\(\int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx\) [16]

Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=2 c^2 e \sqrt {a+b x}+\frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac {2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}-2 \sqrt {a} c^2 e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

2/7*f*(b*x+a)^(3/2)*(d*x+c)^2/b+2/105*(b*x+a)^(3/2)*(8*a^2*d^2*f-14*a*b*d*(2*c*f+d*e)+10*b^2*c*(2*c*f+7*d*e)+3
*b*d*(-4*a*d*f+4*b*c*f+7*b*d*e)*x)/b^3-2*c^2*e*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2)+2*c^2*e*(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {158, 152, 52, 65, 214} \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\frac {2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (2 c f+d e)+5 b^2 c (2 c f+7 d e)\right )+3 b d x (-4 a d f+4 b c f+7 b d e)\right )}{105 b^3}-2 \sqrt {a} c^2 e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 c^2 e \sqrt {a+b x}+\frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b} \]

[In]

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

[Out]

2*c^2*e*Sqrt[a + b*x] + (2*f*(a + b*x)^(3/2)*(c + d*x)^2)/(7*b) + (2*(a + b*x)^(3/2)*(2*(4*a^2*d^2*f - 7*a*b*d
*(d*e + 2*c*f) + 5*b^2*c*(7*d*e + 2*c*f)) + 3*b*d*(7*b*d*e + 4*b*c*f - 4*a*d*f)*x))/(105*b^3) - 2*Sqrt[a]*c^2*
e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> 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] + Dist[(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 158

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 214

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac {2 \int \frac {\sqrt {a+b x} (c+d x) \left (\frac {7 b c e}{2}+\frac {1}{2} (7 b d e+4 b c f-4 a d f) x\right )}{x} \, dx}{7 b} \\ & = \frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac {2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}+\left (c^2 e\right ) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = 2 c^2 e \sqrt {a+b x}+\frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac {2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}+\left (a c^2 e\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = 2 c^2 e \sqrt {a+b x}+\frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac {2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}+\frac {\left (2 a c^2 e\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b} \\ & = 2 c^2 e \sqrt {a+b x}+\frac {2 f (a+b x)^{3/2} (c+d x)^2}{7 b}+\frac {2 (a+b x)^{3/2} \left (2 \left (4 a^2 d^2 f-7 a b d (d e+2 c f)+5 b^2 c (7 d e+2 c f)\right )+3 b d (7 b d e+4 b c f-4 a d f) x\right )}{105 b^3}-2 \sqrt {a} c^2 e \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\frac {2 \sqrt {a+b x} \left (8 a^3 d^2 f-2 a^2 b d (7 d e+14 c f+2 d f x)+a b^2 \left (35 c^2 f+14 c d (5 e+f x)+d^2 x (7 e+3 f x)\right )+b^3 \left (35 c^2 (3 e+f x)+14 c d x (5 e+3 f x)+3 d^2 x^2 (7 e+5 f x)\right )\right )}{105 b^3}-2 \sqrt {a} c^2 e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

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

[Out]

