3.12.92 \(\int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx\) [1192]

Optimal. Leaf size=59 \[ \frac {4}{15} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2} \]

[Out]

4/15*(-4*a*c+b^2)*d^3*(c*x^2+b*x+a)^(3/2)+2/5*d^3*(2*c*x+b)^2*(c*x^2+b*x+a)^(3/2)

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Rubi [A]
time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {706, 643} \begin {gather*} \frac {4}{15} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^3*Sqrt[a + b*x + c*x^2],x]

[Out]

(4*(b^2 - 4*a*c)*d^3*(a + b*x + c*x^2)^(3/2))/15 + (2*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(3/2))/5

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

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)^3 \sqrt {a+b x+c x^2} \, dx &=\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}+\frac {1}{5} \left (2 \left (b^2-4 a c\right ) d^2\right ) \int (b d+2 c d x) \sqrt {a+b x+c x^2} \, dx\\ &=\frac {4}{15} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 80, normalized size = 1.36 \begin {gather*} -\frac {2}{15} d^3 \left (-5 a b^2+8 a^2 c-5 b^3 x-4 a b c x-17 b^2 c x^2-4 a c^2 x^2-24 b c^2 x^3-12 c^3 x^4\right ) \sqrt {a+x (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^3*Sqrt[a + b*x + c*x^2],x]

[Out]

(-2*d^3*(-5*a*b^2 + 8*a^2*c - 5*b^3*x - 4*a*b*c*x - 17*b^2*c*x^2 - 4*a*c^2*x^2 - 24*b*c^2*x^3 - 12*c^3*x^4)*Sq
rt[a + x*(b + c*x)])/15

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Maple [C] Result contains higher order function than in optimal. Order 3 vs. order 2.
time = 0.68, size = 666, normalized size = 11.29

method	result	size
gosper	\(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (-12 c^{2} x^{2}-12 b c x +8 a c -5 b^{2}\right ) d^{3}}{15}\)	\(41\)
trager	\(d^{3} \left (\frac {8}{5} c^{3} x^{4}+\frac {16}{5} b \,c^{2} x^{3}+\frac {8}{15} a \,c^{2} x^{2}+\frac {34}{15} b^{2} c \,x^{2}+\frac {8}{15} a b c x +\frac {2}{3} b^{3} x -\frac {16}{15} a^{2} c +\frac {2}{3} a \,b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\)	\(77\)
risch	\(-\frac {2 d^{3} \left (-12 c^{3} x^{4}-24 b \,c^{2} x^{3}-4 a \,c^{2} x^{2}-17 b^{2} c \,x^{2}-4 a b c x -5 b^{3} x +8 a^{2} c -5 a \,b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{15}\)	\(78\)
default	\(d^{3} \left (8 c^{3} \left (\frac {x^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \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 )}{10 c}-\frac {2 a \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 )}{5 c}\right )+12 b \,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 )+6 b^{2} 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^{3} \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 )\)	\(666\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

d^3*(8*c^3*(1/5*x^2*(c*x^2+b*x+a)^(3/2)/c-7/10*b/c*(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))))-2/5*a/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)))))+12*b*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))))+6*b^2*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^3*(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)^3*(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 = 1.50, size = 91, normalized size = 1.54 \begin {gather*} \frac {2}{15} \, {\left (12 \, c^{3} d^{3} x^{4} + 24 \, b c^{2} d^{3} x^{3} + {\left (17 \, b^{2} c + 4 \, a c^{2}\right )} d^{3} x^{2} + {\left (5 \, b^{3} + 4 \, a b c\right )} d^{3} x + {\left (5 \, a b^{2} - 8 \, a^{2} c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2/15*(12*c^3*d^3*x^4 + 24*b*c^2*d^3*x^3 + (17*b^2*c + 4*a*c^2)*d^3*x^2 + (5*b^3 + 4*a*b*c)*d^3*x + (5*a*b^2 -
8*a^2*c)*d^3)*sqrt(c*x^2 + b*x + a)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (58) = 116\).
time = 0.11, size = 216, normalized size = 3.66 \begin {gather*} - \frac {16 a^{2} c d^{3} \sqrt {a + b x + c x^{2}}}{15} + \frac {2 a b^{2} d^{3} \sqrt {a + b x + c x^{2}}}{3} + \frac {8 a b c d^{3} x \sqrt {a + b x + c x^{2}}}{15} + \frac {8 a c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{15} + \frac {2 b^{3} d^{3} x \sqrt {a + b x + c x^{2}}}{3} + \frac {34 b^{2} c d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{15} + \frac {16 b c^{2} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{5} + \frac {8 c^{3} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**3*(c*x**2+b*x+a)**(1/2),x)

[Out]

-16*a**2*c*d**3*sqrt(a + b*x + c*x**2)/15 + 2*a*b**2*d**3*sqrt(a + b*x + c*x**2)/3 + 8*a*b*c*d**3*x*sqrt(a + b
*x + c*x**2)/15 + 8*a*c**2*d**3*x**2*sqrt(a + b*x + c*x**2)/15 + 2*b**3*d**3*x*sqrt(a + b*x + c*x**2)/3 + 34*b
**2*c*d**3*x**2*sqrt(a + b*x + c*x**2)/15 + 16*b*c**2*d**3*x**3*sqrt(a + b*x + c*x**2)/5 + 8*c**3*d**3*x**4*sq
rt(a + b*x + c*x**2)/5

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Giac [A]
time = 1.57, size = 58, normalized size = 0.98 \begin {gather*} \frac {2}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} b^{2} d^{3} + \frac {8}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} c d^{3} - \frac {8}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} a c d^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.58, size = 89, normalized size = 1.51 \begin {gather*} \sqrt {c\,x^2+b\,x+a}\,\left (\frac {8\,c^3\,d^3\,x^4}{5}-\frac {2\,a\,d^3\,\left (8\,a\,c-5\,b^2\right )}{15}+\frac {2\,c\,d^3\,x^2\,\left (17\,b^2+4\,a\,c\right )}{15}+\frac {16\,b\,c^2\,d^3\,x^3}{5}+\frac {2\,b\,d^3\,x\,\left (5\,b^2+4\,a\,c\right )}{15}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(1/2),x)

[Out]

(a + b*x + c*x^2)^(1/2)*((8*c^3*d^3*x^4)/5 - (2*a*d^3*(8*a*c - 5*b^2))/15 + (2*c*d^3*x^2*(4*a*c + 17*b^2))/15
+ (16*b*c^2*d^3*x^3)/5 + (2*b*d^3*x*(4*a*c + 5*b^2))/15)

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