\(\int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx\) [81]

Optimal result

Integrand size = 22, antiderivative size = 130 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\frac {2 (d e-c f) (d g-c h) \sqrt {e+f x}}{d^3}+\frac {2 (d g-c h) (e+f x)^{3/2}}{3 d^2}+\frac {2 h (e+f x)^{5/2}}{5 d f}-\frac {2 (d e-c f)^{3/2} (d g-c h) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}} \] Output:

2*(-c*f+d*e)*(-c*h+d*g)*(f*x+e)^(1/2)/d^3+2/3*(-c*h+d*g)*(f*x+e)^(3/2)/d^2 
+2/5*h*(f*x+e)^(5/2)/d/f-2*(-c*f+d*e)^(3/2)*(-c*h+d*g)*arctanh(d^(1/2)*(f* 
x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\frac {2 \sqrt {e+f x} \left (15 c^2 f^2 h-5 c d f (3 f g+4 e h+f h x)+d^2 \left (3 e^2 h+f^2 x (5 g+3 h x)+e f (20 g+6 h x)\right )\right )}{15 d^3 f}+\frac {2 (-d e+c f)^{3/2} (d g-c h) \arctan \left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2}} \] Input:

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

Output:

(2*Sqrt[e + f*x]*(15*c^2*f^2*h - 5*c*d*f*(3*f*g + 4*e*h + f*h*x) + d^2*(3* 
e^2*h + f^2*x*(5*g + 3*h*x) + e*f*(20*g + 6*h*x))))/(15*d^3*f) + (2*(-(d*e 
) + c*f)^(3/2)*(d*g - c*h)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c* 
f]])/d^(7/2)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {90, 60, 60, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {(d g-c h) \int \frac {(e+f x)^{3/2}}{c+d x}dx}{d}+\frac {2 h (e+f x)^{5/2}}{5 d f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(d g-c h) \left (\frac {(d e-c f) \int \frac {\sqrt {e+f x}}{c+d x}dx}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}+\frac {2 h (e+f x)^{5/2}}{5 d f}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(d g-c h) \left (\frac {(d e-c f) \left (\frac {(d e-c f) \int \frac {1}{(c+d x) \sqrt {e+f x}}dx}{d}+\frac {2 \sqrt {e+f x}}{d}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}+\frac {2 h (e+f x)^{5/2}}{5 d f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(d g-c h) \left (\frac {(d e-c f) \left (\frac {2 (d e-c f) \int \frac {1}{c+\frac {d (e+f x)}{f}-\frac {d e}{f}}d\sqrt {e+f x}}{d f}+\frac {2 \sqrt {e+f x}}{d}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}+\frac {2 h (e+f x)^{5/2}}{5 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(d g-c h) \left (\frac {(d e-c f) \left (\frac {2 \sqrt {e+f x}}{d}-\frac {2 \sqrt {d e-c f} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2}}\right )}{d}+\frac {2 (e+f x)^{3/2}}{3 d}\right )}{d}+\frac {2 h (e+f x)^{5/2}}{5 d f}\)

Input:

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

Output:

