\(\int \frac {A+B x}{(d+e x)^{5/2} (b x+c x^2)} \, dx\) [74]

Optimal result

Integrand size = 26, antiderivative size = 164 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx=\frac {2 (B d-A e)}{3 d (c d-b e) (d+e x)^{3/2}}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{d^2 (c d-b e)^2 \sqrt {d+e x}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{5/2}}-\frac {2 c^{3/2} (b B-A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b (c d-b e)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx=\frac {2 \left (B d^2 (4 c d-b e+3 c e x)+A e (b e (4 d+3 e x)-c d (7 d+6 e x))\right )}{3 d^2 (c d-b e)^2 (d+e x)^{3/2}}-\frac {2 c^{3/2} (-b B+A c) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b (-c d+b e)^{5/2}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{5/2}} \] Input:

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

Output:

(2*(B*d^2*(4*c*d - b*e + 3*c*e*x) + A*e*(b*e*(4*d + 3*e*x) - c*d*(7*d + 6* 
e*x))))/(3*d^2*(c*d - b*e)^2*(d + e*x)^(3/2)) - (2*c^(3/2)*(-(b*B) + A*c)* 
ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + b*e]])/(b*(-(c*d) + b*e)^(5/2 
)) - (2*A*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*d^(5/2))
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1198, 1198, 1197, 25, 1480, 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.

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

\(\Big \downarrow \) 1198

\(\displaystyle \frac {\int \frac {A (c d-b e)+c (B d-A e) x}{(d+e x)^{3/2} \left (c x^2+b x\right )}dx}{d (c d-b e)}+\frac {2 (B d-A e)}{3 d (d+e x)^{3/2} (c d-b e)}\)

\(\Big \downarrow \) 1198

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

\(\Big \downarrow \) 1197

\(\displaystyle \frac {\frac {2 \int -\frac {B c^2 d^3-A e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )-c \left (B c d^2-A e (2 c d-b e)\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{3 d (d+e x)^{3/2} (c d-b e)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{d \sqrt {d+e x} (c d-b e)}-\frac {2 \int \frac {B c^2 d^3-A e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )-c \left (B c d^2-A e (2 c d-b e)\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{3 d (d+e x)^{3/2} (c d-b e)}\)

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*(B*d - A*e))/(3*d*(c*d - b*e)*(d + e*x)^(3/2)) + ((2*(B*c*d^2 - A*e*(2* 
c*d - b*e)))/(d*(c*d - b*e)*Sqrt[d + e*x]) + (2*(-((A*(c*d - b*e)^2*ArcTan 
h[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d])) - (c^(3/2)*(b*B - A*c)*d^2*ArcTanh[ 
(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*Sqrt[c*d - b*e])))/(d*(c*d - 
b*e)))/(d*(c*d - b*e))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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]
 

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1198

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c 
*d^2 - b*d*e + a*e^2))), x] + Simp[1/(c*d^2 - b*d*e + a*e^2)   Int[(d + e*x 
)^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^ 
2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] && LtQ[m, -1 
]
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {\frac {2 A e}{3}-\frac {2 B d}{3}}{d \left (b e -c d \right ) \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 A b \,e^{2}-4 A c d e +2 B c \,d^{2}}{\sqrt {e x +d}\, d^{2} \left (b e -c d \right )^{2}}-\frac {2 c^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{2} b \sqrt {c \left (b e -c d \right )}}-\frac {2 A \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {5}{2}}}\)	\(155\)
derivativedivides	\(-\frac {2 \left (-A e +B d \right )}{3 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-A b \,e^{2}+2 A c d e -B c \,d^{2}\right )}{d^{2} \left (b e -c d \right )^{2} \sqrt {e x +d}}-\frac {2 c^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{2} b \sqrt {c \left (b e -c d \right )}}-\frac {2 A \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {5}{2}}}\)	\(156\)
default	\(-\frac {2 \left (-A e +B d \right )}{3 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-A b \,e^{2}+2 A c d e -B c \,d^{2}\right )}{d^{2} \left (b e -c d \right )^{2} \sqrt {e x +d}}-\frac {2 c^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{2} b \sqrt {c \left (b e -c d \right )}}-\frac {2 A \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {5}{2}}}\)	\(156\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (142) = 284\).

