\(\int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx\) [1382]

Optimal result

Integrand size = 17, antiderivative size = 70 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=-\frac {\sqrt {c+d x}}{b (a+b x)}-\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {43, 65, 214} \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=-\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x}}{b (a+b x)} \]

[In]

Int[Sqrt[c + d*x]/(a + b*x)^2,x]

[Out]

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

Rule 43

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

Rule 65

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 214

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=-\frac {\sqrt {c+d x}}{b (a+b x)}+\frac {d \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{3/2} \sqrt {-b c+a d}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {d x +c}}{b x +a}+\frac {d \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{b}\)	\(58\)
derivativedivides	\(2 d \left (-\frac {\sqrt {d x +c}}{2 b \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 b \sqrt {\left (a d -b c \right ) b}}\right )\)	\(73\)
default	\(2 d \left (-\frac {\sqrt {d x +c}}{2 b \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 b \sqrt {\left (a d -b c \right ) b}}\right )\)	\(73\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 232, normalized size of antiderivative = 3.31 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=\left [\frac {\sqrt {b^{2} c - a b d} {\left (b d x + a d\right )} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) - 2 \, {\left (b^{2} c - a b d\right )} \sqrt {d x + c}}{2 \, {\left (a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x\right )}}, \frac {\sqrt {-b^{2} c + a b d} {\left (b d x + a d\right )} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) - {\left (b^{2} c - a b d\right )} \sqrt {d x + c}}{a b^{3} c - a^{2} b^{2} d + {\left (b^{4} c - a b^{3} d\right )} x}\right ] \]

[In]

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

[Out]

[1/2*(sqrt(b^2*c - a*b*d)*(b*d*x + a*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x +
a)) - 2*(b^2*c - a*b*d)*sqrt(d*x + c))/(a*b^3*c - a^2*b^2*d + (b^4*c - a*b^3*d)*x), (sqrt(-b^2*c + a*b*d)*(b*
d*x + a*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (b^2*c - a*b*d)*sqrt(d*x + c))/(a*b^3*c
- a^2*b^2*d + (b^4*c - a*b^3*d)*x)]

Sympy [F]

\[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=\int \frac {\sqrt {c + d x}}{\left (a + b x\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(sqrt(c + d*x)/(a + b*x)**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=\frac {d \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b} - \frac {\sqrt {d x + c} d}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {c+d x}}{(a+b x)^2} \, dx=\frac {d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{b^{3/2}\,\sqrt {a\,d-b\,c}}-\frac {d\,\sqrt {c+d\,x}}{d\,x\,b^2+a\,d\,b} \]

[In]

int((c + d*x)^(1/2)/(a + b*x)^2,x)

[Out]

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