\(\int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx\) [70]

Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 (B c d-b B e+A c e) \sqrt {d+e x}}{c^2}+\frac {2 B (d+e x)^{3/2}}{3 c}-\frac {2 A d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}-\frac {2 (b B-A c) (c d-b e)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 \sqrt {d+e x} (3 A c e+B (4 c d-3 b e+c e x))}{3 c^2}-\frac {2 (-b B+A c) (-c d+b e)^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b c^{5/2}}-\frac {2 A d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1196, 1196, 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.

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

\(\Big \downarrow \) 1196

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

\(\Big \downarrow \) 1196

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

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 1196

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c   Int 
[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + 
 b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & 
& GtQ[m, 0]
 

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 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 = 1.41 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(\frac {-2 \left (b e -c d \right )^{2} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )+2 \sqrt {c \left (b e -c d \right )}\, \left (-A \,d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right ) c^{2}+\left (\left (\frac {4 B d}{3}+e \left (\frac {B x}{3}+A \right )\right ) c -B b e \right ) b \sqrt {e x +d}\right )}{b \,c^{2} \sqrt {c \left (b e -c d \right )}}\)	\(129\)
derivativedivides	\(\frac {\frac {2 B c \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A c e \sqrt {e x +d}-2 B b e \sqrt {e x +d}+2 B c d \sqrt {e x +d}}{c^{2}}-\frac {2 A \,d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \left (-A \,b^{2} c \,e^{2}+2 A b \,c^{2} d e -A \,c^{3} d^{2}+b^{3} B \,e^{2}-2 B \,b^{2} c d e +B b \,c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{b \,c^{2} \sqrt {c \left (b e -c d \right )}}\)	\(171\)
default	\(\frac {\frac {2 B c \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A c e \sqrt {e x +d}-2 B b e \sqrt {e x +d}+2 B c d \sqrt {e x +d}}{c^{2}}-\frac {2 A \,d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \left (-A \,b^{2} c \,e^{2}+2 A b \,c^{2} d e -A \,c^{3} d^{2}+b^{3} B \,e^{2}-2 B \,b^{2} c d e +B b \,c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{b \,c^{2} \sqrt {c \left (b e -c d \right )}}\)	\(171\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 640, normalized size of antiderivative = 4.89 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\left [\frac {3 \, A c^{2} d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 3 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {3 \, A c^{2} d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - 6 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + 2 \, {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {6 \, A c^{2} \sqrt {-d} d \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x + d}}\right ) + 3 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {2 \, {\left (3 \, A c^{2} \sqrt {-d} d \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x + d}}\right ) - 3 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}\right )}}{3 \, b c^{2}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 11.70 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\begin {cases} \frac {2 \left (\frac {A d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} + \frac {B e \left (d + e x\right )^{\frac {3}{2}}}{3 c} + \frac {\sqrt {d + e x} \left (A c e^{2} - B b e^{2} + B c d e\right )}{c^{2}} + \frac {e \left (- A c + B b\right ) \left (b e - c d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{3} \sqrt {\frac {b e - c d}{c}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (\frac {B \log {\left (b x + c x^{2} \right )}}{2 c} + \left (A - \frac {B b}{2 c}\right ) \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right )\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [F(-2)]

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

integrate((B*x+A)*(e*x+d)^(3/2)/(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 = 0.26 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 \, A d^{2} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, {\left (B b c^{2} d^{2} - A c^{3} d^{2} - 2 \, B b^{2} c d e + 2 \, A b c^{2} d e + B b^{3} e^{2} - A b^{2} c e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c^{2}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B c^{2} + 3 \, \sqrt {e x + d} B c^{2} d - 3 \, \sqrt {e x + d} B b c e + 3 \, \sqrt {e x + d} A c^{2} e\right )}}{3 \, c^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 3810, normalized size of antiderivative = 29.08 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*A*e - 2*B*d)/c - (2*B*(b*e - 2*c*d))/c^2)*(d + e*x)^(1/2) + (2*B*(d + 
e*x)^(3/2))/(3*c) - (A*atan(((A*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4 
*c^2*e^6 + 2*A^2*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 
 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4*B^2*b^5*c*d*e^5 - 4*A 
^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4 
*e^2 + 8*A*B*b^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^ 
4))/c^3 + (A*((8*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + 
B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3 + (8*A*(b^3*c^5*e^3 - 2*b^2* 
c^6*d*e^2)*(d^3)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^3)^(1/2))/b)*(d^3)^(1/ 
2)*1i)/b + (A*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2*A^2*c 
^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d 
^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 
 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A*B*b^4*c 
^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 - (A*((8*( 
B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 
- 2*B*b^3*c^4*d^2*e^3))/c^3 - (8*A*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^3)^( 
1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^3)^(1/2))/b)*(d^3)^(1/2)*1i)/b)/((16*(2* 
A^3*c^5*d^5*e^3 + 4*A^3*b^2*c^3*d^3*e^5 - A^3*b^3*c^2*d^2*e^6 - A*B^2*b^5* 
d^2*e^6 + A^2*B*c^5*d^6*e^2 - 5*A^3*b*c^4*d^4*e^4 + 4*A*B^2*b^2*c^3*d^5*e^ 
3 - 6*A*B^2*b^3*c^2*d^4*e^4 + 11*A^2*B*b^2*c^3*d^4*e^4 - 8*A^2*B*b^3*c^...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.08 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {-6 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) a b c e +6 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) a \,c^{2} d +6 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{3} e -6 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b^{2} c d +6 \sqrt {e x +d}\, a b \,c^{2} e -6 \sqrt {e x +d}\, b^{3} c e +8 \sqrt {e x +d}\, b^{2} c^{2} d +2 \sqrt {e x +d}\, b^{2} c^{2} e x +3 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) a \,c^{3} d -3 \sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) a \,c^{3} d}{3 b \,c^{3}} \] Input:

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

Output:

( - 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c 
*d)))*a*b*c*e + 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)* 
sqrt(b*e - c*d)))*a*c**2*d + 6*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x) 
*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*e - 6*sqrt(c)*sqrt(b*e - c*d)*atan((sq 
rt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c*d + 6*sqrt(d + e*x)*a*b*c 
**2*e - 6*sqrt(d + e*x)*b**3*c*e + 8*sqrt(d + e*x)*b**2*c**2*d + 2*sqrt(d 
+ e*x)*b**2*c**2*e*x + 3*sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*a*c**3*d - 3 
*sqrt(d)*log(sqrt(d + e*x) + sqrt(d))*a*c**3*d)/(3*b*c**3)