### 3.1264 $$\int \frac{a+b x+c x^2}{\sqrt{b d+2 c d x}} \, dx$$

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

[Out]

-((b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x])/(4*c^2*d) + (b*d + 2*c*d*x)^(5/2)/(20*c^2*d^3)

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Rubi [A]  time = 0.0219751, 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} $\frac{(b d+2 c d x)^{5/2}}{20 c^2 d^3}-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x}}{4 c^2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0247459, size = 43, normalized size = 0.78 $\frac{\left (c \left (5 a+c x^2\right )-b^2+b c x\right ) \sqrt{d (b+2 c x)}}{5 c^2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[d*(b + 2*c*x)]*(-b^2 + b*c*x + c*(5*a + c*x^2)))/(5*c^2*d)

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Maple [A]  time = 0.041, size = 44, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,cx+b \right ) \left ({c}^{2}{x}^{2}+bcx+5\,ac-{b}^{2} \right ) }{5\,{c}^{2}}{\frac{1}{\sqrt{2\,cdx+bd}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16641, size = 157, normalized size = 2.85 \begin{align*} \frac{60 \, \sqrt{2 \, c d x + b d} a - \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}}}{60 \, c d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*sqrt(2*c*d*x + b*d)*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))/(c*d)

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Fricas [A]  time = 1.95994, size = 88, normalized size = 1.6 \begin{align*} \frac{{\left (c^{2} x^{2} + b c x - b^{2} + 5 \, a c\right )} \sqrt{2 \, c d x + b d}}{5 \, c^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(c^2*x^2 + b*c*x - b^2 + 5*a*c)*sqrt(2*c*d*x + b*d)/(c^2*d)

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Sympy [A]  time = 12.0852, size = 258, normalized size = 4.69 \begin{align*} \begin{cases} - \frac{\frac{a b}{\sqrt{b d + 2 c d x}} + \frac{a \left (- \frac{b d}{\sqrt{b d + 2 c d x}} - \sqrt{b d + 2 c d x}\right )}{d} + \frac{b^{2} \left (- \frac{b d}{\sqrt{b d + 2 c d x}} - \sqrt{b d + 2 c d x}\right )}{2 c d} + \frac{3 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 )}{4 c d^{2}} + \frac{- \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}}{4 c d^{3}}}{c} & \text{for}\: c \neq 0 \\\frac{a x + \frac{b x^{2}}{2}}{\sqrt{b d}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(a*b/sqrt(b*d + 2*c*d*x) + a*(-b*d/sqrt(b*d + 2*c*d*x) - sqrt(b*d + 2*c*d*x))/d + b**2*(-b*d/sqrt(
b*d + 2*c*d*x) - sqrt(b*d + 2*c*d*x))/(2*c*d) + 3*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)/(4*c*d**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)/(4*c*d**3))/c, Ne(c, 0)), ((a*x + b*x**2/2)/sqrt(b*d),
True))

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Giac [B]  time = 1.14201, size = 157, normalized size = 2.85 \begin{align*} \frac{60 \, \sqrt{2 \, c d x + b d} a - \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}}}{60 \, c d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*sqrt(2*c*d*x + b*d)*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))/(c*d)