\(\int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx\) [391]

Optimal result

Integrand size = 31, antiderivative size = 133 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\frac {2 f \sqrt {e+f x}}{b d}-\frac {2 (b e-a f)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{3/2} (b c-a d)}+\frac {2 (d e-c f)^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (b c-a d)} \] Output:

2*f*(f*x+e)^(1/2)/b/d-2*(-a*f+b*e)^(3/2)*arctanh(b^(1/2)*(f*x+e)^(1/2)/(-a 
*f+b*e)^(1/2))/b^(3/2)/(-a*d+b*c)+2*(-c*f+d*e)^(3/2)*arctanh(d^(1/2)*(f*x+ 
e)^(1/2)/(-c*f+d*e)^(1/2))/d^(3/2)/(-a*d+b*c)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\frac {2 \left (d^{3/2} (-b e+a f)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )+\sqrt {b} \left (\sqrt {d} (b c-a d) f \sqrt {e+f x}-b (-d e+c f)^{3/2} \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )\right )\right )}{b^{3/2} d^{3/2} (b c-a d)} \] Input:

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

Output:

(2*(d^(3/2)*(-(b*e) + a*f)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[e + f*x])/Sqrt[-(b*e 
) + a*f]] + Sqrt[b]*(Sqrt[d]*(b*c - a*d)*f*Sqrt[e + f*x] - b*(-(d*e) + c*f 
)^(3/2)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])))/(b^(3/2)*d^( 
3/2)*(b*c - a*d))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {1146, 1197, 25, 27, 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.

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

\(\Big \downarrow \) 1146

\(\displaystyle \frac {\int \frac {b d e^2-a c f^2+f (2 b d e-b c f-a d f) x}{\sqrt {e+f x} \left (b d x^2+(b c+a d) x+a c\right )}dx}{b d}+\frac {2 f \sqrt {e+f x}}{b d}\)

\(\Big \downarrow \) 1197

\(\displaystyle \frac {2 \int -\frac {f ((b e-a f) (d e-c f)-(2 b d e-b c f-a d f) (e+f x))}{b d (e+f x)^2-(2 b d e-b c f-a d f) (e+f x)+(b e-a f) (d e-c f)}d\sqrt {e+f x}}{b d}+\frac {2 f \sqrt {e+f x}}{b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 f \sqrt {e+f x}}{b d}-\frac {2 \int \frac {f ((b e-a f) (d e-c f)-(2 b d e-b c f-a d f) (e+f x))}{b d (e+f x)^2-(2 b d e-b c f-a d f) (e+f x)+(b e-a f) (d e-c f)}d\sqrt {e+f x}}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 f \sqrt {e+f x}}{b d}-\frac {2 f \int \frac {(b e-a f) (d e-c f)-(2 b d e-b c f-a d f) (e+f x)}{b d (e+f x)^2-(2 b d e-b c f-a d f) (e+f x)+(b e-a f) (d e-c f)}d\sqrt {e+f x}}{b d}\)

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*f*Sqrt[e + f*x])/(b*d) - (2*f*((d*(b*e - a*f)^(3/2)*ArcTanh[(Sqrt[b]*Sq 
rt[e + f*x])/Sqrt[b*e - a*f]])/(Sqrt[b]*(b*c - a*d)*f) - (b*(d*e - c*f)^(3 
/2)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(Sqrt[d]*(b*c - a*d) 
*f)))/(b*d)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 1146

