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

Optimal. Leaf size=201 \[ \frac{b \left (a^2 d^2-8 a b c d+5 b^2 c^2\right )}{a^3 c \sqrt{a+\frac{b}{x}} (b c-a d)^2}-\frac{(2 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2} c^2}+\frac{b (5 b c-3 a d)}{3 a^2 c \left (a+\frac{b}{x}\right )^{3/2} (b c-a d)}-\frac{2 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 (b c-a d)^{5/2}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}} \]

[Out]

(b*(5*b*c - 3*a*d))/(3*a^2*c*(b*c - a*d)*(a + b/x)^(3/2)) + (b*(5*b^2*c^2 - 8*a*b*c*d + a^2*d^2))/(a^3*c*(b*c
- a*d)^2*Sqrt[a + b/x]) + x/(a*c*(a + b/x)^(3/2)) - (2*d^(7/2)*ArcTan[(Sqrt[d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]]
)/(c^2*(b*c - a*d)^(5/2)) - ((5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(a^(7/2)*c^2)

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Rubi [A]  time = 0.315132, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {375, 103, 152, 156, 63, 208, 205} \[ \frac{b \left (a^2 d^2-8 a b c d+5 b^2 c^2\right )}{a^3 c \sqrt{a+\frac{b}{x}} (b c-a d)^2}-\frac{(2 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2} c^2}+\frac{b (5 b c-3 a d)}{3 a^2 c \left (a+\frac{b}{x}\right )^{3/2} (b c-a d)}-\frac{2 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 (b c-a d)^{5/2}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(5/2)*(c + d/x)),x]

[Out]

(b*(5*b*c - 3*a*d))/(3*a^2*c*(b*c - a*d)*(a + b/x)^(3/2)) + (b*(5*b^2*c^2 - 8*a*b*c*d + a^2*d^2))/(a^3*c*(b*c
- a*d)^2*Sqrt[a + b/x]) + x/(a*c*(a + b/x)^(3/2)) - (2*d^(7/2)*ArcTan[(Sqrt[d]*Sqrt[a + b/x])/Sqrt[b*c - a*d]]
)/(c^2*(b*c - a*d)^(5/2)) - ((5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(a^(7/2)*c^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 103

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

Rule 152

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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{1}{\left (a+\frac{b}{x}\right )^{5/2} \left (c+\frac{d}{x}\right )} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{5/2} (c+d x)} \, dx,x,\frac{1}{x}\right )\\ &=\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (5 b c+2 a d)+\frac{5 b d x}{2}}{x (a+b x)^{5/2} (c+d x)} \, dx,x,\frac{1}{x}\right )}{a c}\\ &=\frac{b (5 b c-3 a d)}{3 a^2 c (b c-a d) \left (a+\frac{b}{x}\right )^{3/2}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{\frac{3}{4} (b c-a d) (5 b c+2 a d)+\frac{3}{4} b d (5 b c-3 a d) x}{x (a+b x)^{3/2} (c+d x)} \, dx,x,\frac{1}{x}\right )}{3 a^2 c (b c-a d)}\\ &=\frac{b (5 b c-3 a d)}{3 a^2 c (b c-a d) \left (a+\frac{b}{x}\right )^{3/2}}+\frac{b \left (5 b^2 c^2-8 a b c d+a^2 d^2\right )}{a^3 c (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}}+\frac{4 \operatorname{Subst}\left (\int \frac{\frac{3}{8} (b c-a d)^2 (5 b c+2 a d)+\frac{3}{8} b d \left (5 b^2 c^2-8 a b c d+a^2 d^2\right ) x}{x \sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{3 a^3 c (b c-a d)^2}\\ &=\frac{b (5 b c-3 a d)}{3 a^2 c (b c-a d) \left (a+\frac{b}{x}\right )^{3/2}}+\frac{b \left (5 b^2 c^2-8 a b c d+a^2 d^2\right )}{a^3 c (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}}-\frac{d^4 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{c^2 (b c-a d)^2}+\frac{(5 b c+2 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a^3 c^2}\\ &=\frac{b (5 b c-3 a d)}{3 a^2 c (b c-a d) \left (a+\frac{b}{x}\right )^{3/2}}+\frac{b \left (5 b^2 c^2-8 a b c d+a^2 d^2\right )}{a^3 c (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}}-\frac{\left (2 d^4\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^2 (b c-a d)^2}+\frac{(5 b c+2 a d) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{a^3 b c^2}\\ &=\frac{b (5 b c-3 a d)}{3 a^2 c (b c-a d) \left (a+\frac{b}{x}\right )^{3/2}}+\frac{b \left (5 b^2 c^2-8 a b c d+a^2 d^2\right )}{a^3 c (b c-a d)^2 \sqrt{a+\frac{b}{x}}}+\frac{x}{a c \left (a+\frac{b}{x}\right )^{3/2}}-\frac{2 d^{7/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 (b c-a d)^{5/2}}-\frac{(5 b c+2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2} c^2}\\ \end{align*}

