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

Optimal. Leaf size=84 \[ \frac {e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{3/2}}+\frac {\sqrt {a+b x+c x^2} (-b e+4 c d+2 c e x)}{2 c} \]

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Rubi [A]  time = 0.05, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {779, 621, 206} \begin {gather*} \frac {e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 c^{3/2}}+\frac {\sqrt {a+b x+c x^2} (-b e+4 c d+2 c e x)}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 82, normalized size = 0.98 \begin {gather*} \frac {e \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{4 c^{3/2}}+\frac {\sqrt {a+x (b+c x)} (-b e+4 c d+2 c e x)}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.47, size = 86, normalized size = 1.02 \begin {gather*} \frac {\sqrt {a+b x+c x^2} (-b e+4 c d+2 c e x)}{2 c}-\frac {e \left (b^2-4 a c\right ) \log \left (-2 c^{3/2} \sqrt {a+b x+c x^2}+b c+2 c^2 x\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*c*d - b*e + 2*c*e*x)*Sqrt[a + b*x + c*x^2])/(2*c) - ((b^2 - 4*a*c)*e*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[a
+ b*x + c*x^2]])/(4*c^(3/2))

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fricas [A]  time = 0.75, size = 197, normalized size = 2.35 \begin {gather*} \left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} e \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (2 \, c^{2} e x + 4 \, c^{2} d - b c e\right )} \sqrt {c x^{2} + b x + a}}{8 \, c^{2}}, -\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} e \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (2 \, c^{2} e x + 4 \, c^{2} d - b c e\right )} \sqrt {c x^{2} + b x + a}}{4 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.44, size = 84, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + b x + a} {\left (2 \, x e + \frac {4 \, c d - b e}{c}\right )} - \frac {{\left (b^{2} e - 4 \, a c e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{4 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 117, normalized size = 1.39 \begin {gather*} -\frac {a e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\frac {b^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}+\sqrt {c \,x^{2}+b x +a}\, e x -\frac {\sqrt {c \,x^{2}+b x +a}\, b e}{2 c}+2 \sqrt {c \,x^{2}+b x +a}\, d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)/(c*x^2+b*x+a)^(1/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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