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]

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

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Rubi [A]  time = 0.23, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 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 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 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]

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 \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*}

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Mathematica [A]  time = 0.43, size = 145, normalized size = 0.87 \[ \frac {a^{3/2} (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )+\frac {c x \sqrt {a+\frac {b}{x}} \left (a^2 d (c x+2 d)-2 a b c d+b^2 c^2\right )}{d (c x+d)}-\frac {(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{d^{3/2}}}{c^3} \]

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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fricas [A]  time = 1.23, size = 1001, normalized size = 6.03 \[ \left [-\frac {{\left (5 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (5 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + {\left (b^{2} c^{2} d + 3 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (b^{2} c^{3} + 3 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{d}} \log \left (\frac {2 \, d x \sqrt {-\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}} + b d - {\left (b c - 2 \, a d\right )} x}{c x + d}\right ) - 2 \, {\left (a^{2} c^{2} d x^{2} + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + 2 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (c^{4} d x + c^{3} d^{2}\right )}}, -\frac {2 \, {\left (5 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (5 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (b^{2} c^{2} d + 3 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (b^{2} c^{3} + 3 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {-\frac {b c - a d}{d}} \log \left (\frac {2 \, d x \sqrt {-\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}} + b d - {\left (b c - 2 \, a d\right )} x}{c x + d}\right ) - 2 \, {\left (a^{2} c^{2} d x^{2} + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + 2 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (c^{4} d x + c^{3} d^{2}\right )}}, \frac {2 \, {\left (b^{2} c^{2} d + 3 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (b^{2} c^{3} + 3 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{d}} \arctan \left (-\frac {d \sqrt {\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}}}{b c - a d}\right ) - {\left (5 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (5 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (a^{2} c^{2} d x^{2} + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + 2 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (c^{4} d x + c^{3} d^{2}\right )}}, \frac {{\left (b^{2} c^{2} d + 3 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (b^{2} c^{3} + 3 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {b c - a d}{d}} \arctan \left (-\frac {d \sqrt {\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}}}{b c - a d}\right ) - {\left (5 \, a b c d^{2} - 4 \, a^{2} d^{3} + {\left (5 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (a^{2} c^{2} d x^{2} + {\left (b^{2} c^{3} - 2 \, a b c^{2} d + 2 \, a^{2} c d^{2}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{c^{4} d x + c^{3} d^{2}}\right ] \]

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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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)^(5/2)/(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)]Unable to divide, perhaps due to rounding error%%%{%%{[-2,0]:[1,0,%%%{-1,[1]%%%}]%%},[4,6,4,0]%%%}+%%%{%%
%{8,[1]%%%},[3,5,4,1]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1]%%%}]%%},[2,5,5,1]%%%}+%%%{%%{[%%%{-8,[1]%%%},0]:[1,0,%
%%{-1,[1]%%%}]%%},[2,4,4,2]%%%}+%%%{%%%{8,[1]%%%},[1,4,5,2]%%%}+%%%{%%{[-2,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,4,6,2
]%%%} / %%%{%%%{1,[1]%%%},[4,2,0,0]%%%}+%%%{%%{poly1[%%%{-4,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1,0,1]%%%}+%
%%{%%%{2,[1]%%%},[2,1,1,1]%%%}+%%%{%%%{4,[2]%%%},[2,0,0,2]%%%}+%%%{%%{poly1[%%%{-4,[1]%%%},0]:[1,0,%%%{-1,[1]%
%%}]%%},[1,0,1,2]%%%}+%%%{%%%{1,[1]%%%},[0,0,2,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.06, size = 1323, normalized size = 7.97 \[ \text {result too large to display} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x}\right )}^{\frac {5}{2}}}{{\left (c + \frac {d}{x}\right )}^{2}}\,{d x} \]

