3.642 \(\int \frac{1}{(a+b \sqrt{c+d x})^2} \, dx\)

Optimal. Leaf size=47 \[ \frac{2 a}{b^2 d \left (a+b \sqrt{c+d x}\right )}+\frac{2 \log \left (a+b \sqrt{c+d x}\right )}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0327996, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {247, 190, 43} \[ \frac{2 a}{b^2 d \left (a+b \sqrt{c+d x}\right )}+\frac{2 \log \left (a+b \sqrt{c+d x}\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[c + d*x])^(-2),x]

[Out]

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

Rule 247

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]

Rule 190

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 43

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])

Rubi steps

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

Mathematica [A]  time = 0.0335396, size = 40, normalized size = 0.85 \[ \frac{2 \left (\frac{a}{a+b \sqrt{c+d x}}+\log \left (a+b \sqrt{c+d x}\right )\right )}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[c + d*x])^(-2),x]

[Out]

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

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Maple [B]  time = 0.016, size = 142, normalized size = 3. \begin{align*} -2\,{\frac{{a}^{2}}{ \left ({b}^{2}dx+{b}^{2}c-{a}^{2} \right ){b}^{2}d}}+{\frac{\ln \left ({b}^{2}dx+{b}^{2}c-{a}^{2} \right ) }{{b}^{2}d}}+{\frac{a}{{b}^{2}d} \left ( -a+b\sqrt{dx+c} \right ) ^{-1}}-{\frac{1}{{b}^{2}d}\ln \left ( -a+b\sqrt{dx+c} \right ) }+{\frac{a}{{b}^{2}d} \left ( a+b\sqrt{dx+c} \right ) ^{-1}}+{\frac{1}{{b}^{2}d}\ln \left ( a+b\sqrt{dx+c} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01556, size = 58, normalized size = 1.23 \begin{align*} \frac{2 \,{\left (\frac{a}{\sqrt{d x + c} b^{3} + a b^{2}} + \frac{\log \left (\sqrt{d x + c} b + a\right )}{b^{2}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(a/(sqrt(d*x + c)*b^3 + a*b^2) + log(sqrt(d*x + c)*b + a)/b^2)/d

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Fricas [A]  time = 1.71093, size = 154, normalized size = 3.28 \begin{align*} \frac{2 \,{\left (\sqrt{d x + c} a b - a^{2} +{\left (b^{2} d x + b^{2} c - a^{2}\right )} \log \left (\sqrt{d x + c} b + a\right )\right )}}{b^{4} d^{2} x +{\left (b^{4} c - a^{2} b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.918834, size = 124, normalized size = 2.64 \begin{align*} \begin{cases} \frac{x}{a^{2}} & \text{for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\\frac{x}{\left (a + b \sqrt{c}\right )^{2}} & \text{for}\: d = 0 \\\frac{2 a \log{\left (\frac{a}{b} + \sqrt{c + d x} \right )}}{a b^{2} d + b^{3} d \sqrt{c + d x}} + \frac{2 a}{a b^{2} d + b^{3} d \sqrt{c + d x}} + \frac{2 b \sqrt{c + d x} \log{\left (\frac{a}{b} + \sqrt{c + d x} \right )}}{a b^{2} d + b^{3} d \sqrt{c + d x}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a**2, Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x/(a + b*sqrt(c))**2, Eq(d, 0)), (2*a*log(a/b + sqrt(c
+ d*x))/(a*b**2*d + b**3*d*sqrt(c + d*x)) + 2*a/(a*b**2*d + b**3*d*sqrt(c + d*x)) + 2*b*sqrt(c + d*x)*log(a/b
+ sqrt(c + d*x))/(a*b**2*d + b**3*d*sqrt(c + d*x)), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef