3.11.82 \(\int \sqrt {b d+2 c d x} (a+b x+c x^2)^2 \, dx\)

Optimal. Leaf size=88 \[ -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}+\frac {(b d+2 c d x)^{11/2}}{176 c^3 d^5} \]

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {683} \begin {gather*} -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}+\frac {(b d+2 c d x)^{11/2}}{176 c^3 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2))/(48*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(56*c^3*d^3) + (b*d
+ 2*c*d*x)^(11/2)/(176*c^3*d^5)

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

\begin {align*} \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx &=\int \left (\frac {\left (-b^2+4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^2}+\frac {\left (-b^2+4 a c\right ) (b d+2 c d x)^{5/2}}{8 c^2 d^2}+\frac {(b d+2 c d x)^{9/2}}{16 c^2 d^4}\right ) \, dx\\ &=\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac {(b d+2 c d x)^{11/2}}{176 c^3 d^5}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 92, normalized size = 1.05 \begin {gather*} \frac {\left (c^2 \left (77 a^2+66 a c x^2+21 c^2 x^4\right )+b^2 c \left (15 c x^2-22 a\right )+6 b c^2 x \left (11 a+7 c x^2\right )+2 b^4-6 b^3 c x\right ) (d (b+2 c x))^{3/2}}{231 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*(b + 2*c*x))^(3/2)*(2*b^4 - 6*b^3*c*x + 6*b*c^2*x*(11*a + 7*c*x^2) + b^2*c*(-22*a + 15*c*x^2) + c^2*(77*a^
2 + 66*a*c*x^2 + 21*c^2*x^4)))/(231*c^3*d)

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IntegrateAlgebraic [A]  time = 0.09, size = 96, normalized size = 1.09 \begin {gather*} \frac {\left (77 a^2 c^2-22 a b^2 c+66 a b c^2 x+66 a c^3 x^2+2 b^4-6 b^3 c x+15 b^2 c^2 x^2+42 b c^3 x^3+21 c^4 x^4\right ) (b d+2 c d x)^{3/2}}{231 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d + 2*c*d*x)^(3/2)*(2*b^4 - 22*a*b^2*c + 77*a^2*c^2 - 6*b^3*c*x + 66*a*b*c^2*x + 15*b^2*c^2*x^2 + 66*a*c^3
*x^2 + 42*b*c^3*x^3 + 21*c^4*x^4))/(231*c^3*d)

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fricas [A]  time = 0.40, size = 121, normalized size = 1.38 \begin {gather*} \frac {{\left (42 \, c^{5} x^{5} + 105 \, b c^{4} x^{4} + 2 \, b^{5} - 22 \, a b^{3} c + 77 \, a^{2} b c^{2} + 12 \, {\left (6 \, b^{2} c^{3} + 11 \, a c^{4}\right )} x^{3} + 3 \, {\left (b^{3} c^{2} + 66 \, a b c^{3}\right )} x^{2} - 2 \, {\left (b^{4} c - 11 \, a b^{2} c^{2} - 77 \, a^{2} c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{231 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/231*(42*c^5*x^5 + 105*b*c^4*x^4 + 2*b^5 - 22*a*b^3*c + 77*a^2*b*c^2 + 12*(6*b^2*c^3 + 11*a*c^4)*x^3 + 3*(b^3
*c^2 + 66*a*b*c^3)*x^2 - 2*(b^4*c - 11*a*b^2*c^2 - 77*a^2*c^3)*x)*sqrt(2*c*d*x + b*d)/c^3

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giac [B]  time = 0.19, size = 580, normalized size = 6.59 \begin {gather*} \frac {55440 \, \sqrt {2 \, c d x + b d} a^{2} b - \frac {18480 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} a^{2}}{d} - \frac {18480 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} a b^{2}}{c d} + \frac {924 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} b^{3}}{c^{2} d^{2}} + \frac {5544 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} a b}{c d^{2}} - \frac {792 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} b^{2}}{c^{2} d^{3}} - \frac {792 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} a}{c d^{3}} + \frac {55 \, {\left (315 \, \sqrt {2 \, c d x + b d} b^{4} d^{4} - 420 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{3} d^{3} + 378 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} d^{2} - 180 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b d + 35 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}\right )} b}{c^{2} d^{4}} - \frac {5 \, {\left (693 \, \sqrt {2 \, c d x + b d} b^{5} d^{5} - 1155 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} d^{4} + 1386 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{3} d^{3} - 990 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{2} d^{2} + 385 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} b d - 63 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}\right )}}{c^{2} d^{5}}}{55440 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55440*(55440*sqrt(2*c*d*x + b*d)*a^2*b - 18480*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a^2/d - 1
8480*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a*b^2/(c*d) + 924*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 1
0*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*b^3/(c^2*d^2) + 5544*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 -
10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*a*b/(c*d^2) - 792*(35*sqrt(2*c*d*x + b*d)*b^3*d^3 - 35
*(2*c*d*x + b*d)^(3/2)*b^2*d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x + b*d)^(7/2))*b^2/(c^2*d^3) - 792*(
35*sqrt(2*c*d*x + b*d)*b^3*d^3 - 35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x
+ b*d)^(7/2))*a/(c*d^3) + 55*(315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 420*(2*c*d*x + b*d)^(3/2)*b^3*d^3 + 378*(2*c*d
*x + b*d)^(5/2)*b^2*d^2 - 180*(2*c*d*x + b*d)^(7/2)*b*d + 35*(2*c*d*x + b*d)^(9/2))*b/(c^2*d^4) - 5*(693*sqrt(
2*c*d*x + b*d)*b^5*d^5 - 1155*(2*c*d*x + b*d)^(3/2)*b^4*d^4 + 1386*(2*c*d*x + b*d)^(5/2)*b^3*d^3 - 990*(2*c*d*
x + b*d)^(7/2)*b^2*d^2 + 385*(2*c*d*x + b*d)^(9/2)*b*d - 63*(2*c*d*x + b*d)^(11/2))/(c^2*d^5))/c

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maple [A]  time = 0.05, size = 96, normalized size = 1.09 \begin {gather*} \frac {\left (2 c x +b \right ) \left (21 c^{4} x^{4}+42 b \,c^{3} x^{3}+66 a \,c^{3} x^{2}+15 x^{2} b^{2} c^{2}+66 a b \,c^{2} x -6 x \,b^{3} c +77 a^{2} c^{2}-22 a \,b^{2} c +2 b^{4}\right ) \sqrt {2 c d x +b d}}{231 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/231*(2*c*x+b)*(21*c^4*x^4+42*b*c^3*x^3+66*a*c^3*x^2+15*b^2*c^2*x^2+66*a*b*c^2*x-6*b^3*c*x+77*a^2*c^2-22*a*b^
2*c+2*b^4)*(2*c*d*x+b*d)^(1/2)/c^3

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maxima [A]  time = 1.41, size = 81, normalized size = 0.92 \begin {gather*} -\frac {66 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 77 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d^{4} - 21 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{3696 \, c^{3} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3696*(66*(2*c*d*x + b*d)^(7/2)*(b^2 - 4*a*c)*d^2 - 77*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^(3/2)*
d^4 - 21*(2*c*d*x + b*d)^(11/2))/(c^3*d^5)

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mupad [B]  time = 0.50, size = 99, normalized size = 1.12 \begin {gather*} \frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (21\,{\left (b\,d+2\,c\,d\,x\right )}^4+77\,b^4\,d^4-66\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+1232\,a^2\,c^2\,d^4+264\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-616\,a\,b^2\,c\,d^4\right )}{3696\,c^3\,d^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*d + 2*c*d*x)^(3/2)*(21*(b*d + 2*c*d*x)^4 + 77*b^4*d^4 - 66*b^2*d^2*(b*d + 2*c*d*x)^2 + 1232*a^2*c^2*d^4 +
264*a*c*d^2*(b*d + 2*c*d*x)^2 - 616*a*b^2*c*d^4))/(3696*c^3*d^5)

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sympy [A]  time = 3.63, size = 94, normalized size = 1.07 \begin {gather*} \frac {\frac {\left (b d + 2 c d x\right )^{\frac {3}{2}} \left (16 a^{2} c^{2} - 8 a b^{2} c + b^{4}\right )}{48 c^{2}} + \frac {\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {7}{2}}}{56 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {11}{2}}}{176 c^{2} d^{4}}}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*d + 2*c*d*x)**(3/2)*(16*a**2*c**2 - 8*a*b**2*c + b**4)/(48*c**2) + (4*a*c - b**2)*(b*d + 2*c*d*x)**(7/2)/(
56*c**2*d**2) + (b*d + 2*c*d*x)**(11/2)/(176*c**2*d**4))/(c*d)

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