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

Optimal. Leaf size=126 \[ \frac {c^2 x \left (a+\frac {b}{x}\right )^{5/2}}{a}-\frac {c \left (a+\frac {b}{x}\right )^{3/2} (4 a d+3 b c)}{3 a}-c \sqrt {a+\frac {b}{x}} (4 a d+3 b c)+\sqrt {a} c (4 a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {2 d^2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {c^2 x \left (a+\frac {b}{x}\right )^{5/2}}{a}-\frac {c \left (a+\frac {b}{x}\right )^{3/2} (4 a d+3 b c)}{3 a}-c \sqrt {a+\frac {b}{x}} (4 a d+3 b c)+\sqrt {a} c (4 a d+3 b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )-\frac {2 d^2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)*(c + d/x)^2,x]

[Out]

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

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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 106, normalized size = 0.84 \[ -\frac {c (4 a d+3 b c) \left (\sqrt {a+\frac {b}{x}} (4 a x+b)-3 a^{3/2} x \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\right )}{3 a x}+\frac {c^2 x \left (a+\frac {b}{x}\right )^{5/2}}{a}-\frac {2 d^2 \left (a+\frac {b}{x}\right )^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 268, normalized size = 2.13 \[ \left [\frac {15 \, {\left (3 \, b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {a} x^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (15 \, a b c^{2} x^{3} - 6 \, b^{2} d^{2} - 2 \, {\left (15 \, b^{2} c^{2} + 40 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{2} - 4 \, {\left (5 \, b^{2} c d + 3 \, a b d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{30 \, b x^{2}}, -\frac {15 \, {\left (3 \, b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (15 \, a b c^{2} x^{3} - 6 \, b^{2} d^{2} - 2 \, {\left (15 \, b^{2} c^{2} + 40 \, a b c d + 3 \, a^{2} d^{2}\right )} x^{2} - 4 \, {\left (5 \, b^{2} c d + 3 \, a b d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{15 \, b x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(3*b^2*c^2 + 4*a*b*c*d)*sqrt(a)*x^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(15*a*b*c^2*x
^3 - 6*b^2*d^2 - 2*(15*b^2*c^2 + 40*a*b*c*d + 3*a^2*d^2)*x^2 - 4*(5*b^2*c*d + 3*a*b*d^2)*x)*sqrt((a*x + b)/x))
/(b*x^2), -1/15*(15*(3*b^2*c^2 + 4*a*b*c*d)*sqrt(-a)*x^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (15*a*b*c^2*x^
3 - 6*b^2*d^2 - 2*(15*b^2*c^2 + 40*a*b*c*d + 3*a^2*d^2)*x^2 - 4*(5*b^2*c*d + 3*a*b*d^2)*x)*sqrt((a*x + b)/x))/
(b*x^2)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-2,[1,2,0]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[2,4,0]%%
%}+%%%{-2,[2,2,0]%%%}+%%%{1,[2,0,0]%%%}+%%%{2,[1,3,1]%%%}+%%%{-2,[1,1,1]%%%}+%%%{1,[0,2,2]%%%}] at parameters
values [86,-97,-82]Warning, choosing root of [1,0,%%%{-2,[1,2,0]%%%}+%%%{-2,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%
%%{1,[2,4,0]%%%}+%%%{-2,[2,2,0]%%%}+%%%{1,[2,0,0]%%%}+%%%{2,[1,3,1]%%%}+%%%{-2,[1,1,1]%%%}+%%%{1,[0,2,2]%%%}]
at parameters values [7,-27,26]Warning, choosing root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[0
,2,2]%%%}] at parameters values [18.6420984049,-49,-86]Warning, choosing root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-
2,[0,1,1]%%%},0,%%%{1,[0,2,2]%%%}] at parameters values [78.6493344628,22,42]Warning, choosing root of [1,0,%%
%{-2,[1,0,1]%%%}+%%%{-4,[0,1,0]%%%},0,%%%{1,[2,0,2]%%%}] at parameters values [-13,74.7709350525,24]Sign error
(%%%{-b,0%%%}+%%%{2*sqrt(a)*sqrt(b),1/2%%%}+%%%{-2*a,1%%%}+%%%{a*sqrt(a)*sqrt(b)/b,3/2%%%}+%%%{-a^2*sqrt(a)*s
qrt(b)/(4*b^2),5/2%%%}+%%%{undef,7/2%%%})Evaluation time: 0.43Limit: Max order reached or unable to make serie
s expansion Error: Bad Argument Value

