### 3.1234 $$\int \frac{(b d+2 c d x)^4}{\sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=117 $\frac{3}{4} d^4 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}+\frac{3 d^4 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c}}+\frac{1}{2} d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0547876, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {692, 621, 206} $\frac{3}{4} d^4 \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}+\frac{3 d^4 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 \sqrt{c}}+\frac{1}{2} d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^4/Sqrt[a + b*x + c*x^2],x]

[Out]

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

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

Mathematica [A]  time = 0.106997, size = 100, normalized size = 0.85 $d^4 \left (\frac{1}{4} (b+2 c x) \sqrt{a+x (b+c x)} \left (4 c \left (2 c x^2-3 a\right )+5 b^2+8 b c x\right )+\frac{3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{8 \sqrt{c}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^4/Sqrt[a + b*x + c*x^2],x]

[Out]

d^4*(((b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(5*b^2 + 8*b*c*x + 4*c*(-3*a + 2*c*x^2)))/4 + (3*(b^2 - 4*a*c)^2*ArcTa
nh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8*Sqrt[c]))

________________________________________________________________________________________

Maple [B]  time = 0.053, size = 242, normalized size = 2.1 \begin{align*} 4\,{d}^{4}{c}^{3}{x}^{3}\sqrt{c{x}^{2}+bx+a}+6\,{d}^{4}{c}^{2}b{x}^{2}\sqrt{c{x}^{2}+bx+a}+{\frac{9\,{d}^{4}{b}^{2}cx}{2}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{d}^{4}{b}^{3}}{4}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{d}^{4}{b}^{4}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-3\,{d}^{4}\sqrt{c}{b}^{2}a\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) -3\,{d}^{4}cba\sqrt{c{x}^{2}+bx+a}-6\,{d}^{4}{c}^{2}ax\sqrt{c{x}^{2}+bx+a}+6\,{d}^{4}{c}^{3/2}{a}^{2}\ln \left ({\frac{b/2+cx}{\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)^4/(c*x^2+b*x+a)^(1/2),x)

[Out]

4*d^4*c^3*x^3*(c*x^2+b*x+a)^(1/2)+6*d^4*c^2*b*x^2*(c*x^2+b*x+a)^(1/2)+9/2*d^4*c*b^2*x*(c*x^2+b*x+a)^(1/2)+5/4*
d^4*b^3*(c*x^2+b*x+a)^(1/2)+3/8*d^4*b^4*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-3*d^4*c^(1/2)*b^2*
a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3*d^4*c*b*a*(c*x^2+b*x+a)^(1/2)-6*d^4*c^2*a*x*(c*x^2+b*x+a)^(1/2
)+6*d^4*c^(3/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

________________________________________________________________________________________

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)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 3.3668, size = 710, normalized size = 6.07 \begin{align*} \left [\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{c} d^{4} \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^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \,{\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x +{\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{16 \, c}, -\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-c} d^{4} \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^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \,{\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x +{\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{8 \, c}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c)*d^4*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^4*x^3 + 24*b*c^3*d^4*x^2 + 6*(3*b^2*c^2 - 4*a*c^3)*d^4*x + (5*b^3*c
- 12*a*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a))/c, -1/8*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c)*d^4*arctan(1/2*sq
rt(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^4*x^3 + 24*b*c^3*d^4*x^2 + 6*(
3*b^2*c^2 - 4*a*c^3)*d^4*x + (5*b^3*c - 12*a*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a))/c]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{4} \left (\int \frac{b^{4}}{\sqrt{a + b x + c x^{2}}}\, dx + \int \frac{16 c^{4} x^{4}}{\sqrt{a + b x + c x^{2}}}\, dx + \int \frac{32 b c^{3} x^{3}}{\sqrt{a + b x + c x^{2}}}\, dx + \int \frac{24 b^{2} c^{2} x^{2}}{\sqrt{a + b x + c x^{2}}}\, dx + \int \frac{8 b^{3} 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)**4/(c*x**2+b*x+a)**(1/2),x)

[Out]

d**4*(Integral(b**4/sqrt(a + b*x + c*x**2), x) + Integral(16*c**4*x**4/sqrt(a + b*x + c*x**2), x) + Integral(3
2*b*c**3*x**3/sqrt(a + b*x + c*x**2), x) + Integral(24*b**2*c**2*x**2/sqrt(a + b*x + c*x**2), x) + Integral(8*
b**3*c*x/sqrt(a + b*x + c*x**2), x))

________________________________________________________________________________________

Giac [A]  time = 1.17606, size = 215, normalized size = 1.84 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \, c^{3} d^{4} x + 3 \, b c^{2} d^{4}\right )} x + \frac{3 \,{\left (3 \, b^{2} c^{4} d^{4} - 4 \, a c^{5} d^{4}\right )}}{c^{3}}\right )} x + \frac{5 \, b^{3} c^{3} d^{4} - 12 \, a b c^{4} d^{4}}{c^{3}}\right )} - \frac{3 \,{\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, \sqrt{c}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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