### 3.1261 $$\int (b d+2 c d x)^{5/2} (a+b x+c x^2) \, dx$$

Optimal. Leaf size=55 $\frac{(b d+2 c d x)^{11/2}}{44 c^2 d^3}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{28 c^2 d}$

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(28*c^2*d) + (b*d + 2*c*d*x)^(11/2)/(44*c^2*d^3)

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Rubi [A]  time = 0.0238854, antiderivative size = 55, normalized size of antiderivative = 1., 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} $\frac{(b d+2 c d x)^{11/2}}{44 c^2 d^3}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{28 c^2 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2),x]

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(28*c^2*d) + (b*d + 2*c*d*x)^(11/2)/(44*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)^{5/2} \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)^{5/2}}{4 c}+\frac{(b d+2 c d x)^{9/2}}{4 c d^2}\right ) \, dx\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{28 c^2 d}+\frac{(b d+2 c d x)^{11/2}}{44 c^2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0436917, size = 45, normalized size = 0.82 $\frac{\left (c \left (11 a+7 c x^2\right )-b^2+7 b c x\right ) (d (b+2 c x))^{7/2}}{77 c^2 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^(5/2)*(a + b*x + c*x^2),x]

[Out]

((d*(b + 2*c*x))^(7/2)*(-b^2 + 7*b*c*x + c*(11*a + 7*c*x^2)))/(77*c^2*d)

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Maple [A]  time = 0.041, size = 46, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,cx+b \right ) \left ( 7\,{c}^{2}{x}^{2}+7\,bcx+11\,ac-{b}^{2} \right ) }{77\,{c}^{2}} \left ( 2\,cdx+bd \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a),x)

[Out]

1/77*(2*c*x+b)*(7*c^2*x^2+7*b*c*x+11*a*c-b^2)*(2*c*d*x+b*d)^(5/2)/c^2

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Maxima [A]  time = 1.098, size = 62, normalized size = 1.13 \begin{align*} -\frac{11 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}{\left (b^{2} - 4 \, a c\right )} d^{2} - 7 \,{\left (2 \, c d x + b d\right )}^{\frac{11}{2}}}{308 \, c^{2} d^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

-1/308*(11*(2*c*d*x + b*d)^(7/2)*(b^2 - 4*a*c)*d^2 - 7*(2*c*d*x + b*d)^(11/2))/(c^2*d^3)

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Fricas [B]  time = 1.98296, size = 265, normalized size = 4.82 \begin{align*} \frac{{\left (56 \, c^{5} d^{2} x^{5} + 140 \, b c^{4} d^{2} x^{4} + 2 \,{\left (59 \, b^{2} c^{3} + 44 \, a c^{4}\right )} d^{2} x^{3} +{\left (37 \, b^{3} c^{2} + 132 \, a b c^{3}\right )} d^{2} x^{2} +{\left (b^{4} c + 66 \, a b^{2} c^{2}\right )} d^{2} x -{\left (b^{5} - 11 \, a b^{3} c\right )} d^{2}\right )} \sqrt{2 \, c d x + b d}}{77 \, c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

1/77*(56*c^5*d^2*x^5 + 140*b*c^4*d^2*x^4 + 2*(59*b^2*c^3 + 44*a*c^4)*d^2*x^3 + (37*b^3*c^2 + 132*a*b*c^3)*d^2*
x^2 + (b^4*c + 66*a*b^2*c^2)*d^2*x - (b^5 - 11*a*b^3*c)*d^2)*sqrt(2*c*d*x + b*d)/c^2

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Sympy [A]  time = 5.2453, size = 289, normalized size = 5.25 \begin{align*} \begin{cases} \frac{a b^{3} d^{2} \sqrt{b d + 2 c d x}}{7 c} + \frac{6 a b^{2} d^{2} x \sqrt{b d + 2 c d x}}{7} + \frac{12 a b c d^{2} x^{2} \sqrt{b d + 2 c d x}}{7} + \frac{8 a c^{2} d^{2} x^{3} \sqrt{b d + 2 c d x}}{7} - \frac{b^{5} d^{2} \sqrt{b d + 2 c d x}}{77 c^{2}} + \frac{b^{4} d^{2} x \sqrt{b d + 2 c d x}}{77 c} + \frac{37 b^{3} d^{2} x^{2} \sqrt{b d + 2 c d x}}{77} + \frac{118 b^{2} c d^{2} x^{3} \sqrt{b d + 2 c d x}}{77} + \frac{20 b c^{2} d^{2} x^{4} \sqrt{b d + 2 c d x}}{11} + \frac{8 c^{3} d^{2} x^{5} \sqrt{b d + 2 c d x}}{11} & \text{for}\: c \neq 0 \\\left (b d\right )^{\frac{5}{2}} \left (a x + \frac{b x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(5/2)*(c*x**2+b*x+a),x)

