\(\int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx\) [335]

Optimal result

Integrand size = 22, antiderivative size = 171 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\frac {d (7 b c-3 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}-\frac {2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2}} \] Output:

1/4*d*(-3*a*d+7*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2+1/2*d*(b*x+a)^(1/2)*( 
d*x+c)^(3/2)/b-2*c^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/ 
2))/a^(1/2)+1/4*d^(1/2)*(3*a^2*d^2-10*a*b*c*d+15*b^2*c^2)*arctanh(d^(1/2)* 
(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\frac {1}{4} \left (\frac {d \sqrt {a+b x} \sqrt {c+d x} (9 b c-3 a d+2 b d x)}{b^2}-\frac {8 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {\sqrt {d} \left (15 b^2 c^2-10 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2}}\right ) \] Input:

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

Output:

((d*Sqrt[a + b*x]*Sqrt[c + d*x]*(9*b*c - 3*a*d + 2*b*d*x))/b^2 - (8*c^(5/2 
)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/Sqrt[a] + (Sqr 
t[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x]) 
/(Sqrt[b]*Sqrt[c + d*x])])/b^(5/2))/4
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {113, 27, 171, 27, 175, 66, 104, 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.

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

\(\Big \downarrow \) 113

\(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left (4 b c^2+d (7 b c-3 a d) x\right )}{2 x \sqrt {a+b x}}dx}{2 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {c+d x} \left (4 b c^2+d (7 b c-3 a d) x\right )}{x \sqrt {a+b x}}dx}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 171

\(\displaystyle \frac {\frac {\int \frac {8 b^2 c^3+d \left (15 b^2 c^2-10 a b d c+3 a^2 d^2\right ) x}{2 x \sqrt {a+b x} \sqrt {c+d x}}dx}{b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {8 b^2 c^3+d \left (15 b^2 c^2-10 a b d c+3 a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {\frac {d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx+8 b^2 c^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {2 d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+8 b^2 c^3 \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\frac {2 d \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}+16 b^2 c^3 \int \frac {1}{\frac {c (a+b x)}{c+d x}-a}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {2 \sqrt {d} \left (3 a^2 d^2-10 a b c d+15 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b}}-\frac {16 b^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}}{2 b}+\frac {d \sqrt {a+b x} \sqrt {c+d x} (7 b c-3 a d)}{b}}{4 b}+\frac {d \sqrt {a+b x} (c+d x)^{3/2}}{2 b}\)

Input:

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

Output:

(d*Sqrt[a + b*x]*(c + d*x)^(3/2))/(2*b) + ((d*(7*b*c - 3*a*d)*Sqrt[a + b*x 
]*Sqrt[c + d*x])/b + ((-16*b^2*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sq 
rt[a]*Sqrt[c + d*x])])/Sqrt[a] + (2*Sqrt[d]*(15*b^2*c^2 - 10*a*b*c*d + 3*a 
^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/Sqrt[b]) 
/(2*b))/(4*b)
 

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 66

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

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 113

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
 

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 175

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

Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(133)=266\).

Time = 0.25 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.00

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (-4 b \,d^{2} x \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}+8 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) b^{2} c^{3} \sqrt {d b}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} d^{3} \sqrt {a c}+10 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a b c \,d^{2} \sqrt {a c}-15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{2} c^{2} d \sqrt {a c}+6 a \,d^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}-18 b c d \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}\, \sqrt {a c}\right )}{8 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} \sqrt {d b}\, \sqrt {a c}}\)	\(342\)

Input:

int((d*x+c)^(5/2)/x/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-4*b*d^2*x*((b*x+a)*(d*x+c))^(1/2)*(d*b) 
^(1/2)*(a*c)^(1/2)+8*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2) 
+2*a*c)/x)*b^2*c^3*(d*b)^(1/2)-3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2) 
*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*d^3*(a*c)^(1/2)+10*ln(1/2*(2*b*d*x+ 
2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b*c*d^2*(a*c 
)^(1/2)-15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/ 
(d*b)^(1/2))*b^2*c^2*d*(a*c)^(1/2)+6*a*d^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^( 
1/2)*(a*c)^(1/2)-18*b*c*d*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*(a*c)^(1/2)) 
/((b*x+a)*(d*x+c))^(1/2)/b^2/(d*b)^(1/2)/(a*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 987, normalized size of antiderivative = 5.77 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^(5/2)/x/(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/16*(8*b^2*c^2*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) + (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2 
)*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) + 4*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c) 
)/b^2, 1/8*(4*b^2*c^2*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)*s 
qrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^ 
2*d^2)*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*(2*b*d^2*x + 
9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/b^2, 1/16*(16*b^2*c^2*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)) + (15*b^2*c^2 - 10*a*b*c*d 
 + 3*a^2*d^2)*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) + 4*(2*b*d^2*x + 9*b*c*d - 3*a*d^2)*sqrt(b*x + a)*s 
qrt(d*x + c))/b^2, 1/8*(8*b^2*c^2*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)) - (15*b^2*c^2 - 10*a*b*c*d + 3*a^2*d^2)*sqrt(-d/b)*arctan(1...
 

Sympy [F]

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

integrate((d*x+c)**(5/2)/x/(b*x+a)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)^(5/2)/x/(b*x+a)^(1/2),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 [F(-2)]

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

integrate((d*x+c)^(5/2)/x/(b*x+a)^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

int((c + d*x)^(5/2)/(x*(a + b*x)^(1/2)),x)
 

Output:

int((c + d*x)^(5/2)/(x*(a + b*x)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.07 \[ \int \frac {(c+d x)^{5/2}}{x \sqrt {a+b x}} \, dx=\frac {-3 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{2}+9 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c d +2 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{2} x +4 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b^{3} c^{2}+4 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b^{3} c^{2}-4 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {b}\, \sqrt {d x +c}\, \sqrt {b x +a}+2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+2 b d x \right ) b^{3} c^{2}+3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{2}-10 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c d +15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2}}{4 a \,b^{3}} \] Input:

int((d*x+c)^(5/2)/x/(b*x+a)^(1/2),x)
 

Output:

( - 3*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**2 + 9*sqrt(c + d*x)*sqrt(a + b 
*x)*a*b**2*c*d + 2*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**2*x + 4*sqrt(c)*s 
qrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d 
)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b**3*c**2 + 4*sqrt(c)*sqrt(a)*log 
(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b* 
x) + sqrt(b)*sqrt(c + d*x))*b**3*c**2 - 4*sqrt(c)*sqrt(a)*log(2*sqrt(d)*sq 
rt(b)*sqrt(c + d*x)*sqrt(a + b*x) + 2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2* 
b*d*x)*b**3*c**2 + 3*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)* 
sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*d**2 - 10*sqrt(d)*sqrt(b)*log((sqrt(d 
)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b*c*d + 15* 
sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a 
*d - b*c))*a*b**2*c**2)/(4*a*b**3)