\(\int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx\) [170]

Optimal result

Integrand size = 22, antiderivative size = 222 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx=-\frac {2 e (B d-A e)}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac {2 e (2 b B d-3 A b e+a B e)}{(b d-a e)^4 \sqrt {d+e x}}-\frac {b (A b-a B) \sqrt {d+e x}}{2 (b d-a e)^3 (a+b x)^2}-\frac {b (4 b B d-11 A b e+7 a B e) \sqrt {d+e x}}{4 (b d-a e)^4 (a+b x)}+\frac {5 \sqrt {b} e (4 b B d-7 A b e+3 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2}} \] Output:

-2/3*e*(-A*e+B*d)/(-a*e+b*d)^3/(e*x+d)^(3/2)-2*e*(-3*A*b*e+B*a*e+2*B*b*d)/ 
(-a*e+b*d)^4/(e*x+d)^(1/2)-1/2*b*(A*b-B*a)*(e*x+d)^(1/2)/(-a*e+b*d)^3/(b*x 
+a)^2-1/4*b*(-11*A*b*e+7*B*a*e+4*B*b*d)*(e*x+d)^(1/2)/(-a*e+b*d)^4/(b*x+a) 
+5/4*b^(1/2)*e*(-7*A*b*e+3*B*a*e+4*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(- 
a*e+b*d)^(1/2))/(-a*e+b*d)^(9/2)
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx=\frac {-B \left (8 a^3 e^2 (2 d+3 e x)+4 b^3 d x \left (3 d^2+20 d e x+15 e^2 x^2\right )+a^2 b e \left (83 d^2+134 d e x+75 e^2 x^2\right )+a b^2 \left (6 d^3+145 d^2 e x+160 d e^2 x^2+45 e^3 x^3\right )\right )+A \left (-8 a^3 e^3+8 a^2 b e^2 (10 d+7 e x)+a b^2 e \left (39 d^2+238 d e x+175 e^2 x^2\right )+b^3 \left (-6 d^3+21 d^2 e x+140 d e^2 x^2+105 e^3 x^3\right )\right )}{12 (b d-a e)^4 (a+b x)^2 (d+e x)^{3/2}}-\frac {5 \sqrt {b} e (4 b B d-7 A b e+3 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 (-b d+a e)^{9/2}} \] Input:

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

Output:

(-(B*(8*a^3*e^2*(2*d + 3*e*x) + 4*b^3*d*x*(3*d^2 + 20*d*e*x + 15*e^2*x^2) 
+ a^2*b*e*(83*d^2 + 134*d*e*x + 75*e^2*x^2) + a*b^2*(6*d^3 + 145*d^2*e*x + 
 160*d*e^2*x^2 + 45*e^3*x^3))) + A*(-8*a^3*e^3 + 8*a^2*b*e^2*(10*d + 7*e*x 
) + a*b^2*e*(39*d^2 + 238*d*e*x + 175*e^2*x^2) + b^3*(-6*d^3 + 21*d^2*e*x 
+ 140*d*e^2*x^2 + 105*e^3*x^3)))/(12*(b*d - a*e)^4*(a + b*x)^2*(d + e*x)^( 
3/2)) - (5*Sqrt[b]*e*(4*b*B*d - 7*A*b*e + 3*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d 
+ e*x])/Sqrt[-(b*d) + a*e]])/(4*(-(b*d) + a*e)^(9/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 52, 61, 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.

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {(3 a B e-7 A b e+4 b B d) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}}dx}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(3 a B e-7 A b e+4 b B d) \left (-\frac {5 e \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(3 a B e-7 A b e+4 b B d) \left (-\frac {5 e \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(3 a B e-7 A b e+4 b B d) \left (-\frac {5 e \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(3 a B e-7 A b e+4 b B d) \left (-\frac {5 e \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 a B e-7 A b e+4 b B d) \left (-\frac {5 e \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 b (b d-a e)}-\frac {A b-a B}{2 b (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\)

Input:

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

Output:

