3.1193 \(\int (b d+2 c d x)^2 \sqrt{a+b x+c x^2} \, dx\)

Optimal. Leaf size=123 \[ -\frac{d^2 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{32 c^{3/2}}-\frac{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{16 c}+\frac{d^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{8 c} \]

[Out]

-((b^2 - 4*a*c)*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(16*c) + (d^2*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/(8*c
) - ((b^2 - 4*a*c)^2*d^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(32*c^(3/2))

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Rubi [A]  time = 0.0559653, antiderivative size = 123, normalized size of antiderivative = 1., 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 = {685, 692, 621, 206} \[ -\frac{d^2 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{32 c^{3/2}}-\frac{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{16 c}+\frac{d^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{8 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(16*c) + (d^2*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/(8*c
) - ((b^2 - 4*a*c)^2*d^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(32*c^(3/2))

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(d*p*(b^2 - 4*a*c))/(b*e*(m + 2*p + 1)), Int[(d + e*x)^m*(a +
b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0] &&
NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p]))
&& RationalQ[m] && IntegerQ[2*p]

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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])

Rubi steps

\begin{align*} \int (b d+2 c d x)^2 \sqrt{a+b x+c x^2} \, dx &=\frac{d^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{8 c}-\frac{\left (b^2-4 a c\right ) \int \frac{(b d+2 c d x)^2}{\sqrt{a+b x+c x^2}} \, dx}{16 c}\\ &=-\frac{\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt{a+b x+c x^2}}{16 c}+\frac{d^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{8 c}-\frac{\left (\left (b^2-4 a c\right )^2 d^2\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{32 c}\\ &=-\frac{\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt{a+b x+c x^2}}{16 c}+\frac{d^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{8 c}-\frac{\left (\left (b^2-4 a c\right )^2 d^2\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 )}{16 c}\\ &=-\frac{\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt{a+b x+c x^2}}{16 c}+\frac{d^2 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{8 c}-\frac{\left (b^2-4 a c\right )^2 d^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{32 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.371687, size = 140, normalized size = 1.14 \[ \frac{d^2 \sqrt{a+x (b+c x)} \left (2 (b+2 c x) \left (4 c \left (a+2 c x^2\right )+b^2+8 b c x\right )-\frac{c^{3/2} \sqrt{4 a-\frac{b^2}{c}} (a+x (b+c x)) \sinh ^{-1}\left (\frac{b+2 c x}{\sqrt{c} \sqrt{4 a-\frac{b^2}{c}}}\right )}{\left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/2}}\right )}{32 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*Sqrt[a + x*(b + c*x)]*(2*(b + 2*c*x)*(b^2 + 8*b*c*x + 4*c*(a + 2*c*x^2)) - (Sqrt[4*a - b^2/c]*c^(3/2)*(a
+ x*(b + c*x))*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])])/((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(3/2))
)/(32*c)

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Maple [B]  time = 0.048, size = 230, normalized size = 1.9 \begin{align*}{d}^{2}cx \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}+{\frac{{d}^{2}b}{2} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{2}{d}^{2}x}{8}\sqrt{c{x}^{2}+bx+a}}+{\frac{{d}^{2}{b}^{3}}{16\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{2}{d}^{2}a}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{d}^{2}{b}^{4}}{32}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{c{d}^{2}ax}{2}\sqrt{c{x}^{2}+bx+a}}-{\frac{a{d}^{2}b}{4}\sqrt{c{x}^{2}+bx+a}}-{\frac{{a}^{2}{d}^{2}}{2}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*c*x*(c*x^2+b*x+a)^(3/2)+1/2*d^2*b*(c*x^2+b*x+a)^(3/2)+1/8*d^2*b^2*x*(c*x^2+b*x+a)^(1/2)+1/16*d^2/c*b^3*(c*
x^2+b*x+a)^(1/2)+1/4*d^2*b^2/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/32*d^2*b^4/c^(3/2)*ln((1/
2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/2*d^2*c*a*x*(c*x^2+b*x+a)^(1/2)-1/4*d^2*a*(c*x^2+b*x+a)^(1/2)*b-1/2*d^
2*c^(1/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.29423, size = 702, normalized size = 5.71 \begin{align*} \left [\frac{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{c} d^{2} \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 (16 \, c^{4} d^{2} x^{3} + 24 \, b c^{3} d^{2} x^{2} + 2 \,{\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} d^{2} x +{\left (b^{3} c + 4 \, a b c^{2}\right )} d^{2}\right )} \sqrt{c x^{2} + b x + a}}{64 \, c^{2}}, \frac{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-c} d^{2} \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 (16 \, c^{4} d^{2} x^{3} + 24 \, b c^{3} d^{2} x^{2} + 2 \,{\left (5 \, b^{2} c^{2} + 4 \, a c^{3}\right )} d^{2} x +{\left (b^{3} c + 4 \, a b c^{2}\right )} d^{2}\right )} \sqrt{c x^{2} + b x + a}}{32 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*d^2*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*(16*c^4*d^2*x^3 + 24*b*c^3*d^2*x^2 + 2*(5*b^2*c^2 + 4*a*c^3)*d^2*x + (b^3*c + 4*
a*b*c^2)*d^2)*sqrt(c*x^2 + b*x + a))/c^2, 1/32*((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*d^2*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*(16*c^4*d^2*x^3 + 24*b*c^3*d^2*x^2 + 2*(5*b^2
*c^2 + 4*a*c^3)*d^2*x + (b^3*c + 4*a*b*c^2)*d^2)*sqrt(c*x^2 + b*x + a))/c^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int b^{2} \sqrt{a + b x + c x^{2}}\, dx + \int 4 c^{2} x^{2} \sqrt{a + b x + c x^{2}}\, dx + \int 4 b c x \sqrt{a + b x + c x^{2}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(b**2*sqrt(a + b*x + c*x**2), x) + Integral(4*c**2*x**2*sqrt(a + b*x + c*x**2), x) + Integral(4*
b*c*x*sqrt(a + b*x + c*x**2), x))

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Giac [A]  time = 1.18944, size = 209, normalized size = 1.7 \begin{align*} \frac{1}{16} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \, c^{2} d^{2} x + 3 \, b c d^{2}\right )} x + \frac{5 \, b^{2} c^{3} d^{2} + 4 \, a c^{4} d^{2}}{c^{3}}\right )} x + \frac{b^{3} c^{2} d^{2} + 4 \, a b c^{3} d^{2}}{c^{3}}\right )} + \frac{{\left (b^{4} d^{2} - 8 \, a b^{2} c d^{2} + 16 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{32 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(c*x^2 + b*x + a)*(2*(4*(2*c^2*d^2*x + 3*b*c*d^2)*x + (5*b^2*c^3*d^2 + 4*a*c^4*d^2)/c^3)*x + (b^3*c^2
*d^2 + 4*a*b*c^3*d^2)/c^3) + 1/32*(b^4*d^2 - 8*a*b^2*c*d^2 + 16*a^2*c^2*d^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))*sqrt(c) - b))/c^(3/2)