\(\int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx\) [17]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=2 c e \sqrt {a+b x}-\frac {2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}-2 \sqrt {a} c e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

-2/15*(b*x+a)^(3/2)*(2*a*d*f-5*b*(c*f+d*e)-3*b*d*f*x)/b^2-2*c*e*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2)+2*c*e*(
b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {152, 52, 65, 214} \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=-2 \sqrt {a} c e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )-\frac {2 (a+b x)^{3/2} (2 a d f-5 b (c f+d e)-3 b d f x)}{15 b^2}+2 c e \sqrt {a+b x} \]

[In]

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

[Out]

2*c*e*Sqrt[a + b*x] - (2*(a + b*x)^(3/2)*(2*a*d*f - 5*b*(d*e + c*f) - 3*b*d*f*x))/(15*b^2) - 2*Sqrt[a]*c*e*Arc
Tanh[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 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 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}+(c e) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = 2 c e \sqrt {a+b x}-\frac {2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}+(a c e) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = 2 c e \sqrt {a+b x}-\frac {2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}+\frac {(2 a c e) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{b} \\ & = 2 c e \sqrt {a+b x}-\frac {2 (a+b x)^{3/2} (2 a d f-5 b (d e+c f)-3 b d f x)}{15 b^2}-2 \sqrt {a} c e \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=\frac {2 \sqrt {a+b x} \left (15 b^2 c e+5 b d e (a+b x)+5 b c f (a+b x)-5 a d f (a+b x)+3 d f (a+b x)^2\right )}{15 b^2}-2 \sqrt {a} c e \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(\frac {-2 \sqrt {a}\, b^{2} c e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {4 \sqrt {b x +a}\, \left (\frac {5 \left (-x \left (\frac {3 f x}{5}+e \right ) d -3 \left (\frac {f x}{3}+e \right ) c \right ) b^{2}}{2}-\frac {5 \left (\left (\frac {f x}{5}+e \right ) d +c f \right ) a b}{2}+a^{2} d f \right )}{15}}{b^{2}}\)	\(86\)
derivativedivides	\(\frac {\frac {2 d f \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {2 a d f \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 b c f \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 b d e \left (b x +a \right )^{\frac {3}{2}}}{3}+2 b^{2} c e \sqrt {b x +a}-2 \sqrt {a}\, b^{2} c e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b^{2}}\)	\(89\)
default	\(\frac {\frac {2 d f \left (b x +a \right )^{\frac {5}{2}}}{5}-\frac {2 a d f \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 b c f \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 b d e \left (b x +a \right )^{\frac {3}{2}}}{3}+2 b^{2} c e \sqrt {b x +a}-2 \sqrt {a}\, b^{2} c e \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b^{2}}\)	\(89\)

[In]

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

[Out]

2/15*(-15*a^(1/2)*b^2*c*e*arctanh((b*x+a)^(1/2)/a^(1/2))-2*(b*x+a)^(1/2)*(5/2*(-x*(3/5*f*x+e)*d-3*(1/3*f*x+e)*
c)*b^2-5/2*((1/5*f*x+e)*d+c*f)*a*b+a^2*d*f))/b^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.82 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=\left [\frac {15 \, \sqrt {a} b^{2} c e \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, b^{2} d f x^{2} + 5 \, {\left (3 \, b^{2} c + a b d\right )} e + {\left (5 \, a b c - 2 \, a^{2} d\right )} f + {\left (5 \, b^{2} d e + {\left (5 \, b^{2} c + a b d\right )} f\right )} x\right )} \sqrt {b x + a}}{15 \, b^{2}}, \frac {2 \, {\left (15 \, \sqrt {-a} b^{2} c e \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, b^{2} d f x^{2} + 5 \, {\left (3 \, b^{2} c + a b d\right )} e + {\left (5 \, a b c - 2 \, a^{2} d\right )} f + {\left (5 \, b^{2} d e + {\left (5 \, b^{2} c + a b d\right )} f\right )} x\right )} \sqrt {b x + a}\right )}}{15 \, b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.58 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=\begin {cases} \frac {2 a c e \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 c e \sqrt {a + b x} + \frac {2 d f \left (a + b x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {2 \left (a + b x\right )^{\frac {3}{2}} \left (- a d f + b c f + b d e\right )}{3 b^{2}} & \text {for}\: b \neq 0 \\\sqrt {a} \left (c e \log {\left (x \right )} + c f x + d e x + \frac {d f x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=\sqrt {a} c e \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2 \, {\left (15 \, \sqrt {b x + a} b^{2} c e + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} d f + 5 \, {\left (b d e + {\left (b c - a d\right )} f\right )} {\left (b x + a\right )}^{\frac {3}{2}}\right )}}{15 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=\frac {2 \, a c e \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (15 \, \sqrt {b x + a} b^{10} c e + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{9} d e + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{9} c f + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{8} d f - 5 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{8} d f\right )}}{15 \, b^{10}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.85 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.77 \[ \int \frac {\sqrt {a+b x} (c+d x) (e+f x)}{x} \, dx=\left (a\,\left (\frac {2\,b\,c\,f-4\,a\,d\,f+2\,b\,d\,e}{b^2}+\frac {2\,a\,d\,f}{b^2}\right )+\frac {2\,\left (a\,d-b\,c\right )\,\left (a\,f-b\,e\right )}{b^2}\right )\,\sqrt {a+b\,x}+\left (\frac {2\,b\,c\,f-4\,a\,d\,f+2\,b\,d\,e}{3\,b^2}+\frac {2\,a\,d\,f}{3\,b^2}\right )\,{\left (a+b\,x\right )}^{3/2}+\frac {2\,d\,f\,{\left (a+b\,x\right )}^{5/2}}{5\,b^2}+\sqrt {a}\,c\,e\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

[In]

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

[Out]

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