[Out]

Piecewise((a*b**3*d**2*sqrt(b*d + 2*c*d*x)/(7*c) + 6*a*b**2*d**2*x*sqrt(b*d + 2*c*d*x)/7 + 12*a*b*c*d**2*x**2*
sqrt(b*d + 2*c*d*x)/7 + 8*a*c**2*d**2*x**3*sqrt(b*d + 2*c*d*x)/7 - b**5*d**2*sqrt(b*d + 2*c*d*x)/(77*c**2) + b
**4*d**2*x*sqrt(b*d + 2*c*d*x)/(77*c) + 37*b**3*d**2*x**2*sqrt(b*d + 2*c*d*x)/77 + 118*b**2*c*d**2*x**3*sqrt(b
*d + 2*c*d*x)/77 + 20*b*c**2*d**2*x**4*sqrt(b*d + 2*c*d*x)/11 + 8*c**3*d**2*x**5*sqrt(b*d + 2*c*d*x)/11, Ne(c,
0)), ((b*d)**(5/2)*(a*x + b*x**2/2), True))

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Giac [B]  time = 1.14354, size = 508, normalized size = 9.24 \begin{align*} \frac{4620 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a b^{2} d - 1848 \,{\left (5 \,{\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 - \frac{462 \,{\left (5 \,{\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} + \frac{132 \,{\left (35 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} d^{2} - 42 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b d + 15 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}\right )} a}{d} + \frac{165 \,{\left (35 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} d^{2} - 42 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b d + 15 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}\right )} b^{2}}{c d} - \frac{44 \,{\left (105 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{3} d^{3} - 189 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b^{2} d^{2} + 135 \,{\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 d^{2}} + \frac{1155 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{4} d^{4} - 2772 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b^{3} d^{3} + 2970 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} b^{2} d^{2} - 1540 \,{\left (2 \, c d x + b d\right )}^{\frac{9}{2}} b d + 315 \,{\left (2 \, c d x + b d\right )}^{\frac{11}{2}}}{c d^{3}}}{13860 \, c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)*(c*x^2+b*x+a),x, algorithm="giac")

[Out]

1/13860*(4620*(2*c*d*x + b*d)^(3/2)*a*b^2*d - 1848*(5*(2*c*d*x + b*d)^(3/2)*b*d - 3*(2*c*d*x + b*d)^(5/2))*a*b
- 462*(5*(2*c*d*x + b*d)^(3/2)*b*d - 3*(2*c*d*x + b*d)^(5/2))*b^3/c + 132*(35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 -
42*(2*c*d*x + b*d)^(5/2)*b*d + 15*(2*c*d*x + b*d)^(7/2))*a/d + 165*(35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 - 42*(2*
c*d*x + b*d)^(5/2)*b*d + 15*(2*c*d*x + b*d)^(7/2))*b^2/(c*d) - 44*(105*(2*c*d*x + b*d)^(3/2)*b^3*d^3 - 189*(2*
c*d*x + b*d)^(5/2)*b^2*d^2 + 135*(2*c*d*x + b*d)^(7/2)*b*d - 35*(2*c*d*x + b*d)^(9/2))*b/(c*d^2) + (1155*(2*c*
d*x + b*d)^(3/2)*b^4*d^4 - 2772*(2*c*d*x + b*d)^(5/2)*b^3*d^3 + 2970*(2*c*d*x + b*d)^(7/2)*b^2*d^2 - 1540*(2*c
*d*x + b*d)^(9/2)*b*d + 315*(2*c*d*x + b*d)^(11/2))/(c*d^3))/c