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

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 = 3.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(2 f \left (\frac {\sqrt {f x +e}}{d b}+\frac {\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d f \left (a d -b c \right ) \sqrt {\left (c f -d e \right ) d}}+\frac {\left (-a^{2} f^{2}+2 a b e f -b^{2} e^{2}\right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{b f \left (a d -b c \right ) \sqrt {\left (a f -b e \right ) b}}\right )\)	\(165\)
default	\(2 f \left (\frac {\sqrt {f x +e}}{d b}+\frac {\left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d f \left (a d -b c \right ) \sqrt {\left (c f -d e \right ) d}}+\frac {\left (-a^{2} f^{2}+2 a b e f -b^{2} e^{2}\right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{b f \left (a d -b c \right ) \sqrt {\left (a f -b e \right ) b}}\right )\)	\(165\)
pseudoelliptic	\(\frac {-2 d \left (a f -b e \right )^{2} \sqrt {\left (c f -d e \right ) d}\, \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )+2 \left (b \left (c f -d e \right )^{2} \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )+f \sqrt {f x +e}\, \left (a d -b c \right ) \sqrt {\left (c f -d e \right ) d}\right ) \sqrt {\left (a f -b e \right ) b}}{b d \left (a d -b c \right ) \sqrt {\left (a f -b e \right ) b}\, \sqrt {\left (c f -d e \right ) d}}\)	\(168\)
risch	\(\frac {2 f \sqrt {f x +e}}{b d}-\frac {2 f \left (\frac {d \left (a^{2} f^{2}-2 a b e f +b^{2} e^{2}\right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{f \left (a d -b c \right ) \sqrt {\left (a f -b e \right ) b}}-\frac {b \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{f \left (a d -b c \right ) \sqrt {\left (c f -d e \right ) d}}\right )}{b d}\)	\(169\)

Input:

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

Output:

2*f*(1/d/b*(f*x+e)^(1/2)+1/d*(c^2*f^2-2*c*d*e*f+d^2*e^2)/f/(a*d-b*c)/((c*f 
-d*e)*d)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2))+(-a^2*f^2+2*a*b 
*e*f-b^2*e^2)/b/f/(a*d-b*c)/((a*f-b*e)*b)^(1/2)*arctan(b*(f*x+e)^(1/2)/((a 
*f-b*e)*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 680, normalized size of antiderivative = 5.11 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\left [\frac {2 \, {\left (b c - a d\right )} \sqrt {f x + e} f + {\left (b d e - a d f\right )} \sqrt {\frac {b e - a f}{b}} \log \left (\frac {b f x + 2 \, b e - a f - 2 \, \sqrt {f x + e} b \sqrt {\frac {b e - a f}{b}}}{b x + a}\right ) + {\left (b d e - b c f\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f + 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right )}{b^{2} c d - a b d^{2}}, \frac {2 \, {\left (b c - a d\right )} \sqrt {f x + e} f - 2 \, {\left (b d e - a d f\right )} \sqrt {-\frac {b e - a f}{b}} \arctan \left (-\frac {\sqrt {f x + e} b \sqrt {-\frac {b e - a f}{b}}}{b e - a f}\right ) + {\left (b d e - b c f\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f + 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right )}{b^{2} c d - a b d^{2}}, \frac {2 \, {\left (b c - a d\right )} \sqrt {f x + e} f + 2 \, {\left (b d e - b c f\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) + {\left (b d e - a d f\right )} \sqrt {\frac {b e - a f}{b}} \log \left (\frac {b f x + 2 \, b e - a f - 2 \, \sqrt {f x + e} b \sqrt {\frac {b e - a f}{b}}}{b x + a}\right )}{b^{2} c d - a b d^{2}}, \frac {2 \, {\left ({\left (b c - a d\right )} \sqrt {f x + e} f - {\left (b d e - a d f\right )} \sqrt {-\frac {b e - a f}{b}} \arctan \left (-\frac {\sqrt {f x + e} b \sqrt {-\frac {b e - a f}{b}}}{b e - a f}\right ) + {\left (b d e - b c f\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right )\right )}}{b^{2} c d - a b d^{2}}\right ] \] Input:

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

Output:

