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

Optimal. Leaf size=99 \[ \frac{c^2 x \left (a+\frac{b}{x}\right )^{3/2}}{a}-\frac{c \sqrt{a+\frac{b}{x}} (4 a d+b c)}{a}+\frac{c (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0666848, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {375, 89, 80, 50, 63, 208} \[ \frac{c^2 x \left (a+\frac{b}{x}\right )^{3/2}}{a}-\frac{c \sqrt{a+\frac{b}{x}} (4 a d+b c)}{a}+\frac{c (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]*(c + d/x)^2,x]

[Out]

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

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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )^2 \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x} (c+d x)^2}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c^2 \left (a+\frac{b}{x}\right )^{3/2} x}{a}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x} \left (\frac{1}{2} c (b c+4 a d)+a d^2 x\right )}{x} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{3/2}}{3 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{3/2} x}{a}-\frac{(c (b c+4 a d)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=-\frac{c (b c+4 a d) \sqrt{a+\frac{b}{x}}}{a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{3/2}}{3 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{3/2} x}{a}-\frac{1}{2} (c (b c+4 a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{c (b c+4 a d) \sqrt{a+\frac{b}{x}}}{a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{3/2}}{3 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{3/2} x}{a}-\frac{(c (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 )}{b}\\ &=-\frac{c (b c+4 a d) \sqrt{a+\frac{b}{x}}}{a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{3/2}}{3 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{3/2} x}{a}+\frac{c (b c+4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0882163, size = 84, normalized size = 0.85 \[ \frac{\sqrt{a+\frac{b}{x}} \left (b \left (3 c^2 x^2-12 c d x-2 d^2\right )-2 a d^2 x\right )}{3 b x}+\frac{c (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b/x]*(-2*a*d^2*x + b*(-2*d^2 - 12*c*d*x + 3*c^2*x^2)))/(3*b*x) + (c*(b*c + 4*a*d)*ArcTanh[Sqrt[a + b
/x]/Sqrt[a]])/Sqrt[a]

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Maple [B]  time = 0.01, size = 191, normalized size = 1.9 \begin{align*} -{\frac{1}{6\,b{x}^{2}}\sqrt{{\frac{ax+b}{x}}} \left ( -24\,\sqrt{a{x}^{2}+bx}{a}^{3/2}{x}^{3}cd-6\,\sqrt{a{x}^{2}+bx}\sqrt{a}{x}^{3}b{c}^{2}-12\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}abcd-3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}{b}^{2}{c}^{2}+24\, \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}xcd+4\,{d}^{2} \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a} \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((a*x+b)/x)^(1/2)/x^2*(-24*(a*x^2+b*x)^(1/2)*a^(3/2)*x^3*c*d-6*(a*x^2+b*x)^(1/2)*a^(1/2)*x^3*b*c^2-12*ln(
1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a*b*c*d-3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/
a^(1/2))*x^3*b^2*c^2+24*(a*x^2+b*x)^(3/2)*a^(1/2)*x*c*d+4*d^2*(a*x^2+b*x)^(3/2)*a^(1/2))/((a*x+b)*x)^(1/2)/a^(
1/2)/b

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33427, size = 462, normalized size = 4.67 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt{a} x \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (3 \, a b c^{2} x^{2} - 2 \, a b d^{2} - 2 \,{\left (6 \, a b c d + a^{2} d^{2}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{6 \, a b x}, -\frac{3 \,{\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (3 \, a b c^{2} x^{2} - 2 \, a b d^{2} - 2 \,{\left (6 \, a b c d + a^{2} d^{2}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{3 \, a b x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(b^2*c^2 + 4*a*b*c*d)*sqrt(a)*x*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(3*a*b*c^2*x^2 - 2*
a*b*d^2 - 2*(6*a*b*c*d + a^2*d^2)*x)*sqrt((a*x + b)/x))/(a*b*x), -1/3*(3*(b^2*c^2 + 4*a*b*c*d)*sqrt(-a)*x*arct
an(sqrt(-a)*sqrt((a*x + b)/x)/a) - (3*a*b*c^2*x^2 - 2*a*b*d^2 - 2*(6*a*b*c*d + a^2*d^2)*x)*sqrt((a*x + b)/x))/
(a*b*x)]

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Sympy [A]  time = 17.7495, size = 121, normalized size = 1.22 \begin{align*} - \frac{4 a c d \operatorname{atan}{\left (\frac{\sqrt{a + \frac{b}{x}}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + \sqrt{b} c^{2} \sqrt{x} \sqrt{\frac{a x}{b} + 1} - 4 c d \sqrt{a + \frac{b}{x}} + d^{2} \left (\begin{cases} - \frac{\sqrt{a}}{x} & \text{for}\: b = 0 \\- \frac{2 \left (a + \frac{b}{x}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) + \frac{b c^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )}}{\sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a*c*d*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) + sqrt(b)*c**2*sqrt(x)*sqrt(a*x/b + 1) - 4*c*d*sqrt(a + b/x) +
d**2*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) + b*c**2*asinh(sqrt(a)*sqrt(x)/sqrt(
b))/sqrt(a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError