### 3.1191 $$\int (b d+2 c d x)^4 \sqrt{a+b x+c x^2} \, dx$$

Optimal. Leaf size=165 $-\frac{d^4 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2}}-\frac{d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}-\frac{d^4 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}$

[Out]

-((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c) - ((b^2 - 4*a*c)*d^4*(b + 2*c*x)^3*Sqrt[a + b*
x + c*x^2])/(48*c) + (d^4*(b + 2*c*x)^5*Sqrt[a + b*x + c*x^2])/(12*c) - ((b^2 - 4*a*c)^3*d^4*ArcTanh[(b + 2*c*
x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(3/2))

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Rubi [A]  time = 0.0976735, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.154, Rules used = {685, 692, 621, 206} $-\frac{d^4 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2}}-\frac{d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}-\frac{d^4 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c) - ((b^2 - 4*a*c)*d^4*(b + 2*c*x)^3*Sqrt[a + b*
x + c*x^2])/(48*c) + (d^4*(b + 2*c*x)^5*Sqrt[a + b*x + c*x^2])/(12*c) - ((b^2 - 4*a*c)^3*d^4*ArcTanh[(b + 2*c*
x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*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)^4 \sqrt{a+b x+c x^2} \, dx &=\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}-\frac{\left (b^2-4 a c\right ) \int \frac{(b d+2 c d x)^4}{\sqrt{a+b x+c x^2}} \, dx}{24 c}\\ &=-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}-\frac{\left (\left (b^2-4 a c\right )^2 d^2\right ) \int \frac{(b d+2 c d x)^2}{\sqrt{a+b x+c x^2}} \, dx}{32 c}\\ &=-\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}-\frac{\left (\left (b^2-4 a c\right )^3 d^4\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{64 c}\\ &=-\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}-\frac{\left (\left (b^2-4 a c\right )^3 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 )}{32 c}\\ &=-\frac{\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}-\frac{\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c}-\frac{\left (b^2-4 a c\right )^3 d^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.888537, size = 203, normalized size = 1.23 $d^4 \left (\frac{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \left (2 (b+2 c x) \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \left (4 c \left (a+2 c x^2\right )+b^2+8 b c x\right )-\sqrt{c} \sqrt{4 a-\frac{b^2}{c}} \left (4 a c-b^2\right ) \sinh ^{-1}\left (\frac{b+2 c x}{\sqrt{c} \sqrt{4 a-\frac{b^2}{c}}}\right )\right )}{64 c \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}+\frac{1}{3} (b+2 c x)^3 (a+x (b+c x))^{3/2}\right )$

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)^3*(a + x*(b + c*x))^(3/2))/3 + ((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*(2*(b + 2*c*x)*Sqrt[(c*(
a + x*(b + c*x)))/(-b^2 + 4*a*c)]*(b^2 + 8*b*c*x + 4*c*(a + 2*c*x^2)) - Sqrt[4*a - b^2/c]*Sqrt[c]*(-b^2 + 4*a*
c)*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])]))/(64*c*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]))

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Maple [B]  time = 0.055, size = 413, normalized size = 2.5 \begin{align*}{\frac{7\,{d}^{4}{b}^{3}}{12} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+4\,{d}^{4}{c}^{2}b{x}^{2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}+{\frac{5\,{d}^{4}{b}^{2}cx}{2} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{4}{b}^{6}}{64}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{d}^{4}cba \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}-2\,{d}^{4}{c}^{2}ax \left ( c{x}^{2}+bx+a \right ) ^{3/2}+{d}^{4}{c}^{2}{a}^{2}x\sqrt{c{x}^{2}+bx+a}+{\frac{c{d}^{4}{a}^{2}b}{2}\sqrt{c{x}^{2}+bx+a}}+{d}^{4}{c}^{{\frac{3}{2}}}{a}^{3}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) +{\frac{8\,{d}^{4}{c}^{3}{x}^{3}}{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{{d}^{4}{b}^{4}x}{16}\sqrt{c{x}^{2}+bx+a}}+{\frac{{d}^{4}{b}^{5}}{32\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{{d}^{4}{b}^{3}a}{4}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{d}^{4}{b}^{4}a}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{c{d}^{4}{b}^{2}ax}{2}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{d}^{4}{b}^{2}{a}^{2}}{4}\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)^4*(c*x^2+b*x+a)^(1/2),x)

[Out]