[(2*(b*c - a*d)*sqrt(f*x + e)*f + (b*d*e - a*d*f)*sqrt((b*e - a*f)/b)*log( 
(b*f*x + 2*b*e - a*f - 2*sqrt(f*x + e)*b*sqrt((b*e - a*f)/b))/(b*x + a)) + 
 (b*d*e - b*c*f)*sqrt((d*e - c*f)/d)*log((d*f*x + 2*d*e - c*f + 2*sqrt(f*x 
 + e)*d*sqrt((d*e - c*f)/d))/(d*x + c)))/(b^2*c*d - a*b*d^2), (2*(b*c - a* 
d)*sqrt(f*x + e)*f - 2*(b*d*e - a*d*f)*sqrt(-(b*e - a*f)/b)*arctan(-sqrt(f 
*x + e)*b*sqrt(-(b*e - a*f)/b)/(b*e - a*f)) + (b*d*e - b*c*f)*sqrt((d*e - 
c*f)/d)*log((d*f*x + 2*d*e - c*f + 2*sqrt(f*x + e)*d*sqrt((d*e - c*f)/d))/ 
(d*x + c)))/(b^2*c*d - a*b*d^2), (2*(b*c - a*d)*sqrt(f*x + e)*f + 2*(b*d*e 
 - b*c*f)*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d 
)/(d*e - c*f)) + (b*d*e - a*d*f)*sqrt((b*e - a*f)/b)*log((b*f*x + 2*b*e - 
a*f - 2*sqrt(f*x + e)*b*sqrt((b*e - a*f)/b))/(b*x + a)))/(b^2*c*d - a*b*d^ 
2), 2*((b*c - a*d)*sqrt(f*x + e)*f - (b*d*e - a*d*f)*sqrt(-(b*e - a*f)/b)* 
arctan(-sqrt(f*x + e)*b*sqrt(-(b*e - a*f)/b)/(b*e - a*f)) + (b*d*e - b*c*f 
)*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/(d*e - 
 c*f)))/(b^2*c*d - a*b*d^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (112) = 224\).

Time = 4.53 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.21 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\begin {cases} \frac {2 \left (\frac {f \left (c f - d e\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{2} \sqrt {\frac {c f - d e}{d}} \left (a d - b c\right )} + \frac {f^{2} \sqrt {e + f x}}{b d} - \frac {f \left (a f - b e\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {a f - b e}{b}}} \right )}}{b^{2} \sqrt {\frac {a f - b e}{b}} \left (a d - b c\right )}\right )}{f} & \text {for}\: f \neq 0 \\e^{\frac {3}{2}} \left (- \frac {2 b d \left (\begin {cases} \frac {x + \frac {a d + b c}{2 b d}}{a d} & \text {for}\: b = 0 \\- \frac {x + \frac {a d + b c}{2 b d}}{b c} & \text {for}\: d = 0 \\- \frac {\log {\left (a d - b c - 2 b d \left (x + \frac {a d + b c}{2 b d}\right ) \right )}}{2 b d} & \text {otherwise} \end {cases}\right )}{a d - b c} - \frac {2 b d \left (\begin {cases} \frac {x + \frac {a d + b c}{2 b d}}{a d} & \text {for}\: b = 0 \\- \frac {x + \frac {a d + b c}{2 b d}}{b c} & \text {for}\: d = 0 \\\frac {\log {\left (a d - b c + 2 b d \left (x + \frac {a d + b c}{2 b d}\right ) \right )}}{2 b d} & \text {otherwise} \end {cases}\right )}{a d - b c}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [F(-2)]

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

integrate((f*x+e)^(3/2)/(a*c+(a*d+b*c)*x+b*d*x^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(c*f-d*e>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.21 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\frac {2 \, {\left (b^{2} e^{2} - 2 \, a b e f + a^{2} f^{2}\right )} \arctan \left (\frac {\sqrt {f x + e} b}{\sqrt {-b^{2} e + a b f}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt {-b^{2} e + a b f}} - \frac {2 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{{\left (b c d - a d^{2}\right )} \sqrt {-d^{2} e + c d f}} + \frac {2 \, \sqrt {f x + e} f}{b d} \] Input:

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

Output:

2*(b^2*e^2 - 2*a*b*e*f + a^2*f^2)*arctan(sqrt(f*x + e)*b/sqrt(-b^2*e + a*b 
*f))/((b^2*c - a*b*d)*sqrt(-b^2*e + a*b*f)) - 2*(d^2*e^2 - 2*c*d*e*f + c^2 
*f^2)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/((b*c*d - a*d^2)*sqrt(- 
d^2*e + c*d*f)) + 2*sqrt(f*x + e)*f/(b*d)
 

Mupad [B] (verification not implemented)

