3.1440 \(\int \frac{1}{(a+b x) (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=93 \[ -\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac{2 b}{\sqrt{c+d x} (b c-a d)^2}+\frac{2}{3 (c+d x)^{3/2} (b c-a d)} \]

[Out]

2/(3*(b*c - a*d)*(c + d*x)^(3/2)) + (2*b)/((b*c - a*d)^2*Sqrt[c + d*x]) - (2*b^(3/2)*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 = 0.0387108, antiderivative size = 93, 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} \[ -\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac{2 b}{\sqrt{c+d x} (b c-a d)^2}+\frac{2}{3 (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(5/2)),x]

[Out]

2/(3*(b*c - a*d)*(c + d*x)^(3/2)) + (2*b)/((b*c - a*d)^2*Sqrt[c + d*x]) - (2*b^(3/2)*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) (c+d x)^{5/2}} \, dx &=\frac{2}{3 (b c-a d) (c+d x)^{3/2}}+\frac{b \int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx}{b c-a d}\\ &=\frac{2}{3 (b c-a d) (c+d x)^{3/2}}+\frac{2 b}{(b c-a d)^2 \sqrt{c+d x}}+\frac{b^2 \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{(b c-a d)^2}\\ &=\frac{2}{3 (b c-a d) (c+d x)^{3/2}}+\frac{2 b}{(b c-a d)^2 \sqrt{c+d x}}+\frac{\left (2 b^2\right ) \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)^2}\\ &=\frac{2}{3 (b c-a d) (c+d x)^{3/2}}+\frac{2 b}{(b c-a d)^2 \sqrt{c+d x}}-\frac{2 b^{3/2} \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 = 0.011491, size = 48, normalized size = 0.52 \[ \frac{2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b (c+d x)}{b c-a d}\right )}{3 (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)^(5/2)),x]

[Out]

(2*Hypergeometric2F1[-3/2, 1, -1/2, (b*(c + d*x))/(b*c - a*d)])/(3*(b*c - a*d)*(c + d*x)^(3/2))

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Maple [A]  time = 0.009, size = 90, normalized size = 1. \begin{align*} -{\frac{2}{3\,ad-3\,bc} \left ( dx+c \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{b}{ \left ( ad-bc \right ) ^{2}\sqrt{dx+c}}}+2\,{\frac{{b}^{2}}{ \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)/(d*x+c)^(5/2),x)

[Out]

-2/3/(a*d-b*c)/(d*x+c)^(3/2)+2*b/(a*d-b*c)^2/(d*x+c)^(1/2)+2*b^2/(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)/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14546, size = 841, normalized size = 9.04 \begin{align*} \left [\frac{3 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\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 + 4 \, b c - a d\right )} \sqrt{d x + c}}{3 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x\right )}}, -\frac{2 \,{\left (3 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\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 + 4 \, b c - a d\right )} \sqrt{d x + c}\right )}}{3 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{2} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.62344, size = 83, normalized size = 0.89 \begin{align*} \frac{2 b}{\sqrt{c + d x} \left (a d - b c\right )^{2}} + \frac{2 b \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 )^{2}} - \frac{2}{3 \left (c + d x\right )^{\frac{3}{2}} \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)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.07016, size = 153, normalized size = 1.65 \begin{align*} \frac{2 \, b^{2} \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{2 \,{\left (3 \,{\left (d x + c\right )} b + b c - a d\right )}}{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left (d x + c\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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