(2*Sqrt[a + b*x]*(8*a^3*d^2*f - 2*a^2*b*d*(7*d*e + 14*c*f + 2*d*f*x) + a*b^2*(35*c^2*f + 14*c*d*(5*e + f*x) +
d^2*x*(7*e + 3*f*x)) + b^3*(35*c^2*(3*e + f*x) + 14*c*d*x*(5*e + 3*f*x) + 3*d^2*x^2*(7*e + 5*f*x))))/(105*b^3)
- 2*Sqrt[a]*c^2*e*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {-210 \sqrt {a}\, b^{3} c^{2} e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+16 \sqrt {b x +a}\, \left (\left (\frac {21 \left (\frac {5 f x}{7}+e \right ) x^{2} d^{2}}{8}+\frac {35 x \left (\frac {3 f x}{5}+e \right ) c d}{4}+\frac {105 \left (\frac {f x}{3}+e \right ) c^{2}}{8}\right ) b^{3}+\frac {35 \left (\frac {\left (\frac {3 f x}{7}+e \right ) x \,d^{2}}{5}+2 \left (\frac {f x}{5}+e \right ) c d +c^{2} f \right ) a \,b^{2}}{8}-\frac {7 \left (\left (\frac {f x}{7}+\frac {e}{2}\right ) d +c f \right ) d \,a^{2} b}{2}+a^{3} d^{2} f \right )}{105 b^{3}}\)	\(145\)
derivativedivides	\(\frac {\frac {2 d^{2} f \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {4 a \,d^{2} f \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {4 b c d f \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 b \,d^{2} e \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} d^{2} f \left (b x +a \right )^{\frac {3}{2}}}{3}-\frac {4 a b c d f \left (b x +a \right )^{\frac {3}{2}}}{3}-\frac {2 a b \,d^{2} e \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} c^{2} f \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {4 b^{2} c d e \left (b x +a \right )^{\frac {3}{2}}}{3}+2 b^{3} c^{2} e \sqrt {b x +a}-2 \sqrt {a}\, b^{3} c^{2} e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b^{3}}\)	\(176\)
default	\(\frac {\frac {2 d^{2} f \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {4 a \,d^{2} f \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {4 b c d f \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 b \,d^{2} e \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} d^{2} f \left (b x +a \right )^{\frac {3}{2}}}{3}-\frac {4 a b c d f \left (b x +a \right )^{\frac {3}{2}}}{3}-\frac {2 a b \,d^{2} e \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} c^{2} f \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {4 b^{2} c d e \left (b x +a \right )^{\frac {3}{2}}}{3}+2 b^{3} c^{2} e \sqrt {b x +a}-2 \sqrt {a}\, b^{3} c^{2} e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b^{3}}\)	\(176\)

[In]

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

[Out]

1/105*(-210*a^(1/2)*b^3*c^2*e*arctanh((b*x+a)^(1/2)/a^(1/2))+16*(b*x+a)^(1/2)*((21/8*(5/7*f*x+e)*x^2*d^2+35/4*
x*(3/5*f*x+e)*c*d+105/8*(1/3*f*x+e)*c^2)*b^3+35/8*(1/5*(3/7*f*x+e)*x*d^2+2*(1/5*f*x+e)*c*d+c^2*f)*a*b^2-7/2*((
1/7*f*x+1/2*e)*d+c*f)*d*a^2*b+a^3*d^2*f))/b^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.78 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\left [\frac {105 \, \sqrt {a} b^{3} c^{2} e \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, b^{3} d^{2} f x^{3} + 3 \, {\left (7 \, b^{3} d^{2} e + {\left (14 \, b^{3} c d + a b^{2} d^{2}\right )} f\right )} x^{2} + 7 \, {\left (15 \, b^{3} c^{2} + 10 \, a b^{2} c d - 2 \, a^{2} b d^{2}\right )} e + {\left (35 \, a b^{2} c^{2} - 28 \, a^{2} b c d + 8 \, a^{3} d^{2}\right )} f + {\left (7 \, {\left (10 \, b^{3} c d + a b^{2} d^{2}\right )} e + {\left (35 \, b^{3} c^{2} + 14 \, a b^{2} c d - 4 \, a^{2} b d^{2}\right )} f\right )} x\right )} \sqrt {b x + a}}{105 \, b^{3}}, \frac {2 \, {\left (105 \, \sqrt {-a} b^{3} c^{2} e \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, b^{3} d^{2} f x^{3} + 3 \, {\left (7 \, b^{3} d^{2} e + {\left (14 \, b^{3} c d + a b^{2} d^{2}\right )} f\right )} x^{2} + 7 \, {\left (15 \, b^{3} c^{2} + 10 \, a b^{2} c d - 2 \, a^{2} b d^{2}\right )} e + {\left (35 \, a b^{2} c^{2} - 28 \, a^{2} b c d + 8 \, a^{3} d^{2}\right )} f + {\left (7 \, {\left (10 \, b^{3} c d + a b^{2} d^{2}\right )} e + {\left (35 \, b^{3} c^{2} + 14 \, a b^{2} c d - 4 \, a^{2} b d^{2}\right )} f\right )} x\right )} \sqrt {b x + a}\right )}}{105 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/105*(105*sqrt(a)*b^3*c^2*e*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*b^3*d^2*f*x^3 + 3*(7*b^3*d^
2*e + (14*b^3*c*d + a*b^2*d^2)*f)*x^2 + 7*(15*b^3*c^2 + 10*a*b^2*c*d - 2*a^2*b*d^2)*e + (35*a*b^2*c^2 - 28*a^2
*b*c*d + 8*a^3*d^2)*f + (7*(10*b^3*c*d + a*b^2*d^2)*e + (35*b^3*c^2 + 14*a*b^2*c*d - 4*a^2*b*d^2)*f)*x)*sqrt(b
*x + a))/b^3, 2/105*(105*sqrt(-a)*b^3*c^2*e*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (15*b^3*d^2*f*x^3 + 3*(7*b^3*d^
2*e + (14*b^3*c*d + a*b^2*d^2)*f)*x^2 + 7*(15*b^3*c^2 + 10*a*b^2*c*d - 2*a^2*b*d^2)*e + (35*a*b^2*c^2 - 28*a^2
*b*c*d + 8*a^3*d^2)*f + (7*(10*b^3*c*d + a*b^2*d^2)*e + (35*b^3*c^2 + 14*a*b^2*c*d - 4*a^2*b*d^2)*f)*x)*sqrt(b
*x + a))/b^3]

