\(\int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx\) [228]

Optimal result

Integrand size = 17, antiderivative size = 172 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx=-\frac {d \sqrt {c+d x}}{8 b^2 (a+b x)^3}-\frac {d^2 \sqrt {c+d x}}{32 b^2 (b c-a d) (a+b x)^2}+\frac {3 d^3 \sqrt {c+d x}}{64 b^2 (b c-a d)^2 (a+b x)}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}-\frac {3 d^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{64 b^{5/2} (b c-a d)^{5/2}} \] Output:

-1/8*d*(d*x+c)^(1/2)/b^2/(b*x+a)^3-1/32*d^2*(d*x+c)^(1/2)/b^2/(-a*d+b*c)/( 
b*x+a)^2+3/64*d^3*(d*x+c)^(1/2)/b^2/(-a*d+b*c)^2/(b*x+a)-1/4*(d*x+c)^(3/2) 
/b/(b*x+a)^4-3/64*d^4*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5 
/2)/(-a*d+b*c)^(5/2)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx=-\frac {\sqrt {c+d x} \left (3 a^3 d^3+a^2 b d^2 (2 c+11 d x)-a b^2 d \left (24 c^2+44 c d x+11 d^2 x^2\right )+b^3 \left (16 c^3+24 c^2 d x+2 c d^2 x^2-3 d^3 x^3\right )\right )}{64 b^2 (b c-a d)^2 (a+b x)^4}+\frac {3 d^4 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{64 b^{5/2} (-b c+a d)^{5/2}} \] Input:

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

Output:

-1/64*(Sqrt[c + d*x]*(3*a^3*d^3 + a^2*b*d^2*(2*c + 11*d*x) - a*b^2*d*(24*c 
^2 + 44*c*d*x + 11*d^2*x^2) + b^3*(16*c^3 + 24*c^2*d*x + 2*c*d^2*x^2 - 3*d 
^3*x^3)))/(b^2*(b*c - a*d)^2*(a + b*x)^4) + (3*d^4*ArcTan[(Sqrt[b]*Sqrt[c 
+ d*x])/Sqrt[-(b*c) + a*d]])/(64*b^(5/2)*(-(b*c) + a*d)^(5/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {51, 51, 52, 52, 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.

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

\(\displaystyle \frac {3 d \left (\frac {d \int \frac {1}{(a+b x)^3 \sqrt {c+d x}}dx}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {3 d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}}dx}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {3 d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {3 d \left (-\frac {\int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b c-a d}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {3 d \left (\frac {d \left (-\frac {3 d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 (b c-a d)}-\frac {\sqrt {c+d x}}{2 (a+b x)^2 (b c-a d)}\right )}{6 b}-\frac {\sqrt {c+d x}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(c+d x)^{3/2}}{4 b (a+b x)^4}\)

Input:

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

Output:

-1/4*(c + d*x)^(3/2)/(b*(a + b*x)^4) + (3*d*(-1/3*Sqrt[c + d*x]/(b*(a + b* 
x)^3) + (d*(-1/2*Sqrt[c + d*x]/((b*c - a*d)*(a + b*x)^2) - (3*d*(-(Sqrt[c 
+ d*x]/((b*c - a*d)*(a + b*x))) + (d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[ 
b*c - a*d]])/(Sqrt[b]*(b*c - a*d)^(3/2))))/(4*(b*c - a*d))))/(6*b)))/(8*b)
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

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 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 = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {\frac {3 d^{4} \left (b x +a \right )^{4} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{64}-\frac {3 \left (\left (d^{2} x^{2}-\frac {8}{3} c d x -\frac {8}{3} c^{2}\right ) b^{2}+\frac {8 a \left (\frac {7 x d}{4}+c \right ) d b}{3}+a^{2} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, \left (\left (-x d -2 c \right ) b +a d \right ) \sqrt {x d +c}}{64}}{\sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{4} \left (a d -b c \right )^{2} b^{2}}\)	\(145\)
derivativedivides	\(2 d^{4} \left (\frac {\frac {3 b \left (x d +c \right )^{\frac {7}{2}}}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {11 \left (x d +c \right )^{\frac {5}{2}}}{128 \left (a d -b c \right )}-\frac {11 \left (x d +c \right )^{\frac {3}{2}}}{128 b}-\frac {3 \left (a d -b c \right ) \sqrt {x d +c}}{128 b^{2}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{4}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(172\)
default	\(2 d^{4} \left (\frac {\frac {3 b \left (x d +c \right )^{\frac {7}{2}}}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {11 \left (x d +c \right )^{\frac {5}{2}}}{128 \left (a d -b c \right )}-\frac {11 \left (x d +c \right )^{\frac {3}{2}}}{128 b}-\frac {3 \left (a d -b c \right ) \sqrt {x d +c}}{128 b^{2}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{4}}+\frac {3 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{128 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(172\)

