\(\int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx\) [5]

Optimal result

Integrand size = 27, antiderivative size = 156 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx=\frac {2 (b c-a d) (b g-a h) \sqrt {e+f x}}{b^3}-\frac {2 (d (b e+a f) h-b f (d g+c h)) (e+f x)^{3/2}}{3 b^2 f^2}+\frac {2 d h (e+f x)^{5/2}}{5 b f^2}-\frac {2 (b c-a d) \sqrt {b e-a f} (b g-a h) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx=\frac {2 \sqrt {e+f x} \left (15 a^2 d f^2 h-5 a b f (3 c f h+d (3 f g+e h+f h x))+b^2 (-d (e+f x) (-5 f g+2 e h-3 f h x)+5 c f (3 f g+e h+f h x))\right )}{15 b^3 f^2}-\frac {2 (b c-a d) \sqrt {-b e+a f} (b g-a h) \arctan \left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {-b e+a f}}\right )}{b^{7/2}} \] Input:

Integrate[((c + d*x)*Sqrt[e + f*x]*(g + h*x))/(a + b*x),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {164, 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.

\(\displaystyle \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {(b c-a d) (b g-a h) \int \frac {\sqrt {e+f x}}{a+b x}dx}{b^2}-\frac {2 (e+f x)^{3/2} (5 a d f h-5 b f (c h+d g)+2 b d e h-3 b d f h x)}{15 b^2 f^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) (b g-a h) \left (\frac {(b e-a f) \int \frac {1}{(a+b x) \sqrt {e+f x}}dx}{b}+\frac {2 \sqrt {e+f x}}{b}\right )}{b^2}-\frac {2 (e+f x)^{3/2} (5 a d f h-5 b f (c h+d g)+2 b d e h-3 b d f h x)}{15 b^2 f^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(b c-a d) (b g-a h) \left (\frac {2 (b e-a f) \int \frac {1}{a+\frac {b (e+f x)}{f}-\frac {b e}{f}}d\sqrt {e+f x}}{b f}+\frac {2 \sqrt {e+f x}}{b}\right )}{b^2}-\frac {2 (e+f x)^{3/2} (5 a d f h-5 b f (c h+d g)+2 b d e h-3 b d f h x)}{15 b^2 f^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(b c-a d) (b g-a h) \left (\frac {2 \sqrt {e+f x}}{b}-\frac {2 \sqrt {b e-a f} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e+f x}}{\sqrt {b e-a f}}\right )}{b^{3/2}}\right )}{b^2}-\frac {2 (e+f x)^{3/2} (5 a d f h-5 b f (c h+d g)+2 b d e h-3 b d f h x)}{15 b^2 f^2}\)

Input:

Int[((c + d*x)*Sqrt[e + f*x]*(g + h*x))/(a + b*x),x]
 

Output:

(-2*(e + f*x)^(3/2)*(2*b*d*e*h + 5*a*d*f*h - 5*b*f*(d*g + c*h) - 3*b*d*f*h 
*x))/(15*b^2*f^2) + ((b*c - a*d)*(b*g - a*h)*((2*Sqrt[e + f*x])/b - (2*Sqr 
t[b*e - a*f]*ArcTanh[(Sqrt[b]*Sqrt[e + f*x])/Sqrt[b*e - a*f]])/b^(3/2)))/b 
^2
 

