\(\int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx\) [124]

Optimal result

Integrand size = 20, antiderivative size = 128 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}-\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {-\frac {a \sqrt {c+d x} \left (b c^2-2 a d^2 x\right )}{b x}-\frac {2 (-b c+a d)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{3/2}}+c^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {109, 27, 171, 27, 174, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 109

\(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (2 b c-5 a d)-d (b c+2 a d) x)}{2 x (a+b x)}dx}{a}-\frac {c (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (2 b c-5 a d)-d (b c+2 a d) x)}{x (a+b x)}dx}{2 a}-\frac {c (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 171

\(\displaystyle -\frac {\frac {2 \int \frac {b (2 b c-5 a d) c^2+d \left (b^2 c^2-6 a b d c+2 a^2 d^2\right ) x}{2 x (a+b x) \sqrt {c+d x}}dx}{b}-\frac {2 d \sqrt {c+d x} (2 a d+b c)}{b}}{2 a}-\frac {c (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {b (2 b c-5 a d) c^2+d \left (b^2 c^2-6 a b d c+2 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}}dx}{b}-\frac {2 d \sqrt {c+d x} (2 a d+b c)}{b}}{2 a}-\frac {c (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 174

\(\displaystyle -\frac {\frac {\frac {b c^2 (2 b c-5 a d) \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {2 (b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{b}-\frac {2 d \sqrt {c+d x} (2 a d+b c)}{b}}{2 a}-\frac {c (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {\frac {2 b c^2 (2 b c-5 a d) \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {4 (b c-a d)^3 \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{b}-\frac {2 d \sqrt {c+d x} (2 a d+b c)}{b}}{2 a}-\frac {c (c+d x)^{3/2}}{a x}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\frac {\frac {4 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {2 b c^{3/2} (2 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}}{b}-\frac {2 d \sqrt {c+d x} (2 a d+b c)}{b}}{2 a}-\frac {c (c+d x)^{3/2}}{a x}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> 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] + Simp[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 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[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] /; Fre 
eQ[{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 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(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 221

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]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {c}\, \left (-a d +b c \right )^{3} x \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\sqrt {\left (a d -b c \right ) b}\, \left (x \left (5 a b \,c^{2} d -2 b^{2} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )+\sqrt {c}\, a \left (-2 x a \,d^{2}+b \,c^{2}\right ) \sqrt {x d +c}\right )}{b \,a^{2} \sqrt {\left (a d -b c \right ) b}\, x \sqrt {c}}\)	\(136\)
derivativedivides	\(2 d^{2} \left (\frac {\sqrt {x d +c}}{b}+\frac {\left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \,a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {c^{2} \left (\frac {a \sqrt {x d +c}}{2 x}+\frac {\left (5 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{a^{2} d^{2}}\right )\)	\(149\)
default	\(2 d^{2} \left (\frac {\sqrt {x d +c}}{b}+\frac {\left (-a^{3} d^{3}+3 a^{2} b c \,d^{2}-3 a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{b \,a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {c^{2} \left (\frac {a \sqrt {x d +c}}{2 x}+\frac {\left (5 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{a^{2} d^{2}}\right )\)	\(149\)
risch	\(-\frac {c^{2} \sqrt {x d +c}}{a x}+\frac {d \left (\frac {2 a d \sqrt {x d +c}}{b}+\frac {\left (-2 a^{3} d^{3}+6 a^{2} b c \,d^{2}-6 a \,b^{2} c^{2} d +2 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{a b d \sqrt {\left (a d -b c \right ) b}}-\frac {c^{\frac {3}{2}} \left (5 a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )}{a d}\right )}{a}\)	\(153\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 632, normalized size of antiderivative = 4.94 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\left [\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{2 \, a^{2} b x}, -\frac {4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{2 \, a^{2} b x}, -\frac {{\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{a^{2} b x}, -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{a^{2} b x}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {2 \, \sqrt {d x + c} d^{2}}{b} - \frac {\sqrt {d x + c} c^{2}}{a x} - \frac {{\left (2 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2} b} \] Input:

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

Output:

2*sqrt(d*x + c)*d^2/b - sqrt(d*x + c)*c^2/(a*x) - (2*b*c^3 - 5*a*c^2*d)*ar 
ctan(sqrt(d*x + c)/sqrt(-c))/(a^2*sqrt(-c)) + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 
 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sq 
rt(-b^2*c + a*b*d)*a^2*b)
 

Mupad [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 1402, normalized size of antiderivative = 10.95 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.04 \[ \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx=\frac {-4 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{2} x +8 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b c d x -4 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} c^{2} x +4 \sqrt {d x +c}\, a^{2} b \,d^{2} x -2 \sqrt {d x +c}\, a \,b^{2} c^{2}+5 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,b^{2} c d x -2 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) b^{3} c^{2} x -5 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,b^{2} c d x +2 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) b^{3} c^{2} x}{2 a^{2} b^{2} x} \] Input:

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

Output:

( - 4*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b 
*c)))*a**2*d**2*x + 8*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt 
(b)*sqrt(a*d - b*c)))*a*b*c*d*x - 4*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + 
 d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**2*c**2*x + 4*sqrt(c + d*x)*a**2*b*d 
**2*x - 2*sqrt(c + d*x)*a*b**2*c**2 + 5*sqrt(c)*log(sqrt(c + d*x) - sqrt(c 
))*a*b**2*c*d*x - 2*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*b**3*c**2*x - 5*s 
qrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*b**2*c*d*x + 2*sqrt(c)*log(sqrt(c + 
d*x) + sqrt(c))*b**3*c**2*x)/(2*a**2*b**2*x)