Mathematica [C]  time = 0.0585514, size = 118, normalized size = 0.59 \[ \frac{x \left ((a d-b c) \left ((2 a d+5 b c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b}{a x}+1\right )+3 a c x\right )-2 a^2 d^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{d \left (a+\frac{b}{x}\right )}{a d-b c}\right )\right )}{3 a^2 c^2 \sqrt{a+\frac{b}{x}} (a x+b) (a d-b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-2*a^2*d^2*Hypergeometric2F1[-3/2, 1, -1/2, (d*(a + b/x))/(-(b*c) + a*d)] + (-(b*c) + a*d)*(3*a*c*x + (5*b
*c + 2*a*d)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + b/(a*x)])))/(3*a^2*c^2*(-(b*c) + a*d)*Sqrt[a + b/x]*(b + a*x)
)

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Maple [B]  time = 0.013, size = 1767, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(5/2)/(c+d/x),x)

[Out]

-1/6*(-32*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*b^2*c^3*d+18*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2
*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x*a^4*b^2*c*d^3+9*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2
))*((a*d-b*c)*d/c^2)^(1/2)*x*a^3*b^3*c^2*d^2-72*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*
c)*d/c^2)^(1/2)*x*a^2*b^4*c^3*d+3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2
)*x^3*a^5*b*c^2*d^2-24*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x^3*a^4*b
^2*c^3*d+18*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x^2*a^5*b*c*d^3+9*ln
(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x^2*a^4*b^2*c^2*d^2-72*ln(1/2*(2*(
(a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x^2*a^3*b^3*c^3*d+144*a^(5/2)*((a*d-b*c)*d/
c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^3*c^3*d-18*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b*c^2*d^2+14
4*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b^2*c^3*d-18*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*
x)^(1/2)*x*b^2*c^2*d^2+48*a^(9/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*b*c^3*d-36*a^(7/2)*((a*d-b*c)*
d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*x*b*c^3*d-30*a^(7/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*b^2*c^4+45*l
n(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x^2*a^2*b^4*c^4+45*ln(1/2*(2*((a*
x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x*a*b^5*c^4+6*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1
/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*a^3*b^3*c*d^3+3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(
1/2))*((a*d-b*c)*d/c^2)^(1/2)*a^2*b^4*c^2*d^2-24*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b
*c)*d/c^2)^(1/2)*a*b^5*c^3*d+6*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x
^3*a^6*c*d^3+15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*x^3*a^3*b^3*c^4-
90*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x*b^4*c^4-6*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(
1/2)*b^3*c^2*d^2+48*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^4*c^3*d+24*a^(5/2)*((a*d-b*c)*d/c^2)^(
1/2)*((a*x+b)*x)^(3/2)*x*b^2*c^4-90*a^(5/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^2*b^3*c^4-6*a^(11/2)*(
(a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*x^3*c^2*d^2+15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*
((a*d-b*c)*d/c^2)^(1/2)*b^6*c^4+6*a^(13/2)*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^3*d^4+6*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^3
*d^4+18*a^(11/2)*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^2*b*d^4+20*a^
(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(3/2)*b^3*c^4+18*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*b^2*d^4-30*a^(1/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b^5*c^4)/a^(