(2*h*(e + f*x)^(5/2))/(5*d*f) + ((d*g - c*h)*((2*(e + f*x)^(3/2))/(3*d) + 
((d*e - c*f)*((2*Sqrt[e + f*x])/d - (2*Sqrt[d*e - c*f]*ArcTanh[(Sqrt[d]*Sq 
rt[e + f*x])/Sqrt[d*e - c*f]])/d^(3/2)))/d))/d
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 = 0.39 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {2 \left (\left (\frac {\left (f x +e \right )^{2} h}{5}+\frac {4 f g \left (\frac {f x}{4}+e \right )}{3}\right ) d^{2}-\frac {4 c f \left (\left (\frac {f x}{4}+e \right ) h +\frac {3 f g}{4}\right ) d}{3}+c^{2} f^{2} h \right ) \sqrt {\left (c f -d e \right ) d}\, \sqrt {f x +e}-2 f \left (c f -d e \right )^{2} \left (c h -d g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{f \,d^{3} \sqrt {\left (c f -d e \right ) d}}\)	\(137\)
risch	\(\frac {2 \left (3 f^{2} d^{2} h \,x^{2}-5 f^{2} c d h x +6 e f x h \,d^{2}+5 f^{2} d^{2} g x +15 c^{2} f^{2} h -20 c d e h f -15 c d \,f^{2} g +3 d^{2} e^{2} h +20 d^{2} e g f \right ) \sqrt {f x +e}}{15 f \,d^{3}}-\frac {2 \left (c^{3} f^{2} h -2 c^{2} d e f h -c^{2} d \,f^{2} g +c \,d^{2} e^{2} h +2 c \,d^{2} e f g -d^{3} e^{2} g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}\)	\(193\)
derivativedivides	\(\frac {\frac {2 \left (\frac {h \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}-\frac {c d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {d^{2} f g \left (f x +e \right )^{\frac {3}{2}}}{3}+c^{2} f^{2} h \sqrt {f x +e}-c d e f h \sqrt {f x +e}-c d \,f^{2} g \sqrt {f x +e}+d^{2} e f g \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f \left (c^{3} f^{2} h -2 c^{2} d e f h -c^{2} d \,f^{2} g +c \,d^{2} e^{2} h +2 c \,d^{2} e f g -d^{3} e^{2} g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f}\)	\(205\)
default	\(\frac {\frac {2 \left (\frac {h \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}-\frac {c d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {d^{2} f g \left (f x +e \right )^{\frac {3}{2}}}{3}+c^{2} f^{2} h \sqrt {f x +e}-c d e f h \sqrt {f x +e}-c d \,f^{2} g \sqrt {f x +e}+d^{2} e f g \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f \left (c^{3} f^{2} h -2 c^{2} d e f h -c^{2} d \,f^{2} g +c \,d^{2} e^{2} h +2 c \,d^{2} e f g -d^{3} e^{2} g \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f}\)	\(205\)

Input:

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

Output:

2*(((1/5*(f*x+e)^2*h+4/3*f*g*(1/4*f*x+e))*d^2-4/3*c*f*((1/4*f*x+e)*h+3/4*f 
*g)*d+c^2*f^2*h)*((c*f-d*e)*d)^(1/2)*(f*x+e)^(1/2)-f*(c*f-d*e)^2*(c*h-d*g) 
*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)))/((c*f-d*e)*d)^(1/2)/f/d^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.00 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\left [\frac {15 \, {\left ({\left (d^{2} e f - c d f^{2}\right )} g - {\left (c d e f - c^{2} f^{2}\right )} h\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) + 2 \, {\left (3 \, d^{2} f^{2} h x^{2} + 5 \, {\left (4 \, d^{2} e f - 3 \, c d f^{2}\right )} g + {\left (3 \, d^{2} e^{2} - 20 \, c d e f + 15 \, c^{2} f^{2}\right )} h + {\left (5 \, d^{2} f^{2} g + {\left (6 \, d^{2} e f - 5 \, c d f^{2}\right )} h\right )} x\right )} \sqrt {f x + e}}{15 \, d^{3} f}, -\frac {2 \, {\left (15 \, {\left ({\left (d^{2} e f - c d f^{2}\right )} g - {\left (c d e f - c^{2} f^{2}\right )} h\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) - {\left (3 \, d^{2} f^{2} h x^{2} + 5 \, {\left (4 \, d^{2} e f - 3 \, c d f^{2}\right )} g + {\left (3 \, d^{2} e^{2} - 20 \, c d e f + 15 \, c^{2} f^{2}\right )} h + {\left (5 \, d^{2} f^{2} g + {\left (6 \, d^{2} e f - 5 \, c d f^{2}\right )} h\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, d^{3} f}\right ] \] Input:

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

Output:

[1/15*(15*((d^2*e*f - c*d*f^2)*g - (c*d*e*f - c^2*f^2)*h)*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)) + 2*(3*d^2*f^2*h*x^2 + 5*(4*d^2*e*f - 3*c*d*f^2)*g + (3*d^2*e^2 - 2 
0*c*d*e*f + 15*c^2*f^2)*h + (5*d^2*f^2*g + (6*d^2*e*f - 5*c*d*f^2)*h)*x)*s 
qrt(f*x + e))/(d^3*f), -2/15*(15*((d^2*e*f - c*d*f^2)*g - (c*d*e*f - c^2*f 
^2)*h)*sqrt(-(d*e - c*f)/d)*arctan(-sqrt(f*x + e)*d*sqrt(-(d*e - c*f)/d)/( 
d*e - c*f)) - (3*d^2*f^2*h*x^2 + 5*(4*d^2*e*f - 3*c*d*f^2)*g + (3*d^2*e^2 
- 20*c*d*e*f + 15*c^2*f^2)*h + (5*d^2*f^2*g + (6*d^2*e*f - 5*c*d*f^2)*h)*x 
)*sqrt(f*x + e))/(d^3*f)]
 

