3.1431 \(\int \frac{1}{(a+b x)^2 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=99 \[ -\frac{3 d}{\sqrt{c+d x} (b c-a d)^2}-\frac{1}{(a+b x) \sqrt{c+d x} (b c-a d)}+\frac{3 \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}} \]

[Out]

(-3*d)/((b*c - a*d)^2*Sqrt[c + d*x]) - 1/((b*c - a*d)*(a + b*x)*Sqrt[c + d*x]) + (3*Sqrt[b]*d*ArcTanh[(Sqrt[b]
*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

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Rubi [A]  time = 0.0384196, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ -\frac{3 d}{\sqrt{c+d x} (b c-a d)^2}-\frac{1}{(a+b x) \sqrt{c+d x} (b c-a d)}+\frac{3 \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d)/((b*c - a*d)^2*Sqrt[c + d*x]) - 1/((b*c - a*d)*(a + b*x)*Sqrt[c + d*x]) + (3*Sqrt[b]*d*ArcTanh[(Sqrt[b]
*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(5/2)

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 (c+d x)^{3/2}} \, dx &=-\frac{1}{(b c-a d) (a+b x) \sqrt{c+d x}}-\frac{(3 d) \int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx}{2 (b c-a d)}\\ &=-\frac{3 d}{(b c-a d)^2 \sqrt{c+d x}}-\frac{1}{(b c-a d) (a+b x) \sqrt{c+d x}}-\frac{(3 b d) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 (b c-a d)^2}\\ &=-\frac{3 d}{(b c-a d)^2 \sqrt{c+d x}}-\frac{1}{(b c-a d) (a+b x) \sqrt{c+d x}}-\frac{(3 b) \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)^2}\\ &=-\frac{3 d}{(b c-a d)^2 \sqrt{c+d x}}-\frac{1}{(b c-a d) (a+b x) \sqrt{c+d x}}+\frac{3 \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0121071, size = 48, normalized size = 0.48 \[ -\frac{2 d \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{b (c+d x)}{a d-b c}\right )}{\sqrt{c+d x} (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 101, normalized size = 1. \begin{align*} -2\,{\frac{d}{ \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}-{\frac{bd}{ \left ( ad-bc \right ) ^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}-3\,{\frac{bd}{ \left ( ad-bc \right ) ^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.12702, size = 883, normalized size = 8.92 \begin{align*} \left [\frac{3 \,{\left (b d^{2} x^{2} + a c d +{\left (b c d + a d^{2}\right )} x\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 \,{\left (3 \, b d x + b c + 2 \, a d\right )} \sqrt{d x + c}}{2 \,{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )}}, \frac{3 \,{\left (b d^{2} x^{2} + a c d +{\left (b c d + a d^{2}\right )} x\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 ) -{\left (3 \, b d x + b c + 2 \, a d\right )} \sqrt{d x + c}}{a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [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)**2/(d*x+c)**(3/2),x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.05468, size = 193, normalized size = 1.95 \begin{align*} -\frac{3 \, b d \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} - \frac{3 \,{\left (d x + c\right )} b d - 2 \, b c d + 2 \, a d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left ({\left (d x + c\right )}^{\frac{3}{2}} b - \sqrt{d x + c} b c + \sqrt{d x + c} a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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