-1/2*(A*b - a*B)/(b*(b*d - a*e)*(a + b*x)^2*(d + e*x)^(3/2)) + ((4*b*B*d - 
 7*A*b*e + 3*a*B*e)*(-(1/((b*d - a*e)*(a + b*x)*(d + e*x)^(3/2))) - (5*e*( 
2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(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)))/(b*d - a*e)))/(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 = 229, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(2 e \left (\frac {b \left (\frac {\left (\frac {11}{8} A \,b^{2} e -\frac {7}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (\frac {13}{8} A a b \,e^{2}-\frac {13}{8} A \,b^{2} d e -\frac {9}{8} B \,a^{2} e^{2}+\frac {5}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {5 \left (7 A b e -3 B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}\right )}{\left (a e -d b \right )^{4}}-\frac {A e -B d}{3 \left (a e -d b \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {-3 A b e +B a e +2 B b d}{\left (a e -d b \right )^{4} \sqrt {e x +d}}\right )\)	\(229\)
default	\(2 e \left (\frac {b \left (\frac {\left (\frac {11}{8} A \,b^{2} e -\frac {7}{8} B a b e -\frac {1}{2} b^{2} B d \right ) \left (e x +d \right )^{\frac {3}{2}}+\left (\frac {13}{8} A a b \,e^{2}-\frac {13}{8} A \,b^{2} d e -\frac {9}{8} B \,a^{2} e^{2}+\frac {5}{8} B a b d e +\frac {1}{2} b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -d b \right )^{2}}+\frac {5 \left (7 A b e -3 B a e -4 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8 \sqrt {\left (a e -d b \right ) b}}\right )}{\left (a e -d b \right )^{4}}-\frac {A e -B d}{3 \left (a e -d b \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {-3 A b e +B a e +2 B b d}{\left (a e -d b \right )^{4} \sqrt {e x +d}}\right )\)	\(229\)
pseudoelliptic	\(-\frac {2 \left (-\frac {105 \left (\left (A e -\frac {4 B d}{7}\right ) b -\frac {3 B a e}{7}\right ) \left (e x +d \right )^{\frac {3}{2}} b \left (b x +a \right )^{2} e \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{8}+\sqrt {\left (a e -d b \right ) b}\, \left (\left (-\frac {105 A \,e^{3} x^{3}}{8}-\frac {35 \left (-\frac {3 B x}{7}+A \right ) x^{2} d \,e^{2}}{2}-\frac {21 \left (-\frac {80 B x}{21}+A \right ) x \,d^{2} e}{8}+\frac {3 d^{3} \left (2 B x +A \right )}{4}\right ) b^{3}-\frac {39 \left (\frac {175 \left (-\frac {9 B x}{35}+A \right ) x^{2} e^{3}}{39}+\frac {238 \left (-\frac {80 B x}{119}+A \right ) x d \,e^{2}}{39}+d^{2} \left (-\frac {145 B x}{39}+A \right ) e -\frac {2 B \,d^{3}}{13}\right ) a \,b^{2}}{8}-10 a^{2} \left (\frac {7 \left (-\frac {75 B x}{56}+A \right ) x \,e^{2}}{10}+d \left (-\frac {67 B x}{40}+A \right ) e -\frac {83 B \,d^{2}}{80}\right ) e b +a^{3} e^{2} \left (\left (3 B x +A \right ) e +2 B d \right )\right )\right )}{3 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -d b \right ) b}\, \left (b x +a \right )^{2} \left (a e -d b \right )^{4}}\)	\(269\)

Input:

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

Output:

2*e*(1/(a*e-b*d)^4*b*(((11/8*A*b^2*e-7/8*B*a*b*e-1/2*b^2*B*d)*(e*x+d)^(3/2 
)+(13/8*A*a*b*e^2-13/8*A*b^2*d*e-9/8*B*a^2*e^2+5/8*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+5/8*(7*A*b*e-3*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)))-1/3/(a*e-b*d)^ 
3*(A*e-B*d)/(e*x+d)^(3/2)-1/(a*e-b*d)^4*(-3*A*b*e+B*a*e+2*B*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. 863 vs. \(2 (196) = 392\).

Time = 0.19 (sec) , antiderivative size = 1756, normalized size of antiderivative = 7.91 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/24*(15*(4*B*a^2*b*d^3*e + (3*B*a^3 - 7*A*a^2*b)*d^2*e^2 + (4*B*b^3*d*e 
^3 + (3*B*a*b^2 - 7*A*b^3)*e^4)*x^4 + 2*(4*B*b^3*d^2*e^2 + 7*(B*a*b^2 - A* 
b^3)*d*e^3 + (3*B*a^2*b - 7*A*a*b^2)*e^4)*x^3 + (4*B*b^3*d^3*e + (19*B*a*b 
^2 - 7*A*b^3)*d^2*e^2 + 4*(4*B*a^2*b - 7*A*a*b^2)*d*e^3 + (3*B*a^3 - 7*A*a 
^2*b)*e^4)*x^2 + 2*(4*B*a*b^2*d^3*e + 7*(B*a^2*b - A*a*b^2)*d^2*e^2 + (3*B 
*a^3 - 7*A*a^2*b)*d*e^3)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 
 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(8*A*a^3* 
e^3 + 6*(B*a*b^2 + A*b^3)*d^3 + (83*B*a^2*b - 39*A*a*b^2)*d^2*e + 16*(B*a^ 
3 - 5*A*a^2*b)*d*e^2 + 15*(4*B*b^3*d*e^2 + (3*B*a*b^2 - 7*A*b^3)*e^3)*x^3 
+ 5*(16*B*b^3*d^2*e + 4*(8*B*a*b^2 - 7*A*b^3)*d*e^2 + 5*(3*B*a^2*b - 7*A*a 
*b^2)*e^3)*x^2 + (12*B*b^3*d^3 + (145*B*a*b^2 - 21*A*b^3)*d^2*e + 2*(67*B* 
a^2*b - 119*A*a*b^2)*d*e^2 + 8*(3*B*a^3 - 7*A*a^2*b)*e^3)*x)*sqrt(e*x + d) 
)/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a 
^6*d^2*e^4 + (b^6*d^4*e^2 - 4*a*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^ 
3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e - 3*a*b^5*d^4*e^2 + 2*a^2*b^4*d^ 
3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 
9*a^2*b^4*d^4*e^2 + 16*a^3*b^3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 
+ 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3*b^3*d^4*e^2 + 2*a^4*b^2*d^3*e^3 - 
 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x), -1/12*(15*(4*B*a^2*b*d^3*e + (3*B*a^3 - 
7*A*a^2*b)*d^2*e^2 + (4*B*b^3*d*e^3 + (3*B*a*b^2 - 7*A*b^3)*e^4)*x^4 + ...
 

