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

Optimal. Leaf size=76 \[ \frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)} \]

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Rubi [A]  time = 0.03, antiderivative size = 76, 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 = {51, 63, 208} \begin {gather*} \frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^2*Sqrt[c + d*x]),x]

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 {1}{(a+b x)^2 \sqrt {c+d x}} \, dx &=-\frac {\sqrt {c+d x}}{(b c-a d) (a+b x)}-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 (b c-a d)}\\ &=-\frac {\sqrt {c+d x}}{(b c-a d) (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 c-a d}\\ &=-\frac {\sqrt {c+d x}}{(b c-a d) (a+b x)}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 76, normalized size = 1.00 \begin {gather*} \frac {d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {a d-b c}}\right )}{\sqrt {b} (a d-b c)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^2*Sqrt[c + d*x]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 98, normalized size = 1.29 \begin {gather*} \frac {d \sqrt {c+d x}}{(b c-a d) (-a d-b (c+d x)+b c)}-\frac {d \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{\sqrt {b} (a d-b c)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^2*Sqrt[c + d*x]),x]

[Out]

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

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fricas [B]  time = 1.29, size = 280, normalized size = 3.68 \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^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\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^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(d*x+c)^(1/2),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^2 - 2*a^2*b^2*c*d + a^3*b*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^2
*b^2*d^2)*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^2 - 2*a^2*b^2*c*d + a^3*b*d^2 + (b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*
x)]

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giac [A]  time = 1.03, size = 87, normalized size = 1.14 \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} {\left (b c - a d\right )}} - \frac {\sqrt {d x + c} d}{{\left ({\left (d x + c\right )} b - b c + a d\right )} {\left (b c - a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.01, size = 77, normalized size = 1.01 \begin {gather*} \frac {d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) \sqrt {\left (a d -b c \right ) b}}+\frac {\sqrt {d x +c}\, d}{\left (a d -b c \right ) \left (b d x +a d \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(d*x+c)^(1/2)/(a*d-b*c)/(b*d*x+a*d)+d/(a*d-b*c)/((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(1/(b*x+a)^2/(d*x+c)^(1/2),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 = 0.09, size = 74, normalized size = 0.97 \begin {gather*} \frac {d\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{\sqrt {b}\,{\left (a\,d-b\,c\right )}^{3/2}}+\frac {d\,\sqrt {c+d\,x}}{\left (a\,d-b\,c\right )\,\left (a\,d-b\,c+b\,\left (c+d\,x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{2} \sqrt {c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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