Sympy [A] (verification not implemented)

Time = 3.54 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\begin {cases} \frac {2 \left (\frac {h \left (e + f x\right )^{\frac {5}{2}}}{5 d} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (- c f h + d f g\right )}{3 d^{2}} + \frac {\sqrt {e + f x} \left (c^{2} f^{2} h - c d e f h - c d f^{2} g + d^{2} e f g\right )}{d^{3}} - \frac {f \left (c f - d e\right )^{2} \left (c h - d g\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{4} \sqrt {\frac {c f - d e}{d}}}\right )}{f} & \text {for}\: f \neq 0 \\e^{\frac {3}{2}} \left (\frac {h x}{d} - \frac {\left (c h - d g\right ) \left (\begin {cases} \frac {x}{c} & \text {for}\: d = 0 \\\frac {\log {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right )}{d}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((f*x+e)**(3/2)*(h*x+g)/(d*x+c),x)
 

Output:

Piecewise((2*(h*(e + f*x)**(5/2)/(5*d) + (e + f*x)**(3/2)*(-c*f*h + d*f*g) 
/(3*d**2) + sqrt(e + f*x)*(c**2*f**2*h - c*d*e*f*h - c*d*f**2*g + d**2*e*f 
*g)/d**3 - f*(c*f - d*e)**2*(c*h - d*g)*atan(sqrt(e + f*x)/sqrt((c*f - d*e 
)/d))/(d**4*sqrt((c*f - d*e)/d)))/f, Ne(f, 0)), (e**(3/2)*(h*x/d - (c*h - 
d*g)*Piecewise((x/c, Eq(d, 0)), (log(c + d*x)/d, True))/d), True))
 

Maxima [F(-2)]

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

integrate((f*x+e)^(3/2)*(h*x+g)/(d*x+c),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (110) = 220\).

Time = 0.13 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.74 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\frac {2 \, {\left (d^{3} e^{2} g - 2 \, c d^{2} e f g + c^{2} d f^{2} g - c d^{2} e^{2} h + 2 \, c^{2} d e f h - c^{3} f^{2} h\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {-d^{2} e + c d f}}\right )}{\sqrt {-d^{2} e + c d f} d^{3}} + \frac {2 \, {\left (5 \, {\left (f x + e\right )}^{\frac {3}{2}} d^{4} f^{5} g + 15 \, \sqrt {f x + e} d^{4} e f^{5} g - 15 \, \sqrt {f x + e} c d^{3} f^{6} g + 3 \, {\left (f x + e\right )}^{\frac {5}{2}} d^{4} f^{4} h - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} c d^{3} f^{5} h - 15 \, \sqrt {f x + e} c d^{3} e f^{5} h + 15 \, \sqrt {f x + e} c^{2} d^{2} f^{6} h\right )}}{15 \, d^{5} f^{5}} \] Input:

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

Output:

2*(d^3*e^2*g - 2*c*d^2*e*f*g + c^2*d*f^2*g - c*d^2*e^2*h + 2*c^2*d*e*f*h - 
 c^3*f^2*h)*arctan(sqrt(f*x + e)*d/sqrt(-d^2*e + c*d*f))/(sqrt(-d^2*e + c* 
d*f)*d^3) + 2/15*(5*(f*x + e)^(3/2)*d^4*f^5*g + 15*sqrt(f*x + e)*d^4*e*f^5 
*g - 15*sqrt(f*x + e)*c*d^3*f^6*g + 3*(f*x + e)^(5/2)*d^4*f^4*h - 5*(f*x + 
 e)^(3/2)*c*d^3*f^5*h - 15*sqrt(f*x + e)*c*d^3*e*f^5*h + 15*sqrt(f*x + e)* 
c^2*d^2*f^6*h)/(d^5*f^5)
 

Mupad [B] (verification not implemented)