Sympy [F(-1)]

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

integrate((B*x+A)/(b*x+a)**3/(e*x+d)**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)/(b*x+a)^3/(e*x+d)^(5/2),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. 446 vs. \(2 (196) = 392\).

Time = 0.13 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.01 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx=-\frac {5 \, {\left (4 \, B b^{2} d e + 3 \, B a b e^{2} - 7 \, A b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (6 \, {\left (e x + d\right )} B b d e + B b d^{2} e + 3 \, {\left (e x + d\right )} B a e^{2} - 9 \, {\left (e x + d\right )} A b e^{2} - B a d e^{2} - A b d e^{2} + A a e^{3}\right )}}{3 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left (e x + d\right )}^{\frac {3}{2}}} - \frac {4 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d e - 4 \, \sqrt {e x + d} B b^{3} d^{2} e + 7 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} e^{2} - 11 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} e^{2} - 5 \, \sqrt {e x + d} B a b^{2} d e^{2} + 13 \, \sqrt {e x + d} A b^{3} d e^{2} + 9 \, \sqrt {e x + d} B a^{2} b e^{3} - 13 \, \sqrt {e x + d} A a b^{2} e^{3}}{4 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} \] Input:

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

Output:

-5/4*(4*B*b^2*d*e + 3*B*a*b*e^2 - 7*A*b^2*e^2)*arctan(sqrt(e*x + d)*b/sqrt 
(-b^2*d + a*b*e))/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b* 
d*e^3 + a^4*e^4)*sqrt(-b^2*d + a*b*e)) - 2/3*(6*(e*x + d)*B*b*d*e + B*b*d^ 
2*e + 3*(e*x + d)*B*a*e^2 - 9*(e*x + d)*A*b*e^2 - B*a*d*e^2 - A*b*d*e^2 + 
A*a*e^3)/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a 
^4*e^4)*(e*x + d)^(3/2)) - 1/4*(4*(e*x + d)^(3/2)*B*b^3*d*e - 4*sqrt(e*x + 
 d)*B*b^3*d^2*e + 7*(e*x + d)^(3/2)*B*a*b^2*e^2 - 11*(e*x + d)^(3/2)*A*b^3 
*e^2 - 5*sqrt(e*x + d)*B*a*b^2*d*e^2 + 13*sqrt(e*x + d)*A*b^3*d*e^2 + 9*sq 
rt(e*x + d)*B*a^2*b*e^3 - 13*sqrt(e*x + d)*A*a*b^2*e^3)/((b^4*d^4 - 4*a*b^ 
3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*((e*x + d)*b - b*d 
+ a*e)^2)
 

Mupad [B] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.63 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx=-\frac {\frac {2\,\left (A\,e^2-B\,d\,e\right )}{3\,\left (a\,e-b\,d\right )}+\frac {2\,\left (d+e\,x\right )\,\left (3\,B\,a\,e^2-7\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {25\,{\left (d+e\,x\right )}^2\,\left (-7\,A\,b^2\,e^2+4\,B\,d\,b^2\,e+3\,B\,a\,b\,e^2\right )}{12\,{\left (a\,e-b\,d\right )}^3}+\frac {5\,b^2\,{\left (d+e\,x\right )}^3\,\left (3\,B\,a\,e^2-7\,A\,b\,e^2+4\,B\,b\,d\,e\right )}{4\,{\left (a\,e-b\,d\right )}^4}}{b^2\,{\left (d+e\,x\right )}^{7/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{5/2}+{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}-\frac {5\,\sqrt {b}\,e\,\mathrm {atan}\left (\frac {\sqrt {b}\,e\,\sqrt {d+e\,x}\,\left (3\,B\,a\,e-7\,A\,b\,e+4\,B\,b\,d\right )\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^{9/2}\,\left (3\,B\,a\,e^2-7\,A\,b\,e^2+4\,B\,b\,d\,e\right )}\right )\,\left (3\,B\,a\,e-7\,A\,b\,e+4\,B\,b\,d\right )}{4\,{\left (a\,e-b\,d\right )}^{9/2}} \] Input:

int((A + B*x)/((a + b*x)^3*(d + e*x)^(5/2)),x)
 