Sympy [A] (verification not implemented)

Time = 8.86 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\begin {cases} \frac {2 a c^{2} e \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 c^{2} e \sqrt {a + b x} + \frac {2 d^{2} f \left (a + b x\right )^{\frac {7}{2}}}{7 b^{3}} + \frac {2 \left (a + b x\right )^{\frac {5}{2}} \left (- 2 a d^{2} f + 2 b c d f + b d^{2} e\right )}{5 b^{3}} + \frac {2 \left (a + b x\right )^{\frac {3}{2}} \left (a^{2} d^{2} f - 2 a b c d f - a b d^{2} e + b^{2} c^{2} f + 2 b^{2} c d e\right )}{3 b^{3}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (c^{2} e \log {\left (x \right )} + c^{2} f x + 2 c d e x + \frac {d^{2} f x^{3}}{3} + \frac {x^{2} \cdot \left (2 c d f + d^{2} e\right )}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a*c**2*e*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*c**2*e*sqrt(a + b*x) + 2*d**2*f*(a + b*x)**(7/
2)/(7*b**3) + 2*(a + b*x)**(5/2)*(-2*a*d**2*f + 2*b*c*d*f + b*d**2*e)/(5*b**3) + 2*(a + b*x)**(3/2)*(a**2*d**2
*f - 2*a*b*c*d*f - a*b*d**2*e + b**2*c**2*f + 2*b**2*c*d*e)/(3*b**3), Ne(b, 0)), (sqrt(a)*(c**2*e*log(x) + c**
2*f*x + 2*c*d*e*x + d**2*f*x**3/3 + x**2*(2*c*d*f + d**2*e)/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\sqrt {a} c^{2} e \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (105 \, \sqrt {b x + a} b^{3} c^{2} e + 15 \, {\left (b x + a\right )}^{\frac {7}{2}} d^{2} f + 21 \, {\left (b d^{2} e + 2 \, {\left (b c d - a d^{2}\right )} f\right )} {\left (b x + a\right )}^{\frac {5}{2}} + 35 \, {\left ({\left (2 \, b^{2} c d - a b d^{2}\right )} e + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f\right )} {\left (b x + a\right )}^{\frac {3}{2}}\right )}}{105 \, b^{3}} \]

[In]