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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mupad [B]  time = 2.31, size = 1153, normalized size = 6.95 \[ \frac {\frac {\sqrt {a+\frac {b}{x}}\,\left (2\,a^3\,b\,d^2-3\,a^2\,b^2\,c\,d+a\,b^3\,c^2\right )}{c^2\,d}-\frac {b\,{\left (a+\frac {b}{x}\right )}^{3/2}\,\left (2\,a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{c^2\,d}}{\left (a+\frac {b}{x}\right )\,\left (2\,a\,d-b\,c\right )-d\,{\left (a+\frac {b}{x}\right )}^2-a^2\,d+a\,b\,c}-\frac {\mathrm {atanh}\left (\frac {10\,b^9\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}}{10\,a^2\,b^9+\frac {32\,a^3\,b^8\,d}{c}-\frac {132\,a^4\,b^7\,d^2}{c^2}+\frac {130\,a^5\,b^6\,d^3}{c^3}-\frac {40\,a^6\,b^5\,d^4}{c^4}}+\frac {32\,a\,b^8\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}}{32\,a^3\,b^8+\frac {10\,a^2\,b^9\,c}{d}-\frac {132\,a^4\,b^7\,d}{c}+\frac {130\,a^5\,b^6\,d^2}{c^2}-\frac {40\,a^6\,b^5\,d^3}{c^3}}-\frac {132\,a^2\,b^7\,d\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}}{32\,a^3\,b^8\,c-132\,a^4\,b^7\,d+\frac {10\,a^2\,b^9\,c^2}{d}+\frac {130\,a^5\,b^6\,d^2}{c}-\frac {40\,a^6\,b^5\,d^3}{c^2}}+\frac {130\,a^3\,b^6\,d^2\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}}{32\,a^3\,b^8\,c^2+130\,a^5\,b^6\,d^2+\frac {10\,a^2\,b^9\,c^3}{d}-\frac {40\,a^6\,b^5\,d^3}{c}-132\,a^4\,b^7\,c\,d}-\frac {40\,a^4\,b^5\,d^3\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}}{32\,a^3\,b^8\,c^3-40\,a^6\,b^5\,d^3-132\,a^4\,b^7\,c^2\,d+130\,a^5\,b^6\,c\,d^2+\frac {10\,a^2\,b^9\,c^4}{d}}\right )\,\left (4\,a\,d-5\,b\,c\right )\,\sqrt {a^3}}{c^3}+\frac {\mathrm {atanh}\left (\frac {30\,a^3\,b^6\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3\,d^6-3\,a^2\,b\,c\,d^5+3\,a\,b^2\,c^2\,d^4-b^3\,c^3\,d^3}}{14\,a^2\,b^9\,c^3+110\,a^5\,b^6\,d^3-4\,a^3\,b^8\,c^2\,d-82\,a^4\,b^7\,c\,d^2+\frac {2\,a\,b^{10}\,c^4}{d}-\frac {40\,a^6\,b^5\,d^4}{c}}+\frac {18\,a^2\,b^7\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3\,d^6-3\,a^2\,b\,c\,d^5+3\,a\,b^2\,c^2\,d^4-b^3\,c^3\,d^3}}{2\,a\,b^{10}\,c^3-82\,a^4\,b^7\,d^3+14\,a^2\,b^9\,c^2\,d-4\,a^3\,b^8\,c\,d^2+\frac {110\,a^5\,b^6\,d^4}{c}-\frac {40\,a^6\,b^5\,d^5}{c^2}}+\frac {40\,a^4\,b^5\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3\,d^6-3\,a^2\,b\,c\,d^5+3\,a\,b^2\,c^2\,d^4-b^3\,c^3\,d^3}}{4\,a^3\,b^8\,c^3+40\,a^6\,b^5\,d^3+82\,a^4\,b^7\,c^2\,d-110\,a^5\,b^6\,c\,d^2-\frac {2\,a\,b^{10}\,c^5}{d^2}-\frac {14\,a^2\,b^9\,c^4}{d}}-\frac {2\,a\,b^8\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3\,d^6-3\,a^2\,b\,c\,d^5+3\,a\,b^2\,c^2\,d^4-b^3\,c^3\,d^3}}{4\,a^3\,b^8\,d^3-14\,a^2\,b^9\,c\,d^2+\frac {82\,a^4\,b^7\,d^4}{c}-\frac {110\,a^5\,b^6\,d^5}{c^2}+\frac {40\,a^6\,b^5\,d^6}{c^3}-2\,a\,b^{10}\,c^2\,d}\right )\,\sqrt {d^3\,{\left (a\,d-b\,c\right )}^3}\,\left (4\,a\,d+b\,c\right )}{c^3\,d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(5/2)/(c + d/x)^2,x)