Time = 6.54 (sec) , antiderivative size = 4619, normalized size of antiderivative = 34.73 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((-b^3*(a*f - b*e)^3)^(1/2)*((8*(e + f*x)^(1/2)*(a^4*d^4*f^6 + b^4* 
c^4*f^6 + 2*b^4*d^4*e^4*f^2 + 6*a^2*b^2*d^4*e^2*f^4 + 6*b^4*c^2*d^2*e^2*f^ 
4 - 4*a^3*b*d^4*e*f^5 - 4*b^4*c^3*d*e*f^5 - 4*a*b^3*d^4*e^3*f^3 - 4*b^4*c* 
d^3*e^3*f^3))/(b*d) + (((8*(a^2*b^3*d^5*e^2*f^3 - 2*a^2*b^3*c^2*d^3*f^5 + 
b^5*c^2*d^3*e^2*f^3 + a*b^4*c^3*d^2*f^5 + a^3*b^2*c*d^4*f^5 - a^3*b^2*d^5* 
e*f^4 - b^5*c^3*d^2*e*f^4 - 2*a*b^4*c*d^4*e^2*f^3 + a*b^4*c^2*d^3*e*f^4 + 
a^2*b^3*c*d^4*e*f^4))/(b*d) - (8*(e + f*x)^(1/2)*(-b^3*(a*f - b*e)^3)^(1/2 
)*(a*b^5*c^2*d^4*f^3 - b^6*c^3*d^3*f^3 - a^3*b^3*d^6*f^3 + a^2*b^4*c*d^5*f 
^3 + 2*a^2*b^4*d^6*e*f^2 + 2*b^6*c^2*d^4*e*f^2 - 4*a*b^5*c*d^5*e*f^2))/(b^ 
4*d*(a*d - b*c)))*(-b^3*(a*f - b*e)^3)^(1/2))/(b^3*(a*d - b*c)))*1i)/(b^3* 
(a*d - b*c)) + ((-b^3*(a*f - b*e)^3)^(1/2)*((8*(e + f*x)^(1/2)*(a^4*d^4*f^ 
6 + b^4*c^4*f^6 + 2*b^4*d^4*e^4*f^2 + 6*a^2*b^2*d^4*e^2*f^4 + 6*b^4*c^2*d^ 
2*e^2*f^4 - 4*a^3*b*d^4*e*f^5 - 4*b^4*c^3*d*e*f^5 - 4*a*b^3*d^4*e^3*f^3 - 
4*b^4*c*d^3*e^3*f^3))/(b*d) - (((8*(a^2*b^3*d^5*e^2*f^3 - 2*a^2*b^3*c^2*d^ 
3*f^5 + b^5*c^2*d^3*e^2*f^3 + a*b^4*c^3*d^2*f^5 + a^3*b^2*c*d^4*f^5 - a^3* 
b^2*d^5*e*f^4 - b^5*c^3*d^2*e*f^4 - 2*a*b^4*c*d^4*e^2*f^3 + a*b^4*c^2*d^3* 
e*f^4 + a^2*b^3*c*d^4*e*f^4))/(b*d) + (8*(e + f*x)^(1/2)*(-b^3*(a*f - b*e) 
^3)^(1/2)*(a*b^5*c^2*d^4*f^3 - b^6*c^3*d^3*f^3 - a^3*b^3*d^6*f^3 + a^2*b^4 
*c*d^5*f^3 + 2*a^2*b^4*d^6*e*f^2 + 2*b^6*c^2*d^4*e*f^2 - 4*a*b^5*c*d^5*e*f 
^2))/(b^4*d*(a*d - b*c)))*(-b^3*(a*f - b*e)^3)^(1/2))/(b^3*(a*d - b*c))...
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.59 \[ \int \frac {(e+f x)^{3/2}}{a c+(b c+a d) x+b d x^2} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a \,d^{2} f +2 \sqrt {b}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) b \,d^{2} e +2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) b^{2} c f -2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) b^{2} d e +2 \sqrt {f x +e}\, a b \,d^{2} f -2 \sqrt {f x +e}\, b^{2} c d f}{b^{2} d^{2} \left (a d -b c \right )} \] Input:

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

Output:

(2*( - sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - 
b*e)))*a*d**2*f + sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)* 
sqrt(a*f - b*e)))*b*d**2*e + sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d 
)/(sqrt(d)*sqrt(c*f - d*e)))*b**2*c*f - sqrt(d)*sqrt(c*f - d*e)*atan((sqrt 
(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*b**2*d*e + sqrt(e + f*x)*a*b*d**2* 
f - sqrt(e + f*x)*b**2*c*d*f))/(b**2*d**2*(a*d - b*c))