3.7.35 \(\int \frac {1}{a+b \sqrt {c+d x}} \, dx\) [635]

Optimal. Leaf size=41 \[ \frac {2 \sqrt {c+d x}}{b d}-\frac {2 a \log \left (a+b \sqrt {c+d x}\right )}{b^2 d} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {253, 196, 45} \begin {gather*} \frac {2 \sqrt {c+d x}}{b d}-\frac {2 a \log \left (a+b \sqrt {c+d x}\right )}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[c + d*x])^(-1),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 40, normalized size = 0.98 \begin {gather*} \frac {2 b \sqrt {c+d x}-2 a \log \left (b d \left (a+b \sqrt {c+d x}\right )\right )}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(86\) vs. \(2(37)=74\).
time = 0.02, size = 87, normalized size = 2.12

method	result	size
derivativedivides	\(\frac {\frac {2 \sqrt {d x +c}}{b}-\frac {2 a \ln \left (a +b \sqrt {d x +c}\right )}{b^{2}}}{d}\)	\(36\)
default	\(\frac {2 \sqrt {d x +c}}{b d}-\frac {a \ln \left (a +b \sqrt {d x +c}\right )}{b^{2} d}+\frac {a \ln \left (-a +b \sqrt {d x +c}\right )}{b^{2} d}-\frac {a \ln \left (b^{2} d x +b^{2} c -a^{2}\right )}{b^{2} d}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 35, normalized size = 0.85 \begin {gather*} -\frac {2 \, {\left (\frac {a \log \left (\sqrt {d x + c} b + a\right )}{b^{2}} - \frac {\sqrt {d x + c}}{b}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 33, normalized size = 0.80 \begin {gather*} -\frac {2 \, {\left (a \log \left (\sqrt {d x + c} b + a\right ) - \sqrt {d x + c} b\right )}}{b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.26, size = 49, normalized size = 1.20 \begin {gather*} \begin {cases} \frac {x}{a} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\\frac {x}{a + b \sqrt {c}} & \text {for}\: d = 0 \\- \frac {2 a \log {\left (\frac {a}{b} + \sqrt {c + d x} \right )}}{b^{2} d} + \frac {2 \sqrt {c + d x}}{b d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a, Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x/(a + b*sqrt(c)), Eq(d, 0)), (-2*a*log(a/b + sqrt(c + d*x
))/(b**2*d) + 2*sqrt(c + d*x)/(b*d), True))

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Giac [A]
time = 3.89, size = 38, normalized size = 0.93 \begin {gather*} -\frac {2 \, a \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{2} d} + \frac {2 \, \sqrt {d x + c}}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 33, normalized size = 0.80 \begin {gather*} -\frac {2\,\left (a\,\ln \left (a+b\,\sqrt {c+d\,x}\right )-b\,\sqrt {c+d\,x}\right )}{b^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*(c + d*x)^(1/2)),x)

[Out]

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

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