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

Optimal. Leaf size=122 \[ -\frac {e \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{5/2}}+\frac {e \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^2}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c} \]

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Rubi [A]  time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {779, 612, 621, 206} \begin {gather*} \frac {e \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^2}-\frac {e \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{5/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 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]

((b^2 - 4*a*c)*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c^2) + ((8*c*d - b*e + 6*c*e*x)*(a + b*x + c*x^2)^(3/2
))/(12*c) - ((b^2 - 4*a*c)^2*e*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(5/2))

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 612

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

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 (b+2 c x) (d+e x) \sqrt {a+b x+c x^2} \, dx &=\frac {(8 c d-b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (\left (b^2-4 a c\right ) e\right ) \int \sqrt {a+b x+c x^2} \, dx}{8 c}\\ &=\frac {\left (b^2-4 a c\right ) e (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^2}+\frac {(8 c d-b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 c}-\frac {\left (\left (b^2-4 a c\right )^2 e\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{64 c^2}\\ &=\frac {\left (b^2-4 a c\right ) e (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^2}+\frac {(8 c d-b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 c}-\frac {\left (\left (b^2-4 a c\right )^2 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 )}{32 c^2}\\ &=\frac {\left (b^2-4 a c\right ) e (b+2 c x) \sqrt {a+b x+c x^2}}{32 c^2}+\frac {(8 c d-b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 c}-\frac {\left (b^2-4 a c\right )^2 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 134, normalized size = 1.10 \begin {gather*} \frac {\sqrt {a+x (b+c x)} \left (4 a c (-5 b e+16 c d+6 c e x)+3 b^3 e-2 b^2 c e x+8 b c^2 x (8 d+5 e x)+16 c^3 x^2 (4 d+3 e x)\right )}{96 c^2}-\frac {e \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{64 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*(3*b^3*e - 2*b^2*c*e*x + 16*c^3*x^2*(4*d + 3*e*x) + 8*b*c^2*x*(8*d + 5*e*x) + 4*a*c*(16
*c*d - 5*b*e + 6*c*e*x)))/(96*c^2) - ((b^2 - 4*a*c)^2*e*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]
)/(64*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.63, size = 151, normalized size = 1.24 \begin {gather*} \frac {e \left (16 a^2 c^2-8 a b^2 c+b^4\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{64 c^{5/2}}+\frac {\sqrt {a+b x+c x^2} \left (-20 a b c e+64 a c^2 d+24 a c^2 e x+3 b^3 e-2 b^2 c e x+64 b c^2 d x+40 b c^2 e x^2+64 c^3 d x^2+48 c^3 e x^3\right )}{96 c^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]

(Sqrt[a + b*x + c*x^2]*(64*a*c^2*d + 3*b^3*e - 20*a*b*c*e + 64*b*c^2*d*x - 2*b^2*c*e*x + 24*a*c^2*e*x + 64*c^3
*d*x^2 + 40*b*c^2*e*x^2 + 48*c^3*e*x^3))/(96*c^2) + ((b^4 - 8*a*b^2*c + 16*a^2*c^2)*e*Log[b + 2*c*x - 2*Sqrt[c
]*Sqrt[a + b*x + c*x^2]])/(64*c^(5/2))

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fricas [A]  time = 0.45, size = 343, normalized size = 2.81 \begin {gather*} \left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 (48 \, c^{4} e x^{3} + 64 \, a c^{3} d + 8 \, {\left (8 \, c^{4} d + 5 \, b c^{3} e\right )} x^{2} + {\left (3 \, b^{3} c - 20 \, a b c^{2}\right )} e + 2 \, {\left (32 \, b c^{3} d - {\left (b^{2} c^{2} - 12 \, a c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{3}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\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 (48 \, c^{4} e x^{3} + 64 \, a c^{3} d + 8 \, {\left (8 \, c^{4} d + 5 \, b c^{3} e\right )} x^{2} + {\left (3 \, b^{3} c - 20 \, a b c^{2}\right )} e + 2 \, {\left (32 \, b c^{3} d - {\left (b^{2} c^{2} - 12 \, a c^{3}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{192 \, c^{3}}\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]

