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

Optimal. Leaf size=70 \[ -\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x}}{b (a+b x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 70, 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 = {47, 63, 208} \begin {gather*} -\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}}-\frac {\sqrt {c+d x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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)^2} \, dx &=-\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 {\operatorname {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*}

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Mathematica [A]  time = 0.08, size = 69, normalized size = 0.99 \begin {gather*} \frac {d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {a d-b c}}\right )}{b^{3/2} \sqrt {a d-b c}}-\frac {\sqrt {c+d x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[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])

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IntegrateAlgebraic [A]  time = 0.22, size = 91, normalized size = 1.30 \begin {gather*} -\frac {d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{b^{3/2} \sqrt {a d-b c}}-\frac {d \sqrt {c+d x}}{b (a d+b (c+d x)-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.44, size = 232, normalized size = 3.31 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 1.38, size = 72, normalized size = 1.03 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 64, normalized size = 0.91 \begin {gather*} \frac {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 {\sqrt {d x +c}\, d}{\left (b d x +a d \right ) b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^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 details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.24, size = 61, normalized size = 0.87 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [B]  time = 58.58, size = 573, normalized size = 8.19 \begin {gather*} - \frac {2 a d^{2} \sqrt {c + d x}}{2 a^{2} b d^{2} - 2 a b^{2} c d + 2 a b^{2} d^{2} x - 2 b^{3} c d x} + \frac {a d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (- a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b} - \frac {a d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b} - \frac {c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (- a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2} + \frac {c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2} + \frac {2 c d \sqrt {c + d x}}{2 a^{2} d^{2} - 2 a b c d + 2 a b d^{2} x - 2 b^{2} c d x} + \frac {2 d \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{b^{2} \sqrt {\frac {a d}{b} - c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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