3.2.49 \(\int x^{7/2} (A+B x) (b x+c x^2)^2 \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{13} A b^2 x^{13/2}+\frac {2}{17} c x^{17/2} (A c+2 b B)+\frac {2}{15} b x^{15/2} (2 A c+b B)+\frac {2}{19} B c^2 x^{19/2} \]

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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}{13} A b^2 x^{13/2}+\frac {2}{17} c x^{17/2} (A c+2 b B)+\frac {2}{15} b x^{15/2} (2 A c+b B)+\frac {2}{19} B c^2 x^{19/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*A*b^2*x^(13/2))/13 + (2*b*(b*B + 2*A*c)*x^(15/2))/15 + (2*c*(2*b*B + A*c)*x^(17/2))/17 + (2*B*c^2*x^(19/2))
/19

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^{7/2} (A+B x) \left (b x+c x^2\right )^2 \, dx &=\int \left (A b^2 x^{11/2}+b (b B+2 A c) x^{13/2}+c (2 b B+A c) x^{15/2}+B c^2 x^{17/2}\right ) \, dx\\ &=\frac {2}{13} A b^2 x^{13/2}+\frac {2}{15} b (b B+2 A c) x^{15/2}+\frac {2}{17} c (2 b B+A c) x^{17/2}+\frac {2}{19} B c^2 x^{19/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.87 \begin {gather*} \frac {2 x^{13/2} \left (19 A \left (255 b^2+442 b c x+195 c^2 x^2\right )+13 B x \left (323 b^2+570 b c x+255 c^2 x^2\right )\right )}{62985} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*x^(13/2)*(19*A*(255*b^2 + 442*b*c*x + 195*c^2*x^2) + 13*B*x*(323*b^2 + 570*b*c*x + 255*c^2*x^2)))/62985

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IntegrateAlgebraic [A]  time = 0.04, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (4845 A b^2 x^{13/2}+8398 A b c x^{15/2}+3705 A c^2 x^{17/2}+4199 b^2 B x^{15/2}+7410 b B c x^{17/2}+3315 B c^2 x^{19/2}\right )}{62985} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(7/2)*(A + B*x)*(b*x + c*x^2)^2,x]

[Out]

(2*(4845*A*b^2*x^(13/2) + 4199*b^2*B*x^(15/2) + 8398*A*b*c*x^(15/2) + 7410*b*B*c*x^(17/2) + 3705*A*c^2*x^(17/2
) + 3315*B*c^2*x^(19/2)))/62985

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fricas [A]  time = 0.38, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{62985} \, {\left (3315 \, B c^{2} x^{9} + 4845 \, A b^{2} x^{6} + 3705 \, {\left (2 \, B b c + A c^{2}\right )} x^{8} + 4199 \, {\left (B b^{2} + 2 \, A b c\right )} x^{7}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

2/62985*(3315*B*c^2*x^9 + 4845*A*b^2*x^6 + 3705*(2*B*b*c + A*c^2)*x^8 + 4199*(B*b^2 + 2*A*b*c)*x^7)*sqrt(x)

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giac [A]  time = 0.15, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{19} \, B c^{2} x^{\frac {19}{2}} + \frac {4}{17} \, B b c x^{\frac {17}{2}} + \frac {2}{17} \, A c^{2} x^{\frac {17}{2}} + \frac {2}{15} \, B b^{2} x^{\frac {15}{2}} + \frac {4}{15} \, A b c x^{\frac {15}{2}} + \frac {2}{13} \, A b^{2} x^{\frac {13}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

2/19*B*c^2*x^(19/2) + 4/17*B*b*c*x^(17/2) + 2/17*A*c^2*x^(17/2) + 2/15*B*b^2*x^(15/2) + 4/15*A*b*c*x^(15/2) +
2/13*A*b^2*x^(13/2)

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maple [A]  time = 0.05, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 \left (3315 B \,c^{2} x^{3}+3705 A \,c^{2} x^{2}+7410 B b c \,x^{2}+8398 A b c x +4199 B \,b^{2} x +4845 A \,b^{2}\right ) x^{\frac {13}{2}}}{62985} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)*(c*x^2+b*x)^2,x)

[Out]

2/62985*x^(13/2)*(3315*B*c^2*x^3+3705*A*c^2*x^2+7410*B*b*c*x^2+8398*A*b*c*x+4199*B*b^2*x+4845*A*b^2)

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maxima [A]  time = 0.51, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{19} \, B c^{2} x^{\frac {19}{2}} + \frac {2}{13} \, A b^{2} x^{\frac {13}{2}} + \frac {2}{17} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {17}{2}} + \frac {2}{15} \, {\left (B b^{2} + 2 \, A b c\right )} x^{\frac {15}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

2/19*B*c^2*x^(19/2) + 2/13*A*b^2*x^(13/2) + 2/17*(2*B*b*c + A*c^2)*x^(17/2) + 2/15*(B*b^2 + 2*A*b*c)*x^(15/2)

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mupad [B]  time = 1.03, size = 51, normalized size = 0.81 \begin {gather*} x^{15/2}\,\left (\frac {2\,B\,b^2}{15}+\frac {4\,A\,c\,b}{15}\right )+x^{17/2}\,\left (\frac {2\,A\,c^2}{17}+\frac {4\,B\,b\,c}{17}\right )+\frac {2\,A\,b^2\,x^{13/2}}{13}+\frac {2\,B\,c^2\,x^{19/2}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(b*x + c*x^2)^2*(A + B*x),x)

[Out]

x^(15/2)*((2*B*b^2)/15 + (4*A*b*c)/15) + x^(17/2)*((2*A*c^2)/17 + (4*B*b*c)/17) + (2*A*b^2*x^(13/2))/13 + (2*B
*c^2*x^(19/2))/19

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sympy [A]  time = 14.98, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A b^{2} x^{\frac {13}{2}}}{13} + \frac {4 A b c x^{\frac {15}{2}}}{15} + \frac {2 A c^{2} x^{\frac {17}{2}}}{17} + \frac {2 B b^{2} x^{\frac {15}{2}}}{15} + \frac {4 B b c x^{\frac {17}{2}}}{17} + \frac {2 B c^{2} x^{\frac {19}{2}}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(B*x+A)*(c*x**2+b*x)**2,x)

[Out]

2*A*b**2*x**(13/2)/13 + 4*A*b*c*x**(15/2)/15 + 2*A*c**2*x**(17/2)/17 + 2*B*b**2*x**(15/2)/15 + 4*B*b*c*x**(17/
2)/17 + 2*B*c**2*x**(19/2)/19

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