Input:

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

Output:

3/64*(d^4*(b*x+a)^4*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-((d^2*x^2- 
8/3*c*d*x-8/3*c^2)*b^2+8/3*a*(7/4*x*d+c)*d*b+a^2*d^2)*((a*d-b*c)*b)^(1/2)* 
((-d*x-2*c)*b+a*d)*(d*x+c)^(1/2))/((a*d-b*c)*b)^(1/2)/(b*x+a)^4/(a*d-b*c)^ 
2/b^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (144) = 288\).

Time = 0.10 (sec) , antiderivative size = 1043, normalized size of antiderivative = 6.06 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/128*(3*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4 
*x + a^4*d^4)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c 
- a*b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(16*b^5*c^4 - 40*a*b^4*c^3*d + 26*a 
^2*b^3*c^2*d^2 + a^3*b^2*c*d^3 - 3*a^4*b*d^4 - 3*(b^5*c*d^3 - a*b^4*d^4)*x 
^3 + (2*b^5*c^2*d^2 - 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 + (24*b^5*c^3*d 
 - 68*a*b^4*c^2*d^2 + 55*a^2*b^3*c*d^3 - 11*a^3*b^2*d^4)*x)*sqrt(d*x + c)) 
/(a^4*b^6*c^3 - 3*a^5*b^5*c^2*d + 3*a^6*b^4*c*d^2 - a^7*b^3*d^3 + (b^10*c^ 
3 - 3*a*b^9*c^2*d + 3*a^2*b^8*c*d^2 - a^3*b^7*d^3)*x^4 + 4*(a*b^9*c^3 - 3* 
a^2*b^8*c^2*d + 3*a^3*b^7*c*d^2 - a^4*b^6*d^3)*x^3 + 6*(a^2*b^8*c^3 - 3*a^ 
3*b^7*c^2*d + 3*a^4*b^6*c*d^2 - a^5*b^5*d^3)*x^2 + 4*(a^3*b^7*c^3 - 3*a^4* 
b^6*c^2*d + 3*a^5*b^5*c*d^2 - a^6*b^4*d^3)*x), 1/64*(3*(b^4*d^4*x^4 + 4*a* 
b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*sqrt(-b^2*c + a 
*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (16*b^5*c 
^4 - 40*a*b^4*c^3*d + 26*a^2*b^3*c^2*d^2 + a^3*b^2*c*d^3 - 3*a^4*b*d^4 - 3 
*(b^5*c*d^3 - a*b^4*d^4)*x^3 + (2*b^5*c^2*d^2 - 13*a*b^4*c*d^3 + 11*a^2*b^ 
3*d^4)*x^2 + (24*b^5*c^3*d - 68*a*b^4*c^2*d^2 + 55*a^2*b^3*c*d^3 - 11*a^3* 
b^2*d^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^3 - 3*a^5*b^5*c^2*d + 3*a^6*b^4*c*d^ 
2 - a^7*b^3*d^3 + (b^10*c^3 - 3*a*b^9*c^2*d + 3*a^2*b^8*c*d^2 - a^3*b^7*d^ 
3)*x^4 + 4*(a*b^9*c^3 - 3*a^2*b^8*c^2*d + 3*a^3*b^7*c*d^2 - a^4*b^6*d^3)*x 
^3 + 6*(a^2*b^8*c^3 - 3*a^3*b^7*c^2*d + 3*a^4*b^6*c*d^2 - a^5*b^5*d^3)*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((d*x+c)^(3/2)/(b*x+a)^5,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 = 285, normalized size of antiderivative = 1.66 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx=\frac {3 \, d^{4} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{64 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {3 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d^{4} - 11 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} c d^{4} - 11 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c^{2} d^{4} + 3 \, \sqrt {d x + c} b^{3} c^{3} d^{4} + 11 \, {\left (d x + c\right )}^{\frac {5}{2}} a b^{2} d^{5} + 22 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} c d^{5} - 9 \, \sqrt {d x + c} a b^{2} c^{2} d^{5} - 11 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b d^{6} + 9 \, \sqrt {d x + c} a^{2} b c d^{6} - 3 \, \sqrt {d x + c} a^{3} d^{7}}{64 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{4}} \] Input:

