3.3.36 \(\int \frac {(a+\frac {b}{x})^{3/2}}{(c+\frac {d}{x})^2} \, dx\) [236]

Optimal. Leaf size=156 \[ -\frac {(b c-2 a d) \sqrt {a+\frac {b}{x}}}{c^2 \left (c+\frac {d}{x}\right )}+\frac {a \sqrt {a+\frac {b}{x}} x}{c \left (c+\frac {d}{x}\right )}-\frac {(b c-4 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^3 \sqrt {d}}+\frac {\sqrt {a} (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^3} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 156, 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 = {382, 100, 156, 162, 65, 214, 211} \begin {gather*} -\frac {(b c-4 a d) \sqrt {b c-a d} \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^3 \sqrt {d}}+\frac {\sqrt {a} (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^3}-\frac {\sqrt {a+\frac {b}{x}} (b c-2 a d)}{c^2 \left (c+\frac {d}{x}\right )}+\frac {a x \sqrt {a+\frac {b}{x}}}{c \left (c+\frac {d}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)/(c + d/x)^2,x]

[Out]

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

Rule 65

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 100

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 156

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] && ILtQ[m, -1]

Rule 162

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 211

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

Rule 214

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

Rule 382

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 )^{3/2}}{\left (c+\frac {d}{x}\right )^2} \, dx &=-\text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2 (c+d x)^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {a \sqrt {a+\frac {b}{x}} x}{c \left (c+\frac {d}{x}\right )}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} a (3 b c-4 a d)-\frac {1}{2} b (2 b c-3 a d) x}{x \sqrt {a+b x} (c+d x)^2} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {(b c-2 a d) \sqrt {a+\frac {b}{x}}}{c^2 \left (c+\frac {d}{x}\right )}+\frac {a \sqrt {a+\frac {b}{x}} x}{c \left (c+\frac {d}{x}\right )}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (3 b c-4 a d) (b c-a d)+\frac {1}{2} b (b c-2 a d) (b c-a d) x}{x \sqrt {a+b x} (c+d x)} \, dx,x,\frac {1}{x}\right )}{c^2 (b c-a d)}\\ &=-\frac {(b c-2 a d) \sqrt {a+\frac {b}{x}}}{c^2 \left (c+\frac {d}{x}\right )}+\frac {a \sqrt {a+\frac {b}{x}} x}{c \left (c+\frac {d}{x}\right )}-\frac {(a (3 b c-4 a d)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{2 c^3}-\frac {((b c-4 a d) (b c-a d)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} (c+d x)} \, dx,x,\frac {1}{x}\right )}{2 c^3}\\ &=-\frac {(b c-2 a d) \sqrt {a+\frac {b}{x}}}{c^2 \left (c+\frac {d}{x}\right )}+\frac {a \sqrt {a+\frac {b}{x}} x}{c \left (c+\frac {d}{x}\right )}-\frac {(a (3 b c-4 a d)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{b c^3}-\frac {((b c-4 a d) (b c-a d)) \text {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}\\ &=-\frac {(b c-2 a d) \sqrt {a+\frac {b}{x}}}{c^2 \left (c+\frac {d}{x}\right )}+\frac {a \sqrt {a+\frac {b}{x}} x}{c \left (c+\frac {d}{x}\right )}-\frac {(b c-4 a d) \sqrt {b c-a d} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{c^3 \sqrt {d}}+\frac {\sqrt {a} (3 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 = 0.36, size = 144, normalized size = 0.92 \begin {gather*} \frac {\frac {c \sqrt {a+\frac {b}{x}} x (-b c+2 a d+a c x)}{d+c x}-\frac {\left (b^2 c^2-5 a b c d+4 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {a+\frac {b}{x}}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \sqrt {b c-a d}}-\sqrt {a} (-3 b c+4 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(835\) vs. \(2(136)=272\).
time = 0.08, size = 836, normalized size = 5.36