[1/384*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*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*(48*c^4*e*x^3 + 64*a*c^3*d + 8*(8*c^4*d + 5*b*c^3*e)*x^2 + (3*b^3*c - 20*a*b*c^
2)*e + 2*(32*b*c^3*d - (b^2*c^2 - 12*a*c^3)*e)*x)*sqrt(c*x^2 + b*x + a))/c^3, 1/192*(3*(b^4 - 8*a*b^2*c + 16*a
^2*c^2)*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*(48*c^4*
e*x^3 + 64*a*c^3*d + 8*(8*c^4*d + 5*b*c^3*e)*x^2 + (3*b^3*c - 20*a*b*c^2)*e + 2*(32*b*c^3*d - (b^2*c^2 - 12*a*
c^3)*e)*x)*sqrt(c*x^2 + b*x + a))/c^3]

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giac [A]  time = 0.22, size = 170, normalized size = 1.39 \begin {gather*} \frac {1}{96} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, c x e + \frac {8 \, c^{4} d + 5 \, b c^{3} e}{c^{3}}\right )} x + \frac {32 \, b c^{3} d - b^{2} c^{2} e + 12 \, a c^{3} e}{c^{3}}\right )} x + \frac {64 \, a c^{3} d + 3 \, b^{3} c e - 20 \, a b c^{2} e}{c^{3}}\right )} + \frac {{\left (b^{4} e - 8 \, a b^{2} c e + 16 \, a^{2} c^{2} e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{64 \, c^{\frac {5}{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]

1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(6*c*x*e + (8*c^4*d + 5*b*c^3*e)/c^3)*x + (32*b*c^3*d - b^2*c^2*e + 12*a*c^3*
e)/c^3)*x + (64*a*c^3*d + 3*b^3*c*e - 20*a*b*c^2*e)/c^3) + 1/64*(b^4*e - 8*a*b^2*c*e + 16*a^2*c^2*e)*log(abs(-
2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.05, size = 235, normalized size = 1.93 \begin {gather*} -\frac {a^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 \sqrt {c}}+\frac {a \,b^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}-\frac {b^{4} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{64 c^{\frac {5}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a e x}{4}+\frac {\sqrt {c \,x^{2}+b x +a}\, b^{2} e x}{16 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, a b e}{8 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, b^{3} e}{32 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} e x}{2}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b e}{12 c}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d}{3} \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]

1/2*e*x*(c*x^2+b*x+a)^(3/2)-1/12/c*e*b*(c*x^2+b*x+a)^(3/2)+1/16/c*e*b^2*x*(c*x^2+b*x+a)^(1/2)+1/32/c^2*e*b^3*(
c*x^2+b*x+a)^(1/2)+1/8/c^(3/2)*e*b^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/64/c^(5/2)*e*b^4*ln((c*x+
1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/4*e*a*x*(c*x^2+b*x+a)^(1/2)-1/8/c*e*a*(c*x^2+b*x+a)^(1/2)*b-1/4/c^(1/2)*
e*a^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2/3*(c*x^2+b*x+a)^(3/2)*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 positive, negative or zero?

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mupad [B]  time = 2.79, size = 395, normalized size = 3.24 \begin {gather*} \frac {e\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{2}-\frac {a\,e\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{2}-\frac {5\,b\,e\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{4}+\frac {d\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{8\,c^{3/2}}+\frac {d\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{12\,c}+b\,d\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {b\,e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}+\frac {b\,d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}+\frac {b\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}} \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]

(e*x*(a + b*x + c*x^2)^(3/2))/2 - (a*e*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (
a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/2 - (5*b*e*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2
)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2
)))/4 + (d*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(8*c^(3/2)) + (d*(8*c*(a + c*
x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(12*c) + b*d*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (b*e*(
8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2) + (b*d*log((b/2 + c*x)/c^(1/2) + (a + b*x
+ c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)) + (b*e*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 -
4*a*b*c))/(16*c^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \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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