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]

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

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.11, 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]))

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fricas [A]  time = 1.39, size = 313, normalized size = 2.68 \[ \left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} d^{4} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (16 \, c^{4} d^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x + {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{16 \, c}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} d^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (16 \, c^{4} d^{4} x^{3} + 24 \, b c^{3} d^{4} x^{2} + 6 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} d^{4} x + {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{8 \, c}\right ] \]

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]

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giac [A]  time = 0.28, size = 159, normalized size = 1.36 \[ \frac {1}{4} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, c^{3} d^{4} x + 3 \, b c^{2} d^{4}\right )} x + \frac {3 \, {\left (3 \, b^{2} c^{4} d^{4} - 4 \, a c^{5} d^{4}\right )}}{c^{3}}\right )} x + \frac {5 \, b^{3} c^{3} d^{4} - 12 \, a b c^{4} d^{4}}{c^{3}}\right )} - \frac {3 \, {\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{8 \, \sqrt {c}} \]

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)

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maple [B]  time = 0.05, size = 242, normalized size = 2.07 \[ 4 \sqrt {c \,x^{2}+b x +a}\, c^{3} d^{4} x^{3}+6 \sqrt {c \,x^{2}+b x +a}\, b \,c^{2} d^{4} x^{2}+6 a^{2} c^{\frac {3}{2}} d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )-3 a \,b^{2} \sqrt {c}\, d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )+\frac {3 b^{4} d^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-6 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2} d^{4} x +\frac {9 \sqrt {c \,x^{2}+b x +a}\, b^{2} c \,d^{4} x}{2}-3 \sqrt {c \,x^{2}+b x +a}\, a b c \,d^{4}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{3} d^{4}}{4} \]

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

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

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 >> 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 \[ \int \frac {{\left (b\,d+2\,c\,d\,x\right )}^4}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(1/2),x)

[Out]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{4} \left (\int \frac {b^{4}}{\sqrt {a + b x + c x^{2}}}\, dx + \int \frac {16 c^{4} x^{4}}{\sqrt {a + b x + c x^{2}}}\, dx + \int \frac {32 b c^{3} x^{3}}{\sqrt {a + b x + c x^{2}}}\, dx + \int \frac {24 b^{2} c^{2} x^{2}}{\sqrt {a + b x + c x^{2}}}\, dx + \int \frac {8 b^{3} c x}{\sqrt {a + b x + c x^{2}}}\, dx\right ) \]

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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