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

Optimal. Leaf size=134 \[ \frac {a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^2}+\frac {2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^2 d^{3/2}}-\frac {b \sqrt {a+\frac {b}{x}} (a d+2 b c)}{c d}+\frac {a x \left (a+\frac {b}{x}\right )^{3/2}}{c} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 134, 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, 154, 156, 63, 208, 205} \[ \frac {a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^2}+\frac {2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^2 d^{3/2}}-\frac {b \sqrt {a+\frac {b}{x}} (a d+2 b c)}{c d}+\frac {a x \left (a+\frac {b}{x}\right )^{3/2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/(c + d/x),x]

[Out]

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

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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && 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 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}}{c+\frac {d}{x}} \, dx &=-\operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^2 (c+d x)} \, dx,x,\frac {1}{x}\right )\\ &=\frac {a \left (a+\frac {b}{x}\right )^{3/2} x}{c}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x} \left (-\frac {1}{2} a (5 b c-2 a d)-\frac {1}{2} b (2 b c+a d) x\right )}{x (c+d x)} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {b (2 b c+a d) \sqrt {a+\frac {b}{x}}}{c d}+\frac {a \left (a+\frac {b}{x}\right )^{3/2} x}{c}+\frac {2 \operatorname {Subst}\left (\int \frac {-\frac {1}{4} a^2 d (5 b c-2 a d)+\frac {1}{4} b \left (2 b^2 c^2-6 a b c d+a^2 d^2\right ) x}{x \sqrt {a+b x} (c+d x)} \, dx,x,\frac {1}{x}\right )}{c d}\\ &=-\frac {b (2 b c+a d) \sqrt {a+\frac {b}{x}}}{c d}+\frac {a \left (a+\frac {b}{x}\right )^{3/2} x}{c}-\frac {\left (a^2 (5 b c-2 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{2 c^2}+\frac {(b c-a d)^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} (c+d x)} \, dx,x,\frac {1}{x}\right )}{c^2 d}\\ &=-\frac {b (2 b c+a d) \sqrt {a+\frac {b}{x}}}{c d}+\frac {a \left (a+\frac {b}{x}\right )^{3/2} x}{c}-\frac {\left (a^2 (5 b c-2 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^2}+\frac {\left (2 (b c-a d)^3\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 d}\\ &=-\frac {b (2 b c+a d) \sqrt {a+\frac {b}{x}}}{c d}+\frac {a \left (a+\frac {b}{x}\right )^{3/2} x}{c}+\frac {2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^2 d^{3/2}}+\frac {a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^2}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 116, normalized size = 0.87 \[ \frac {a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )+\frac {c \sqrt {a+\frac {b}{x}} \left (a^2 d x-2 b^2 c\right )}{d}+\frac {2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{d^{3/2}}}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.31, size = 659, normalized size = 4.92 \[ \left [-\frac {{\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{d}} \log \left (\frac {2 \, d x \sqrt {-\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}} + b d - {\left (b c - 2 \, a d\right )} x}{c x + d}\right ) - 2 \, {\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt {\frac {a x + b}{x}}}{2 \, c^{2} d}, -\frac {{\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{d}} \log \left (\frac {2 \, d x \sqrt {-\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}} + b d - {\left (b c - 2 \, a d\right )} x}{c x + d}\right ) - {\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt {\frac {a x + b}{x}}}{c^{2} d}, -\frac {4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{d}} \arctan \left (-\frac {d \sqrt {\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}}}{b c - a d}\right ) + {\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt {\frac {a x + b}{x}}}{2 \, c^{2} d}, -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{d}} \arctan \left (-\frac {d \sqrt {\frac {b c - a d}{d}} \sqrt {\frac {a x + b}{x}}}{b c - a d}\right ) + {\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt {\frac {a x + b}{x}}}{c^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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),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)]Error: Bad Argument Type

