3.460 \(\int (\frac{b^2}{4 c}+b \sqrt{x}+c x)^2 \, dx\)

Optimal. Leaf size=40 \[ \frac{\left (b+2 c \sqrt{x}\right )^6}{192 c^4}-\frac{b \left (b+2 c \sqrt{x}\right )^5}{160 c^4} \]

[Out]

-(b*(b + 2*c*Sqrt[x])^5)/(160*c^4) + (b + 2*c*Sqrt[x])^6/(192*c^4)

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Rubi [A]  time = 0.019456, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {28, 190, 43} \[ \frac{\left (b+2 c \sqrt{x}\right )^6}{192 c^4}-\frac{b \left (b+2 c \sqrt{x}\right )^5}{160 c^4} \]

Antiderivative was successfully verified.

[In]

Int[(b^2/(4*c) + b*Sqrt[x] + c*x)^2,x]

[Out]

-(b*(b + 2*c*Sqrt[x])^5)/(160*c^4) + (b + 2*c*Sqrt[x])^6/(192*c^4)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (\frac{b^2}{4 c}+b \sqrt{x}+c x\right )^2 \, dx &=\frac{\int \left (\frac{b}{2}+c \sqrt{x}\right )^4 \, dx}{c^2}\\ &=\frac{2 \operatorname{Subst}\left (\int x \left (\frac{b}{2}+c x\right )^4 \, dx,x,\sqrt{x}\right )}{c^2}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (-\frac{b \left (\frac{b}{2}+c x\right )^4}{2 c}+\frac{\left (\frac{b}{2}+c x\right )^5}{c}\right ) \, dx,x,\sqrt{x}\right )}{c^2}\\ &=-\frac{b \left (b+2 c \sqrt{x}\right )^5}{160 c^4}+\frac{\left (b+2 c \sqrt{x}\right )^6}{192 c^4}\\ \end{align*}

Mathematica [A]  time = 0.0257015, size = 29, normalized size = 0.72 \[ -\frac{\left (b-10 c \sqrt{x}\right ) \left (b+2 c \sqrt{x}\right )^5}{960 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(b^2/(4*c) + b*Sqrt[x] + c*x)^2,x]

[Out]

-((b - 10*c*Sqrt[x])*(b + 2*c*Sqrt[x])^5)/(960*c^4)

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Maple [A]  time = 0.003, size = 52, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}{x}^{2}}{2}}+{\frac{b}{2\,c} \left ({\frac{8\,{c}^{2}}{5}{x}^{{\frac{5}{2}}}}+{\frac{2\,{b}^{2}}{3}{x}^{{\frac{3}{2}}}} \right ) }+{\frac{1}{3\,c} \left ({\frac{{b}^{2}}{4\,c}}+cx \right ) ^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/4*b^2/c+c*x+b*x^(1/2))^2,x)

[Out]

1/2*b^2*x^2+1/2*b/c*(8/5*c^2*x^(5/2)+2/3*x^(3/2)*b^2)+1/3*(1/4*b^2/c+c*x)^3/c

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Maxima [A]  time = 1.0648, size = 73, normalized size = 1.82 \begin{align*} \frac{1}{3} \, c^{2} x^{3} + \frac{4}{5} \, b c x^{\frac{5}{2}} + \frac{1}{2} \, b^{2} x^{2} + \frac{b^{4} x}{16 \, c^{2}} + \frac{{\left (3 \, c x^{2} + 4 \, b x^{\frac{3}{2}}\right )} b^{2}}{12 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4/c*b^2+c*x+b*x^(1/2))^2,x, algorithm="maxima")

[Out]

1/3*c^2*x^3 + 4/5*b*c*x^(5/2) + 1/2*b^2*x^2 + 1/16*b^4*x/c^2 + 1/12*(3*c*x^2 + 4*b*x^(3/2))*b^2/c

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Fricas [A]  time = 1.99539, size = 126, normalized size = 3.15 \begin{align*} \frac{80 \, c^{4} x^{3} + 180 \, b^{2} c^{2} x^{2} + 15 \, b^{4} x + 16 \,{\left (12 \, b c^{3} x^{2} + 5 \, b^{3} c x\right )} \sqrt{x}}{240 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4/c*b^2+c*x+b*x^(1/2))^2,x, algorithm="fricas")

[Out]

1/240*(80*c^4*x^3 + 180*b^2*c^2*x^2 + 15*b^4*x + 16*(12*b*c^3*x^2 + 5*b^3*c*x)*sqrt(x))/c^2

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Sympy [A]  time = 0.411806, size = 51, normalized size = 1.27 \begin{align*} \frac{b^{4} x}{16 c^{2}} + \frac{b^{3} x^{\frac{3}{2}}}{3 c} + \frac{3 b^{2} x^{2}}{4} + \frac{4 b c x^{\frac{5}{2}}}{5} + \frac{c^{2} x^{3}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4/c*b**2+c*x+b*x**(1/2))**2,x)

[Out]

b**4*x/(16*c**2) + b**3*x**(3/2)/(3*c) + 3*b**2*x**2/4 + 4*b*c*x**(5/2)/5 + c**2*x**3/3

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Giac [A]  time = 1.11856, size = 66, normalized size = 1.65 \begin{align*} \frac{80 \, c^{4} x^{3} + 192 \, b c^{3} x^{\frac{5}{2}} + 180 \, b^{2} c^{2} x^{2} + 80 \, b^{3} c x^{\frac{3}{2}} + 15 \, b^{4} x}{240 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4/c*b^2+c*x+b*x^(1/2))^2,x, algorithm="giac")

[Out]

1/240*(80*c^4*x^3 + 192*b*c^3*x^(5/2) + 180*b^2*c^2*x^2 + 80*b^3*c*x^(3/2) + 15*b^4*x)/c^2