\(\int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx\) [856]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F(-1)]
   Sympy [F]
   Maxima [F(-2)]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 27, antiderivative size = 219 \[ \int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx=\frac {(4 c e f-4 c d g+b e g+2 c e g x) \sqrt {a+b x+c x^2}}{4 c e^2}-\frac {\left (b^2 e^2 g+8 c^2 d (e f-d g)-4 c e (b e f-b d g+a e g)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2} e^3}+\frac {\sqrt {c d^2-b d e+a e^2} (e f-d g) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {828, 857, 635, 212, 738} \[ \int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx=-\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e (a e g-b d g+b e f)+b^2 e^2 g+8 c^2 d (e f-d g)\right )}{8 c^{3/2} e^3}+\frac {(e f-d g) \sqrt {a e^2-b d e+c d^2} \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^3}+\frac {\sqrt {a+b x+c x^2} (b e g-4 c d g+4 c e f+2 c e g x)}{4 c e^2} \]

[In]

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

[Out]

((4*c*e*f - 4*c*d*g + b*e*g + 2*c*e*g*x)*Sqrt[a + b*x + c*x^2])/(4*c*e^2) - ((b^2*e^2*g + 8*c^2*d*(e*f - d*g)
- 4*c*e*(b*e*f - b*d*g + a*e*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(3/2)*e^3) + (Sq
rt[c*d^2 - b*d*e + a*e^2]*(e*f - d*g)*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*S
qrt[a + b*x + c*x^2])])/e^3

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
 + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
 c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(4 c e f-4 c d g+b e g+2 c e g x) \sqrt {a+b x+c x^2}}{4 c e^2}-\frac {\int \frac {\frac {1}{2} (4 c e (b d-2 a e) f+4 a c d e g-b d (4 c d-b e) g)+\frac {1}{2} \left (b^2 e^2 g+8 c^2 d (e f-d g)-4 c e (b e f-b d g+a e g)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{4 c e^2} \\ & = \frac {(4 c e f-4 c d g+b e g+2 c e g x) \sqrt {a+b x+c x^2}}{4 c e^2}+\frac {\left (\left (c d^2-b d e+a e^2\right ) (e f-d g)\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^3}-\frac {\left (b^2 e^2 g+8 c^2 d (e f-d g)-4 c e (b e f-b d g+a e g)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c e^3} \\ & = \frac {(4 c e f-4 c d g+b e g+2 c e g x) \sqrt {a+b x+c x^2}}{4 c e^2}-\frac {\left (2 \left (c d^2-b d e+a e^2\right ) (e f-d g)\right ) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{e^3}-\frac {\left (b^2 e^2 g+8 c^2 d (e f-d g)-4 c e (b e f-b d g+a e g)\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{4 c e^3} \\ & = \frac {(4 c e f-4 c d g+b e g+2 c e g x) \sqrt {a+b x+c x^2}}{4 c e^2}-\frac {\left (b^2 e^2 g+8 c^2 d (e f-d g)-4 c e (b e f-b d g+a e g)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2} e^3}+\frac {\sqrt {c d^2-b d e+a e^2} (e f-d g) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.97 \[ \int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx=\frac {2 \sqrt {c} \left (e \sqrt {a+x (b+c x)} (b e g+2 c (2 e f-2 d g+e g x))-8 c \sqrt {-c d^2+b d e-a e^2} (-e f+d g) \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )\right )+\left (-b^2 e^2 g+8 c^2 d (-e f+d g)+4 c e (b e f-b d g+a e g)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{8 c^{3/2} e^3} \]

[In]

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

[Out]

(2*Sqrt[c]*(e*Sqrt[a + x*(b + c*x)]*(b*e*g + 2*c*(2*e*f - 2*d*g + e*g*x)) - 8*c*Sqrt[-(c*d^2) + b*d*e - a*e^2]
*(-(e*f) + d*g)*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c*x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)]]) + (-(b^2
*e^2*g) + 8*c^2*d*(-(e*f) + d*g) + 4*c*e*(b*e*f - b*d*g + a*e*g))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b
 + c*x)])])/(8*c^(3/2)*e^3)

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.53

method result size
risch \(\frac {\left (2 c e g x +b e g -4 c d g +4 c e f \right ) \sqrt {c \,x^{2}+b x +a}}{4 c \,e^{2}}+\frac {\frac {\left (4 a c \,e^{2} g -b^{2} e^{2} g -4 b c d e g +4 b c \,e^{2} f +8 c^{2} d^{2} g -8 c^{2} d e f \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e \sqrt {c}}+\frac {8 \left (a \,e^{2} g d -a \,e^{3} f -b \,d^{2} e g +b \,e^{2} f d +c \,d^{3} g -c \,d^{2} e f \right ) c \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}}{8 c \,e^{2}}\) \(335\)
default \(\frac {g \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{e}+\frac {\left (-d g +e f \right ) \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (e^{2} a -b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}\) \(407\)

[In]

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

[Out]

1/4*(2*c*e*g*x+b*e*g-4*c*d*g+4*c*e*f)*(c*x^2+b*x+a)^(1/2)/c/e^2+1/8/c/e^2*((4*a*c*e^2*g-b^2*e^2*g-4*b*c*d*e*g+
4*b*c*e^2*f+8*c^2*d^2*g-8*c^2*d*e*f)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+8*(a*d*e^2*g-a*e^3*
f-b*d^2*e*g+b*d*e^2*f+c*d^3*g-c*d^2*e*f)*c/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(
b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/
e^2)^(1/2))/(x+d/e)))

Fricas [F(-1)]

Timed out. \[ \int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

Integral((f + g*x)*sqrt(a + b*x + c*x**2)/(d + e*x), x)

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F(-2)]

Exception generated. \[ \int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

Timed out. \[ \int \frac {(f+g x) \sqrt {a+b x+c x^2}}{d+e x} \, dx=\int \frac {\left (f+g\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{d+e\,x} \,d x \]

[In]

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

[Out]

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