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maple [B]  time = 0.06, size = 859, normalized size = 6.41 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (2 a^{\frac {7}{2}} d^{4} x^{2} \ln \left (\frac {-2 a d x +b c x -b d +2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, c}{c x +d}\right )-6 a^{\frac {5}{2}} b c \,d^{3} x^{2} \ln \left (\frac {-2 a d x +b c x -b d +2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, c}{c x +d}\right )+6 a^{\frac {3}{2}} b^{2} c^{2} d^{2} x^{2} \ln \left (\frac {-2 a d x +b c x -b d +2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, c}{c x +d}\right )-2 \sqrt {a}\, b^{3} c^{3} d \,x^{2} \ln \left (\frac {-2 a d x +b c x -b d +2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, c}{c x +d}\right )+2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, a^{3} c \,d^{3} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-5 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, a^{2} b \,c^{2} d^{2} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, a \,b^{2} c^{3} d \,x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-4 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, a \,b^{2} c^{3} d \,x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-\sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, b^{3} c^{4} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+\sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, b^{3} c^{4} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} c^{2} d^{2} x^{2}-8 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b \,c^{3} d \,x^{2}+4 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b \,c^{3} d \,x^{2}+2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {a \,x^{2}+b x}\, \sqrt {a}\, b^{2} c^{4} x^{2}-2 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {\left (a x +b \right ) x}\, \sqrt {a}\, b^{2} c^{4} x^{2}+4 \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b \,c^{3} d \right )}{2 \sqrt {\left (a x +b \right ) x}\, \sqrt {\frac {\left (a d -b c \right ) d}{c^{2}}}\, \sqrt {a}\, c^{3} d^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*x+b)/x)^(1/2)/x*(2*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^2*
a^3*c*d^3-5*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^2*a^2*b*c^2*d^2+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^2*a*b^2*c^3*d-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^2*b^3*c^4-8*((a*d-b*c)/c^2*d)^(1/2)*(a*x^2+b*x
)^(1/2)*a^(3/2)*x^2*b*c^3*d+2*((a*d-b*c)/c^2*d)^(1/2)*(a*x^2+b*x)^(1/2)*a^(1/2)*x^2*b^2*c^4-4*((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^2*a*b^2*c^3*d+((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^2*b^3*c^4-2*((a*d-b*c)/c^2*d)^(1/2)*a^(5/2)*((a*x+b)*x)^(1/2)*
x^2*c^2*d^2+4*((a*d-b*c)/c^2*d)^(1/2)*a^(3/2)*((a*x+b)*x)^(1/2)*x^2*b*c^3*d-2*((a*d-b*c)/c^2*d)^(1/2)*a^(1/2)*
((a*x+b)*x)^(1/2)*x^2*b^2*c^4+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))
*a^(7/2)*x^2*d^4-6*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))*a^(5/2)*x^2*
b*c*d^3+6*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))*a^(3/2)*x^2*b^2*c^2*d
^2-2*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))*a^(1/2)*x^2*b^3*c^3*d+4*((
a*d-b*c)/c^2*d)^(1/2)*(a*x^2+b*x)^(3/2)*a^(1/2)*b*c^3*d)/((a*x+b)*x)^(1/2)/d^2/c^3/a^(1/2)/((a*d-b*c)/c^2*d)^(
