3.12.93 \(\int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx\) [1193]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 123, 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 = {699, 706, 635, 212} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((b^2 - 4*a*c)*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 699

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 706

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 (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 ) \text {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*}

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Mathematica [A]
time = 0.31, size = 97, normalized size = 0.79 \begin {gather*} \frac {d^2 \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (b^2+8 b c x+4 c \left (a+2 c x^2\right )\right )+\left (b^2-4 a c\right )^2 \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{32 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(353\) vs. \(2(105)=210\).
time = 0.68, size = 354, normalized size = 2.88

method	result	size
risch	\(\frac {\left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+8 a \,c^{2} x +10 b^{2} c x +4 a b c +b^{3}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{2}}{16 c}+\left (-\frac {\sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a^{2}}{2}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) a \,b^{2}}{4 \sqrt {c}}-\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) b^{4}}{32 c^{\frac {3}{2}}}\right ) d^{2}\)	\(167\)
default	\(d^{2} \left (4 c^{2} \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+4 b c \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )+b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )\right )\)	\(354\)

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,method=_RETURNVERBOSE)

[Out]

d^2*(4*c^2*(1/4*x*(c*x^2+b*x+a)^(3/2)/c-5/8*b/c*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a
)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))-1/4*a/c*(1/4*(2*c*x+b)*(c*x^2+
b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))+4*b*c*(1/3*(c*x^2+b*x+a)^
(3/2)/c-1/2*b/c*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))))+b^2*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*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 = 0.00, 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*d*x+b*d)^2*(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 deta

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Fricas [A]
time = 1.91, size = 309, normalized size = 2.51 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 {gather*}

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 = 1.13, size = 155, normalized size = 1.26 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 1.07, size = 335, normalized size = 2.72 \begin {gather*} b^2\,d^2\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}-a\,c\,d^2\,\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 )-\frac {5\,b\,c\,d^2\,\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 )}{2}+c\,d^2\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}+\frac {b\,d^2\,\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 )}{4\,c^{3/2}}+\frac {b\,d^2\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{6\,c}+\frac {b^2\,d^2\,\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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