[Out]

(((a + b/x)^(1/2)*(a*b^3*c^2 + 2*a^3*b*d^2 - 3*a^2*b^2*c*d))/(c^2*d) - (b*(a + b/x)^(3/2)*(2*a^2*d^2 + b^2*c^2
- 2*a*b*c*d))/(c^2*d))/((a + b/x)*(2*a*d - b*c) - d*(a + b/x)^2 - a^2*d + a*b*c) - (atanh((10*b^9*(a + b/x)^(
1/2)*(a^3)^(1/2))/(10*a^2*b^9 + (32*a^3*b^8*d)/c - (132*a^4*b^7*d^2)/c^2 + (130*a^5*b^6*d^3)/c^3 - (40*a^6*b^5
*d^4)/c^4) + (32*a*b^8*(a + b/x)^(1/2)*(a^3)^(1/2))/(32*a^3*b^8 + (10*a^2*b^9*c)/d - (132*a^4*b^7*d)/c + (130*
a^5*b^6*d^2)/c^2 - (40*a^6*b^5*d^3)/c^3) - (132*a^2*b^7*d*(a + b/x)^(1/2)*(a^3)^(1/2))/(32*a^3*b^8*c - 132*a^4
*b^7*d + (10*a^2*b^9*c^2)/d + (130*a^5*b^6*d^2)/c - (40*a^6*b^5*d^3)/c^2) + (130*a^3*b^6*d^2*(a + b/x)^(1/2)*(
a^3)^(1/2))/(32*a^3*b^8*c^2 + 130*a^5*b^6*d^2 + (10*a^2*b^9*c^3)/d - (40*a^6*b^5*d^3)/c - 132*a^4*b^7*c*d) - (
40*a^4*b^5*d^3*(a + b/x)^(1/2)*(a^3)^(1/2))/(32*a^3*b^8*c^3 - 40*a^6*b^5*d^3 - 132*a^4*b^7*c^2*d + 130*a^5*b^6
*c*d^2 + (10*a^2*b^9*c^4)/d))*(4*a*d - 5*b*c)*(a^3)^(1/2))/c^3 + (atanh((30*a^3*b^6*(a + b/x)^(1/2)*(a^3*d^6 -
b^3*c^3*d^3 + 3*a*b^2*c^2*d^4 - 3*a^2*b*c*d^5)^(1/2))/(14*a^2*b^9*c^3 + 110*a^5*b^6*d^3 - 4*a^3*b^8*c^2*d - 8
2*a^4*b^7*c*d^2 + (2*a*b^10*c^4)/d - (40*a^6*b^5*d^4)/c) + (18*a^2*b^7*(a + b/x)^(1/2)*(a^3*d^6 - b^3*c^3*d^3
+ 3*a*b^2*c^2*d^4 - 3*a^2*b*c*d^5)^(1/2))/(2*a*b^10*c^3 - 82*a^4*b^7*d^3 + 14*a^2*b^9*c^2*d - 4*a^3*b^8*c*d^2
+ (110*a^5*b^6*d^4)/c - (40*a^6*b^5*d^5)/c^2) + (40*a^4*b^5*(a + b/x)^(1/2)*(a^3*d^6 - b^3*c^3*d^3 + 3*a*b^2*c
^2*d^4 - 3*a^2*b*c*d^5)^(1/2))/(4*a^3*b^8*c^3 + 40*a^6*b^5*d^3 + 82*a^4*b^7*c^2*d - 110*a^5*b^6*c*d^2 - (2*a*b
^10*c^5)/d^2 - (14*a^2*b^9*c^4)/d) - (2*a*b^8*(a + b/x)^(1/2)*(a^3*d^6 - b^3*c^3*d^3 + 3*a*b^2*c^2*d^4 - 3*a^2
*b*c*d^5)^(1/2))/(4*a^3*b^8*d^3 - 14*a^2*b^9*c*d^2 + (82*a^4*b^7*d^4)/c - (110*a^5*b^6*d^5)/c^2 + (40*a^6*b^5*
d^6)/c^3 - 2*a*b^10*c^2*d))*(d^3*(a*d - b*c)^3)^(1/2)*(4*a*d + b*c))/(c^3*d^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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