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

Optimal. Leaf size=94 \[ \frac{2 a^2 d^2+b c (3 b c-4 a d)}{a^2 b \sqrt{a+\frac{b}{x}}}-\frac{c (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{c^2 x}{a \sqrt{a+\frac{b}{x}}} \]

[Out]

(2*a^2*d^2 + b*c*(3*b*c - 4*a*d))/(a^2*b*Sqrt[a + b/x]) + (c^2*x)/(a*Sqrt[a + b/x]) - (c*(3*b*c - 4*a*d)*ArcTa
nh[Sqrt[a + b/x]/Sqrt[a]])/a^(5/2)

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Rubi [A]  time = 0.0787857, antiderivative size = 90, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {375, 89, 78, 63, 208} \[ \frac{\frac{c (3 b c-4 a d)}{a^2}+\frac{2 d^2}{b}}{\sqrt{a+\frac{b}{x}}}-\frac{c (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{c^2 x}{a \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d/x)^2/(a + b/x)^(3/2),x]

[Out]

((2*d^2)/b + (c*(3*b*c - 4*a*d))/a^2)/Sqrt[a + b/x] + (c^2*x)/(a*Sqrt[a + b/x]) - (c*(3*b*c - 4*a*d)*ArcTanh[S
qrt[a + b/x]/Sqrt[a]])/a^(5/2)

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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{\left (c+\frac{d}{x}\right )^2}{\left (a+\frac{b}{x}\right )^{3/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{(c+d x)^2}{x^2 (a+b x)^{3/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c^2 x}{a \sqrt{a+\frac{b}{x}}}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} c (3 b c-4 a d)+a d^2 x}{x (a+b x)^{3/2}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{\frac{2 d^2}{b}+\frac{c (3 b c-4 a d)}{a^2}}{\sqrt{a+\frac{b}{x}}}+\frac{c^2 x}{a \sqrt{a+\frac{b}{x}}}+\frac{(c (3 b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a^2}\\ &=\frac{\frac{2 d^2}{b}+\frac{c (3 b c-4 a d)}{a^2}}{\sqrt{a+\frac{b}{x}}}+\frac{c^2 x}{a \sqrt{a+\frac{b}{x}}}+\frac{(c (3 b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{a^2 b}\\ &=\frac{\frac{2 d^2}{b}+\frac{c (3 b c-4 a d)}{a^2}}{\sqrt{a+\frac{b}{x}}}+\frac{c^2 x}{a \sqrt{a+\frac{b}{x}}}-\frac{c (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0857521, size = 81, normalized size = 0.86 \[ \frac{2 a^2 d^2+a b c (c x-4 d)+3 b^2 c^2}{a^2 b \sqrt{a+\frac{b}{x}}}+\frac{c (4 a d-3 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*c^2 + 2*a^2*d^2 + a*b*c*(-4*d + c*x))/(a^2*b*Sqrt[a + b/x]) + (c*(-3*b*c + 4*a*d)*ArcTanh[Sqrt[a + b/x]
/Sqrt[a]])/a^(5/2)

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Maple [B]  time = 0.012, size = 789, normalized size = 8.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x+b)/x)^(1/2)*x/a^(5/2)*(4*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^3*b^2*c*d+8*ln(
1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^2*b^3*c*d-8*a^(7/2)*((a*x+b)*x)^(1/2)*x^2*b*c*d-3*ln(1/
2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*b^5*c^2-4*a^(7/2)*((a*x+b)*x)^(3/2)*d^2-16*a^(5/2)*((a*x+b)*x
)^(1/2)*x*b^2*c*d+2*(a*x^2+b*x)^(1/2)*a^(9/2)*x^2*d^2+2*a^(9/2)*((a*x+b)*x)^(1/2)*x^2*d^2-4*a^(3/2)*((a*x+b)*x
)^(3/2)*b^2*c^2+2*(a*x^2+b*x)^(1/2)*a^(5/2)*b^2*d^2+2*a^(5/2)*((a*x+b)*x)^(1/2)*b^2*d^2+ln(1/2*(2*(a*x^2+b*x)^
(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^2*b^3*d^2+6*a^(1/2)*((a*x+b)*x)^(1/2)*b^4*c^2-ln(1/2*(2*((a*x+b)*x)^(1/2)*a^
(1/2)+2*a*x+b)/a^(1/2))*a^2*b^3*d^2+4*(a*x^2+b*x)^(1/2)*a^(7/2)*x*b*d^2+8*a^(5/2)*((a*x+b)*x)^(3/2)*b*c*d+4*a^
(7/2)*((a*x+b)*x)^(1/2)*x*b*d^2+12*a^(3/2)*((a*x+b)*x)^(1/2)*x*b^3*c^2-8*a^(3/2)*((a*x+b)*x)^(1/2)*b^3*c*d+6*a
^(5/2)*((a*x+b)*x)^(1/2)*x^2*b^2*c^2+ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^4*b*d^2-ln(1/
2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^4*b*d^2-3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/
a^(1/2))*x^2*a^2*b^3*c^2+2*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^3*b^2*d^2-2*ln(1/2*(2*((a
*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^3*b^2*d^2-6*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))
*x*a*b^4*c^2+4*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*b^4*c*d)/((a*x+b)*x)^(1/2)/b^2/(a*x+b)^
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((c+d/x)^2/(a+b/x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.30421, size = 586, normalized size = 6.23 \begin{align*} \left [-\frac{{\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d +{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x\right )} \sqrt{a} \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) - 2 \,{\left (a^{2} b c^{2} x^{2} +{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{2 \,{\left (a^{4} b x + a^{3} b^{2}\right )}}, \frac{{\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d +{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (a^{2} b c^{2} x^{2} +{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{a^{4} b x + a^{3} b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x + d\right )^{2}}{x^{2} \left (a + \frac{b}{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.71262, size = 217, normalized size = 2.31 \begin{align*} b{\left (\frac{{\left (3 \, b c^{2} - 4 \, a c d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2} b} + \frac{2 \, a b^{2} c^{2} - 4 \, a^{2} b c d + 2 \, a^{3} d^{2} - \frac{3 \,{\left (a x + b\right )} b^{2} c^{2}}{x} + \frac{4 \,{\left (a x + b\right )} a b c d}{x} - \frac{2 \,{\left (a x + b\right )} a^{2} d^{2}}{x}}{{\left (a \sqrt{\frac{a x + b}{x}} - \frac{{\left (a x + b\right )} \sqrt{\frac{a x + b}{x}}}{x}\right )} a^{2} b^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*((3*b*c^2 - 4*a*c*d)*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^2*b) + (2*a*b^2*c^2 - 4*a^2*b*c*d + 2*a^
3*d^2 - 3*(a*x + b)*b^2*c^2/x + 4*(a*x + b)*a*b*c*d/x - 2*(a*x + b)*a^2*d^2/x)/((a*sqrt((a*x + b)/x) - (a*x +
b)*sqrt((a*x + b)/x)/x)*a^2*b^2))