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maple [B]  time = 0.06, size = 260, normalized size = 2.06 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-60 a^{2} b c d \,x^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-45 a \,b^{2} c^{2} x^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-120 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} c d \,x^{4}-90 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b \,c^{2} x^{4}+120 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} c d \,x^{2}+60 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b \,c^{2} x^{2}+12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} d^{2} x +40 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b c d x +12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b \,d^{2}\right )}{30 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((a*x+b)/x)^(1/2)/x^3/b*(-120*(a*x^2+b*x)^(1/2)*a^(5/2)*x^4*c*d-90*(a*x^2+b*x)^(1/2)*a^(3/2)*x^4*b*c^2-6
0*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*x^4*a^2*b*c*d-45*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a
^(1/2))/a^(1/2))*x^4*a*b^2*c^2+120*(a*x^2+b*x)^(3/2)*a^(3/2)*x^2*c*d+60*(a*x^2+b*x)^(3/2)*a^(1/2)*x^2*b*c^2+12
*(a*x^2+b*x)^(3/2)*a^(3/2)*x*d^2+40*(a*x^2+b*x)^(3/2)*a^(1/2)*x*b*c*d+12*(a*x^2+b*x)^(3/2)*a^(1/2)*b*d^2)/((a*
x+b)*x)^(1/2)/a^(1/2)

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maxima [A]  time = 1.16, size = 152, normalized size = 1.21 \[ -\frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} d^{2}}{5 \, b} + \frac {1}{2} \, {\left (2 \, \sqrt {a + \frac {b}{x}} a x - 3 \, \sqrt {a} b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - 4 \, \sqrt {a + \frac {b}{x}} b\right )} c^{2} - \frac {2}{3} \, {\left (3 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) + 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} + 6 \, \sqrt {a + \frac {b}{x}} a\right )} c d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.58, size = 197, normalized size = 1.56 \[ \sqrt {a+\frac {b}{x}}\,\left (2\,a\,\left (\frac {4\,a\,d^2-4\,b\,c\,d}{b}-\frac {4\,a\,d^2}{b}\right )-\frac {2\,{\left (a\,d-b\,c\right )}^2}{b}+\frac {2\,a^2\,d^2}{b}\right )+\left (\frac {4\,a\,d^2-4\,b\,c\,d}{3\,b}-\frac {4\,a\,d^2}{3\,b}\right )\,{\left (a+\frac {b}{x}\right )}^{3/2}-\frac {2\,d^2\,{\left (a+\frac {b}{x}\right )}^{5/2}}{5\,b}+a\,c^2\,x\,\sqrt {a+\frac {b}{x}}-2\,c\,\mathrm {atan}\left (\frac {2\,c\,\sqrt {a+\frac {b}{x}}\,\left (4\,a\,d+3\,b\,c\right )\,\sqrt {-\frac {a}{4}}}{4\,d\,a^2\,c+3\,b\,a\,c^2}\right )\,\left (4\,a\,d+3\,b\,c\right )\,\sqrt {-\frac {a}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(3/2)*(c + d/x)^2,x)

[Out]

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

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sympy [A]  time = 94.07, size = 534, normalized size = 4.24 \[ \frac {4 a^{\frac {11}{2}} b^{\frac {5}{2}} d^{2} x^{3} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} + \frac {2 a^{\frac {9}{2}} b^{\frac {7}{2}} d^{2} x^{2} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {8 a^{\frac {7}{2}} b^{\frac {9}{2}} d^{2} x \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {6 a^{\frac {5}{2}} b^{\frac {11}{2}} d^{2} \sqrt {\frac {a x}{b} + 1}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} + \sqrt {a} b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )} - \frac {4 a^{6} b^{2} d^{2} x^{\frac {7}{2}}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {4 a^{5} b^{3} d^{2} x^{\frac {5}{2}}}{15 a^{\frac {7}{2}} b^{3} x^{\frac {7}{2}} + 15 a^{\frac {5}{2}} b^{4} x^{\frac {5}{2}}} - \frac {4 a^{2} c d \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + a \sqrt {b} c^{2} \sqrt {x} \sqrt {\frac {a x}{b} + 1} - \frac {2 a b c^{2} \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} - 4 a c d \sqrt {a + \frac {b}{x}} + a d^{2} \left (\begin {cases} - \frac {\sqrt {a}}{x} & \text {for}\: b = 0 \\- \frac {2 \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) - 2 b c^{2} \sqrt {a + \frac {b}{x}} + 2 b c d \left (\begin {cases} - \frac {\sqrt {a}}{x} & \text {for}\: b = 0 \\- \frac {2 \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**(11/2)*b**(5/2)*d**2*x**3*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + 2*a**
(9/2)*b**(7/2)*d**2*x**2*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**(7/2)*
b**(9/2)*d**2*x*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 6*a**(5/2)*b**(11/2)
*d**2*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + sqrt(a)*b*c**2*asinh(sqrt(a)*s
qrt(x)/sqrt(b)) - 4*a**6*b**2*d**2*x**(7/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**5*b
**3*d**2*x**(5/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**2*c*d*atan(sqrt(a + b/x)/sqrt
(-a))/sqrt(-a) + a*sqrt(b)*c**2*sqrt(x)*sqrt(a*x/b + 1) - 2*a*b*c**2*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) - 4
*a*c*d*sqrt(a + b/x) + a*d**2*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) - 2*b*c**2*
sqrt(a + b/x) + 2*b*c*d*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True))

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