3.1430 \(\int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac{2}{\sqrt{c+d x} (b c-a d)}-\frac{2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0273585, antiderivative size = 69, normalized size of antiderivative = 1., 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} \[ \frac{2}{\sqrt{c+d x} (b c-a d)}-\frac{2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(3/2)),x]

[Out]

2/((b*c - a*d)*Sqrt[c + d*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(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) (c+d x)^{3/2}} \, dx &=\frac{2}{(b c-a d) \sqrt{c+d x}}+\frac{b \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{b c-a d}\\ &=\frac{2}{(b c-a d) \sqrt{c+d x}}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d (b c-a d)}\\ &=\frac{2}{(b c-a d) \sqrt{c+d x}}-\frac{2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.009799, size = 46, normalized size = 0.67 \[ -\frac{2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b (c+d x)}{b c-a d}\right )}{\sqrt{c+d x} (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)^(3/2)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x))/(b*c - a*d)])/((-(b*c) + a*d)*Sqrt[c + d*x])

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Maple [A]  time = 0.009, size = 68, normalized size = 1. \begin{align*} -2\,{\frac{b}{ \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-2\,{\frac{1}{ \left ( ad-bc \right ) \sqrt{dx+c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(d*x+c)^(3/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10208, size = 456, normalized size = 6.61 \begin{align*} \left [-\frac{{\left (d x + c\right )} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x + 2 \, b c - a d + 2 \,{\left (b c - a d\right )} \sqrt{d x + c} \sqrt{\frac{b}{b c - a d}}}{b x + a}\right ) - 2 \, \sqrt{d x + c}}{b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x}, -\frac{2 \,{\left ({\left (d x + c\right )} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{{\left (b c - a d\right )} \sqrt{d x + c} \sqrt{-\frac{b}{b c - a d}}}{b d x + b c}\right ) - \sqrt{d x + c}\right )}}{b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 5.06421, size = 60, normalized size = 0.87 \begin{align*} - \frac{2}{\sqrt{c + d x} \left (a d - b c\right )} - \frac{2 \operatorname{atan}{\left (\frac{\sqrt{c + d x}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{\sqrt{\frac{a d - b c}{b}} \left (a d - b c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.06208, size = 93, normalized size = 1.35 \begin{align*} \frac{2 \, b \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{2}{{\left (b c - a d\right )} \sqrt{d x + c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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