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

Optimal. Leaf size=166 \[ \frac{a^{3/2} (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^3}-\frac{(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 d^{3/2}}+\frac{\sqrt{a+\frac{b}{x}} (b c-2 a d) (b c-a d)}{c^2 d \left (c+\frac{d}{x}\right )}+\frac{a x \left (a+\frac{b}{x}\right )^{3/2}}{c \left (c+\frac{d}{x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.233858, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {375, 98, 149, 156, 63, 208, 205} \[ \frac{a^{3/2} (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^3}-\frac{(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^3 d^{3/2}}+\frac{\sqrt{a+\frac{b}{x}} (b c-2 a d) (b c-a d)}{c^2 d \left (c+\frac{d}{x}\right )}+\frac{a x \left (a+\frac{b}{x}\right )^{3/2}}{c \left (c+\frac{d}{x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/(c + d/x)^2,x]

[Out]

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

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 98

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

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.340474, size = 145, normalized size = 0.87 \[ \frac{\frac{c x \sqrt{a+\frac{b}{x}} \left (a^2 d (c x+2 d)-2 a b c d+b^2 c^2\right )}{d (c x+d)}+a^{3/2} (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )-\frac{(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{d^{3/2}}}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.034, size = 1323, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)/(c+d/x)^2,x)

[Out]

-1/2*(ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*((a*d-b*c)*d/c^2)^(1/2)*x*b^3*c^5+4*a^(9/2)*ln((
2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*d^5+2*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2
)*((a*x+b)*x)^(1/2)*x*b*c^4*d-5*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^3*((a*d-b*c)*d/c^2)^(1
/2)*x*b*c^3*d^2+2*a^(5/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*b^2*c^
2*d^3+2*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*b*c^5+a^(3/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)
*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*b^3*c^3*d^2-2*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*c^4*d+
4*a^(9/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*x*c*d^4+4*ln(1/2*(2*((
a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^4*((a*d-b*c)*d/c^2)^(1/2)*c*d^4-4*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*
((a*x+b)*x)^(1/2)*c^2*d^3-7*a^(7/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+
d))*b*c*d^4-7*a^(7/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*x*b*c^2*d^
3+2*a^(5/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*x*b^2*c^3*d^2-5*ln(1
/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^3*((a*d-b*c)*d/c^2)^(1/2)*b*c^2*d^3+2*a^(7/2)*((a*d-b*c)*d
/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*c^4*d+4*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b*c^3*d^2-2*a^(3/2
)*((a*d-b*c)*d/c^2)^(1/2)*(a*x^2+b*x)^(1/2)*x*b^2*c^5+a^(3/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*
c-2*a*d*x+b*c*x-b*d)/(c*x+d))*x*b^3*c^4*d-a*((a*d-b*c)*d/c^2)^(1/2)*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+
b)/a^(1/2))*x*b^3*c^5+ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*((a*d-b*c)*d/c^2)^(1/2)*b^3*c^4*
d-2*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*(a*x^2+b*x)^(1/2)*b^2*c^4*d-a*((a*d-b*c)*d/c^2)^(1/2)*ln(1/2*(2*(a*x^2+b*x
)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*b^3*c^4*d-2*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b*c^5+4*ln
(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^4*((a*d-b*c)*d/c^2)^(1/2)*x*c^2*d^3-2*a^(7/2)*((a*d-b*c)
*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*c^3*d^2)*x*((a*x+b)/x)^(1/2)/c^4/((a*d-b*c)*d/c^2)^(1/2)/a^(3/2)/(c*x+d)/d^2
/((a*x+b)*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x}\right )}^{\frac{5}{2}}}{{\left (c + \frac{d}{x}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x)^(5/2)/(c + d/x)^2, x)

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Fricas [A]  time = 1.95058, size = 2125, normalized size = 12.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((5*a*b*c*d^2 - 4*a^2*d^3 + (5*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b
)/x) + b) + (b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-(b*c - a*d)/
d)*log((2*d*x*sqrt(-(b*c - a*d)/d)*sqrt((a*x + b)/x) + b*d - (b*c - 2*a*d)*x)/(c*x + d)) - 2*(a^2*c^2*d*x^2 +
(b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(c^4*d*x + c^3*d^2), -1/2*(2*(5*a*b*c*d^2 - 4*a^2*
d^3 + (5*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (b^2*c^2*d + 3*a*b*c*d^2
- 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-(b*c - a*d)/d)*log((2*d*x*sqrt(-(b*c - a*d)/d)*sq
rt((a*x + b)/x) + b*d - (b*c - 2*a*d)*x)/(c*x + d)) - 2*(a^2*c^2*d*x^2 + (b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)
*x)*sqrt((a*x + b)/x))/(c^4*d*x + c^3*d^2), 1/2*(2*(b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2
*d - 4*a^2*c*d^2)*x)*sqrt((b*c - a*d)/d)*arctan(-d*sqrt((b*c - a*d)/d)*sqrt((a*x + b)/x)/(b*c - a*d)) - (5*a*b
*c*d^2 - 4*a^2*d^3 + (5*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2
*(a^2*c^2*d*x^2 + (b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)*x)*sqrt((a*x + b)/x))/(c^4*d*x + c^3*d^2), ((b^2*c^2*d
+ 3*a*b*c*d^2 - 4*a^2*d^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt((b*c - a*d)/d)*arctan(-d*sqrt((b*c
- a*d)/d)*sqrt((a*x + b)/x)/(b*c - a*d)) - (5*a*b*c*d^2 - 4*a^2*d^3 + (5*a*b*c^2*d - 4*a^2*c*d^2)*x)*sqrt(-a)*
arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (a^2*c^2*d*x^2 + (b^2*c^3 - 2*a*b*c^2*d + 2*a^2*c*d^2)*x)*sqrt((a*x + b
)/x))/(c^4*d*x + c^3*d^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)/(c+d/x)**2,x)

[Out]

Timed out

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

[Out]

Exception raised: TypeError