3.13.75 \(\int \frac {\sqrt {c+d x}}{a+b x} \, dx\)

Optimal. Leaf size=62 \[ \frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {50, 63, 208} \begin {gather*} \frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x),x]

[Out]

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

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 62, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {c+d x}}{b}-\frac {2 \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]/(a + b*x),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.07, size = 72, normalized size = 1.16 \begin {gather*} \frac {2 \sqrt {a d-b c} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{b^{3/2}}+\frac {2 \sqrt {c+d x}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[c + d*x]/(a + b*x),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.36, size = 143, normalized size = 2.31 \begin {gather*} \left [\frac {\sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + 2 \, \sqrt {d x + c}}{b}, -\frac {2 \, {\left (\sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - \sqrt {d x + c}\right )}}{b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.27, size = 62, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b} + \frac {2 \, \sqrt {d x + c}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*c - a*d)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b) + 2*sqrt(d*x + c)/b

________________________________________________________________________________________

maple [A]  time = 0.01, size = 92, normalized size = 1.48 \begin {gather*} -\frac {2 a d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}+\frac {2 c \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}+\frac {2 \sqrt {d x +c}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 50, normalized size = 0.81 \begin {gather*} \frac {2\,\sqrt {c+d\,x}}{b}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )\,\sqrt {a\,d-b\,c}}{b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/2)/(a + b*x),x)

[Out]

(2*(c + d*x)^(1/2))/b - (2*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2))*(a*d - b*c)^(1/2))/b^(3/2)

________________________________________________________________________________________

sympy [A]  time = 4.39, size = 61, normalized size = 0.98 \begin {gather*} \frac {2 \left (\frac {d \sqrt {c + d x}}{b} - \frac {d \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{2} \sqrt {\frac {a d - b c}{b}}}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*sqrt(c + d*x)/b - d*(a*d - b*c)*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**2*sqrt((a*d - b*c)/b)))/d

________________________________________________________________________________________