\(\int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx\) [169]

Optimal result

Integrand size = 22, antiderivative size = 180 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx=-\frac {2 e (B d-A e)}{(b d-a e)^3 \sqrt {d+e x}}-\frac {(A b-a B) \sqrt {d+e x}}{2 (b d-a e)^2 (a+b x)^2}-\frac {(4 b B d-7 A b e+3 a B e) \sqrt {d+e x}}{4 (b d-a e)^3 (a+b x)}+\frac {3 e (4 b B d-5 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2}} \] Output:

-2*e*(-A*e+B*d)/(-a*e+b*d)^3/(e*x+d)^(1/2)-1/2*(A*b-B*a)*(e*x+d)^(1/2)/(-a 
*e+b*d)^2/(b*x+a)^2-1/4*(-7*A*b*e+3*B*a*e+4*B*b*d)*(e*x+d)^(1/2)/(-a*e+b*d 
)^3/(b*x+a)+3/4*e*(-5*A*b*e+B*a*e+4*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/( 
-a*e+b*d)^(1/2))/b^(1/2)/(-a*e+b*d)^(7/2)
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx=\frac {1}{4} \left (\frac {-B \left (4 b^2 d x (d+3 e x)+a^2 e (13 d+5 e x)+a b \left (2 d^2+21 d e x+3 e^2 x^2\right )\right )+A \left (8 a^2 e^2+a b e (9 d+25 e x)+b^2 \left (-2 d^2+5 d e x+15 e^2 x^2\right )\right )}{(b d-a e)^3 (a+b x)^2 \sqrt {d+e x}}+\frac {3 e (4 b B d-5 A b e+a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{7/2}}\right ) \] Input:

Integrate[(A + B*x)/((a + b*x)^3*(d + e*x)^(3/2)),x]
 

Output:

((-(B*(4*b^2*d*x*(d + 3*e*x) + a^2*e*(13*d + 5*e*x) + a*b*(2*d^2 + 21*d*e* 
x + 3*e^2*x^2))) + A*(8*a^2*e^2 + a*b*e*(9*d + 25*e*x) + b^2*(-2*d^2 + 5*d 
*e*x + 15*e^2*x^2)))/((b*d - a*e)^3*(a + b*x)^2*Sqrt[d + e*x]) + (3*e*(4*b 
*B*d - 5*A*b*e + a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]] 
)/(Sqrt[b]*(-(b*d) + a*e)^(7/2)))/4
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {87, 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.

\(\displaystyle \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(A + B*x)/((a + b*x)^3*(d + e*x)^(3/2)),x]
 

Output:

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

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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 = 0.52 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(2 e \left (-\frac {\frac {\left (\frac {7}{8} A \,b^{2} e -\frac {3}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (\frac {9}{8} A a b \,e^{2}-\frac {9}{8} A \,b^{2} d e -\frac {5}{8} B \,a^{2} e^{2}+\frac {1}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {3 \left (5 A b e -B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}}{\left (a e -d b \right )^{3}}-\frac {A e -B d}{\left (a e -d b \right )^{3} \sqrt {e x +d}}\right )\)	\(195\)
default	\(2 e \left (-\frac {\frac {\left (\frac {7}{8} A \,b^{2} e -\frac {3}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (\frac {9}{8} A a b \,e^{2}-\frac {9}{8} A \,b^{2} d e -\frac {5}{8} B \,a^{2} e^{2}+\frac {1}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {3 \left (5 A b e -B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}}{\left (a e -d b \right )^{3}}-\frac {A e -B d}{\left (a e -d b \right )^{3} \sqrt {e x +d}}\right )\)	\(195\)
pseudoelliptic	\(-\frac {2 \left (\frac {15 \left (b x +a \right )^{2} e \sqrt {e x +d}\, \left (\left (A e -\frac {4 B d}{5}\right ) b -\frac {B a e}{5}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8}+\sqrt {\left (a e -d b \right ) b}\, \left (\left (\frac {15 A \,e^{2} x^{2}}{8}+\frac {5 x d \left (-\frac {12 B x}{5}+A \right ) e}{8}-\frac {d^{2} \left (2 B x +A \right )}{4}\right ) b^{2}+\frac {9 a \left (\left (-\frac {1}{3} x^{2} B +\frac {25}{9} x A \right ) e^{2}+d \left (-\frac {7 B x}{3}+A \right ) e -\frac {2 B \,d^{2}}{9}\right ) b}{8}+a^{2} \left (\left (-\frac {5 B x}{8}+A \right ) e -\frac {13 B d}{8}\right ) e \right )\right )}{\sqrt {e x +d}\, \sqrt {\left (a e -d b \right ) b}\, \left (b x +a \right )^{2} \left (a e -d b \right )^{3}}\)	\(198\)