1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.16, size = 1427, normalized size = 10.65 \[ \frac {a^2\,b\,d\,\sqrt {a+\frac {b}{x}}}{c\,\left (d\,\left (a+\frac {b}{x}\right )-a\,d\right )}-\frac {2\,b^2\,\sqrt {a+\frac {b}{x}}}{d}+\frac {\mathrm {atan}\left (\frac {a^3\,b^5\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^5\,d^8-5\,a^4\,b\,c\,d^7+10\,a^3\,b^2\,c^2\,d^6-10\,a^2\,b^3\,c^3\,d^5+5\,a\,b^4\,c^4\,d^4-b^5\,c^5\,d^3}\,160{}\mathrm {i}}{448\,a^3\,b^8\,c^3\,d-340\,a^6\,b^5\,d^4-128\,a^2\,b^9\,c^4+740\,a^5\,b^6\,c\,d^3+\frac {16\,a\,b^{10}\,c^5}{d}-796\,a^4\,b^7\,c^2\,d^2+\frac {60\,a^7\,b^4\,d^5}{c}}-\frac {a^2\,b^6\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^5\,d^8-5\,a^4\,b\,c\,d^7+10\,a^3\,b^2\,c^2\,d^6-10\,a^2\,b^3\,c^3\,d^5+5\,a\,b^4\,c^4\,d^4-b^5\,c^5\,d^3}\,80{}\mathrm {i}}{16\,a\,b^{10}\,c^4+740\,a^5\,b^6\,d^4-128\,a^2\,b^9\,c^3\,d-796\,a^4\,b^7\,c\,d^3+448\,a^3\,b^8\,c^2\,d^2-\frac {340\,a^6\,b^5\,d^5}{c}+\frac {60\,a^7\,b^4\,d^6}{c^2}}-\frac {a^4\,b^4\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^5\,d^8-5\,a^4\,b\,c\,d^7+10\,a^3\,b^2\,c^2\,d^6-10\,a^2\,b^3\,c^3\,d^5+5\,a\,b^4\,c^4\,d^4-b^5\,c^5\,d^3}\,60{}\mathrm {i}}{448\,a^3\,b^8\,c^4+60\,a^7\,b^4\,d^4-796\,a^4\,b^7\,c^3\,d-340\,a^6\,b^5\,c\,d^3+\frac {16\,a\,b^{10}\,c^6}{d^2}+740\,a^5\,b^6\,c^2\,d^2-\frac {128\,a^2\,b^9\,c^5}{d}}+\frac {a\,b^7\,c\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^5\,d^8-5\,a^4\,b\,c\,d^7+10\,a^3\,b^2\,c^2\,d^6-10\,a^2\,b^3\,c^3\,d^5+5\,a\,b^4\,c^4\,d^4-b^5\,c^5\,d^3}\,16{}\mathrm {i}}{740\,a^5\,b^6\,d^5-796\,a^4\,b^7\,c\,d^4-128\,a^2\,b^9\,c^3\,d^2+448\,a^3\,b^8\,c^2\,d^3-\frac {340\,a^6\,b^5\,d^6}{c}+\frac {60\,a^7\,b^4\,d^7}{c^2}+16\,a\,b^{10}\,c^4\,d}\right )\,\sqrt {d^3\,{\left (a\,d-b\,c\right )}^5}\,2{}\mathrm {i}}{c^2\,d^3}+\frac {\mathrm {atan}\left (\frac {b^9\,c^3\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}\,40{}\mathrm {i}}{40\,a^2\,b^9\,c^3-790\,a^5\,b^6\,d^3-256\,a^3\,b^8\,c^2\,d+696\,a^4\,b^7\,c\,d^2+\frac {370\,a^6\,b^5\,d^4}{c}-\frac {60\,a^7\,b^4\,d^5}{c^2}}+\frac {a\,b^8\,c^2\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}\,256{}\mathrm {i}}{256\,a^3\,b^8\,c^2+790\,a^5\,b^6\,d^2-\frac {40\,a^2\,b^9\,c^3}{d}-\frac {370\,a^6\,b^5\,d^3}{c}+\frac {60\,a^7\,b^4\,d^4}{c^2}-696\,a^4\,b^7\,c\,d}+\frac {a^3\,b^6\,d^2\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}\,790{}\mathrm {i}}{256\,a^3\,b^8\,c^2+790\,a^5\,b^6\,d^2-\frac {40\,a^2\,b^9\,c^3}{d}-\frac {370\,a^6\,b^5\,d^3}{c}+\frac {60\,a^7\,b^4\,d^4}{c^2}-696\,a^4\,b^7\,c\,d}-\frac {a^4\,b^5\,d^3\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}\,370{}\mathrm {i}}{256\,a^3\,b^8\,c^3-370\,a^6\,b^5\,d^3-696\,a^4\,b^7\,c^2\,d+790\,a^5\,b^6\,c\,d^2-\frac {40\,a^2\,b^9\,c^4}{d}+\frac {60\,a^7\,b^4\,d^4}{c}}+\frac {a^5\,b^4\,d^4\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}\,60{}\mathrm {i}}{256\,a^3\,b^8\,c^4+60\,a^7\,b^4\,d^4-696\,a^4\,b^7\,c^3\,d-370\,a^6\,b^5\,c\,d^3+790\,a^5\,b^6\,c^2\,d^2-\frac {40\,a^2\,b^9\,c^5}{d}}-\frac {a^2\,b^7\,c\,d\,\sqrt {a+\frac {b}{x}}\,\sqrt {a^3}\,696{}\mathrm {i}}{256\,a^3\,b^8\,c^2+790\,a^5\,b^6\,d^2-\frac {40\,a^2\,b^9\,c^3}{d}-\frac {370\,a^6\,b^5\,d^3}{c}+\frac {60\,a^7\,b^4\,d^4}{c^2}-696\,a^4\,b^7\,c\,d}\right )\,\left (2\,a\,d-5\,b\,c\right )\,\sqrt {a^3}\,1{}\mathrm {i}}{c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(5/2)/(c + d/x),x)

