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

Optimal. Leaf size=88 \[ -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{40 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^3 d}+\frac {(b d+2 c d x)^{9/2}}{144 c^3 d^5} \]

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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)^{5/2}}{40 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^3 d}+\frac {(b d+2 c d x)^{9/2}}{144 c^3 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2 - 4*a*c)^2*Sqrt[b*d + 2*c*d*x])/(16*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2))/(40*c^3*d^3) + (b*d +
2*c*d*x)^(9/2)/(144*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 \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx &=\int \left (\frac {\left (-b^2+4 a c\right )^2}{16 c^2 \sqrt {b d+2 c d x}}+\frac {\left (-b^2+4 a c\right ) (b d+2 c d x)^{3/2}}{8 c^2 d^2}+\frac {(b d+2 c d x)^{7/2}}{16 c^2 d^4}\right ) \, dx\\ &=\frac {\left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{40 c^3 d^3}+\frac {(b d+2 c d x)^{9/2}}{144 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 (45 a^2+18 a c x^2+5 c^2 x^4\right )+3 b^2 c \left (c x^2-6 a\right )+2 b c^2 x \left (9 a+5 c x^2\right )+2 b^4-2 b^3 c x\right ) \sqrt {d (b+2 c x)}}{45 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*(b + 2*c*x)]*(2*b^4 - 2*b^3*c*x + 3*b^2*c*(-6*a + c*x^2) + 2*b*c^2*x*(9*a + 5*c*x^2) + c^2*(45*a^2 + 1
8*a*c*x^2 + 5*c^2*x^4)))/(45*c^3*d)

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IntegrateAlgebraic [A]  time = 0.10, size = 96, normalized size = 1.09 \begin {gather*} \frac {\left (45 a^2 c^2-18 a b^2 c+18 a b c^2 x+18 a c^3 x^2+2 b^4-2 b^3 c x+3 b^2 c^2 x^2+10 b c^3 x^3+5 c^4 x^4\right ) \sqrt {b d+2 c d x}}{45 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*d + 2*c*d*x]*(2*b^4 - 18*a*b^2*c + 45*a^2*c^2 - 2*b^3*c*x + 18*a*b*c^2*x + 3*b^2*c^2*x^2 + 18*a*c^3*x^
2 + 10*b*c^3*x^3 + 5*c^4*x^4))/(45*c^3*d)

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fricas [A]  time = 0.39, size = 92, normalized size = 1.05 \begin {gather*} \frac {{\left (5 \, c^{4} x^{4} + 10 \, b c^{3} x^{3} + 2 \, b^{4} - 18 \, a b^{2} c + 45 \, a^{2} c^{2} + 3 \, {\left (b^{2} c^{2} + 6 \, a c^{3}\right )} x^{2} - 2 \, {\left (b^{3} c - 9 \, a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{45 \, c^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(5*c^4*x^4 + 10*b*c^3*x^3 + 2*b^4 - 18*a*b^2*c + 45*a^2*c^2 + 3*(b^2*c^2 + 6*a*c^3)*x^2 - 2*(b^3*c - 9*a*
b*c^2)*x)*sqrt(2*c*d*x + b*d)/(c^3*d)

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giac [B]  time = 0.18, size = 349, normalized size = 3.97 \begin {gather*} \frac {5040 \, \sqrt {2 \, c d x + b d} a^{2} - \frac {1680 \, {\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}{c d} + \frac {84 \, {\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^{2}}{c^{2} d^{2}} + \frac {168 \, {\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}{c d^{2}} - \frac {36 \, {\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}{c^{2} d^{3}} + \frac {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}}}{c^{2} d^{4}}}{5040 \, c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5040*(5040*sqrt(2*c*d*x + b*d)*a^2 - 1680*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a*b/(c*d) + 84
*(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))*b^2/(c^2*d^2) + 168
*(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/(c*d^2) - 36*(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/(c^2*d^3) + (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))/(c^2*d^4))/(c*d)

