3.2.50 \(\int x^{5/2} (A+B x) (b x+c x^2)^2 \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{11} A b^2 x^{11/2}+\frac {2}{15} c x^{15/2} (A c+2 b B)+\frac {2}{13} b x^{13/2} (2 A c+b B)+\frac {2}{17} B c^2 x^{17/2} \]

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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}{11} A b^2 x^{11/2}+\frac {2}{15} c x^{15/2} (A c+2 b B)+\frac {2}{13} b x^{13/2} (2 A c+b B)+\frac {2}{17} B c^2 x^{17/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*A*b^2*x^(11/2))/11 + (2*b*(b*B + 2*A*c)*x^(13/2))/13 + (2*c*(2*b*B + A*c)*x^(15/2))/15 + (2*B*c^2*x^(17/2))
/17

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 x^{5/2} (A+B x) \left (b x+c x^2\right )^2 \, dx &=\int \left (A b^2 x^{9/2}+b (b B+2 A c) x^{11/2}+c (2 b B+A c) x^{13/2}+B c^2 x^{15/2}\right ) \, dx\\ &=\frac {2}{11} A b^2 x^{11/2}+\frac {2}{13} b (b B+2 A c) x^{13/2}+\frac {2}{15} c (2 b B+A c) x^{15/2}+\frac {2}{17} B c^2 x^{17/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.87 \begin {gather*} \frac {2 x^{11/2} \left (17 A \left (195 b^2+330 b c x+143 c^2 x^2\right )+11 B x \left (255 b^2+442 b c x+195 c^2 x^2\right )\right )}{36465} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*x^(11/2)*(17*A*(195*b^2 + 330*b*c*x + 143*c^2*x^2) + 11*B*x*(255*b^2 + 442*b*c*x + 195*c^2*x^2)))/36465

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IntegrateAlgebraic [A]  time = 0.04, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (3315 A b^2 x^{11/2}+5610 A b c x^{13/2}+2431 A c^2 x^{15/2}+2805 b^2 B x^{13/2}+4862 b B c x^{15/2}+2145 B c^2 x^{17/2}\right )}{36465} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*(3315*A*b^2*x^(11/2) + 2805*b^2*B*x^(13/2) + 5610*A*b*c*x^(13/2) + 4862*b*B*c*x^(15/2) + 2431*A*c^2*x^(15/2
) + 2145*B*c^2*x^(17/2)))/36465

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fricas [A]  time = 0.39, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{36465} \, {\left (2145 \, B c^{2} x^{8} + 3315 \, A b^{2} x^{5} + 2431 \, {\left (2 \, B b c + A c^{2}\right )} x^{7} + 2805 \, {\left (B b^{2} + 2 \, A b c\right )} x^{6}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

2/36465*(2145*B*c^2*x^8 + 3315*A*b^2*x^5 + 2431*(2*B*b*c + A*c^2)*x^7 + 2805*(B*b^2 + 2*A*b*c)*x^6)*sqrt(x)

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giac [A]  time = 0.16, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{17} \, B c^{2} x^{\frac {17}{2}} + \frac {4}{15} \, B b c x^{\frac {15}{2}} + \frac {2}{15} \, A c^{2} x^{\frac {15}{2}} + \frac {2}{13} \, B b^{2} x^{\frac {13}{2}} + \frac {4}{13} \, A b c x^{\frac {13}{2}} + \frac {2}{11} \, A b^{2} x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

2/17*B*c^2*x^(17/2) + 4/15*B*b*c*x^(15/2) + 2/15*A*c^2*x^(15/2) + 2/13*B*b^2*x^(13/2) + 4/13*A*b*c*x^(13/2) +
2/11*A*b^2*x^(11/2)

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maple [A]  time = 0.05, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 \left (2145 B \,c^{2} x^{3}+2431 A \,c^{2} x^{2}+4862 B b c \,x^{2}+5610 A b c x +2805 B \,b^{2} x +3315 A \,b^{2}\right ) x^{\frac {11}{2}}}{36465} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)*(c*x^2+b*x)^2,x)

[Out]

2/36465*x^(11/2)*(2145*B*c^2*x^3+2431*A*c^2*x^2+4862*B*b*c*x^2+5610*A*b*c*x+2805*B*b^2*x+3315*A*b^2)

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maxima [A]  time = 0.57, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{17} \, B c^{2} x^{\frac {17}{2}} + \frac {2}{11} \, A b^{2} x^{\frac {11}{2}} + \frac {2}{15} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {15}{2}} + \frac {2}{13} \, {\left (B b^{2} + 2 \, A b c\right )} x^{\frac {13}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

2/17*B*c^2*x^(17/2) + 2/11*A*b^2*x^(11/2) + 2/15*(2*B*b*c + A*c^2)*x^(15/2) + 2/13*(B*b^2 + 2*A*b*c)*x^(13/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \begin {gather*} x^{13/2}\,\left (\frac {2\,B\,b^2}{13}+\frac {4\,A\,c\,b}{13}\right )+x^{15/2}\,\left (\frac {2\,A\,c^2}{15}+\frac {4\,B\,b\,c}{15}\right )+\frac {2\,A\,b^2\,x^{11/2}}{11}+\frac {2\,B\,c^2\,x^{17/2}}{17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x + c*x^2)^2*(A + B*x),x)

[Out]

x^(13/2)*((2*B*b^2)/13 + (4*A*b*c)/13) + x^(15/2)*((2*A*c^2)/15 + (4*B*b*c)/15) + (2*A*b^2*x^(11/2))/11 + (2*B
*c^2*x^(17/2))/17

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sympy [A]  time = 8.36, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A b^{2} x^{\frac {11}{2}}}{11} + \frac {4 A b c x^{\frac {13}{2}}}{13} + \frac {2 A c^{2} x^{\frac {15}{2}}}{15} + \frac {2 B b^{2} x^{\frac {13}{2}}}{13} + \frac {4 B b c x^{\frac {15}{2}}}{15} + \frac {2 B c^{2} x^{\frac {17}{2}}}{17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x+A)*(c*x**2+b*x)**2,x)

[Out]

2*A*b**2*x**(11/2)/11 + 4*A*b*c*x**(13/2)/13 + 2*A*c**2*x**(15/2)/15 + 2*B*b**2*x**(13/2)/13 + 4*B*b*c*x**(15/
2)/15 + 2*B*c**2*x**(17/2)/17

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