3.2.54 \(\int \frac {(A+B x) (b x+c x^2)^2}{x^{3/2}} \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{3} A b^2 x^{3/2}+\frac {2}{7} c x^{7/2} (A c+2 b B)+\frac {2}{5} b x^{5/2} (2 A c+b B)+\frac {2}{9} B c^2 x^{9/2} \]

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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}{3} A b^2 x^{3/2}+\frac {2}{7} c x^{7/2} (A c+2 b B)+\frac {2}{5} b x^{5/2} (2 A c+b B)+\frac {2}{9} B c^2 x^{9/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^2)/x^(3/2),x]

[Out]

(2*A*b^2*x^(3/2))/3 + (2*b*(b*B + 2*A*c)*x^(5/2))/5 + (2*c*(2*b*B + A*c)*x^(7/2))/7 + (2*B*c^2*x^(9/2))/9

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

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Mathematica [A]  time = 0.02, size = 54, normalized size = 0.86 \begin {gather*} \frac {2}{315} x^{3/2} \left (3 A \left (35 b^2+42 b c x+15 c^2 x^2\right )+B x \left (63 b^2+90 b c x+35 c^2 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^2)/x^(3/2),x]

[Out]

(2*x^(3/2)*(3*A*(35*b^2 + 42*b*c*x + 15*c^2*x^2) + B*x*(63*b^2 + 90*b*c*x + 35*c^2*x^2)))/315

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IntegrateAlgebraic [A]  time = 0.04, size = 69, normalized size = 1.10 \begin {gather*} \frac {2}{315} \left (105 A b^2 x^{3/2}+126 A b c x^{5/2}+45 A c^2 x^{7/2}+63 b^2 B x^{5/2}+90 b B c x^{7/2}+35 B c^2 x^{9/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/x^(3/2),x]

[Out]

(2*(105*A*b^2*x^(3/2) + 63*b^2*B*x^(5/2) + 126*A*b*c*x^(5/2) + 90*b*B*c*x^(7/2) + 45*A*c^2*x^(7/2) + 35*B*c^2*
x^(9/2)))/315

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fricas [A]  time = 0.41, size = 54, normalized size = 0.86 \begin {gather*} \frac {2}{315} \, {\left (35 \, B c^{2} x^{4} + 105 \, A b^{2} x + 45 \, {\left (2 \, B b c + A c^{2}\right )} x^{3} + 63 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}\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^(3/2),x, algorithm="fricas")

[Out]

2/315*(35*B*c^2*x^4 + 105*A*b^2*x + 45*(2*B*b*c + A*c^2)*x^3 + 63*(B*b^2 + 2*A*b*c)*x^2)*sqrt(x)

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giac [A]  time = 0.19, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{9} \, B c^{2} x^{\frac {9}{2}} + \frac {4}{7} \, B b c x^{\frac {7}{2}} + \frac {2}{7} \, A c^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B b^{2} x^{\frac {5}{2}} + \frac {4}{5} \, A b c x^{\frac {5}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{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^(3/2),x, algorithm="giac")

[Out]

2/9*B*c^2*x^(9/2) + 4/7*B*b*c*x^(7/2) + 2/7*A*c^2*x^(7/2) + 2/5*B*b^2*x^(5/2) + 4/5*A*b*c*x^(5/2) + 2/3*A*b^2*
x^(3/2)

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maple [A]  time = 0.05, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 \left (35 B \,c^{2} x^{3}+45 A \,c^{2} x^{2}+90 B b c \,x^{2}+126 A b c x +63 B \,b^{2} x +105 A \,b^{2}\right ) x^{\frac {3}{2}}}{315} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^2/x^(3/2),x)

[Out]

2/315*x^(3/2)*(35*B*c^2*x^3+45*A*c^2*x^2+90*B*b*c*x^2+126*A*b*c*x+63*B*b^2*x+105*A*b^2)

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maxima [A]  time = 0.51, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{9} \, B c^{2} x^{\frac {9}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{2}} + \frac {2}{7} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B b^{2} + 2 \, A b c\right )} 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^(3/2),x, algorithm="maxima")

[Out]

2/9*B*c^2*x^(9/2) + 2/3*A*b^2*x^(3/2) + 2/7*(2*B*b*c + A*c^2)*x^(7/2) + 2/5*(B*b^2 + 2*A*b*c)*x^(5/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{5/2}\,\left (\frac {2\,B\,b^2}{5}+\frac {4\,A\,c\,b}{5}\right )+x^{7/2}\,\left (\frac {2\,A\,c^2}{7}+\frac {4\,B\,b\,c}{7}\right )+\frac {2\,A\,b^2\,x^{3/2}}{3}+\frac {2\,B\,c^2\,x^{9/2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^2*(A + B*x))/x^(3/2),x)

[Out]

x^(5/2)*((2*B*b^2)/5 + (4*A*b*c)/5) + x^(7/2)*((2*A*c^2)/7 + (4*B*b*c)/7) + (2*A*b^2*x^(3/2))/3 + (2*B*c^2*x^(
9/2))/9

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sympy [A]  time = 1.50, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A b^{2} x^{\frac {3}{2}}}{3} + \frac {4 A b c x^{\frac {5}{2}}}{5} + \frac {2 A c^{2} x^{\frac {7}{2}}}{7} + \frac {2 B b^{2} x^{\frac {5}{2}}}{5} + \frac {4 B b c x^{\frac {7}{2}}}{7} + \frac {2 B c^{2} x^{\frac {9}{2}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**2/x**(3/2),x)

[Out]

2*A*b**2*x**(3/2)/3 + 4*A*b*c*x**(5/2)/5 + 2*A*c**2*x**(7/2)/7 + 2*B*b**2*x**(5/2)/5 + 4*B*b*c*x**(7/2)/7 + 2*
B*c**2*x**(9/2)/9

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