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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 (5 c^{4} x^{4}+10 b \,c^{3} x^{3}+18 a \,c^{3} x^{2}+3 x^{2} b^{2} c^{2}+18 a b \,c^{2} x -2 x \,b^{3} c +45 a^{2} c^{2}-18 a \,b^{2} c +2 b^{4}\right )}{45 \sqrt {2 c d x +b d}\, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(2*c*x+b)*(5*c^4*x^4+10*b*c^3*x^3+18*a*c^3*x^2+3*b^2*c^2*x^2+18*a*b*c^2*x-2*b^3*c*x+45*a^2*c^2-18*a*b^2*c
+2*b^4)/c^3/(2*c*d*x+b*d)^(1/2)

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maxima [B]  time = 1.47, size = 351, normalized size = 3.99 \begin {gather*} \frac {5040 \, \sqrt {2 \, c d x + b d} a^{2} - 168 \, a {\left (\frac {10 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b}{c d} - \frac {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}}}{c d^{2}}\right )} + \frac {84 \, {\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^{2}}{c^{2} d^{2}} - \frac {36 \, {\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}{c^{2} d^{3}} + \frac {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}}}{c^{2} d^{4}}}{5040 \, c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5040*(5040*sqrt(2*c*d*x + b*d)*a^2 - 168*a*(10*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b/(c*d) -
(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))/(c*d^2)) + 84*(15*s
qrt(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))*b^2/(c^2*d^2) - 36*(35*sq
rt(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/(c^2*d^3) + (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))/(c^2*d^4))/(c*d)

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mupad [B]  time = 0.49, size = 99, normalized size = 1.12 \begin {gather*} \frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (5\,{\left (b\,d+2\,c\,d\,x\right )}^4+45\,b^4\,d^4-18\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+720\,a^2\,c^2\,d^4+72\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-360\,a\,b^2\,c\,d^4\right )}{720\,c^3\,d^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*d + 2*c*d*x)^(1/2)*(5*(b*d + 2*c*d*x)^4 + 45*b^4*d^4 - 18*b^2*d^2*(b*d + 2*c*d*x)^2 + 720*a^2*c^2*d^4 + 72
*a*c*d^2*(b*d + 2*c*d*x)^2 - 360*a*b^2*c*d^4))/(720*c^3*d^5)

