3.2.53 \(\int \frac {(A+B x) (b x+c x^2)^2}{\sqrt {x}} \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{5} A b^2 x^{5/2}+\frac {2}{9} c x^{9/2} (A c+2 b B)+\frac {2}{7} b x^{7/2} (2 A c+b B)+\frac {2}{11} B c^2 x^{11/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {765} \begin {gather*} \frac {2}{5} A b^2 x^{5/2}+\frac {2}{9} c x^{9/2} (A c+2 b B)+\frac {2}{7} b x^{7/2} (2 A c+b B)+\frac {2}{11} B c^2 x^{11/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b^2*x^(5/2))/5 + (2*b*(b*B + 2*A*c)*x^(7/2))/7 + (2*c*(2*b*B + A*c)*x^(9/2))/9 + (2*B*c^2*x^(11/2))/11

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^2}{\sqrt {x}} \, dx &=\int \left (A b^2 x^{3/2}+b (b B+2 A c) x^{5/2}+c (2 b B+A c) x^{7/2}+B c^2 x^{9/2}\right ) \, dx\\ &=\frac {2}{5} A b^2 x^{5/2}+\frac {2}{7} b (b B+2 A c) x^{7/2}+\frac {2}{9} c (2 b B+A c) x^{9/2}+\frac {2}{11} B c^2 x^{11/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.87 \begin {gather*} \frac {2 x^{5/2} \left (11 A \left (63 b^2+90 b c x+35 c^2 x^2\right )+5 B x \left (99 b^2+154 b c x+63 c^2 x^2\right )\right )}{3465} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(11*A*(63*b^2 + 90*b*c*x + 35*c^2*x^2) + 5*B*x*(99*b^2 + 154*b*c*x + 63*c^2*x^2)))/3465

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (693 A b^2 x^{5/2}+990 A b c x^{7/2}+385 A c^2 x^{9/2}+495 b^2 B x^{7/2}+770 b B c x^{9/2}+315 B c^2 x^{11/2}\right )}{3465} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/Sqrt[x],x]

[Out]

(2*(693*A*b^2*x^(5/2) + 495*b^2*B*x^(7/2) + 990*A*b*c*x^(7/2) + 770*b*B*c*x^(9/2) + 385*A*c^2*x^(9/2) + 315*B*
c^2*x^(11/2)))/3465

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fricas [A]  time = 0.39, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{3465} \, {\left (315 \, B c^{2} x^{5} + 693 \, A b^{2} x^{2} + 385 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 495 \, {\left (B b^{2} + 2 \, A b c\right )} x^{3}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*B*c^2*x^5 + 693*A*b^2*x^2 + 385*(2*B*b*c + A*c^2)*x^4 + 495*(B*b^2 + 2*A*b*c)*x^3)*sqrt(x)

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giac [A]  time = 0.16, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{11} \, B c^{2} x^{\frac {11}{2}} + \frac {4}{9} \, B b c x^{\frac {9}{2}} + \frac {2}{9} \, A c^{2} x^{\frac {9}{2}} + \frac {2}{7} \, B b^{2} x^{\frac {7}{2}} + \frac {4}{7} \, A b c x^{\frac {7}{2}} + \frac {2}{5} \, A b^{2} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*c^2*x^(11/2) + 4/9*B*b*c*x^(9/2) + 2/9*A*c^2*x^(9/2) + 2/7*B*b^2*x^(7/2) + 4/7*A*b*c*x^(7/2) + 2/5*A*b^
2*x^(5/2)

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maple [A]  time = 0.06, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 \left (315 B \,c^{2} x^{3}+385 A \,c^{2} x^{2}+770 B b c \,x^{2}+990 A b c x +495 B \,b^{2} x +693 A \,b^{2}\right ) x^{\frac {5}{2}}}{3465} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*x^(5/2)*(315*B*c^2*x^3+385*A*c^2*x^2+770*B*b*c*x^2+990*A*b*c*x+495*B*b^2*x+693*A*b^2)

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maxima [A]  time = 0.56, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{11} \, B c^{2} x^{\frac {11}{2}} + \frac {2}{5} \, A b^{2} x^{\frac {5}{2}} + \frac {2}{9} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B b^{2} + 2 \, A b c\right )} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*c^2*x^(11/2) + 2/5*A*b^2*x^(5/2) + 2/9*(2*B*b*c + A*c^2)*x^(9/2) + 2/7*(B*b^2 + 2*A*b*c)*x^(7/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{7/2}\,\left (\frac {2\,B\,b^2}{7}+\frac {4\,A\,c\,b}{7}\right )+x^{9/2}\,\left (\frac {2\,A\,c^2}{9}+\frac {4\,B\,b\,c}{9}\right )+\frac {2\,A\,b^2\,x^{5/2}}{5}+\frac {2\,B\,c^2\,x^{11/2}}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(7/2)*((2*B*b^2)/7 + (4*A*b*c)/7) + x^(9/2)*((2*A*c^2)/9 + (4*B*b*c)/9) + (2*A*b^2*x^(5/2))/5 + (2*B*c^2*x^(
11/2))/11

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sympy [A]  time = 1.57, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A b^{2} x^{\frac {5}{2}}}{5} + \frac {4 A b c x^{\frac {7}{2}}}{7} + \frac {2 A c^{2} x^{\frac {9}{2}}}{9} + \frac {2 B b^{2} x^{\frac {7}{2}}}{7} + \frac {4 B b c x^{\frac {9}{2}}}{9} + \frac {2 B c^{2} x^{\frac {11}{2}}}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*b**2*x**(5/2)/5 + 4*A*b*c*x**(7/2)/7 + 2*A*c**2*x**(9/2)/9 + 2*B*b**2*x**(7/2)/7 + 4*B*b*c*x**(9/2)/9 + 2*
B*c**2*x**(11/2)/11

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