\(\int \sqrt {a+\frac {b}{x}} (c+\frac {d}{x})^2 \, dx\) [225]

Optimal result

Integrand size = 21, antiderivative size = 99 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=-\frac {c (b c+4 a d) \sqrt {a+\frac {b}{x}}}{a}-\frac {2 d^2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b}+\frac {c^2 \left (a+\frac {b}{x}\right )^{3/2} x}{a}+\frac {c (b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

-2/3*d^2*(a+b/x)^(3/2)/b+c^2*(a+b/x)^(3/2)*x/a+c*(4*a*d+b*c)*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(1/2)-c*(4*a*d+b
*c)*(a+b/x)^(1/2)/a

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {382, 91, 81, 52, 65, 214} \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\frac {c \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) (4 a d+b c)}{\sqrt {a}}+\frac {c^2 x \left (a+\frac {b}{x}\right )^{3/2}}{a}-\frac {c \sqrt {a+\frac {b}{x}} (4 a d+b c)}{a}-\frac {2 d^2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b} \]

[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 52

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 65

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 81

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 91

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 382

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]

Rubi steps \begin{align*} \text {integral}& = -\text {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 {\text {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)) \text {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)) \text {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)) \text {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] (verified)

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\frac {\sqrt {a+\frac {b}{x}} \left (-2 a d^2 x+b \left (-2 d^2-12 c d x+3 c^2 x^2\right )\right )}{3 b x}+\frac {c (b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[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]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {\left (-3 b \,c^{2} x^{2}+2 x a \,d^{2}+12 b c d x +2 b \,d^{2}\right ) \sqrt {\frac {a x +b}{x}}}{3 x b}+\frac {\left (4 a d +b c \right ) c \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{2 \sqrt {a}\, \left (a x +b \right )}\)	\(115\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, \left (24 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, c d \,x^{3}+6 \sqrt {a}\, \sqrt {a \,x^{2}+b x}\, b \,c^{2} x^{3}+12 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a b c d \,x^{3}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{2} c^{2} x^{3}-24 \sqrt {a}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} c d x -4 d^{2} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\right )}{6 x^{2} \sqrt {x \left (a x +b \right )}\, \sqrt {a}\, b}\)	\(191\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.10 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {a} x \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (3 \, a b c^{2} x^{2} - 2 \, a b d^{2} - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{6 \, a b x}, -\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (3 \, a b c^{2} x^{2} - 2 \, a b d^{2} - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, a b x}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 9.91 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\sqrt {b} c^{2} \sqrt {x} \sqrt {\frac {a x}{b} + 1} - 2 c d \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + \frac {b}{x}} & \text {for}\: b \neq 0 \\- \sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + d^{2} \left (\begin {cases} - \frac {\sqrt {a}}{x} & \text {for}\: b = 0 \\- \frac {2 \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + \frac {b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{\sqrt {a}} \]

[In]

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

[Out]

sqrt(b)*c**2*sqrt(x)*sqrt(a*x/b + 1) - 2*c*d*Piecewise((2*a*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) + 2*sqrt(a +
b/x), Ne(b, 0)), (-sqrt(a)*log(x), True)) + 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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.27 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\frac {1}{2} \, {\left (2 \, \sqrt {a + \frac {b}{x}} x - \frac {b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{\sqrt {a}}\right )} c^{2} - 2 \, {\left (\sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) + 2 \, \sqrt {a + \frac {b}{x}}\right )} c d - \frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} d^{2}}{3 \, b} \]

[In]

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

[Out]

1/2*(2*sqrt(a + b/x)*x - b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a))*c^2 - 2*(sqrt(a)*
log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) + 2*sqrt(a + b/x))*c*d - 2/3*(a + b/x)^(3/2)*d^2/b

Giac [F(-2)]

Exception generated. \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \, dx=\left (\frac {4\,a\,d^2-4\,b\,c\,d}{b}-\frac {4\,a\,d^2}{b}\right )\,\sqrt {a+\frac {b}{x}}+c^2\,x\,\sqrt {a+\frac {b}{x}}-\frac {2\,d^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{3\,b}-\frac {c\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\left (4\,a\,d+b\,c\right )\,1{}\mathrm {i}}{\sqrt {a}} \]

[In]

int((a + b/x)^(1/2)*(c + d/x)^2,x)

[Out]

((4*a*d^2 - 4*b*c*d)/b - (4*a*d^2)/b)*(a + b/x)^(1/2) + c^2*x*(a + b/x)^(1/2) - (2*d^2*(a + b/x)^(3/2))/(3*b)
- (c*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*(4*a*d + b*c)*1i)/a^(1/2)