Output:

- ((2*(A*e^2 - B*d*e))/(3*(a*e - b*d)) + (2*(d + e*x)*(3*B*a*e^2 - 7*A*b*e 
^2 + 4*B*b*d*e))/(3*(a*e - b*d)^2) + (25*(d + e*x)^2*(3*B*a*b*e^2 - 7*A*b^ 
2*e^2 + 4*B*b^2*d*e))/(12*(a*e - b*d)^3) + (5*b^2*(d + e*x)^3*(3*B*a*e^2 - 
 7*A*b*e^2 + 4*B*b*d*e))/(4*(a*e - b*d)^4))/(b^2*(d + e*x)^(7/2) - (2*b^2* 
d - 2*a*b*e)*(d + e*x)^(5/2) + (d + e*x)^(3/2)*(a^2*e^2 + b^2*d^2 - 2*a*b* 
d*e)) - (5*b^(1/2)*e*atan((b^(1/2)*e*(d + e*x)^(1/2)*(3*B*a*e - 7*A*b*e + 
4*B*b*d)*(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b* 
d*e^3))/((a*e - b*d)^(9/2)*(3*B*a*e^2 - 7*A*b*e^2 + 4*B*b*d*e)))*(3*B*a*e 
- 7*A*b*e + 4*B*b*d))/(4*(a*e - b*d)^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.22 \[ \int \frac {A+B x}{(a+b x)^3 (d+e x)^{5/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a b d e +15 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a b \,e^{2} x +15 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{2} d e x +15 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{2} e^{2} x^{2}-2 a^{3} e^{3}+16 a^{2} b d \,e^{2}+10 a^{2} b \,e^{3} x -11 a \,b^{2} d^{2} e +10 a \,b^{2} d \,e^{2} x +15 a \,b^{2} e^{3} x^{2}-3 b^{3} d^{3}-20 b^{3} d^{2} e x -15 b^{3} d \,e^{2} x^{2}}{3 \sqrt {e x +d}\, \left (a^{4} b \,e^{5} x^{2}-4 a^{3} b^{2} d \,e^{4} x^{2}+6 a^{2} b^{3} d^{2} e^{3} x^{2}-4 a \,b^{4} d^{3} e^{2} x^{2}+b^{5} d^{4} e \,x^{2}+a^{5} e^{5} x -3 a^{4} b d \,e^{4} x +2 a^{3} b^{2} d^{2} e^{3} x +2 a^{2} b^{3} d^{3} e^{2} x -3 a \,b^{4} d^{4} e x +b^{5} d^{5} x +a^{5} d \,e^{4}-4 a^{4} b \,d^{2} e^{3}+6 a^{3} b^{2} d^{3} e^{2}-4 a^{2} b^{3} d^{4} e +a \,b^{4} d^{5}\right )} \] Input:

int((B*x+A)/(b*x+a)^3/(e*x+d)^(5/2),x)
 

Output:

(15*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*b*d*e + 15*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*b*e**2*x + 15*sqrt(b)*sqrt( 
d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d))) 
*b**2*d*e*x + 15*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x) 
*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**2*e**2*x**2 - 2*a**3*e**3 + 16*a**2*b*d* 
e**2 + 10*a**2*b*e**3*x - 11*a*b**2*d**2*e + 10*a*b**2*d*e**2*x + 15*a*b** 
2*e**3*x**2 - 3*b**3*d**3 - 20*b**3*d**2*e*x - 15*b**3*d*e**2*x**2)/(3*sqr 
t(d + e*x)*(a**5*d*e**4 + a**5*e**5*x - 4*a**4*b*d**2*e**3 - 3*a**4*b*d*e* 
*4*x + a**4*b*e**5*x**2 + 6*a**3*b**2*d**3*e**2 + 2*a**3*b**2*d**2*e**3*x 
- 4*a**3*b**2*d*e**4*x**2 - 4*a**2*b**3*d**4*e + 2*a**2*b**3*d**3*e**2*x + 
 6*a**2*b**3*d**2*e**3*x**2 + a*b**4*d**5 - 3*a*b**4*d**4*e*x - 4*a*b**4*d 
**3*e**2*x**2 + b**5*d**5*x + b**5*d**4*e*x**2))