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

Output:

3/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^2 - 2*a*b^3* 
c*d + a^2*b^2*d^2)*sqrt(-b^2*c + a*b*d)) + 1/64*(3*(d*x + c)^(7/2)*b^3*d^4 
 - 11*(d*x + c)^(5/2)*b^3*c*d^4 - 11*(d*x + c)^(3/2)*b^3*c^2*d^4 + 3*sqrt( 
d*x + c)*b^3*c^3*d^4 + 11*(d*x + c)^(5/2)*a*b^2*d^5 + 22*(d*x + c)^(3/2)*a 
*b^2*c*d^5 - 9*sqrt(d*x + c)*a*b^2*c^2*d^5 - 11*(d*x + c)^(3/2)*a^2*b*d^6 
+ 9*sqrt(d*x + c)*a^2*b*c*d^6 - 3*sqrt(d*x + c)*a^3*d^7)/((b^4*c^2 - 2*a*b 
^3*c*d + a^2*b^2*d^2)*((d*x + c)*b - b*c + a*d)^4)
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.72 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx=\frac {3\,d^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{64\,b^{5/2}\,{\left (a\,d-b\,c\right )}^{5/2}}-\frac {\frac {11\,d^4\,{\left (c+d\,x\right )}^{3/2}}{64\,b}-\frac {11\,d^4\,{\left (c+d\,x\right )}^{5/2}}{64\,\left (a\,d-b\,c\right )}+\frac {3\,d^4\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{64\,b^2}-\frac {3\,b\,d^4\,{\left (c+d\,x\right )}^{7/2}}{64\,{\left (a\,d-b\,c\right )}^2}}{b^4\,{\left (c+d\,x\right )}^4-\left (4\,b^4\,c-4\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^3-\left (c+d\,x\right )\,\left (-4\,a^3\,b\,d^3+12\,a^2\,b^2\,c\,d^2-12\,a\,b^3\,c^2\,d+4\,b^4\,c^3\right )+a^4\,d^4+b^4\,c^4+{\left (c+d\,x\right )}^2\,\left (6\,a^2\,b^2\,d^2-12\,a\,b^3\,c\,d+6\,b^4\,c^2\right )+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d-4\,a^3\,b\,c\,d^3} \] Input:

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

Output:

(3*d^4*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(64*b^(5/2)*(a*d 
 - b*c)^(5/2)) - ((11*d^4*(c + d*x)^(3/2))/(64*b) - (11*d^4*(c + d*x)^(5/2 
))/(64*(a*d - b*c)) + (3*d^4*(a*d - b*c)*(c + d*x)^(1/2))/(64*b^2) - (3*b* 
d^4*(c + d*x)^(7/2))/(64*(a*d - b*c)^2))/(b^4*(c + d*x)^4 - (4*b^4*c - 4*a 
*b^3*d)*(c + d*x)^3 - (c + d*x)*(4*b^4*c^3 - 4*a^3*b*d^3 + 12*a^2*b^2*c*d^ 
2 - 12*a*b^3*c^2*d) + a^4*d^4 + b^4*c^4 + (c + d*x)^2*(6*b^4*c^2 + 6*a^2*b 
^2*d^2 - 12*a*b^3*c*d) + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^3*d - 4*a^3*b*c*d^3 
)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 742, normalized size of antiderivative = 4.31 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^5} \, dx=\frac {3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{4} d^{4}+12 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{3} b \,d^{4} x +18 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b^{2} d^{4} x^{2}+12 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{3} d^{4} x^{3}+3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{4} d^{4} x^{4}-3 \sqrt {d x +c}\, a^{4} b \,d^{4}+\sqrt {d x +c}\, a^{3} b^{2} c \,d^{3}-11 \sqrt {d x +c}\, a^{3} b^{2} d^{4} x +26 \sqrt {d x +c}\, a^{2} b^{3} c^{2} d^{2}+55 \sqrt {d x +c}\, a^{2} b^{3} c \,d^{3} x +11 \sqrt {d x +c}\, a^{2} b^{3} d^{4} x^{2}-40 \sqrt {d x +c}\, a \,b^{4} c^{3} d -68 \sqrt {d x +c}\, a \,b^{4} c^{2} d^{2} x -13 \sqrt {d x +c}\, a \,b^{4} c \,d^{3} x^{2}+3 \sqrt {d x +c}\, a \,b^{4} d^{4} x^{3}+16 \sqrt {d x +c}\, b^{5} c^{4}+24 \sqrt {d x +c}\, b^{5} c^{3} d x +2 \sqrt {d x +c}\, b^{5} c^{2} d^{2} x^{2}-3 \sqrt {d x +c}\, b^{5} c \,d^{3} x^{3}}{64 b^{3} \left (a^{3} b^{4} d^{3} x^{4}-3 a^{2} b^{5} c \,d^{2} x^{4}+3 a \,b^{6} c^{2} d \,x^{4}-b^{7} c^{3} x^{4}+4 a^{4} b^{3} d^{3} x^{3}-12 a^{3} b^{4} c \,d^{2} x^{3}+12 a^{2} b^{5} c^{2} d \,x^{3}-4 a \,b^{6} c^{3} x^{3}+6 a^{5} b^{2} d^{3} x^{2}-18 a^{4} b^{3} c \,d^{2} x^{2}+18 a^{3} b^{4} c^{2} d \,x^{2}-6 a^{2} b^{5} c^{3} x^{2}+4 a^{6} b \,d^{3} x -12 a^{5} b^{2} c \,d^{2} x +12 a^{4} b^{3} c^{2} d x -4 a^{3} b^{4} c^{3} x +a^{7} d^{3}-3 a^{6} b c \,d^{2}+3 a^{5} b^{2} c^{2} d -a^{4} b^{3} c^{3}\right )} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c) 
))*a**4*d**4 + 12*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)* 
sqrt(a*d - b*c)))*a**3*b*d**4*x + 18*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c 
+ d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b**2*d**4*x**2 + 12*sqrt(b)*sqrt 
(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b**3*d**4* 
x**3 + 3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d 
- b*c)))*b**4*d**4*x**4 - 3*sqrt(c + d*x)*a**4*b*d**4 + sqrt(c + d*x)*a**3 
*b**2*c*d**3 - 11*sqrt(c + d*x)*a**3*b**2*d**4*x + 26*sqrt(c + d*x)*a**2*b 
**3*c**2*d**2 + 55*sqrt(c + d*x)*a**2*b**3*c*d**3*x + 11*sqrt(c + d*x)*a** 
2*b**3*d**4*x**2 - 40*sqrt(c + d*x)*a*b**4*c**3*d - 68*sqrt(c + d*x)*a*b** 
4*c**2*d**2*x - 13*sqrt(c + d*x)*a*b**4*c*d**3*x**2 + 3*sqrt(c + d*x)*a*b* 
*4*d**4*x**3 + 16*sqrt(c + d*x)*b**5*c**4 + 24*sqrt(c + d*x)*b**5*c**3*d*x 
 + 2*sqrt(c + d*x)*b**5*c**2*d**2*x**2 - 3*sqrt(c + d*x)*b**5*c*d**3*x**3) 
/(64*b**3*(a**7*d**3 - 3*a**6*b*c*d**2 + 4*a**6*b*d**3*x + 3*a**5*b**2*c** 
2*d - 12*a**5*b**2*c*d**2*x + 6*a**5*b**2*d**3*x**2 - a**4*b**3*c**3 + 12* 
a**4*b**3*c**2*d*x - 18*a**4*b**3*c*d**2*x**2 + 4*a**4*b**3*d**3*x**3 - 4* 
a**3*b**4*c**3*x + 18*a**3*b**4*c**2*d*x**2 - 12*a**3*b**4*c*d**2*x**3 + a 
**3*b**4*d**3*x**4 - 6*a**2*b**5*c**3*x**2 + 12*a**2*b**5*c**2*d*x**3 - 3* 
a**2*b**5*c*d**2*x**4 - 4*a*b**6*c**3*x**3 + 3*a*b**6*c**2*d*x**4 - b**7*c 
**3*x**4))