Time = 0.91 (sec) , antiderivative size = 1664, normalized size of antiderivative = 10.15 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/3*(3*((B*b*c - A*c^2)*d^3*e^2*x^2 + 2*(B*b*c - A*c^2)*d^4*e*x + (B*b*c 
 - A*c^2)*d^5)*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e 
)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) - 3*(A*c^2*d^4 - 2*A*b*c*d 
^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2*e^2 - 2*A*b*c*d*e^3 + A*b^2*e^4)*x^2 + 2 
*(A*c^2*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(d)*log((e*x - 2*sqr 
t(e*x + d)*sqrt(d) + 2*d)/x) - 2*(4*B*b*c*d^4 + 4*A*b^2*d^2*e^2 - (B*b^2 + 
 7*A*b*c)*d^3*e + 3*(B*b*c*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt( 
e*x + d))/(b*c^2*d^7 - 2*b^2*c*d^6*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^ 
2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2*(b*c^2*d^6*e - 2*b^2*c*d^5*e^2 + b^3*d^ 
4*e^3)*x), 1/3*(6*((B*b*c - A*c^2)*d^3*e^2*x^2 + 2*(B*b*c - A*c^2)*d^4*e*x 
 + (B*b*c - A*c^2)*d^5)*sqrt(-c/(c*d - b*e))*arctan(sqrt(e*x + d)*sqrt(-c/ 
(c*d - b*e))) + 3*(A*c^2*d^4 - 2*A*b*c*d^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2* 
e^2 - 2*A*b*c*d*e^3 + A*b^2*e^4)*x^2 + 2*(A*c^2*d^3*e - 2*A*b*c*d^2*e^2 + 
A*b^2*d*e^3)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*( 
4*B*b*c*d^4 + 4*A*b^2*d^2*e^2 - (B*b^2 + 7*A*b*c)*d^3*e + 3*(B*b*c*d^3*e - 
 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(e*x + d))/(b*c^2*d^7 - 2*b^2*c*d^6 
*e + b^3*d^5*e^2 + (b*c^2*d^5*e^2 - 2*b^2*c*d^4*e^3 + b^3*d^3*e^4)*x^2 + 2 
*(b*c^2*d^6*e - 2*b^2*c*d^5*e^2 + b^3*d^4*e^3)*x), 1/3*(6*(A*c^2*d^4 - 2*A 
*b*c*d^3*e + A*b^2*d^2*e^2 + (A*c^2*d^2*e^2 - 2*A*b*c*d*e^3 + A*b^2*e^4)*x 
^2 + 2*(A*c^2*d^3*e - 2*A*b*c*d^2*e^2 + A*b^2*d*e^3)*x)*sqrt(-d)*arctan...
 

Sympy [A] (verification not implemented)