[Out]

(atan((a^3*b^5*(a + b/x)^(1/2)*(a^5*d^8 - b^5*c^5*d^3 + 5*a*b^4*c^4*d^4 - 10*a^2*b^3*c^3*d^5 + 10*a^3*b^2*c^2*
d^6 - 5*a^4*b*c*d^7)^(1/2)*160i)/(448*a^3*b^8*c^3*d - 340*a^6*b^5*d^4 - 128*a^2*b^9*c^4 + 740*a^5*b^6*c*d^3 +
(16*a*b^10*c^5)/d - 796*a^4*b^7*c^2*d^2 + (60*a^7*b^4*d^5)/c) - (a^2*b^6*(a + b/x)^(1/2)*(a^5*d^8 - b^5*c^5*d^
3 + 5*a*b^4*c^4*d^4 - 10*a^2*b^3*c^3*d^5 + 10*a^3*b^2*c^2*d^6 - 5*a^4*b*c*d^7)^(1/2)*80i)/(16*a*b^10*c^4 + 740
*a^5*b^6*d^4 - 128*a^2*b^9*c^3*d - 796*a^4*b^7*c*d^3 + 448*a^3*b^8*c^2*d^2 - (340*a^6*b^5*d^5)/c + (60*a^7*b^4
*d^6)/c^2) - (a^4*b^4*(a + b/x)^(1/2)*(a^5*d^8 - b^5*c^5*d^3 + 5*a*b^4*c^4*d^4 - 10*a^2*b^3*c^3*d^5 + 10*a^3*b
^2*c^2*d^6 - 5*a^4*b*c*d^7)^(1/2)*60i)/(448*a^3*b^8*c^4 + 60*a^7*b^4*d^4 - 796*a^4*b^7*c^3*d - 340*a^6*b^5*c*d
^3 + (16*a*b^10*c^6)/d^2 + 740*a^5*b^6*c^2*d^2 - (128*a^2*b^9*c^5)/d) + (a*b^7*c*(a + b/x)^(1/2)*(a^5*d^8 - b^
5*c^5*d^3 + 5*a*b^4*c^4*d^4 - 10*a^2*b^3*c^3*d^5 + 10*a^3*b^2*c^2*d^6 - 5*a^4*b*c*d^7)^(1/2)*16i)/(740*a^5*b^6
*d^5 - 796*a^4*b^7*c*d^4 - 128*a^2*b^9*c^3*d^2 + 448*a^3*b^8*c^2*d^3 - (340*a^6*b^5*d^6)/c + (60*a^7*b^4*d^7)/
c^2 + 16*a*b^10*c^4*d))*(d^3*(a*d - b*c)^5)^(1/2)*2i)/(c^2*d^3) - (2*b^2*(a + b/x)^(1/2))/d + (atan((b^9*c^3*(
a + b/x)^(1/2)*(a^3)^(1/2)*40i)/(40*a^2*b^9*c^3 - 790*a^5*b^6*d^3 - 256*a^3*b^8*c^2*d + 696*a^4*b^7*c*d^2 + (3
70*a^6*b^5*d^4)/c - (60*a^7*b^4*d^5)/c^2) + (a*b^8*c^2*(a + b/x)^(1/2)*(a^3)^(1/2)*256i)/(256*a^3*b^8*c^2 + 79
0*a^5*b^6*d^2 - (40*a^2*b^9*c^3)/d - (370*a^6*b^5*d^3)/c + (60*a^7*b^4*d^4)/c^2 - 696*a^4*b^7*c*d) + (a^3*b^6*
d^2*(a + b/x)^(1/2)*(a^3)^(1/2)*790i)/(256*a^3*b^8*c^2 + 790*a^5*b^6*d^2 - (40*a^2*b^9*c^3)/d - (370*a^6*b^5*d
^3)/c + (60*a^7*b^4*d^4)/c^2 - 696*a^4*b^7*c*d) - (a^4*b^5*d^3*(a + b/x)^(1/2)*(a^3)^(1/2)*370i)/(256*a^3*b^8*
c^3 - 370*a^6*b^5*d^3 - 696*a^4*b^7*c^2*d + 790*a^5*b^6*c*d^2 - (40*a^2*b^9*c^4)/d + (60*a^7*b^4*d^4)/c) + (a^
5*b^4*d^4*(a + b/x)^(1/2)*(a^3)^(1/2)*60i)/(256*a^3*b^8*c^4 + 60*a^7*b^4*d^4 - 696*a^4*b^7*c^3*d - 370*a^6*b^5
*c*d^3 + 790*a^5*b^6*c^2*d^2 - (40*a^2*b^9*c^5)/d) - (a^2*b^7*c*d*(a + b/x)^(1/2)*(a^3)^(1/2)*696i)/(256*a^3*b
^8*c^2 + 790*a^5*b^6*d^2 - (40*a^2*b^9*c^3)/d - (370*a^6*b^5*d^3)/c + (60*a^7*b^4*d^4)/c^2 - 696*a^4*b^7*c*d))
*(2*a*d - 5*b*c)*(a^3)^(1/2)*1i)/c^2 + (a^2*b*d*(a + b/x)^(1/2))/(c*(d*(a + b/x) - a*d))

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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),x)

[Out]

Timed out

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