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 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 = 0.59 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\left (\left (\frac {\left (\frac {\left (3 f x -2 e \right ) d}{5}+c f \right ) \left (f x +e \right ) h}{3}+\left (\frac {\left (f x +e \right ) d}{3}+c f \right ) g f \right ) b^{2}-a \left (\left (\frac {\left (f x +e \right ) d}{3}+c f \right ) h +d f g \right ) f b +a^{2} d \,f^{2} h \right ) \sqrt {\left (a f -b e \right ) b}\, \sqrt {f x +e}+f^{2} \left (a h -b g \right ) \left (a f -b e \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )\right )}{\sqrt {\left (a f -b e \right ) b}\, f^{2} b^{3}}\)	\(170\)
risch	\(\frac {2 \left (3 d h \,b^{2} f^{2} x^{2}-5 a b d \,f^{2} h x +5 b^{2} c \,f^{2} h x +b^{2} d e f h x +5 b^{2} d \,f^{2} g x +15 a^{2} d \,f^{2} h -15 a b c \,f^{2} h -5 a b d e f h -15 a b d \,f^{2} g +5 b^{2} c e f h +15 b^{2} c \,f^{2} g -2 b^{2} d \,e^{2} h +5 b^{2} d e f g \right ) \sqrt {f x +e}}{15 f^{2} b^{3}}-\frac {2 \left (a^{3} d f h -a^{2} b c f h -a^{2} b d e h -a^{2} b d f g +a \,b^{2} c e h +a \,b^{2} c f g +a \,b^{2} d e g -c e g \,b^{3}\right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{b^{3} \sqrt {\left (a f -b e \right ) b}}\)	\(252\)
derivativedivides	\(\frac {\frac {2 \left (\frac {d h \left (f x +e \right )^{\frac {5}{2}} b^{2}}{5}-\frac {a b d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} c f h \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} d e h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} d f g \left (f x +e \right )^{\frac {3}{2}}}{3}+a^{2} d \,f^{2} h \sqrt {f x +e}-a b c \,f^{2} h \sqrt {f x +e}-a b d \,f^{2} g \sqrt {f x +e}+b^{2} c \,f^{2} g \sqrt {f x +e}\right )}{b^{3}}-\frac {2 f^{2} \left (a^{3} d f h -a^{2} b c f h -a^{2} b d e h -a^{2} b d f g +a \,b^{2} c e h +a \,b^{2} c f g +a \,b^{2} d e g -c e g \,b^{3}\right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{b^{3} \sqrt {\left (a f -b e \right ) b}}}{f^{2}}\)	\(258\)
default	\(\frac {\frac {2 \left (\frac {d h \left (f x +e \right )^{\frac {5}{2}} b^{2}}{5}-\frac {a b d f h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} c f h \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} d e h \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b^{2} d f g \left (f x +e \right )^{\frac {3}{2}}}{3}+a^{2} d \,f^{2} h \sqrt {f x +e}-a b c \,f^{2} h \sqrt {f x +e}-a b d \,f^{2} g \sqrt {f x +e}+b^{2} c \,f^{2} g \sqrt {f x +e}\right )}{b^{3}}-\frac {2 f^{2} \left (a^{3} d f h -a^{2} b c f h -a^{2} b d e h -a^{2} b d f g +a \,b^{2} c e h +a \,b^{2} c f g +a \,b^{2} d e g -c e g \,b^{3}\right ) \arctan \left (\frac {b \sqrt {f x +e}}{\sqrt {\left (a f -b e \right ) b}}\right )}{b^{3} \sqrt {\left (a f -b e \right ) b}}}{f^{2}}\)	\(258\)

Input:

int((d*x+c)*(f*x+e)^(1/2)*(h*x+g)/(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 468, normalized size of antiderivative = 3.00 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx=\left [\frac {15 \, {\left ({\left (b^{2} c - a b d\right )} f^{2} g - {\left (a b c - a^{2} d\right )} f^{2} h\right )} \sqrt {\frac {b e - a f}{b}} \log \left (\frac {b f x + 2 \, b e - a f - 2 \, \sqrt {f x + e} b \sqrt {\frac {b e - a f}{b}}}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} d f^{2} h x^{2} + 5 \, {\left (b^{2} d e f + 3 \, {\left (b^{2} c - a b d\right )} f^{2}\right )} g - {\left (2 \, b^{2} d e^{2} - 5 \, {\left (b^{2} c - a b d\right )} e f + 15 \, {\left (a b c - a^{2} d\right )} f^{2}\right )} h + {\left (5 \, b^{2} d f^{2} g + {\left (b^{2} d e f + 5 \, {\left (b^{2} c - a b d\right )} f^{2}\right )} h\right )} x\right )} \sqrt {f x + e}}{15 \, b^{3} f^{2}}, -\frac {2 \, {\left (15 \, {\left ({\left (b^{2} c - a b d\right )} f^{2} g - {\left (a b c - a^{2} d\right )} f^{2} h\right )} \sqrt {-\frac {b e - a f}{b}} \arctan \left (-\frac {\sqrt {f x + e} b \sqrt {-\frac {b e - a f}{b}}}{b e - a f}\right ) - {\left (3 \, b^{2} d f^{2} h x^{2} + 5 \, {\left (b^{2} d e f + 3 \, {\left (b^{2} c - a b d\right )} f^{2}\right )} g - {\left (2 \, b^{2} d e^{2} - 5 \, {\left (b^{2} c - a b d\right )} e f + 15 \, {\left (a b c - a^{2} d\right )} f^{2}\right )} h + {\left (5 \, b^{2} d f^{2} g + {\left (b^{2} d e f + 5 \, {\left (b^{2} c - a b d\right )} f^{2}\right )} h\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, b^{3} f^{2}}\right ] \] Input:

integrate((d*x+c)*(f*x+e)^(1/2)*(h*x+g)/(b*x+a),x, algorithm="fricas")
 

Output:

[1/15*(15*((b^2*c - a*b*d)*f^2*g - (a*b*c - a^2*d)*f^2*h)*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)) + 2*(3*b^2*d*f^2*h*x^2 + 5*(b^2*d*e*f + 3*(b^2*c - a*b*d)*f^2)*g - 
(2*b^2*d*e^2 - 5*(b^2*c - a*b*d)*e*f + 15*(a*b*c - a^2*d)*f^2)*h + (5*b^2* 
d*f^2*g + (b^2*d*e*f + 5*(b^2*c - a*b*d)*f^2)*h)*x)*sqrt(f*x + e))/(b^3*f^ 
2), -2/15*(15*((b^2*c - a*b*d)*f^2*g - (a*b*c - a^2*d)*f^2*h)*sqrt(-(b*e - 
 a*f)/b)*arctan(-sqrt(f*x + e)*b*sqrt(-(b*e - a*f)/b)/(b*e - a*f)) - (3*b^ 
2*d*f^2*h*x^2 + 5*(b^2*d*e*f + 3*(b^2*c - a*b*d)*f^2)*g - (2*b^2*d*e^2 - 5 
*(b^2*c - a*b*d)*e*f + 15*(a*b*c - a^2*d)*f^2)*h + (5*b^2*d*f^2*g + (b^2*d 
*e*f + 5*(b^2*c - a*b*d)*f^2)*h)*x)*sqrt(f*x + e))/(b^3*f^2)]
 

Sympy [A] (verification not implemented)

Time = 13.55 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.51 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {d h \left (e + f x\right )^{\frac {5}{2}}}{5 b f} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (- a d f h + b c f h - b d e h + b d f g\right )}{3 b^{2} f} + \frac {\sqrt {e + f x} \left (a^{2} d f h - a b c f h - a b d f g + b^{2} c f g\right )}{b^{3}} - \frac {f \left (a d - b c\right ) \left (a f - b e\right ) \left (a h - b g\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {a f - b e}{b}}} \right )}}{b^{4} \sqrt {\frac {a f - b e}{b}}}\right )}{f} & \text {for}\: f \neq 0 \\\sqrt {e} \left (\frac {d h x^{2}}{2 b} + \frac {x \left (- a d h + b c h + b d g\right )}{b^{2}} + \frac {\left (a d - b c\right ) \left (a h - b g\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{2}}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((d*x+c)*(f*x+e)**(1/2)*(h*x+g)/(b*x+a),x)
 

Output:

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

Maxima [F(-2)]

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

integrate((d*x+c)*(f*x+e)^(1/2)*(h*x+g)/(b*x+a),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(a*f-b*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. 278 vs. \(2 (136) = 272\).