Input:

int((B*x+A)/(b*x+a)^3/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2*e*(-1/(a*e-b*d)^3*(((7/8*A*b^2*e-3/8*B*a*b*e-1/2*b^2*B*d)*(e*x+d)^(3/2)+ 
(9/8*A*a*b*e^2-9/8*A*b^2*d*e-5/8*B*a^2*e^2+1/8*B*a*b*d*e+1/2*b^2*B*d^2)*(e 
*x+d)^(1/2))/((e*x+d)*b+a*e-d*b)^2+3/8*(5*A*b*e-B*a*e-4*B*b*d)/((a*e-b*d)* 
b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))-1/(a*e-b*d)^3*(A*e-B 
*d)/(e*x+d)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 698 vs. \(2 (158) = 316\).

Time = 0.17 (sec) , antiderivative size = 1410, normalized size of antiderivative = 7.83 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*(3*(4*B*a^2*b*d^2*e + (B*a^3 - 5*A*a^2*b)*d*e^2 + (4*B*b^3*d*e^2 + (B 
*a*b^2 - 5*A*b^3)*e^3)*x^3 + (4*B*b^3*d^2*e + (9*B*a*b^2 - 5*A*b^3)*d*e^2 
+ 2*(B*a^2*b - 5*A*a*b^2)*e^3)*x^2 + (8*B*a*b^2*d^2*e + 2*(3*B*a^2*b - 5*A 
*a*b^2)*d*e^2 + (B*a^3 - 5*A*a^2*b)*e^3)*x)*sqrt(b^2*d - a*b*e)*log((b*e*x 
 + 2*b*d - a*e + 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(8*A* 
a^3*b*e^3 + 2*(B*a*b^3 + A*b^4)*d^3 + 11*(B*a^2*b^2 - A*a*b^3)*d^2*e - (13 
*B*a^3*b - A*a^2*b^2)*d*e^2 + 3*(4*B*b^4*d^2*e - (3*B*a*b^3 + 5*A*b^4)*d*e 
^2 - (B*a^2*b^2 - 5*A*a*b^3)*e^3)*x^2 + (4*B*b^4*d^3 + (17*B*a*b^3 - 5*A*b 
^4)*d^2*e - 4*(4*B*a^2*b^2 + 5*A*a*b^3)*d*e^2 - 5*(B*a^3*b - 5*A*a^2*b^2)* 
e^3)*x)*sqrt(e*x + d))/(a^2*b^5*d^5 - 4*a^3*b^4*d^4*e + 6*a^4*b^3*d^3*e^2 
- 4*a^5*b^2*d^2*e^3 + a^6*b*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6*a^2*b 
^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^3 + (b^7*d^5 - 2*a*b^6*d^4*e 
 - 2*a^2*b^5*d^3*e^2 + 8*a^3*b^4*d^2*e^3 - 7*a^4*b^3*d*e^4 + 2*a^5*b^2*e^5 
)*x^2 + (2*a*b^6*d^5 - 7*a^2*b^5*d^4*e + 8*a^3*b^4*d^3*e^2 - 2*a^4*b^3*d^2 
*e^3 - 2*a^5*b^2*d*e^4 + a^6*b*e^5)*x), -1/4*(3*(4*B*a^2*b*d^2*e + (B*a^3 
- 5*A*a^2*b)*d*e^2 + (4*B*b^3*d*e^2 + (B*a*b^2 - 5*A*b^3)*e^3)*x^3 + (4*B* 
b^3*d^2*e + (9*B*a*b^2 - 5*A*b^3)*d*e^2 + 2*(B*a^2*b - 5*A*a*b^2)*e^3)*x^2 
 + (8*B*a*b^2*d^2*e + 2*(3*B*a^2*b - 5*A*a*b^2)*d*e^2 + (B*a^3 - 5*A*a^2*b 
)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/( 
b*e*x + b*d)) + (8*A*a^3*b*e^3 + 2*(B*a*b^3 + A*b^4)*d^3 + 11*(B*a^2*b^...
 

Sympy [F(-1)]

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

integrate((B*x+A)/(b*x+a)**3/(e*x+d)**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(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*e-b*d>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. 340 vs. \(2 (158) = 316\).