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sympy [A]  time = 61.62, size = 668, normalized size = 7.59 \begin {gather*} \begin {cases} \frac {- \frac {a^{2} b}{\sqrt {b d + 2 c d x}} - \frac {a^{2} \left (- \frac {b d}{\sqrt {b d + 2 c d x}} - \sqrt {b d + 2 c d x}\right )}{d} - \frac {a b^{2} \left (- \frac {b d}{\sqrt {b d + 2 c d x}} - \sqrt {b d + 2 c d x}\right )}{c d} - \frac {3 a b \left (\frac {b^{2} d^{2}}{\sqrt {b d + 2 c d x}} + 2 b d \sqrt {b d + 2 c d x} - \frac {\left (b d + 2 c d x\right )^{\frac {3}{2}}}{3}\right )}{2 c d^{2}} - \frac {a \left (- \frac {b^{3} d^{3}}{\sqrt {b d + 2 c d x}} - 3 b^{2} d^{2} \sqrt {b d + 2 c d x} + b d \left (b d + 2 c d x\right )^{\frac {3}{2}} - \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{5}\right )}{2 c d^{3}} - \frac {b^{3} \left (\frac {b^{2} d^{2}}{\sqrt {b d + 2 c d x}} + 2 b d \sqrt {b d + 2 c d x} - \frac {\left (b d + 2 c d x\right )^{\frac {3}{2}}}{3}\right )}{4 c^{2} d^{2}} - \frac {b^{2} \left (- \frac {b^{3} d^{3}}{\sqrt {b d + 2 c d x}} - 3 b^{2} d^{2} \sqrt {b d + 2 c d x} + b d \left (b d + 2 c d x\right )^{\frac {3}{2}} - \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{5}\right )}{2 c^{2} d^{3}} - \frac {5 b \left (\frac {b^{4} d^{4}}{\sqrt {b d + 2 c d x}} + 4 b^{3} d^{3} \sqrt {b d + 2 c d x} - 2 b^{2} d^{2} \left (b d + 2 c d x\right )^{\frac {3}{2}} + \frac {4 b d \left (b d + 2 c d x\right )^{\frac {5}{2}}}{5} - \frac {\left (b d + 2 c d x\right )^{\frac {7}{2}}}{7}\right )}{16 c^{2} d^{4}} - \frac {- \frac {b^{5} d^{5}}{\sqrt {b d + 2 c d x}} - 5 b^{4} d^{4} \sqrt {b d + 2 c d x} + \frac {10 b^{3} d^{3} \left (b d + 2 c d x\right )^{\frac {3}{2}}}{3} - 2 b^{2} d^{2} \left (b d + 2 c d x\right )^{\frac {5}{2}} + \frac {5 b d \left (b d + 2 c d x\right )^{\frac {7}{2}}}{7} - \frac {\left (b d + 2 c d x\right )^{\frac {9}{2}}}{9}}{16 c^{2} d^{5}}}{c} & \text {for}\: c \neq 0 \\\frac {\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}}{\sqrt {b d}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-a**2*b/sqrt(b*d + 2*c*d*x) - a**2*(-b*d/sqrt(b*d + 2*c*d*x) - sqrt(b*d + 2*c*d*x))/d - a*b**2*(-b
*d/sqrt(b*d + 2*c*d*x) - sqrt(b*d + 2*c*d*x))/(c*d) - 3*a*b*(b**2*d**2/sqrt(b*d + 2*c*d*x) + 2*b*d*sqrt(b*d +
2*c*d*x) - (b*d + 2*c*d*x)**(3/2)/3)/(2*c*d**2) - a*(-b**3*d**3/sqrt(b*d + 2*c*d*x) - 3*b**2*d**2*sqrt(b*d + 2
*c*d*x) + b*d*(b*d + 2*c*d*x)**(3/2) - (b*d + 2*c*d*x)**(5/2)/5)/(2*c*d**3) - b**3*(b**2*d**2/sqrt(b*d + 2*c*d
*x) + 2*b*d*sqrt(b*d + 2*c*d*x) - (b*d + 2*c*d*x)**(3/2)/3)/(4*c**2*d**2) - b**2*(-b**3*d**3/sqrt(b*d + 2*c*d*
x) - 3*b**2*d**2*sqrt(b*d + 2*c*d*x) + b*d*(b*d + 2*c*d*x)**(3/2) - (b*d + 2*c*d*x)**(5/2)/5)/(2*c**2*d**3) -
5*b*(b**4*d**4/sqrt(b*d + 2*c*d*x) + 4*b**3*d**3*sqrt(b*d + 2*c*d*x) - 2*b**2*d**2*(b*d + 2*c*d*x)**(3/2) + 4*
b*d*(b*d + 2*c*d*x)**(5/2)/5 - (b*d + 2*c*d*x)**(7/2)/7)/(16*c**2*d**4) - (-b**5*d**5/sqrt(b*d + 2*c*d*x) - 5*
b**4*d**4*sqrt(b*d + 2*c*d*x) + 10*b**3*d**3*(b*d + 2*c*d*x)**(3/2)/3 - 2*b**2*d**2*(b*d + 2*c*d*x)**(5/2) + 5
*b*d*(b*d + 2*c*d*x)**(7/2)/7 - (b*d + 2*c*d*x)**(9/2)/9)/(16*c**2*d**5))/c, Ne(c, 0)), (Piecewise((a**2*x, Eq
(b, 0)), ((a + b*x)**3/(3*b), True))/sqrt(b*d), True))

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