3.5.60 \(\int (\frac {b^2}{4 c}+b \sqrt {x}+c x)^2 \, dx\) [460]

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

[Out]

-1/160*b*(b+2*c*x^(1/2))^5/c^4+1/192*(b+2*c*x^(1/2))^6/c^4

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {28, 196, 45} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/160*(b*(b + 2*c*Sqrt[x])^5)/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 45

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])

Rule 196

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]

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 \text {Subst}\left (\int x \left (\frac {b}{2}+c x\right )^4 \, dx,x,\sqrt {x}\right )}{c^2}\\ &=\frac {2 \text {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*}

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Mathematica [A]
time = 0.01, size = 55, normalized size = 1.38 \begin {gather*} \frac {15 b^4 x+80 b^3 c x^{3/2}+180 b^2 c^2 x^2+192 b c^3 x^{5/2}+80 c^4 x^3}{240 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 52, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {\frac {8 c^{4} x^{3}}{3}+\frac {32 b \,c^{3} x^{\frac {5}{2}}}{5}+6 b^{2} c^{2} x^{2}+\frac {8 b^{3} c \,x^{\frac {3}{2}}}{3}+\frac {b^{4} x}{2}}{8 c^{2}}\)	\(50\)
default	\(\frac {b^{2} x^{2}}{2}+\frac {b \left (\frac {8 c^{2} x^{\frac {5}{2}}}{5}+\frac {2 b^{2} x^{\frac {3}{2}}}{3}\right )}{2 c}+\frac {\left (\frac {b^{2}}{4 c}+c x \right )^{3}}{3 c}\)	\(52\)
trager	\(\frac {\frac {\left (16 c^{4} x^{2}+36 b^{2} x \,c^{2}+16 c^{4} x +3 b^{4}+36 b^{2} c^{2}+16 c^{4}\right ) \left (-1+x \right )}{3}+\frac {16 b c \,x^{\frac {3}{2}} \left (12 c^{2} x +5 b^{2}\right )}{15}}{16 c^{2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 54, normalized size = 1.35 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4*b^2/c+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 = 0.41, size = 53, normalized size = 1.32 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4*b^2/c+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 = 0.09, size = 51, normalized size = 1.28 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4*b**2/c+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 = 3.81, size = 49, normalized size = 1.22 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/4*b^2/c+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

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Mupad [B]
time = 0.04, size = 44, normalized size = 1.10 \begin {gather*} \frac {3\,b^2\,x^2}{4}+\frac {c^2\,x^3}{3}+\frac {b^4\,x}{16\,c^2}+\frac {b^3\,x^{3/2}}{3\,c}+\frac {4\,b\,c\,x^{5/2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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