7/12*d^4*b^3*(c*x^2+b*x+a)^(3/2)+4*d^4*c^2*b*x^2*(c*x^2+b*x+a)^(3/2)+5/2*d^4*c*b^2*x*(c*x^2+b*x+a)^(3/2)-1/64*
d^4*b^6/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-d^4*c*b*a*(c*x^2+b*x+a)^(3/2)-2*d^4*c^2*a*x*(c*x^2
+b*x+a)^(3/2)+d^4*c^2*a^2*x*(c*x^2+b*x+a)^(1/2)+1/2*d^4*c*a^2*(c*x^2+b*x+a)^(1/2)*b+d^4*c^(3/2)*a^3*ln((1/2*b+
c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+8/3*d^4*c^3*x^3*(c*x^2+b*x+a)^(3/2)+1/16*d^4*b^4*x*(c*x^2+b*x+a)^(1/2)+1/32*
d^4/c*b^5*(c*x^2+b*x+a)^(1/2)-1/4*d^4*b^3*a*(c*x^2+b*x+a)^(1/2)+3/16*d^4*b^4/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c
*x^2+b*x+a)^(1/2))*a-1/2*d^4*c*b^2*a*x*(c*x^2+b*x+a)^(1/2)-3/4*d^4*c^(1/2)*b^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)^4*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27941, size = 1071, normalized size = 6.49 \begin{align*} \left [-\frac{3 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\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 (256 \, c^{6} d^{4} x^{5} + 640 \, b c^{5} d^{4} x^{4} + 16 \,{\left (39 \, b^{2} c^{4} + 4 \, a c^{5}\right )} d^{4} x^{3} + 8 \,{\left (37 \, b^{3} c^{3} + 12 \, a b c^{4}\right )} d^{4} x^{2} + 2 \,{\left (31 \, b^{4} c^{2} + 48 \, a b^{2} c^{3} - 48 \, a^{2} c^{4}\right )} d^{4} x +{\left (3 \, b^{5} c + 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{384 \, c^{2}}, \frac{3 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\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 (256 \, c^{6} d^{4} x^{5} + 640 \, b c^{5} d^{4} x^{4} + 16 \,{\left (39 \, b^{2} c^{4} + 4 \, a c^{5}\right )} d^{4} x^{3} + 8 \,{\left (37 \, b^{3} c^{3} + 12 \, a b c^{4}\right )} d^{4} x^{2} + 2 \,{\left (31 \, b^{4} c^{2} + 48 \, a b^{2} c^{3} - 48 \, a^{2} c^{4}\right )} d^{4} x +{\left (3 \, b^{5} c + 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{192 \, c^{2}}\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]

[-1/384*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*d^4*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqr
t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(256*c^6*d^4*x^5 + 640*b*c^5*d^4*x^4 + 16*(39*b^2*c^4 + 4*
a*c^5)*d^4*x^3 + 8*(37*b^3*c^3 + 12*a*b*c^4)*d^4*x^2 + 2*(31*b^4*c^2 + 48*a*b^2*c^3 - 48*a^2*c^4)*d^4*x + (3*b
^5*c + 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^4)*sqrt(c*x^2 + b*x + a))/c^2, 1/192*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c
^2 - 64*a^3*c^3)*sqrt(-c)*d^4*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*(256*c^6*d^4*x^5 + 640*b*c^5*d^4*x^4 + 16*(39*b^2*c^4 + 4*a*c^5)*d^4*x^3 + 8*(37*b^3*c^3 + 12*a*b*c^4)*d^4*
x^2 + 2*(31*b^4*c^2 + 48*a*b^2*c^3 - 48*a^2*c^4)*d^4*x + (3*b^5*c + 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^4)*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^{4} \left (\int b^{4} \sqrt{a + b x + c x^{2}}\, dx + \int 16 c^{4} x^{4} \sqrt{a + b x + c x^{2}}\, dx + \int 32 b c^{3} x^{3} \sqrt{a + b x + c x^{2}}\, dx + \int 24 b^{2} c^{2} x^{2} \sqrt{a + b x + c x^{2}}\, dx + \int 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))

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Giac [A]  time = 1.28851, size = 350, normalized size = 2.12 \begin{align*} \frac{1}{96} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \, c^{4} d^{4} x + 5 \, b c^{3} d^{4}\right )} x + \frac{39 \, b^{2} c^{7} d^{4} + 4 \, a c^{8} d^{4}}{c^{5}}\right )} x + \frac{37 \, b^{3} c^{6} d^{4} + 12 \, a b c^{7} d^{4}}{c^{5}}\right )} x + \frac{31 \, b^{4} c^{5} d^{4} + 48 \, a b^{2} c^{6} d^{4} - 48 \, a^{2} c^{7} d^{4}}{c^{5}}\right )} x + \frac{3 \, b^{5} c^{4} d^{4} + 32 \, a b^{3} c^{5} d^{4} - 48 \, a^{2} b c^{6} d^{4}}{c^{5}}\right )} + \frac{{\left (b^{6} d^{4} - 12 \, a b^{4} c d^{4} + 48 \, a^{2} b^{2} c^{2} d^{4} - 64 \, a^{3} c^{3} d^{4}\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{3}{2}}} \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]

1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^4*d^4*x + 5*b*c^3*d^4)*x + (39*b^2*c^7*d^4 + 4*a*c^8*d^4)/c^5)*x +
(37*b^3*c^6*d^4 + 12*a*b*c^7*d^4)/c^5)*x + (31*b^4*c^5*d^4 + 48*a*b^2*c^6*d^4 - 48*a^2*c^7*d^4)/c^5)*x + (3*b
^5*c^4*d^4 + 32*a*b^3*c^5*d^4 - 48*a^2*b*c^6*d^4)/c^5) + 1/64*(b^6*d^4 - 12*a*b^4*c*d^4 + 48*a^2*b^2*c^2*d^4 -
64*a^3*c^3*d^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(3/2)