Time = 0.13 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.78 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx=\frac {2 \, {\left (b^{3} c e g - a b^{2} d e g - a b^{2} c f g + a^{2} b d f g - a b^{2} c e h + a^{2} b d e h + a^{2} b c f h - a^{3} d f h\right )} \arctan \left (\frac {\sqrt {f x + e} b}{\sqrt {-b^{2} e + a b f}}\right )}{\sqrt {-b^{2} e + a b f} b^{3}} + \frac {2 \, {\left (5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{4} d f^{9} g + 15 \, \sqrt {f x + e} b^{4} c f^{10} g - 15 \, \sqrt {f x + e} a b^{3} d f^{10} g + 3 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{4} d f^{8} h - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{4} d e f^{8} h + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{4} c f^{9} h - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} a b^{3} d f^{9} h - 15 \, \sqrt {f x + e} a b^{3} c f^{10} h + 15 \, \sqrt {f x + e} a^{2} b^{2} d f^{10} h\right )}}{15 \, b^{5} f^{10}} \] Input:

integrate((d*x+c)*(f*x+e)^(1/2)*(h*x+g)/(b*x+a),x, algorithm="giac")
 

Output:

2*(b^3*c*e*g - a*b^2*d*e*g - a*b^2*c*f*g + a^2*b*d*f*g - a*b^2*c*e*h + a^2 
*b*d*e*h + a^2*b*c*f*h - a^3*d*f*h)*arctan(sqrt(f*x + e)*b/sqrt(-b^2*e + a 
*b*f))/(sqrt(-b^2*e + a*b*f)*b^3) + 2/15*(5*(f*x + e)^(3/2)*b^4*d*f^9*g + 
15*sqrt(f*x + e)*b^4*c*f^10*g - 15*sqrt(f*x + e)*a*b^3*d*f^10*g + 3*(f*x + 
 e)^(5/2)*b^4*d*f^8*h - 5*(f*x + e)^(3/2)*b^4*d*e*f^8*h + 5*(f*x + e)^(3/2 
)*b^4*c*f^9*h - 5*(f*x + e)^(3/2)*a*b^3*d*f^9*h - 15*sqrt(f*x + e)*a*b^3*c 
*f^10*h + 15*sqrt(f*x + e)*a^2*b^2*d*f^10*h)/(b^5*f^10)
 

Mupad [B] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.48 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx={\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,c\,f\,h-4\,d\,e\,h+2\,d\,f\,g}{3\,b\,f^2}-\frac {2\,d\,h\,\left (a\,f^3-b\,e\,f^2\right )}{3\,b^2\,f^4}\right )-\sqrt {e+f\,x}\,\left (\frac {2\,\left (c\,f-d\,e\right )\,\left (e\,h-f\,g\right )}{b\,f^2}+\frac {\left (a\,f^3-b\,e\,f^2\right )\,\left (\frac {2\,c\,f\,h-4\,d\,e\,h+2\,d\,f\,g}{b\,f^2}-\frac {2\,d\,h\,\left (a\,f^3-b\,e\,f^2\right )}{b^2\,f^4}\right )}{b\,f^2}\right )+\frac {2\,d\,h\,{\left (e+f\,x\right )}^{5/2}}{5\,b\,f^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e+f\,x}\,1{}\mathrm {i}}{\sqrt {b\,e-a\,f}}\right )\,\left (a\,d-b\,c\right )\,\sqrt {b\,e-a\,f}\,\left (a\,h-b\,g\right )\,2{}\mathrm {i}}{b^{7/2}} \] Input:

int(((e + f*x)^(1/2)*(g + h*x)*(c + d*x))/(a + b*x),x)
 

Output:

(e + f*x)^(3/2)*((2*c*f*h - 4*d*e*h + 2*d*f*g)/(3*b*f^2) - (2*d*h*(a*f^3 - 
 b*e*f^2))/(3*b^2*f^4)) - (e + f*x)^(1/2)*((2*(c*f - d*e)*(e*h - f*g))/(b* 
f^2) + ((a*f^3 - b*e*f^2)*((2*c*f*h - 4*d*e*h + 2*d*f*g)/(b*f^2) - (2*d*h* 
(a*f^3 - b*e*f^2))/(b^2*f^4)))/(b*f^2)) + (2*d*h*(e + f*x)^(5/2))/(5*b*f^2 
) + (atan((b^(1/2)*(e + f*x)^(1/2)*1i)/(b*e - a*f)^(1/2))*(a*d - b*c)*(b*e 
 - a*f)^(1/2)*(a*h - b*g)*2i)/b^(7/2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.58 \[ \int \frac {(c+d x) \sqrt {e+f x} (g+h x)}{a+b x} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a^{2} d \,f^{2} h +2 \sqrt {b}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a b c \,f^{2} h +2 \sqrt {b}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) a b d \,f^{2} g -2 \sqrt {b}\, \sqrt {a f -b e}\, \mathit {atan} \left (\frac {\sqrt {f x +e}\, b}{\sqrt {b}\, \sqrt {a f -b e}}\right ) b^{2} c \,f^{2} g +2 \sqrt {f x +e}\, a^{2} b d \,f^{2} h -2 \sqrt {f x +e}\, a \,b^{2} c \,f^{2} h -\frac {2 \sqrt {f x +e}\, a \,b^{2} d e f h}{3}-2 \sqrt {f x +e}\, a \,b^{2} d \,f^{2} g -\frac {2 \sqrt {f x +e}\, a \,b^{2} d \,f^{2} h x}{3}+\frac {2 \sqrt {f x +e}\, b^{3} c e f h}{3}+2 \sqrt {f x +e}\, b^{3} c \,f^{2} g +\frac {2 \sqrt {f x +e}\, b^{3} c \,f^{2} h x}{3}-\frac {4 \sqrt {f x +e}\, b^{3} d \,e^{2} h}{15}+\frac {2 \sqrt {f x +e}\, b^{3} d e f g}{3}+\frac {2 \sqrt {f x +e}\, b^{3} d e f h x}{15}+\frac {2 \sqrt {f x +e}\, b^{3} d \,f^{2} g x}{3}+\frac {2 \sqrt {f x +e}\, b^{3} d \,f^{2} h \,x^{2}}{5}}{b^{4} f^{2}} \] Input:

int((d*x+c)*(f*x+e)^(1/2)*(h*x+g)/(b*x+a),x)
 

Output:

(2*( - 15*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f 
 - b*e)))*a**2*d*f**2*h + 15*sqrt(b)*sqrt(a*f - b*e)*atan((sqrt(e + f*x)*b 
)/(sqrt(b)*sqrt(a*f - b*e)))*a*b*c*f**2*h + 15*sqrt(b)*sqrt(a*f - b*e)*ata 
n((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*a*b*d*f**2*g - 15*sqrt(b)*s 
qrt(a*f - b*e)*atan((sqrt(e + f*x)*b)/(sqrt(b)*sqrt(a*f - b*e)))*b**2*c*f* 
*2*g + 15*sqrt(e + f*x)*a**2*b*d*f**2*h - 15*sqrt(e + f*x)*a*b**2*c*f**2*h 
 - 5*sqrt(e + f*x)*a*b**2*d*e*f*h - 15*sqrt(e + f*x)*a*b**2*d*f**2*g - 5*s 
qrt(e + f*x)*a*b**2*d*f**2*h*x + 5*sqrt(e + f*x)*b**3*c*e*f*h + 15*sqrt(e 
+ f*x)*b**3*c*f**2*g + 5*sqrt(e + f*x)*b**3*c*f**2*h*x - 2*sqrt(e + f*x)*b 
**3*d*e**2*h + 5*sqrt(e + f*x)*b**3*d*e*f*g + sqrt(e + f*x)*b**3*d*e*f*h*x 
 + 5*sqrt(e + f*x)*b**3*d*f**2*g*x + 3*sqrt(e + f*x)*b**3*d*f**2*h*x**2))/ 
(15*b**4*f**2)