3.11.74 \(\int (b d+2 c d x)^{3/2} (a+b x+c x^2) \, dx\)

Optimal. Leaf size=55 \[ \frac {(b d+2 c d x)^{9/2}}{36 c^2 d^3}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{20 c^2 d} \]

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {683} \begin {gather*} \frac {(b d+2 c d x)^{9/2}}{36 c^2 d^3}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{20 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2))/(20*c^2*d) + (b*d + 2*c*d*x)^(9/2)/(36*c^2*d^3)

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 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx &=\int \left (\frac {\left (-b^2+4 a c\right ) (b d+2 c d x)^{3/2}}{4 c}+\frac {(b d+2 c d x)^{7/2}}{4 c d^2}\right ) \, dx\\ &=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{20 c^2 d}+\frac {(b d+2 c d x)^{9/2}}{36 c^2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 45, normalized size = 0.82 \begin {gather*} \frac {\left (c \left (9 a+5 c x^2\right )-b^2+5 b c x\right ) (d (b+2 c x))^{5/2}}{45 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*(b + 2*c*x))^(5/2)*(-b^2 + 5*b*c*x + c*(9*a + 5*c*x^2)))/(45*c^2*d)

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IntegrateAlgebraic [A]  time = 0.08, size = 46, normalized size = 0.84 \begin {gather*} \frac {\left (9 a c-b^2+5 b c x+5 c^2 x^2\right ) (b d+2 c d x)^{5/2}}{45 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^(3/2)*(a + b*x + c*x^2),x]

[Out]

((b*d + 2*c*d*x)^(5/2)*(-b^2 + 9*a*c + 5*b*c*x + 5*c^2*x^2))/(45*c^2*d)

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fricas [A]  time = 0.41, size = 87, normalized size = 1.58 \begin {gather*} \frac {{\left (20 \, c^{4} d x^{4} + 40 \, b c^{3} d x^{3} + 3 \, {\left (7 \, b^{2} c^{2} + 12 \, a c^{3}\right )} d x^{2} + {\left (b^{3} c + 36 \, a b c^{2}\right )} d x - {\left (b^{4} - 9 \, a b^{2} c\right )} d\right )} \sqrt {2 \, c d x + b d}}{45 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(20*c^4*d*x^4 + 40*b*c^3*d*x^3 + 3*(7*b^2*c^2 + 12*a*c^3)*d*x^2 + (b^3*c + 36*a*b*c^2)*d*x - (b^4 - 9*a*b
^2*c)*d)*sqrt(2*c*d*x + b*d)/c^2

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giac [B]  time = 0.18, size = 376, normalized size = 6.84 \begin {gather*} \frac {1260 \, \sqrt {2 \, c d x + b d} a b^{2} d - 840 \, {\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 - \frac {210 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b^{3}}{c} + \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 )} a}{d} + \frac {105 \, {\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 d} - \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 d^{2}} + \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 d^{3}}}{1260 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1260*(1260*sqrt(2*c*d*x + b*d)*a*b^2*d - 840*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a*b - 210*(
3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b^3/c + 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))*a/d + 105*(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*d) - 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*d^2) + (315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 4
20*(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*d^3))/c

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maple [A]  time = 0.04, size = 46, normalized size = 0.84 \begin {gather*} \frac {\left (2 c x +b \right ) \left (5 c^{2} x^{2}+5 b c x +9 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{45 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(2*c*x+b)*(5*c^2*x^2+5*b*c*x+9*a*c-b^2)*(2*c*d*x+b*d)^(3/2)/c^2

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maxima [A]  time = 1.36, size = 46, normalized size = 0.84 \begin {gather*} -\frac {9 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{180 \, c^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*(9*(2*c*d*x + b*d)^(5/2)*(b^2 - 4*a*c)*d^2 - 5*(2*c*d*x + b*d)^(9/2))/(c^2*d^3)

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mupad [B]  time = 0.05, size = 39, normalized size = 0.71 \begin {gather*} \frac {{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (36\,a\,c+5\,{\left (b+2\,c\,x\right )}^2-9\,b^2\right )}{180\,c^2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*d + 2*c*d*x)^(5/2)*(36*a*c + 5*(b + 2*c*x)^2 - 9*b^2))/(180*c^2*d)

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sympy [A]  time = 9.76, size = 274, normalized size = 4.98 \begin {gather*} a b d \left (\begin {cases} x \sqrt {b d} & \text {for}\: c = 0 \\0 & \text {for}\: d = 0 \\\frac {\left (b d + 2 c d x\right )^{\frac {3}{2}}}{3 c d} & \text {otherwise} \end {cases}\right ) + \frac {a \left (- \frac {b d \left (b d + 2 c d x\right )^{\frac {3}{2}}}{3} + \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{5}\right )}{c d} + \frac {b^{2} \left (- \frac {b d \left (b d + 2 c d x\right )^{\frac {3}{2}}}{3} + \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{5}\right )}{2 c^{2} d} + \frac {3 b \left (\frac {b^{2} d^{2} \left (b d + 2 c d x\right )^{\frac {3}{2}}}{3} - \frac {2 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 )}{4 c^{2} d^{2}} + \frac {- \frac {b^{3} d^{3} \left (b d + 2 c d x\right )^{\frac {3}{2}}}{3} + \frac {3 b^{2} d^{2} \left (b d + 2 c d x\right )^{\frac {5}{2}}}{5} - \frac {3 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}}{4 c^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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