antiderivative = 1.25, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {742, 640, 621, 206} \begin {gather*} \frac {\sqrt {a x+b} \sqrt {b (a x+b)+\sqrt {a x+b}-b^2}}{a b}-\frac {3 \sqrt {b (a x+b)+\sqrt {a x+b}-b^2}}{2 a b^2}+\frac {\left (4 b^3+3\right ) \tanh ^{-1}\left (\frac {2 b \sqrt {a x+b}+1}{2 \sqrt {b} \sqrt {b (a x+b)+\sqrt {a x+b}-b^2}}\right )}{4 a b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rubi steps

\begin {align*} \int \frac {\sqrt {b+a x}}{\sqrt {a b x+\sqrt {b+a x}}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {-b^2+x+b x^2}} \, dx,x,\sqrt {b+a x}\right )}{a}\\ &=\frac {\sqrt {b+a x} \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{a b}+\frac {\operatorname {Subst}\left (\int \frac {b^2-\frac {3 x}{2}}{\sqrt {-b^2+x+b x^2}} \, dx,x,\sqrt {b+a x}\right )}{a b}\\ &=-\frac {3 \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{2 a b^2}+\frac {\sqrt {b+a x} \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{a b}+\frac {\left (3+4 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b^2+x+b x^2}} \, dx,x,\sqrt {b+a x}\right )}{4 a b^2}\\ &=-\frac {3 \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{2 a b^2}+\frac {\sqrt {b+a x} \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{a b}+\frac {\left (3+4 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{4 b-x^2} \, dx,x,\frac {1+2 b \sqrt {b+a x}}{\sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}\right )}{2 a b^2}\\ &=-\frac {3 \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{2 a b^2}+\frac {\sqrt {b+a x} \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{a b}+\frac {\left (3+4 b^3\right ) \tanh ^{-1}\left (\frac {1+2 b \sqrt {b+a x}}{2 \sqrt {b} \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}\right )}{4 a b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 102, normalized size = 0.86 \begin {gather*} \frac {\left (4 b^3+3\right ) \tanh ^{-1}\left (\frac {2 b \sqrt {a x+b}+1}{2 \sqrt {b} \sqrt {a b x+\sqrt {a x+b}}}\right )+2 \sqrt {b} \sqrt {a b x+\sqrt {a x+b}} \left (2 b \sqrt {a x+b}-3\right )}{4 a b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.27, size = 126, normalized size = 1.07 \begin {gather*} \frac {\left (-3+2 b \sqrt {b+a x}\right ) \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}}{2 a b^2}+\frac {\left (-3-4 b^3\right ) \log \left (a b^2+2 a b^3 \sqrt {b+a x}-2 a b^{5/2} \sqrt {-b^2+\sqrt {b+a x}+b (b+a x)}\right )}{4 a b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out

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maple [A]  time = 0.08, size = 161, normalized size = 1.36

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {a x +b}\, \sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}}{b}-\frac {3 \left (\frac {\sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}}{b}-\frac {\ln \left (\frac {\frac {1}{2}+b \sqrt {a x +b}}{\sqrt {b}}+\sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}\right )}{2 b^{\frac {3}{2}}}\right )}{2 b}+\sqrt {b}\, \ln \left (\frac {\frac {1}{2}+b \sqrt {a x +b}}{\sqrt {b}}+\sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}\right )}{a}\)	\(161\)
default	\(\frac {\frac {\sqrt {a x +b}\, \sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}}{b}-\frac {3 \left (\frac {\sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}}{b}-\frac {\ln \left (\frac {\frac {1}{2}+b \sqrt {a x +b}}{\sqrt {b}}+\sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}\right )}{2 b^{\frac {3}{2}}}\right )}{2 b}+\sqrt {b}\, \ln \left (\frac {\frac {1}{2}+b \sqrt {a x +b}}{\sqrt {b}}+\sqrt {-b^{2}+\sqrt {a x +b}+b \left (a x +b \right )}\right )}{a}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a*(1/2*(a*x+b)^(1/2)/b*(-b^2+(a*x+b)^(1/2)+b*(a*x+b))^(1/2)-3/4/b*(1/b*(-b^2+(a*x+b)^(1/2)+b*(a*x+b))^(1/2)-
1/2/b^(3/2)*ln((1/2+b*(a*x+b)^(1/2))/b^(1/2)+(-b^2+(a*x+b)^(1/2)+b*(a*x+b))^(1/2)))+1/2*b^(1/2)*ln((1/2+b*(a*x
+b)^(1/2))/b^(1/2)+(-b^2+(a*x+b)^(1/2)+b*(a*x+b))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {b+a\,x}}{\sqrt {\sqrt {b+a\,x}+a\,b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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