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

[Out]

sqrt(a)*c^2*e*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/105*(105*sqrt(b*x + a)*b^3*c^2*e +
15*(b*x + a)^(7/2)*d^2*f + 21*(b*d^2*e + 2*(b*c*d - a*d^2)*f)*(b*x + a)^(5/2) + 35*((2*b^2*c*d - a*b*d^2)*e +
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f)*(b*x + a)^(3/2))/b^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\frac {2 \, a c^{2} e \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (105 \, \sqrt {b x + a} b^{21} c^{2} e + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{20} c d e + 21 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{19} d^{2} e - 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{19} d^{2} e + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{20} c^{2} f + 42 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{19} c d f - 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{19} c d f + 15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{18} d^{2} f - 42 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{18} d^{2} f + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{18} d^{2} f\right )}}{105 \, b^{21}} \]

[In]

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

[Out]

2*a*c^2*e*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 2/105*(105*sqrt(b*x + a)*b^21*c^2*e + 70*(b*x + a)^(3/2)*b
^20*c*d*e + 21*(b*x + a)^(5/2)*b^19*d^2*e - 35*(b*x + a)^(3/2)*a*b^19*d^2*e + 35*(b*x + a)^(3/2)*b^20*c^2*f +
42*(b*x + a)^(5/2)*b^19*c*d*f - 70*(b*x + a)^(3/2)*a*b^19*c*d*f + 15*(b*x + a)^(7/2)*b^18*d^2*f - 42*(b*x + a)
^(5/2)*a*b^18*d^2*f + 35*(b*x + a)^(3/2)*a^2*b^18*d^2*f)/b^21

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.81 \[ \int \frac {\sqrt {a+b x} (c+d x)^2 (e+f x)}{x} \, dx=\left (\frac {2\,b\,d^2\,e-6\,a\,d^2\,f+4\,b\,c\,d\,f}{5\,b^3}+\frac {2\,a\,d^2\,f}{5\,b^3}\right )\,{\left (a+b\,x\right )}^{5/2}+\left (a\,\left (a\,\left (\frac {2\,b\,d^2\,e-6\,a\,d^2\,f+4\,b\,c\,d\,f}{b^3}+\frac {2\,a\,d^2\,f}{b^3}\right )-\frac {2\,\left (a\,d-b\,c\right )\,\left (b\,c\,f-3\,a\,d\,f+2\,b\,d\,e\right )}{b^3}\right )-\frac {2\,{\left (a\,d-b\,c\right )}^2\,\left (a\,f-b\,e\right )}{b^3}\right )\,\sqrt {a+b\,x}+\left (\frac {a\,\left (\frac {2\,b\,d^2\,e-6\,a\,d^2\,f+4\,b\,c\,d\,f}{b^3}+\frac {2\,a\,d^2\,f}{b^3}\right )}{3}-\frac {2\,\left (a\,d-b\,c\right )\,\left (b\,c\,f-3\,a\,d\,f+2\,b\,d\,e\right )}{3\,b^3}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {2\,d^2\,f\,{\left (a+b\,x\right )}^{7/2}}{7\,b^3}+\sqrt {a}\,c^2\,e\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

[In]

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

[Out]

((2*b*d^2*e - 6*a*d^2*f + 4*b*c*d*f)/(5*b^3) + (2*a*d^2*f)/(5*b^3))*(a + b*x)^(5/2) + (a*(a*((2*b*d^2*e - 6*a*
d^2*f + 4*b*c*d*f)/b^3 + (2*a*d^2*f)/b^3) - (2*(a*d - b*c)*(b*c*f - 3*a*d*f + 2*b*d*e))/b^3) - (2*(a*d - b*c)^
2*(a*f - b*e))/b^3)*(a + b*x)^(1/2) + ((a*((2*b*d^2*e - 6*a*d^2*f + 4*b*c*d*f)/b^3 + (2*a*d^2*f)/b^3))/3 - (2*
(a*d - b*c)*(b*c*f - 3*a*d*f + 2*b*d*e))/(3*b^3))*(a + b*x)^(3/2) + a^(1/2)*c^2*e*atan(((a + b*x)^(1/2)*1i)/a^
(1/2))*2i + (2*d^2*f*(a + b*x)^(7/2))/(7*b^3)