7/2)*x*((a*x+b)/x)^(1/2)/((a*d-b*c)*d/c^2)^(1/2)/c^3/(a*x+b)^3/(a*d-b*c)^2/((a*x+b)*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x}\right )}^{\frac{5}{2}}{\left (c + \frac{d}{x}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 4.55797, size = 4049, normalized size = 20.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(5*b^5*c^3 - 8*a*b^4*c^2*d + a^2*b^3*c*d^2 + 2*a^3*b^2*d^3 + (5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d + a^4*b*
c*d^2 + 2*a^5*d^3)*x^2 + 2*(5*a*b^4*c^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2 + 2*a^4*b*d^3)*x)*sqrt(a)*log(2*a*x
- 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 6*(a^6*d^3*x^2 + 2*a^5*b*d^3*x + a^4*b^2*d^3)*sqrt(-d/(b*c - a*d))*log(
-(2*(b*c - a*d)*x*sqrt(-d/(b*c - a*d))*sqrt((a*x + b)/x) - b*d + (b*c - 2*a*d)*x)/(c*x + d)) + 2*(3*(a^3*b^2*c
^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x^3 + 2*(10*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 3*a^4*b*c*d^2)*x^2 + 3*(5*a*b^4*c
^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^4*b^4*c^4 - 2*a^5*b^3*c^3*d + a^6*b^2*c^2*d^2 +
(a^6*b^2*c^4 - 2*a^7*b*c^3*d + a^8*c^2*d^2)*x^2 + 2*(a^5*b^3*c^4 - 2*a^6*b^2*c^3*d + a^7*b*c^2*d^2)*x), 1/3*(
3*(5*b^5*c^3 - 8*a*b^4*c^2*d + a^2*b^3*c*d^2 + 2*a^3*b^2*d^3 + (5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d + a^4*b*c*d^2
+ 2*a^5*d^3)*x^2 + 2*(5*a*b^4*c^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2 + 2*a^4*b*d^3)*x)*sqrt(-a)*arctan(sqrt(-a)
*sqrt((a*x + b)/x)/a) + 3*(a^6*d^3*x^2 + 2*a^5*b*d^3*x + a^4*b^2*d^3)*sqrt(-d/(b*c - a*d))*log(-(2*(b*c - a*d)
*x*sqrt(-d/(b*c - a*d))*sqrt((a*x + b)/x) - b*d + (b*c - 2*a*d)*x)/(c*x + d)) + (3*(a^3*b^2*c^3 - 2*a^4*b*c^2*
d + a^5*c*d^2)*x^3 + 2*(10*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 3*a^4*b*c*d^2)*x^2 + 3*(5*a*b^4*c^3 - 8*a^2*b^3*c^
2*d + a^3*b^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^4*b^4*c^4 - 2*a^5*b^3*c^3*d + a^6*b^2*c^2*d^2 + (a^6*b^2*c^4 - 2
*a^7*b*c^3*d + a^8*c^2*d^2)*x^2 + 2*(a^5*b^3*c^4 - 2*a^6*b^2*c^3*d + a^7*b*c^2*d^2)*x), -1/6*(12*(a^6*d^3*x^2
+ 2*a^5*b*d^3*x + a^4*b^2*d^3)*sqrt(d/(b*c - a*d))*arctan(-(b*c - a*d)*x*sqrt(d/(b*c - a*d))*sqrt((a*x + b)/x)
/(a*d*x + b*d)) - 3*(5*b^5*c^3 - 8*a*b^4*c^2*d + a^2*b^3*c*d^2 + 2*a^3*b^2*d^3 + (5*a^2*b^3*c^3 - 8*a^3*b^2*c^
2*d + a^4*b*c*d^2 + 2*a^5*d^3)*x^2 + 2*(5*a*b^4*c^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2 + 2*a^4*b*d^3)*x)*sqrt(a
)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(3*(a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x^3 + 2*(10*
a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 3*a^4*b*c*d^2)*x^2 + 3*(5*a*b^4*c^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2)*x)*sqr
t((a*x + b)/x))/(a^4*b^4*c^4 - 2*a^5*b^3*c^3*d + a^6*b^2*c^2*d^2 + (a^6*b^2*c^4 - 2*a^7*b*c^3*d + a^8*c^2*d^2)
*x^2 + 2*(a^5*b^3*c^4 - 2*a^6*b^2*c^3*d + a^7*b*c^2*d^2)*x), -1/3*(6*(a^6*d^3*x^2 + 2*a^5*b*d^3*x + a^4*b^2*d^
3)*sqrt(d/(b*c - a*d))*arctan(-(b*c - a*d)*x*sqrt(d/(b*c - a*d))*sqrt((a*x + b)/x)/(a*d*x + b*d)) - 3*(5*b^5*c
^3 - 8*a*b^4*c^2*d + a^2*b^3*c*d^2 + 2*a^3*b^2*d^3 + (5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d + a^4*b*c*d^2 + 2*a^5*d^
3)*x^2 + 2*(5*a*b^4*c^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2 + 2*a^4*b*d^3)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x
+ b)/x)/a) - (3*(a^3*b^2*c^3 - 2*a^4*b*c^2*d + a^5*c*d^2)*x^3 + 2*(10*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 3*a^4*
b*c*d^2)*x^2 + 3*(5*a*b^4*c^3 - 8*a^2*b^3*c^2*d + a^3*b^2*c*d^2)*x)*sqrt((a*x + b)/x))/(a^4*b^4*c^4 - 2*a^5*b^
3*c^3*d + a^6*b^2*c^2*d^2 + (a^6*b^2*c^4 - 2*a^7*b*c^3*d + a^8*c^2*d^2)*x^2 + 2*(a^5*b^3*c^4 - 2*a^6*b^2*c^3*d
+ a^7*b*c^2*d^2)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(5/2)/(c+d/x),x)