Time = 0.13 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.89 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B b d e + B a e^{2} - 5 \, A b e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (B d e - A e^{2}\right )}}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {e x + d}} - \frac {4 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{2} d e - 4 \, \sqrt {e x + d} B b^{2} d^{2} e + 3 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b e^{2} - 7 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{2} e^{2} - \sqrt {e x + d} B a b d e^{2} + 9 \, \sqrt {e x + d} A b^{2} d e^{2} + 5 \, \sqrt {e x + d} B a^{2} e^{3} - 9 \, \sqrt {e x + d} A a b e^{3}}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} \] Input:

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(3/2),x, algorithm="giac")
 

Output:

-3/4*(4*B*b*d*e + B*a*e^2 - 5*A*b*e^2)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d 
+ a*b*e))/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d 
 + a*b*e)) - 2*(B*d*e - A*e^2)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - 
 a^3*e^3)*sqrt(e*x + d)) - 1/4*(4*(e*x + d)^(3/2)*B*b^2*d*e - 4*sqrt(e*x + 
 d)*B*b^2*d^2*e + 3*(e*x + d)^(3/2)*B*a*b*e^2 - 7*(e*x + d)^(3/2)*A*b^2*e^ 
2 - sqrt(e*x + d)*B*a*b*d*e^2 + 9*sqrt(e*x + d)*A*b^2*d*e^2 + 5*sqrt(e*x + 
 d)*B*a^2*e^3 - 9*sqrt(e*x + d)*A*a*b*e^3)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a 
^2*b*d*e^2 - a^3*e^3)*((e*x + d)*b - b*d + a*e)^2)
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.64 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx=\frac {\frac {5\,\left (d+e\,x\right )\,\left (B\,a\,e^2-5\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^2}-\frac {2\,\left (A\,e^2-B\,d\,e\right )}{a\,e-b\,d}+\frac {3\,b\,{\left (d+e\,x\right )}^2\,\left (B\,a\,e^2-5\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^3}}{b^2\,{\left (d+e\,x\right )}^{5/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{3/2}+\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}+\frac {3\,e\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (B\,a\,e-5\,A\,b\,e+4\,B\,b\,d\right )\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}\,\left (3\,B\,a\,e^2-15\,A\,b\,e^2+12\,B\,b\,d\,e\right )}\right )\,\left (B\,a\,e-5\,A\,b\,e+4\,B\,b\,d\right )}{4\,\sqrt {b}\,{\left (a\,e-b\,d\right )}^{7/2}} \] Input:

int((A + B*x)/((a + b*x)^3*(d + e*x)^(3/2)),x)
 

Output:

((5*(d + e*x)*(B*a*e^2 - 5*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^2) - (2*(A 
*e^2 - B*d*e))/(a*e - b*d) + (3*b*(d + e*x)^2*(B*a*e^2 - 5*A*b*e^2 + 4*B*b 
*d*e))/(4*(a*e - b*d)^3))/(b^2*(d + e*x)^(5/2) - (2*b^2*d - 2*a*b*e)*(d + 
e*x)^(3/2) + (d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e)) + (3*e*atan( 
(3*b^(1/2)*e*(d + e*x)^(1/2)*(B*a*e - 5*A*b*e + 4*B*b*d)*(a^3*e^3 - b^3*d^ 
3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/((a*e - b*d)^(7/2)*(3*B*a*e^2 - 15*A*b 
*e^2 + 12*B*b*d*e)))*(B*a*e - 5*A*b*e + 4*B*b*d))/(4*b^(1/2)*(a*e - b*d)^( 
7/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{3/2}} \, dx=\frac {-3 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a e -3 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b e x -2 a^{2} e^{2}+a b d e -3 a b \,e^{2} x +b^{2} d^{2}+3 b^{2} d e x}{\sqrt {e x +d}\, \left (a^{3} b \,e^{3} x -3 a^{2} b^{2} d \,e^{2} x +3 a \,b^{3} d^{2} e x -b^{4} d^{3} x +a^{4} e^{3}-3 a^{3} b d \,e^{2}+3 a^{2} b^{2} d^{2} e -a \,b^{3} d^{3}\right )} \] Input:

int((B*x+A)/(b*x+a)^3/(e*x+d)^(3/2),x)
 

Output:

( - 3*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b 
)*sqrt(a*e - b*d)))*a*e - 3*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sq 
rt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b*e*x - 2*a**2*e**2 + a*b*d*e - 
3*a*b*e**2*x + b**2*d**2 + 3*b**2*d*e*x)/(sqrt(d + e*x)*(a**4*e**3 - 3*a** 
3*b*d*e**2 + a**3*b*e**3*x + 3*a**2*b**2*d**2*e - 3*a**2*b**2*d*e**2*x - a 
*b**3*d**3 + 3*a*b**3*d**2*e*x - b**4*d**3*x))