3.856 \(\int \frac{(f+g x) \sqrt{a+b x+c x^2}}{d+e x} \, dx\)
Optimal. Leaf size=219 \[ -\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-4 c e (a e g-b d g+b e f)+b^2 e^2 g+8 c^2 d (e f-d g)\right )}{8 c^{3/2} e^3}+\frac{(e f-d g) \sqrt{a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^3}+\frac{\sqrt{a+b x+c x^2} (b e g-4 c d g+4 c e f+2 c e g x)}{4 c e^2} \]
[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
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Rubi [A] time = 0.324312, antiderivative size = 219, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used =
{814, 843, 621, 206, 724} \[ -\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-4 c e (a e g-b d g+b e f)+b^2 e^2 g+8 c^2 d (e f-d g)\right )}{8 c^{3/2} e^3}+\frac{(e f-d g) \sqrt{a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^3}+\frac{\sqrt{a+b x+c x^2} (b e g-4 c d g+4 c e f+2 c e g x)}{4 c e^2} \]
Antiderivative was successfully verified.
[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 814
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 843
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]
Rule 621
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 724
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]
Rubi steps
\begin{align*} \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{\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 ) \operatorname{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 ) \operatorname{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] time = 0.369378, size = 216, normalized size = 0.99 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (4 c e (a e g-b d g+b e f)-b^2 e^2 g+8 c^2 d (d g-e f)\right )+2 \sqrt{c} \left (4 c (d g-e f) \sqrt{e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )+e \sqrt{a+x (b+c x)} (b e g+2 c (-2 d g+2 e f+e g x))\right )}{8 c^{3/2} e^3} \]
Antiderivative was successfully verified.
[In]
Integrate[((f + g*x)*Sqrt[a + b*x + c*x^2])/(d + e*x),x]
[Out]
((-(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)])] + 2*Sqrt[c]*(e*Sqrt[a + x*(b + c*x)]*(b*e*g + 2*c*(2*e*f - 2*d*g + e*g*x)) + 4*c*Sqrt[c*d^2
+ e*(-(b*d) + a*e)]*(-(e*f) + d*g)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)
]*Sqrt[a + x*(b + c*x)])]))/(8*c^(3/2)*e^3)
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Maple [B] time = 0.299, size = 1559, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*x+f)*(c*x^2+b*x+a)^(1/2)/(e*x+d),x)
[Out]
1/2/e*g*(c*x^2+b*x+a)^(1/2)*x+1/4/e*g/c*(c*x^2+b*x+a)^(1/2)*b+1/2/e*g/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*
x+a)^(1/2))*a-1/8/e*g/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2-1/e^2*((d/e+x)^2*c+(b*e-2*c*d)/e
*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*d*g+1/e*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1
/2)*f-1/2/e^2*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/
e^2)^(1/2))/c^(1/2)*b*d*g+1/2/e*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a
*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)*b*f+1/e^3*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c
*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d^2*g-1/e^2*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d
/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d*f+1/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*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*
c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*a*d*g-1/e/((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*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(
a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*a*f-1/e^3/((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*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^
2)/e^2)^(1/2))/(d/e+x))*b*d^2*g+1/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*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2
))/(d/e+x))*b*d*f+1/e^4/((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*(d/e+x)+2*
((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*c*
d^3*g-1/e^3/((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*(d/e+x)+2*((a*e^2-b*d*
e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*c*d^2*f
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(c*x^2+b*x+a)^(1/2)/(e*x+d),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(c*x^2+b*x+a)^(1/2)/(e*x+d),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (f + g x\right ) \sqrt{a + b x + c x^{2}}}{d + e x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*x+f)*(c*x^2+b*x+a)^(1/2)/(e*x+d),x, algorithm="giac")
[Out]
Exception raised: TypeError