Time = 10.89 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.57 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {A e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b d^{2} \sqrt {- d}} - \frac {e \left (- A e + B d\right )}{3 d \left (d + e x\right )^{\frac {3}{2}} \left (b e - c d\right )} + \frac {e \left (A b e^{2} - 2 A c d e + B c d^{2}\right )}{d^{2} \sqrt {d + e x} \left (b e - c d\right )^{2}} + \frac {c e \left (- A c + B b\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b \sqrt {\frac {b e - c d}{c}} \left (b e - c d\right )^{2}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\frac {B \log {\left (b x + c x^{2} \right )}}{2 c} + \left (A - \frac {B b}{2 c}\right ) \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right )}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(A*e*atan(sqrt(d + e*x)/sqrt(-d))/(b*d**2*sqrt(-d)) - e*(-A*e 
 + B*d)/(3*d*(d + e*x)**(3/2)*(b*e - c*d)) + e*(A*b*e**2 - 2*A*c*d*e + B*c 
*d**2)/(d**2*sqrt(d + e*x)*(b*e - c*d)**2) + c*e*(-A*c + B*b)*atan(sqrt(d 
+ e*x)/sqrt((b*e - c*d)/c))/(b*sqrt((b*e - c*d)/c)*(b*e - c*d)**2))/e, Ne( 
e, 0)), ((B*log(b*x + c*x**2)/(2*c) + (A - B*b/(2*c))*(-2*c*Piecewise(((b/ 
(2*c) + x)/b, Eq(c, 0)), (-log(b - 2*c*(b/(2*c) + x))/(2*c), True))/b - 2* 
c*Piecewise(((b/(2*c) + x)/b, Eq(c, 0)), (log(b + 2*c*(b/(2*c) + x))/(2*c) 
, True))/b))/d**(5/2), True))
 

Maxima [F(-2)]

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

integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x),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(b*e-c*d>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx=\frac {2 \, {\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{{\left (b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}\right )} \sqrt {-c^{2} d + b c e}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )} B c d^{2} + B c d^{3} - 6 \, {\left (e x + d\right )} A c d e - B b d^{2} e - A c d^{2} e + 3 \, {\left (e x + d\right )} A b e^{2} + A b d e^{2}\right )}}{3 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}}} + \frac {2 \, A \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d} d^{2}} \] Input:

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

Output:

2*(B*b*c^2 - A*c^3)*arctan(sqrt(e*x + d)*c/sqrt(-c^2*d + b*c*e))/((b*c^2*d 
^2 - 2*b^2*c*d*e + b^3*e^2)*sqrt(-c^2*d + b*c*e)) + 2/3*(3*(e*x + d)*B*c*d 
^2 + B*c*d^3 - 6*(e*x + d)*A*c*d*e - B*b*d^2*e - A*c*d^2*e + 3*(e*x + d)*A 
*b*e^2 + A*b*d*e^2)/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*(e*x + d)^(3/2) 
) + 2*A*arctan(sqrt(e*x + d)/sqrt(-d))/(b*sqrt(-d)*d^2)
 

Mupad [B] (verification not implemented)

