\(\int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx\) [10]

Optimal result

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

[Out]

-2/15*(d*x+c)^(3/2)*(2*b*c*f-5*d*(a*f+b*e)-3*b*d*f*x)/d^2-2*a*e*arctanh((d*x+c)^(1/2)/c^(1/2))*c^(1/2)+2*a*e*(
d*x+c)^(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} \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=-2 a \sqrt {c} e \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )-\frac {2 (c+d x)^{3/2} (-5 d (a f+b e)+2 b c f-3 b d f x)}{15 d^2}+2 a e \sqrt {c+d x} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=\frac {2 \sqrt {c+d x} (-b (c+d x) (-5 d e+2 c f-3 d f x)+5 a d (3 d e+c f+d f x))}{15 d^2}-2 a \sqrt {c} e \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {-2 a \sqrt {c}\, d^{2} e \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+\frac {2 \sqrt {d x +c}\, \left (\left (x \left (\frac {3 f x}{5}+e \right ) b +3 \left (\frac {f x}{3}+e \right ) a \right ) d^{2}+\left (\left (\frac {f x}{5}+e \right ) b +a f \right ) c d -\frac {2 c^{2} b f}{5}\right )}{3}}{d^{2}}\)	\(83\)
derivativedivides	\(\frac {\frac {2 f b \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a d f \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 b c f \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {2 b d e \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} e \sqrt {d x +c}-2 a \sqrt {c}\, d^{2} e \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{2}}\)	\(89\)
default	\(\frac {\frac {2 f b \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a d f \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 b c f \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {2 b d e \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} e \sqrt {d x +c}-2 a \sqrt {c}\, d^{2} e \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{2}}\)	\(89\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.84 \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=\left [\frac {15 \, a \sqrt {c} d^{2} e \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (3 \, b d^{2} f x^{2} + 5 \, {\left (b c d + 3 \, a d^{2}\right )} e - {\left (2 \, b c^{2} - 5 \, a c d\right )} f + {\left (5 \, b d^{2} e + {\left (b c d + 5 \, a d^{2}\right )} f\right )} x\right )} \sqrt {d x + c}}{15 \, d^{2}}, \frac {2 \, {\left (15 \, a \sqrt {-c} d^{2} e \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + {\left (3 \, b d^{2} f x^{2} + 5 \, {\left (b c d + 3 \, a d^{2}\right )} e - {\left (2 \, b c^{2} - 5 \, a c d\right )} f + {\left (5 \, b d^{2} e + {\left (b c d + 5 \, a d^{2}\right )} f\right )} x\right )} \sqrt {d x + c}\right )}}{15 \, d^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=\begin {cases} \frac {2 a c e \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a e \sqrt {c + d x} + \frac {2 b f \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {2 \left (c + d x\right )^{\frac {3}{2}} \left (a d f - b c f + b d e\right )}{3 d^{2}} & \text {for}\: d \neq 0 \\\sqrt {c} \left (a e \log {\left (x \right )} + a f x + b e x + \frac {b f x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=a \sqrt {c} e \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) + \frac {2 \, {\left (15 \, \sqrt {d x + c} a d^{2} e + 3 \, {\left (d x + c\right )}^{\frac {5}{2}} b f + 5 \, {\left (b d e - {\left (b c - a d\right )} f\right )} {\left (d x + c\right )}^{\frac {3}{2}}\right )}}{15 \, d^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=\frac {2 \, a c e \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, {\left (5 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{9} e + 15 \, \sqrt {d x + c} a d^{10} e + 3 \, {\left (d x + c\right )}^{\frac {5}{2}} b d^{8} f - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b c d^{8} f + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} a d^{9} f\right )}}{15 \, d^{10}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.77 \[ \int \frac {(a+b x) \sqrt {c+d x} (e+f x)}{x} \, dx=\left (c\,\left (\frac {2\,a\,d\,f-4\,b\,c\,f+2\,b\,d\,e}{d^2}+\frac {2\,b\,c\,f}{d^2}\right )-\frac {2\,\left (a\,d-b\,c\right )\,\left (c\,f-d\,e\right )}{d^2}\right )\,\sqrt {c+d\,x}+\left (\frac {2\,a\,d\,f-4\,b\,c\,f+2\,b\,d\,e}{3\,d^2}+\frac {2\,b\,c\,f}{3\,d^2}\right )\,{\left (c+d\,x\right )}^{3/2}+\frac {2\,b\,f\,{\left (c+d\,x\right )}^{5/2}}{5\,d^2}+a\,\sqrt {c}\,e\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,2{}\mathrm {i} \]

[In]

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

[Out]

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