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

Optimal. Leaf size=97 \[ \frac {2 \sqrt {a+b x}}{b-c}-\frac {2 \sqrt {a+c x}}{b-c}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{b-c}+\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{b-c} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6690, 50, 63, 208} \[ \frac {2 \sqrt {a+b x}}{b-c}-\frac {2 \sqrt {a+c x}}{b-c}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{b-c}+\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{b-c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 6690

Int[(u_.)*((e_.)*Sqrt[(a_.) + (b_.)*(x_)^(n_.)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)^(n_.)])^(m_), x_Symbol] :> Dis
t[(b*e^2 - d*f^2)^m, Int[ExpandIntegrand[(u*x^(m*n))/(e*Sqrt[a + b*x^n] - f*Sqrt[c + d*x^n])^m, x], x], x] /;
FreeQ[{a, b, c, d, e, f, n}, x] && ILtQ[m, 0] && EqQ[a*e^2 - c*f^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 75, normalized size = 0.77 \[ \frac {2 \left (\sqrt {a+b x}-\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\sqrt {a+c x}+\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )\right )}{b-c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 158, normalized size = 1.63 \[ \left [-\frac {\sqrt {a} \log \left (\frac {b x + 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \sqrt {a} \log \left (\frac {c x - 2 \, \sqrt {c x + a} \sqrt {a} + 2 \, a}{x}\right ) - 2 \, \sqrt {b x + a} + 2 \, \sqrt {c x + a}}{b - c}, \frac {2 \, {\left (\sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - \sqrt {-a} \arctan \left (\frac {\sqrt {c x + a} \sqrt {-a}}{a}\right ) + \sqrt {b x + a} - \sqrt {c x + a}\right )}}{b - c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(sqrt(a)*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + sqrt(a)*log((c*x - 2*sqrt(c*x + a)*sqrt(a) + 2*a)/x)
- 2*sqrt(b*x + a) + 2*sqrt(c*x + a))/(b - c), 2*(sqrt(-a)*arctan(sqrt(b*x + a)*sqrt(-a)/a) - sqrt(-a)*arctan(
sqrt(c*x + a)*sqrt(-a)/a) + sqrt(b*x + a) - sqrt(c*x + a))/(b - c)]

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giac [B]  time = 1.00, size = 1093, normalized size = 11.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(a*b^2 + (b*x + a)*b*c - a*b*c)*abs(b)/(b^3 - b^2*c) + 2*a*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*(b
- c)) + 2*sqrt(b*x + a)/(b - c) - 2*(2*(a*b^3*c - a*b^2*c^2)*(a*b^2 - a*b*c)^2*sqrt(-a)*abs(b)*sgn(b - c) + 2*
(a*b^3 - a*b^2*c)*(a*b^2 - a*b*c)^2*sqrt(-a*b*c)*abs(b) + (a^2*b^5 - 3*a^2*b^4*c + 3*a^2*b^3*c^2 - a^2*b^2*c^3
)*sqrt(-a*b*c)*abs(a*b^2 - a*b*c)*abs(b)*sgn(b - c) + (a^2*b^6 - 3*a^2*b^5*c + 3*a^2*b^4*c^2 - a^2*b^3*c^3)*sq
rt(-a)*abs(a*b^2 - a*b*c)*abs(b) + (a^3*b^7*c - 2*a^3*b^6*c^2 + 2*a^3*b^4*c^4 - a^3*b^3*c^5)*sqrt(-a)*abs(b)*s
gn(b - c) + (a^3*b^7 - 2*a^3*b^6*c + 2*a^3*b^4*c^3 - a^3*b^3*c^4)*sqrt(-a*b*c)*abs(b))*arctan(-(sqrt(b*c)*sqrt
(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))/sqrt(-(a*b^3 - a*b*c^2 + sqrt((a*b^3 - a*b*c^2)^2 - (a^2*b^5
- 3*a^2*b^4*c + 3*a^2*b^3*c^2 - a^2*b^2*c^3)*(b - c)))/(b - c)))/((b^8 - 5*b^7*c + 10*b^6*c^2 - 10*b^5*c^3 + 5
*b^4*c^4 - b^3*c^5)*a^2*abs(a*b^2 - a*b*c)) + 2*(2*(a*b^3*c - a*b^2*c^2)*(a*b^2 - a*b*c)^2*sqrt(-a)*abs(b)*sgn
(b - c) + 2*(a*b^3 - a*b^2*c)*(a*b^2 - a*b*c)^2*sqrt(-a*b*c)*abs(b) + (a^2*b^5 - 3*a^2*b^4*c + 3*a^2*b^3*c^2 -
a^2*b^2*c^3)*sqrt(-a*b*c)*abs(a*b^2 - a*b*c)*abs(b)*sgn(b - c) + (a^2*b^6 - 3*a^2*b^5*c + 3*a^2*b^4*c^2 - a^2
*b^3*c^3)*sqrt(-a)*abs(a*b^2 - a*b*c)*abs(b) + (a^3*b^7*c - 2*a^3*b^6*c^2 + 2*a^3*b^4*c^4 - a^3*b^3*c^5)*sqrt(
-a)*abs(b)*sgn(b - c) + (a^3*b^7 - 2*a^3*b^6*c + 2*a^3*b^4*c^3 - a^3*b^3*c^4)*sqrt(-a*b*c)*abs(b))*arctan(-(sq
rt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))/sqrt(-(a*b^3 - a*b*c^2 - sqrt((a*b^3 - a*b*c^2)^2
- (a^2*b^5 - 3*a^2*b^4*c + 3*a^2*b^3*c^2 - a^2*b^2*c^3)*(b - c)))/(b - c)))/((b^8 - 5*b^7*c + 10*b^6*c^2 - 10
*b^5*c^3 + 5*b^4*c^4 - b^3*c^5)*a^2*abs(a*b^2 - a*b*c))

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maple [A]  time = 0.01, size = 73, normalized size = 0.75 \[ \frac {-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 \sqrt {b x +a}}{b -c}-\frac {-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )+2 \sqrt {c x +a}}{b -c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.33, size = 213, normalized size = 2.20 \[ -\frac {2\,\sqrt {a}\,c\,\left (\frac {2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}+\frac {\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}\right )-2\,\sqrt {a}\,b\,\left (\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )-\frac {2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}+4\right )}{\left (b-c\right )\,\left (b-\frac {c\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^(1/2)*c*((2*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2) - a^(1/2)) + (log(((a + b*x)^(1/2) - a^(1/2))/
((a + c*x)^(1/2) - a^(1/2)))*((a + b*x)^(1/2) - a^(1/2))^2)/((a + c*x)^(1/2) - a^(1/2))^2) - 2*a^(1/2)*b*(log(
((a + b*x)^(1/2) - a^(1/2))/((a + c*x)^(1/2) - a^(1/2))) - (2*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2) -
a^(1/2)) + 4))/((b - c)*(b - (c*((a + b*x)^(1/2) - a^(1/2))^2)/((a + c*x)^(1/2) - a^(1/2))^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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