3.1551 \(\int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{d+e x} \, dx\)

Optimal. Leaf size=199 \[ \frac {\left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 \sqrt {c} e^3}-\frac {(2 c d-b e) \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^3}-\frac {\sqrt {a+b x+c x^2} (-3 b e+4 c d-2 c e x)}{2 e^2} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {814, 843, 621, 206, 724} \[ \frac {\left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 \sqrt {c} e^3}-\frac {(2 c d-b e) \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^3}-\frac {\sqrt {a+b x+c x^2} (-3 b e+4 c d-2 c e x)}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{d+e x} \, dx &=-\frac {(4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{2 e^2}-\frac {\int \frac {c \left (3 b^2 d e+4 a c d e-4 b \left (c d^2+a e^2\right )\right )-c \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{4 c e^2}\\ &=-\frac {(4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{2 e^2}-\frac {\left ((2 c d-b e) \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^3}+\frac {\left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{4 e^3}\\ &=-\frac {(4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{2 e^2}+\frac {\left (2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )\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 (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\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 )}{2 e^3}\\ &=-\frac {(4 c d-3 b e-2 c e x) \sqrt {a+b x+c x^2}}{2 e^2}+\frac {\left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 \sqrt {c} e^3}-\frac {(2 c d-b e) \sqrt {c d^2-b d e+a e^2} \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*}

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Mathematica [A]  time = 0.25, size = 195, normalized size = 0.98 \[ \frac {\left (4 c e (a e-2 b d)+b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+2 \sqrt {c} \left (2 (2 c d-b e) \sqrt {e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )+e \sqrt {a+x (b+c x)} (3 b e-4 c d+2 c e x)\right )}{4 \sqrt {c} e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 17.69, size = 1196, normalized size = 6.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x
+ a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(2*c^2*d - b*c*e)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2
- (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*
x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 +
2*d*e*x + d^2)) + 4*(2*c^2*e^2*x - 4*c^2*d*e + 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/(c*e^3), -1/8*(8*(2*c^2*d -
b*c*e)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*
e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a
*b*e^2)*x)) - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^
2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(2*c^2*e^2*x - 4*c^2*d*e + 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/(c*
e^3), -1/4*((8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*
sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(2*c^2*d - b*c*e)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 8*a^2*e^2
- (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*
x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 +
2*d*e*x + d^2)) - 2*(2*c^2*e^2*x - 4*c^2*d*e + 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/(c*e^3), -1/4*(4*(2*c^2*d -
b*c*e)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*
e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a
*b*e^2)*x)) + (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b
)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(2*c^2*e^2*x - 4*c^2*d*e + 3*b*c*e^2)*sqrt(c*x^2 + b*x + a))/(c*e^3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(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,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.06, size = 1302, normalized size = 6.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(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(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b + 2 c x\right ) \sqrt {a + b x + c x^{2}}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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