\(\int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx\) [107]

Optimal result

Integrand size = 21, antiderivative size = 92 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 e \sqrt {d+e x}}{c}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 (c d-b e)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 \left (b \sqrt {c} e \sqrt {d+e x}-(-c d+b e)^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )-c^{3/2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{b c^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, 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.

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

\(\Big \downarrow \) 1146

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

\(\Big \downarrow \) 1197

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 = 0.52 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07

method	result	size
pseudoelliptic	\(\frac {-2 \left (b e -c d \right )^{2} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )+2 \left (-d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right ) c +b e \sqrt {e x +d}\right ) \sqrt {c \left (b e -c d \right )}}{b c \sqrt {c \left (b e -c d \right )}}\)	\(98\)
derivativedivides	\(2 e \left (\frac {\sqrt {e x +d}}{c}+\frac {\left (-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{c b e \sqrt {c \left (b e -c d \right )}}-\frac {d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}\right )\)	\(106\)
default	\(2 e \left (\frac {\sqrt {e x +d}}{c}+\frac {\left (-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{c b e \sqrt {c \left (b e -c d \right )}}-\frac {d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}\right )\)	\(106\)

Input:

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

Output:

2*(-(b*e-c*d)^2*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2))+(-d^(3/2)*arct 
anh((e*x+d)^(1/2)/d^(1/2))*c+b*e*(e*x+d)^(1/2))*(c*(b*e-c*d))^(1/2))/(c*(b 
*e-c*d))^(1/2)/c/b
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 441, normalized size of antiderivative = 4.79 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\left [\frac {c d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} b e - {\left (c d - b e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {c d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} b e + 2 \, {\left (c d - b e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )}{b c}, \frac {2 \, c \sqrt {-d} d \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x + d}}\right ) + 2 \, \sqrt {e x + d} b e - {\left (c d - b e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {2 \, {\left (c \sqrt {-d} d \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x + d}}\right ) + \sqrt {e x + d} b e + {\left (c d - b e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )\right )}}{b c}\right ] \] Input:

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

Output:

[(c*d^(3/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d) 
*b*e - (c*d - b*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)))/(b*c), (c*d^(3/2)*log((e*x - 2* 
sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d)*b*e + 2*(c*d - b*e)*sqrt 
(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) 
)/(b*c), (2*c*sqrt(-d)*d*arctan(sqrt(-d)/sqrt(e*x + d)) + 2*sqrt(e*x + d)* 
b*e - (c*d - b*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)))/(b*c), 2*(c*sqrt(-d)*d*arctan(sq 
rt(-d)/sqrt(e*x + d)) + sqrt(e*x + d)*b*e + (c*d - b*e)*sqrt(-(c*d - b*e)/ 
c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)))/(b*c)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (80) = 160\).

Time = 2.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.83 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\begin {cases} \frac {2 \left (\frac {e^{2} \sqrt {d + e x}}{c} + \frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} - \frac {e \left (b e - c d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{2} \sqrt {\frac {b e - c d}{c}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(e**2*sqrt(d + e*x)/c + d**2*e*atan(sqrt(d + e*x)/sqrt(-d))/( 
b*sqrt(-d)) - e*(b*e - c*d)**2*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b* 
c**2*sqrt((b*e - c*d)/c)))/e, Ne(e, 0)), (d**(3/2)*(-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*P 
iecewise(((b/(2*c) + x)/b, Eq(c, 0)), (log(b + 2*c*(b/(2*c) + x))/(2*c), T 
rue))/b), True))
 

Maxima [F(-2)]

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

integrate((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 = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 \, d^{2} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, \sqrt {e x + d} e}{c} - \frac {2 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 697, normalized size of antiderivative = 7.58 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2\,e\,\sqrt {d+e\,x}}{c}-\frac {2\,\mathrm {atanh}\left (\frac {16\,b^3\,e^6\,\sqrt {d^3}\,\sqrt {d+e\,x}}{16\,b^3\,d^2\,e^6-64\,b^2\,c\,d^3\,e^5+96\,b\,c^2\,d^4\,e^4-48\,c^3\,d^5\,e^3}+\frac {48\,c^2\,d^3\,e^3\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}+\frac {64\,b^2\,d\,e^5\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}-\frac {96\,b\,c\,d^2\,e^4\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}\right )\,\sqrt {d^3}}{b}+\frac {2\,\mathrm {atanh}\left (\frac {48\,d^3\,e^3\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{48\,c^3\,d^5\,e^3-80\,b^3\,d^2\,e^6-144\,b\,c^2\,d^4\,e^4+160\,b^2\,c\,d^3\,e^5+\frac {16\,b^4\,d\,e^7}{c}}+\frac {16\,b^2\,d\,e^5\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{16\,b^4\,c\,d\,e^7-80\,b^3\,c^2\,d^2\,e^6+160\,b^2\,c^3\,d^3\,e^5-144\,b\,c^4\,d^4\,e^4+48\,c^5\,d^5\,e^3}-\frac {48\,b\,d^2\,e^4\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{16\,b^4\,d\,e^7-80\,b^3\,c\,d^2\,e^6+160\,b^2\,c^2\,d^3\,e^5-144\,b\,c^3\,d^4\,e^4+48\,c^4\,d^5\,e^3}\right )\,\sqrt {-c^3\,{\left (b\,e-c\,d\right )}^3}}{b\,c^3} \] Input:

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

Output:

(2*e*(d + e*x)^(1/2))/c - (2*atanh((16*b^3*e^6*(d^3)^(1/2)*(d + e*x)^(1/2) 
)/(16*b^3*d^2*e^6 - 48*c^3*d^5*e^3 + 96*b*c^2*d^4*e^4 - 64*b^2*c*d^3*e^5) 
+ (48*c^2*d^3*e^3*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d^3*e^5 + 48*c^2*d^ 
5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4) + (64*b^2*d*e^5*(d^3)^(1/2)*( 
d + e*x)^(1/2))/(64*b^2*d^3*e^5 + 48*c^2*d^5*e^3 - (16*b^3*d^2*e^6)/c - 96 
*b*c*d^4*e^4) - (96*b*c*d^2*e^4*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d^3*e 
^5 + 48*c^2*d^5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4))*(d^3)^(1/2))/b 
 + (2*atanh((48*d^3*e^3*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4 
*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(48*c^3*d^5*e^3 - 80*b^3*d^2*e^6 - 144*b*c^ 
2*d^4*e^4 + 160*b^2*c*d^3*e^5 + (16*b^4*d*e^7)/c) + (16*b^2*d*e^5*(d + e*x 
)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/( 
48*c^5*d^5*e^3 - 144*b*c^4*d^4*e^4 + 160*b^2*c^3*d^3*e^5 - 80*b^3*c^2*d^2* 
e^6 + 16*b^4*c*d*e^7) - (48*b*d^2*e^4*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e 
^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(16*b^4*d*e^7 + 48*c^4*d^5*e^ 
3 - 144*b*c^3*d^4*e^4 - 80*b^3*c*d^2*e^6 + 160*b^2*c^2*d^3*e^5))*(-c^3*(b* 
e - c*d)^3)^(1/2))/(b*c^3)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.46 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {-2 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) b e +2 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) c d +2 \sqrt {e x +d}\, b c e +\sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) c^{2} d -\sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) c^{2} d}{b \,c^{2}} \] Input:

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

Output:

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