Time = 12.95 (sec) , antiderivative size = 6340, normalized size of antiderivative = 38.66 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((-c^3*(b*e - c*d)^5)^(1/2)*(A*c - B*b)*((d + e*x)^(1/2)*(16*A^2*c^ 
13*d^16*e^2 + 480*A^2*b^2*c^11*d^14*e^4 - 1120*A^2*b^3*c^10*d^13*e^5 + 180 
0*A^2*b^4*c^9*d^12*e^6 - 2064*A^2*b^5*c^8*d^11*e^7 + 1688*A^2*b^6*c^7*d^10 
*e^8 - 960*A^2*b^7*c^6*d^9*e^9 + 360*A^2*b^8*c^5*d^8*e^10 - 80*A^2*b^9*c^4 
*d^7*e^11 + 8*A^2*b^10*c^3*d^6*e^12 + 8*B^2*b^2*c^11*d^16*e^2 - 48*B^2*b^3 
*c^10*d^15*e^3 + 120*B^2*b^4*c^9*d^14*e^4 - 160*B^2*b^5*c^8*d^13*e^5 + 120 
*B^2*b^6*c^7*d^12*e^6 - 48*B^2*b^7*c^6*d^11*e^7 + 8*B^2*b^8*c^5*d^10*e^8 - 
 128*A^2*b*c^12*d^15*e^3 - 16*A*B*b*c^12*d^16*e^2 + 96*A*B*b^2*c^11*d^15*e 
^3 - 240*A*B*b^3*c^10*d^14*e^4 + 320*A*B*b^4*c^9*d^13*e^5 - 240*A*B*b^5*c^ 
8*d^12*e^6 + 96*A*B*b^6*c^7*d^11*e^7 - 16*A*B*b^7*c^6*d^10*e^8) - ((-c^3*( 
b*e - c*d)^5)^(1/2)*(A*c - B*b)*(((-c^3*(b*e - c*d)^5)^(1/2)*(A*c - B*b)*( 
d + e*x)^(1/2)*(16*b^2*c^13*d^21*e^2 - 168*b^3*c^12*d^20*e^3 + 800*b^4*c^1 
1*d^19*e^4 - 2280*b^5*c^10*d^18*e^5 + 4320*b^6*c^9*d^17*e^6 - 5712*b^7*c^8 
*d^16*e^7 + 5376*b^8*c^7*d^15*e^8 - 3600*b^9*c^6*d^14*e^9 + 1680*b^10*c^5* 
d^13*e^10 - 520*b^11*c^4*d^12*e^11 + 96*b^12*c^3*d^11*e^12 - 8*b^13*c^2*d^ 
10*e^13))/(b^6*e^5 - b*c^5*d^5 + 5*b^2*c^4*d^4*e - 10*b^3*c^3*d^3*e^2 + 10 
*b^4*c^2*d^2*e^3 - 5*b^5*c*d*e^4) - 24*A*b^2*c^12*d^18*e^3 + 216*A*b^3*c^1 
1*d^17*e^4 - 872*A*b^4*c^10*d^16*e^5 + 2080*A*b^5*c^9*d^15*e^6 - 3248*A*b^ 
6*c^8*d^14*e^7 + 3472*A*b^7*c^7*d^13*e^8 - 2576*A*b^8*c^6*d^12*e^9 + 1312* 
A*b^9*c^5*d^11*e^10 - 440*A*b^10*c^4*d^10*e^11 + 88*A*b^11*c^3*d^9*e^12...
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 921, normalized size of antiderivative = 5.62 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c 
)*sqrt(b*e - c*d)))*a*c**2*d**4 - 6*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)* 
atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*c**2*d**3*e*x + 6*sqrt 
(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e 
 - c*d)))*b**2*c*d**4 + 6*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt 
(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c*d**3*e*x + 3*sqrt(d)*sqrt(d 
 + e*x)*log(sqrt(d + e*x) - sqrt(d))*a*b**3*d*e**3 + 3*sqrt(d)*sqrt(d + e* 
x)*log(sqrt(d + e*x) - sqrt(d))*a*b**3*e**4*x - 9*sqrt(d)*sqrt(d + e*x)*lo 
g(sqrt(d + e*x) - sqrt(d))*a*b**2*c*d**2*e**2 - 9*sqrt(d)*sqrt(d + e*x)*lo 
g(sqrt(d + e*x) - sqrt(d))*a*b**2*c*d*e**3*x + 9*sqrt(d)*sqrt(d + e*x)*log 
(sqrt(d + e*x) - sqrt(d))*a*b*c**2*d**3*e + 9*sqrt(d)*sqrt(d + e*x)*log(sq 
rt(d + e*x) - sqrt(d))*a*b*c**2*d**2*e**2*x - 3*sqrt(d)*sqrt(d + e*x)*log( 
sqrt(d + e*x) - sqrt(d))*a*c**3*d**4 - 3*sqrt(d)*sqrt(d + e*x)*log(sqrt(d 
+ e*x) - sqrt(d))*a*c**3*d**3*e*x - 3*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e 
*x) + sqrt(d))*a*b**3*d*e**3 - 3*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + 
 sqrt(d))*a*b**3*e**4*x + 9*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + sqrt 
(d))*a*b**2*c*d**2*e**2 + 9*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + sqrt 
(d))*a*b**2*c*d*e**3*x - 9*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + sqrt( 
d))*a*b*c**2*d**3*e - 9*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + sqrt(d)) 
*a*b*c**2*d**2*e**2*x + 3*sqrt(d)*sqrt(d + e*x)*log(sqrt(d + e*x) + sqr...