\(\int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx\) [262]

Optimal result

Integrand size = 17, antiderivative size = 140 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {15 d^2}{4 (b c-a d)^3 \sqrt {c+d x}}-\frac {1}{2 (b c-a d) (a+b x)^2 \sqrt {c+d x}}+\frac {5 d}{4 (b c-a d)^2 (a+b x) \sqrt {c+d x}}-\frac {15 \sqrt {b} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 (b c-a d)^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {1}{4} \left (\frac {8 a^2 d^2+a b d (9 c+25 d x)+b^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )}{(b c-a d)^3 (a+b x)^2 \sqrt {c+d x}}-\frac {15 \sqrt {b} d^2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}}\right ) \] Input:

Integrate[1/((a + b*x)^3*(c + d*x)^(3/2)),x]
 

Output:

((8*a^2*d^2 + a*b*d*(9*c + 25*d*x) + b^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2))/ 
((b*c - a*d)^3*(a + b*x)^2*Sqrt[c + d*x]) - (15*Sqrt[b]*d^2*ArcTan[(Sqrt[b 
]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2))/4
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {52, 52, 61, 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.

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[1/((a + b*x)^3*(c + d*x)^(3/2)),x]
 

Output:

-1/2*1/((b*c - a*d)*(a + b*x)^2*Sqrt[c + d*x]) - (5*d*(-(1/((b*c - a*d)*(a 
 + b*x)*Sqrt[c + d*x])) - (3*d*(2/((b*c - a*d)*Sqrt[c + d*x]) - (2*Sqrt[b] 
*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2)))/(2* 
(b*c - a*d))))/(4*(b*c - a*d))
 

Defintions of rubi rules used

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 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 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 = 0.34 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(2 d^{2} \left (-\frac {1}{\left (a d -b c \right )^{3} \sqrt {x d +c}}-\frac {b \left (\frac {\frac {7 b \left (x d +c \right )^{\frac {3}{2}}}{8}+\left (\frac {9 a d}{8}-\frac {9 b c}{8}\right ) \sqrt {x d +c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\)	\(122\)
default	\(2 d^{2} \left (-\frac {1}{\left (a d -b c \right )^{3} \sqrt {x d +c}}-\frac {b \left (\frac {\frac {7 b \left (x d +c \right )^{\frac {3}{2}}}{8}+\left (\frac {9 a d}{8}-\frac {9 b c}{8}\right ) \sqrt {x d +c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}\right )\)	\(122\)
pseudoelliptic	\(-\frac {15 \left (\sqrt {x d +c}\, b \,d^{2} \left (b x +a \right )^{2} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\frac {8 \sqrt {\left (a d -b c \right ) b}\, \left (\left (\frac {15}{8} d^{2} x^{2}+\frac {5}{8} c d x -\frac {1}{4} c^{2}\right ) b^{2}+\frac {9 a d \left (\frac {25 x d}{9}+c \right ) b}{8}+a^{2} d^{2}\right )}{15}\right )}{4 \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}\, \left (b x +a \right )^{2} \left (a d -b c \right )^{3}}\)	\(137\)

Input:

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

Output:

2*d^2*(-1/(a*d-b*c)^3/(d*x+c)^(1/2)-1/(a*d-b*c)^3*b*((7/8*b*(d*x+c)^(3/2)+ 
(9/8*a*d-9/8*b*c)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)^2+15/8/((a*d-b*c)*b)^ 
(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (116) = 232\).

Time = 0.13 (sec) , antiderivative size = 762, normalized size of antiderivative = 5.44 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\left [-\frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, {\left (b c - a d\right )} \sqrt {d x + c} \sqrt {\frac {b}{b c - a d}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{8 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}, \frac {15 \, {\left (b^{2} d^{3} x^{3} + a^{2} c d^{2} + {\left (b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{2} + {\left (2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\sqrt {d x + c} \sqrt {-\frac {b}{b c - a d}}\right ) + {\left (15 \, b^{2} d^{2} x^{2} - 2 \, b^{2} c^{2} + 9 \, a b c d + 8 \, a^{2} d^{2} + 5 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{4 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}}\right ] \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

[-1/8*(15*(b^2*d^3*x^3 + a^2*c*d^2 + (b^2*c*d^2 + 2*a*b*d^3)*x^2 + (2*a*b* 
c*d^2 + a^2*d^3)*x)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d + 2*(b*c 
- a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) - 2*(15*b^2*d^2*x^2 - 
 2*b^2*c^2 + 9*a*b*c*d + 8*a^2*d^2 + 5*(b^2*c*d + 5*a*b*d^2)*x)*sqrt(d*x + 
 c))/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c 
^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a 
*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a 
*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x) 
, 1/4*(15*(b^2*d^3*x^3 + a^2*c*d^2 + (b^2*c*d^2 + 2*a*b*d^3)*x^2 + (2*a*b* 
c*d^2 + a^2*d^3)*x)*sqrt(-b/(b*c - a*d))*arctan(sqrt(d*x + c)*sqrt(-b/(b*c 
 - a*d))) + (15*b^2*d^2*x^2 - 2*b^2*c^2 + 9*a*b*c*d + 8*a^2*d^2 + 5*(b^2*c 
*d + 5*a*b*d^2)*x)*sqrt(d*x + c))/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b 
*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*d^3 - a^ 
3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c* 
d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^ 
2 + a^4*b*c*d^3 - a^5*d^4)*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)**3/(d*x+c)**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(3/2),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (116) = 232\).