method	result	size
default	\(\frac {\left (-2 \sqrt {x \left (a x +b \right )}\, a^{\frac {5}{2}} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c^{4} x^{2}-4 a^{\frac {7}{2}} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c -2 a d x +b c x -b d}{c x +d}\right ) c \,d^{3} x +2 \sqrt {x \left (a x +b \right )}\, a^{\frac {5}{2}} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c^{3} d x -4 a^{\frac {7}{2}} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c -2 a d x +b c x -b d}{c x +d}\right ) d^{4}+5 a^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c -2 a d x +b c x -b d}{c x +d}\right ) b \,c^{2} d^{2} x +2 c^{4} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} a^{\frac {3}{2}} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}+4 \sqrt {x \left (a x +b \right )}\, a^{\frac {5}{2}} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c^{2} d^{2}-2 \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, b \,c^{4} x +5 a^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c -2 a d x +b c x -b d}{c x +d}\right ) b c \,d^{3}-a^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c -2 a d x +b c x -b d}{c x +d}\right ) b^{2} c^{3} d x -2 \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, b \,c^{3} d -4 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c^{2} d^{2} x +3 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, b \,c^{3} d x -a^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c -2 a d x +b c x -b d}{c x +d}\right ) b^{2} c^{2} d^{2}-4 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, c \,d^{3}+3 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, b \,c^{2} d^{2}\right ) x \sqrt {\frac {a x +b}{x}}}{2 c^{4} \sqrt {\frac {d \left (a d -b c \right )}{c^{2}}}\, a^{\frac {3}{2}} \left (c x +d \right ) d \sqrt {x \left (a x +b \right )}}\)	\(836\)
risch	\(\text {Expression too large to display}\)	\(1336\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+1/x*b)^(3/2)/(c+d/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [A]
time = 3.27, size = 769, normalized size = 4.93 \begin {gather*} \left [-\frac {{\left (3 \, b c d - 4 \, a d^{2} + {\left (3 \, b c^{2} - 4 \, a c d\right )} x\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + {\left (b c d - 4 \, a d^{2} + {\left (b c^{2} - 4 \, a c d\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 c^{2} x^{2} - {\left (b c^{2} - 2 \, a c d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (c^{4} x + c^{3} d\right )}}, \frac {2 \, {\left (b c d - 4 \, a d^{2} + {\left (b c^{2} - 4 \, a c d\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 (3 \, b c d - 4 \, a d^{2} + {\left (3 \, b c^{2} - 4 \, a c d\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 c^{2} x^{2} - {\left (b c^{2} - 2 \, a c d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (c^{4} x + c^{3} d\right )}}, -\frac {2 \, {\left (3 \, b c d - 4 \, a d^{2} + {\left (3 \, b c^{2} - 4 \, a c d\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (b c d - 4 \, a d^{2} + {\left (b c^{2} - 4 \, a c d\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 c^{2} x^{2} - {\left (b c^{2} - 2 \, a c d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{2 \, {\left (c^{4} x + c^{3} d\right )}}, \frac {{\left (b c d - 4 \, a d^{2} + {\left (b c^{2} - 4 \, a c d\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 (3 \, b c d - 4 \, a d^{2} + {\left (3 \, b c^{2} - 4 \, a c d\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (a c^{2} x^{2} - {\left (b c^{2} - 2 \, a c d\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{c^{4} x + c^{3} d}\right ] \end {gather*}

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]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (a + \frac {b}{x}\right )^{\frac {3}{2}}}{\left (c x + d\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 2.16, size = 448, normalized size = 2.87 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {8\,a^2\,b^5\,d^2\,\sqrt {a+\frac {b}{x}}\,\sqrt {a\,d^2-b\,c\,d}}{8\,a^3\,b^5\,d^3-10\,a^2\,b^6\,c\,d^2+2\,a\,b^7\,c^2\,d}-\frac {2\,a\,b^6\,d\,\sqrt {a+\frac {b}{x}}\,\sqrt {a\,d^2-b\,c\,d}}{2\,a\,b^7\,c\,d-10\,a^2\,b^6\,d^2+\frac {8\,a^3\,b^5\,d^3}{c}}\right )\,\sqrt {d\,\left (a\,d-b\,c\right )}\,\left (4\,a\,d-b\,c\right )}{c^3\,d}-\frac {\sqrt {a}\,\mathrm {atanh}\left (\frac {6\,\sqrt {a}\,b^7\,d\,\sqrt {a+\frac {b}{x}}}{6\,a\,b^7\,d-\frac {14\,a^2\,b^6\,d^2}{c}+\frac {8\,a^3\,b^5\,d^3}{c^2}}-\frac {14\,a^{3/2}\,b^6\,d^2\,\sqrt {a+\frac {b}{x}}}{6\,a\,b^7\,c\,d-14\,a^2\,b^6\,d^2+\frac {8\,a^3\,b^5\,d^3}{c}}+\frac {8\,a^{5/2}\,b^5\,d^3\,\sqrt {a+\frac {b}{x}}}{8\,a^3\,b^5\,d^3-14\,a^2\,b^6\,c\,d^2+6\,a\,b^7\,c^2\,d}\right )\,\left (4\,a\,d-3\,b\,c\right )}{c^3}-\frac {\frac {2\,\left (a\,b^2\,c-a^2\,b\,d\right )\,\sqrt {a+\frac {b}{x}}}{c^2}+\frac {b\,{\left (a+\frac {b}{x}\right )}^{3/2}\,\left (2\,a\,d-b\,c\right )}{c^2}}{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(3/2)/(c + d/x)^2,x)

[Out]

(atanh((8*a^2*b^5*d^2*(a + b/x)^(1/2)*(a*d^2 - b*c*d)^(1/2))/(8*a^3*b^5*d^3 - 10*a^2*b^6*c*d^2 + 2*a*b^7*c^2*d
) - (2*a*b^6*d*(a + b/x)^(1/2)*(a*d^2 - b*c*d)^(1/2))/(2*a*b^7*c*d - 10*a^2*b^6*d^2 + (8*a^3*b^5*d^3)/c))*(d*(
a*d - b*c))^(1/2)*(4*a*d - b*c))/(c^3*d) - (a^(1/2)*atanh((6*a^(1/2)*b^7*d*(a + b/x)^(1/2))/(6*a*b^7*d - (14*a
^2*b^6*d^2)/c + (8*a^3*b^5*d^3)/c^2) - (14*a^(3/2)*b^6*d^2*(a + b/x)^(1/2))/(6*a*b^7*c*d - 14*a^2*b^6*d^2 + (8
*a^3*b^5*d^3)/c) + (8*a^(5/2)*b^5*d^3*(a + b/x)^(1/2))/(8*a^3*b^5*d^3 - 14*a^2*b^6*c*d^2 + 6*a*b^7*c^2*d))*(4*
a*d - 3*b*c))/c^3 - ((2*(a*b^2*c - a^2*b*d)*(a + b/x)^(1/2))/c^2 + (b*(a + b/x)^(3/2)*(2*a*d - b*c))/c^2)/((a
+ b/x)*(2*a*d - b*c) - d*(a + b/x)^2 - a^2*d + a*b*c)

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