\(\int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx\) [772]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=-\frac {(3 b c-2 a d) (b c-a d) \sqrt {c+d x}}{a^2 b \sqrt {a+b x}}-\frac {c (c+d x)^{3/2}}{a x \sqrt {a+b x}}+\frac {c^{3/2} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {100, 155, 163, 65, 223, 212, 95, 214} \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\frac {c^{3/2} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{a^{5/2}}-\frac {\sqrt {c+d x} (3 b c-2 a d) (b c-a d)}{a^2 b \sqrt {a+b x}}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {c (c+d x)^{3/2}}{a x \sqrt {a+b x}} \]

[In]

Int[(c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)),x]

[Out]

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

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 155

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.93 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=-\frac {\sqrt {c+d x} \left (3 b^2 c^2 x+2 a^2 d^2 x+a b c (c-4 d x)\right )}{a^2 b x \sqrt {a+b x}}+\frac {c^{3/2} (3 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{a^{5/2}}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{3/2}} \]

[In]

Integrate[(c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(501\) vs. \(2(131)=262\).

Time = 0.59 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.08

method	result	size
default	\(\frac {\sqrt {d x +c}\, \left (2 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b \,d^{3} x^{2} \sqrt {a c}-5 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{2} d \,x^{2} \sqrt {b d}+3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) b^{3} c^{3} x^{2} \sqrt {b d}+2 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}\, a^{3} d^{3} x -5 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a^{2} b \,c^{2} d x +3 \sqrt {b d}\, \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}+2 a c}{x}\right ) a \,b^{2} c^{3} x -4 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2} x +8 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d x -6 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2} x -2 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{2}\right )}{2 a^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, x \sqrt {b d}\, \sqrt {a c}\, \sqrt {b x +a}\, b}\)	\(502\)

[In]

int((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (131) = 262\).

Time = 1.01 (sec) , antiderivative size = 1234, normalized size of antiderivative = 7.57 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{x^{2} \left (a + b x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((d*x+c)**(5/2)/x**2/(b*x+a)**(3/2),x)

[Out]

Integral((c + d*x)**(5/2)/(x**2*(a + b*x)**(3/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{x^2\,{\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int((c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)),x)

[Out]

int((c + d*x)^(5/2)/(x^2*(a + b*x)^(3/2)), x)