Time = 0.14 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {15 \, b d^{2} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, d^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {d x + c}} + \frac {7 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 9 \, \sqrt {d x + c} b^{2} c d^{2} + 9 \, \sqrt {d x + c} a b d^{3}}{4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

15/4*b*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^ 
2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b^2*c + a*b*d)) + 2*d^2/((b^3*c^3 
 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(d*x + c)) + 1/4*(7*(d*x + 
 c)^(3/2)*b^2*d^2 - 9*sqrt(d*x + c)*b^2*c*d^2 + 9*sqrt(d*x + c)*a*b*d^3)/( 
(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c + a 
*d)^2)
 

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {\frac {2\,d^2}{a\,d-b\,c}+\frac {15\,b^2\,d^2\,{\left (c+d\,x\right )}^2}{4\,{\left (a\,d-b\,c\right )}^3}+\frac {25\,b\,d^2\,\left (c+d\,x\right )}{4\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^{5/2}-\left (2\,b^2\,c-2\,a\,b\,d\right )\,{\left (c+d\,x\right )}^{3/2}+\sqrt {c+d\,x}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {15\,\sqrt {b}\,d^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{{\left (a\,d-b\,c\right )}^{7/2}}\right )}{4\,{\left (a\,d-b\,c\right )}^{7/2}} \] Input:

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

Output:

- ((2*d^2)/(a*d - b*c) + (15*b^2*d^2*(c + d*x)^2)/(4*(a*d - b*c)^3) + (25* 
b*d^2*(c + d*x))/(4*(a*d - b*c)^2))/(b^2*(c + d*x)^(5/2) - (2*b^2*c - 2*a* 
b*d)*(c + d*x)^(3/2) + (c + d*x)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - 
(15*b^(1/2)*d^2*atan((b^(1/2)*(c + d*x)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^2 
*c^2*d - 3*a^2*b*c*d^2))/(a*d - b*c)^(7/2)))/(4*(a*d - b*c)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.12 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {-15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} d^{2}-30 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a b \,d^{2} x -15 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{2} d^{2} x^{2}-8 a^{3} d^{3}-a^{2} b c \,d^{2}-25 a^{2} b \,d^{3} x +11 a \,b^{2} c^{2} d +20 a \,b^{2} c \,d^{2} x -15 a \,b^{2} d^{3} x^{2}-2 b^{3} c^{3}+5 b^{3} c^{2} d x +15 b^{3} c \,d^{2} x^{2}}{4 \sqrt {d x +c}\, \left (a^{4} b^{2} d^{4} x^{2}-4 a^{3} b^{3} c \,d^{3} x^{2}+6 a^{2} b^{4} c^{2} d^{2} x^{2}-4 a \,b^{5} c^{3} d \,x^{2}+b^{6} c^{4} x^{2}+2 a^{5} b \,d^{4} x -8 a^{4} b^{2} c \,d^{3} x +12 a^{3} b^{3} c^{2} d^{2} x -8 a^{2} b^{4} c^{3} d x +2 a \,b^{5} c^{4} x +a^{6} d^{4}-4 a^{5} b c \,d^{3}+6 a^{4} b^{2} c^{2} d^{2}-4 a^{3} b^{3} c^{3} d +a^{2} b^{4} c^{4}\right )} \] Input:

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

Output:

( - 15*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt( 
b)*sqrt(a*d - b*c)))*a**2*d**2 - 30*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)* 
atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b*d**2*x - 15*sqrt(b)* 
sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b 
*c)))*b**2*d**2*x**2 - 8*a**3*d**3 - a**2*b*c*d**2 - 25*a**2*b*d**3*x + 11 
*a*b**2*c**2*d + 20*a*b**2*c*d**2*x - 15*a*b**2*d**3*x**2 - 2*b**3*c**3 + 
5*b**3*c**2*d*x + 15*b**3*c*d**2*x**2)/(4*sqrt(c + d*x)*(a**6*d**4 - 4*a** 
5*b*c*d**3 + 2*a**5*b*d**4*x + 6*a**4*b**2*c**2*d**2 - 8*a**4*b**2*c*d**3* 
x + a**4*b**2*d**4*x**2 - 4*a**3*b**3*c**3*d + 12*a**3*b**3*c**2*d**2*x - 
4*a**3*b**3*c*d**3*x**2 + a**2*b**4*c**4 - 8*a**2*b**4*c**3*d*x + 6*a**2*b 
**4*c**2*d**2*x**2 + 2*a*b**5*c**4*x - 4*a*b**5*c**3*d*x**2 + b**6*c**4*x* 
*2))