3.5.59 \(\int \frac {\sqrt {a+b \sqrt {x}+c x}}{x} \, dx\) [459]

Optimal. Leaf size=106 \[ 2 \sqrt {a+b \sqrt {x}+c x}-2 \sqrt {a} \tanh ^{-1}\left (\frac {2 a+b \sqrt {x}}{2 \sqrt {a} \sqrt {a+b \sqrt {x}+c x}}\right )+\frac {b \tanh ^{-1}\left (\frac {b+2 c \sqrt {x}}{2 \sqrt {c} \sqrt {a+b \sqrt {x}+c x}}\right )}{\sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1371, 748, 857, 635, 212, 738} \begin {gather*} 2 \sqrt {a+b \sqrt {x}+c x}-2 \sqrt {a} \tanh ^{-1}\left (\frac {2 a+b \sqrt {x}}{2 \sqrt {a} \sqrt {a+b \sqrt {x}+c x}}\right )+\frac {b \tanh ^{-1}\left (\frac {b+2 c \sqrt {x}}{2 \sqrt {c} \sqrt {a+b \sqrt {x}+c x}}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[x] + c*x]/x,x]

[Out]

2*Sqrt[a + b*Sqrt[x] + c*x] - 2*Sqrt[a]*ArcTanh[(2*a + b*Sqrt[x])/(2*Sqrt[a]*Sqrt[a + b*Sqrt[x] + c*x])] + (b*
ArcTanh[(b + 2*c*Sqrt[x])/(2*Sqrt[c]*Sqrt[a + b*Sqrt[x] + c*x])])/Sqrt[c]

Rule 212

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

Rule 635

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 738

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

Rule 748

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

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 105, normalized size = 0.99 \begin {gather*} 2 \sqrt {a+b \sqrt {x}+c x}+4 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}-\sqrt {a+b \sqrt {x}+c x}}{\sqrt {a}}\right )-\frac {b \log \left (b+2 c \sqrt {x}-2 \sqrt {c} \sqrt {a+b \sqrt {x}+c x}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[a + b*Sqrt[x] + c*x] + 4*Sqrt[a]*ArcTanh[(Sqrt[c]*Sqrt[x] - Sqrt[a + b*Sqrt[x] + c*x])/Sqrt[a]] - (b*Lo
g[b + 2*c*Sqrt[x] - 2*Sqrt[c]*Sqrt[a + b*Sqrt[x] + c*x]])/Sqrt[c]

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Maple [A]
time = 0.02, size = 84, normalized size = 0.79

method	result	size
derivativedivides	\(2 \sqrt {a +c x +b \sqrt {x}}+\frac {b \ln \left (\frac {\frac {b}{2}+c \sqrt {x}}{\sqrt {c}}+\sqrt {a +c x +b \sqrt {x}}\right )}{\sqrt {c}}-2 \sqrt {a}\, \ln \left (\frac {2 a +b \sqrt {x}+2 \sqrt {a}\, \sqrt {a +c x +b \sqrt {x}}}{\sqrt {x}}\right )\)	\(84\)
default	\(2 \sqrt {a +c x +b \sqrt {x}}+\frac {b \ln \left (\frac {\frac {b}{2}+c \sqrt {x}}{\sqrt {c}}+\sqrt {a +c x +b \sqrt {x}}\right )}{\sqrt {c}}-2 \sqrt {a}\, \ln \left (\frac {2 a +b \sqrt {x}+2 \sqrt {a}\, \sqrt {a +c x +b \sqrt {x}}}{\sqrt {x}}\right )\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(sqrt(c*x + b*sqrt(x) + a)/x, 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((a+c*x+b*x^(1/2))^(1/2)/x,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 {a + b \sqrt {x} + c x}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m operator + Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x + b*x^(1/2))^(1/2)/x,x)

[Out]

int((a + c*x + b*x^(1/2))^(1/2)/x, x)

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