Time = 2.39 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.82 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\frac {2\,h\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f}-{\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,e\,h-2\,f\,g}{3\,d\,f}+\frac {2\,h\,\left (c\,f^2-d\,e\,f\right )}{3\,d^2\,f^2}\right )+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (c\,f-d\,e\right )}^{3/2}\,\left (c\,h-d\,g\right )}{-h\,c^3\,f^2+2\,h\,c^2\,d\,e\,f+g\,c^2\,d\,f^2-h\,c\,d^2\,e^2-2\,g\,c\,d^2\,e\,f+g\,d^3\,e^2}\right )\,{\left (c\,f-d\,e\right )}^{3/2}\,\left (c\,h-d\,g\right )}{d^{7/2}}+\frac {\sqrt {e+f\,x}\,\left (c\,f^2-d\,e\,f\right )\,\left (\frac {2\,e\,h-2\,f\,g}{d\,f}+\frac {2\,h\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{d\,f} \] Input:

int(((e + f*x)^(3/2)*(g + h*x))/(c + d*x),x)
 

Output:

(2*h*(e + f*x)^(5/2))/(5*d*f) - (e + f*x)^(3/2)*((2*e*h - 2*f*g)/(3*d*f) + 
 (2*h*(c*f^2 - d*e*f))/(3*d^2*f^2)) + (2*atan((d^(1/2)*(e + f*x)^(1/2)*(c* 
f - d*e)^(3/2)*(c*h - d*g))/(d^3*e^2*g - c^3*f^2*h - c*d^2*e^2*h + c^2*d*f 
^2*g - 2*c*d^2*e*f*g + 2*c^2*d*e*f*h))*(c*f - d*e)^(3/2)*(c*h - d*g))/d^(7 
/2) + ((e + f*x)^(1/2)*(c*f^2 - d*e*f)*((2*e*h - 2*f*g)/(d*f) + (2*h*(c*f^ 
2 - d*e*f))/(d^2*f^2)))/(d*f)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.48 \[ \int \frac {(e+f x)^{3/2} (g+h x)}{c+d x} \, dx=\frac {-2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c^{2} f^{2} h +2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c d e f h +2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) c d \,f^{2} g -2 \sqrt {d}\, \sqrt {c f -d e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, d}{\sqrt {d}\, \sqrt {c f -d e}}\right ) d^{2} e f g +2 \sqrt {f x +e}\, c^{2} d \,f^{2} h -\frac {8 \sqrt {f x +e}\, c \,d^{2} e f h}{3}-2 \sqrt {f x +e}\, c \,d^{2} f^{2} g -\frac {2 \sqrt {f x +e}\, c \,d^{2} f^{2} h x}{3}+\frac {2 \sqrt {f x +e}\, d^{3} e^{2} h}{5}+\frac {8 \sqrt {f x +e}\, d^{3} e f g}{3}+\frac {4 \sqrt {f x +e}\, d^{3} e f h x}{5}+\frac {2 \sqrt {f x +e}\, d^{3} f^{2} g x}{3}+\frac {2 \sqrt {f x +e}\, d^{3} f^{2} h \,x^{2}}{5}}{d^{4} f} \] Input:

int((f*x+e)^(3/2)*(h*x+g)/(d*x+c),x)
 

Output:

(2*( - 15*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f 
 - d*e)))*c**2*f**2*h + 15*sqrt(d)*sqrt(c*f - d*e)*atan((sqrt(e + f*x)*d)/ 
(sqrt(d)*sqrt(c*f - d*e)))*c*d*e*f*h + 15*sqrt(d)*sqrt(c*f - d*e)*atan((sq 
rt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*c*d*f**2*g - 15*sqrt(d)*sqrt(c*f 
 - d*e)*atan((sqrt(e + f*x)*d)/(sqrt(d)*sqrt(c*f - d*e)))*d**2*e*f*g + 15* 
sqrt(e + f*x)*c**2*d*f**2*h - 20*sqrt(e + f*x)*c*d**2*e*f*h - 15*sqrt(e + 
f*x)*c*d**2*f**2*g - 5*sqrt(e + f*x)*c*d**2*f**2*h*x + 3*sqrt(e + f*x)*d** 
3*e**2*h + 20*sqrt(e + f*x)*d**3*e*f*g + 6*sqrt(e + f*x)*d**3*e*f*h*x + 5* 
sqrt(e + f*x)*d**3*f**2*g*x + 3*sqrt(e + f*x)*d**3*f**2*h*x**2))/(15*d**4* 
f)