[Out]

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

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Giac [A]  time = 1.18169, size = 332, normalized size = 1.65 \begin{align*} -\frac{1}{3} \,{\left (\frac{6 \, d^{4} \arctan \left (\frac{d \sqrt{\frac{a x + b}{x}}}{\sqrt{b c d - a d^{2}}}\right )}{{\left (b^{3} c^{4} - 2 \, a b^{2} c^{3} d + a^{2} b c^{2} d^{2}\right )} \sqrt{b c d - a d^{2}}} - \frac{2 \,{\left (a b^{2} c - a^{2} b d + \frac{6 \,{\left (a x + b\right )} b^{2} c}{x} - \frac{9 \,{\left (a x + b\right )} a b d}{x}\right )} x}{{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )}{\left (a x + b\right )} \sqrt{\frac{a x + b}{x}}} + \frac{3 \, \sqrt{\frac{a x + b}{x}}}{{\left (a - \frac{a x + b}{x}\right )} a^{3} c} - \frac{3 \,{\left (5 \, b c + 2 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3} b c^{2}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*d^4*arctan(d*sqrt((a*x + b)/x)/sqrt(b*c*d - a*d^2))/((b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^2)*sqrt(b*
c*d - a*d^2)) - 2*(a*b^2*c - a^2*b*d + 6*(a*x + b)*b^2*c/x - 9*(a*x + b)*a*b*d/x)*x/((a^3*b^2*c^2 - 2*a^4*b*c*
d + a^5*d^2)*(a*x + b)*sqrt((a*x + b)/x)) + 3*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a^3*c) - 3*(5*b*c + 2*a*d)*
arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^3*b*c^2))*b