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

Optimal. Leaf size=63 \[ \frac {2}{9} a^2 A x^{9/2}+\frac {2}{13} b x^{13/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{15} b^2 B x^{15/2} \]

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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 = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {76} \begin {gather*} \frac {2}{9} a^2 A x^{9/2}+\frac {2}{13} b x^{13/2} (2 a B+A b)+\frac {2}{11} a x^{11/2} (a B+2 A b)+\frac {2}{15} b^2 B x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*(a + b*x)^2*(A + B*x),x]

[Out]

(2*a^2*A*x^(9/2))/9 + (2*a*(2*A*b + a*B)*x^(11/2))/11 + (2*b*(A*b + 2*a*B)*x^(13/2))/13 + (2*b^2*B*x^(15/2))/1
5

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin {align*} \int x^{7/2} (a+b x)^2 (A+B x) \, dx &=\int \left (a^2 A x^{7/2}+a (2 A b+a B) x^{9/2}+b (A b+2 a B) x^{11/2}+b^2 B x^{13/2}\right ) \, dx\\ &=\frac {2}{9} a^2 A x^{9/2}+\frac {2}{11} a (2 A b+a B) x^{11/2}+\frac {2}{13} b (A b+2 a B) x^{13/2}+\frac {2}{15} b^2 B x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 x^{9/2} \left (65 a^2 (11 A+9 B x)+90 a b x (13 A+11 B x)+33 b^2 x^2 (15 A+13 B x)\right )}{6435} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(9/2)*(65*a^2*(11*A + 9*B*x) + 90*a*b*x*(13*A + 11*B*x) + 33*b^2*x^2*(15*A + 13*B*x)))/6435

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 1.10 \begin {gather*} \frac {2 \left (715 a^2 A x^{9/2}+585 a^2 B x^{11/2}+1170 a A b x^{11/2}+990 a b B x^{13/2}+495 A b^2 x^{13/2}+429 b^2 B x^{15/2}\right )}{6435} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(715*a^2*A*x^(9/2) + 1170*a*A*b*x^(11/2) + 585*a^2*B*x^(11/2) + 495*A*b^2*x^(13/2) + 990*a*b*B*x^(13/2) + 4
29*b^2*B*x^(15/2)))/6435

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fricas [A]  time = 0.73, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{6435} \, {\left (429 \, B b^{2} x^{7} + 715 \, A a^{2} x^{4} + 495 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + 585 \, {\left (B a^{2} + 2 \, A a b\right )} x^{5}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/6435*(429*B*b^2*x^7 + 715*A*a^2*x^4 + 495*(2*B*a*b + A*b^2)*x^6 + 585*(B*a^2 + 2*A*a*b)*x^5)*sqrt(x)

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giac [A]  time = 1.21, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{15} \, B b^{2} x^{\frac {15}{2}} + \frac {4}{13} \, B a b x^{\frac {13}{2}} + \frac {2}{13} \, A b^{2} x^{\frac {13}{2}} + \frac {2}{11} \, B a^{2} x^{\frac {11}{2}} + \frac {4}{11} \, A a b x^{\frac {11}{2}} + \frac {2}{9} \, A a^{2} x^{\frac {9}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*b^2*x^(15/2) + 4/13*B*a*b*x^(13/2) + 2/13*A*b^2*x^(13/2) + 2/11*B*a^2*x^(11/2) + 4/11*A*a*b*x^(11/2) +
2/9*A*a^2*x^(9/2)

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maple [A]  time = 0.01, size = 52, normalized size = 0.83 \begin {gather*} \frac {2 \left (429 B \,b^{2} x^{3}+495 A \,b^{2} x^{2}+990 B a b \,x^{2}+1170 A a b x +585 B \,a^{2} x +715 a^{2} A \right ) x^{\frac {9}{2}}}{6435} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/6435*x^(9/2)*(429*B*b^2*x^3+495*A*b^2*x^2+990*B*a*b*x^2+1170*A*a*b*x+585*B*a^2*x+715*A*a^2)

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maxima [A]  time = 0.88, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{15} \, B b^{2} x^{\frac {15}{2}} + \frac {2}{9} \, A a^{2} x^{\frac {9}{2}} + \frac {2}{13} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {13}{2}} + \frac {2}{11} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*b^2*x^(15/2) + 2/9*A*a^2*x^(9/2) + 2/13*(2*B*a*b + A*b^2)*x^(13/2) + 2/11*(B*a^2 + 2*A*a*b)*x^(11/2)

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mupad [B]  time = 0.37, size = 51, normalized size = 0.81 \begin {gather*} x^{11/2}\,\left (\frac {2\,B\,a^2}{11}+\frac {4\,A\,b\,a}{11}\right )+x^{13/2}\,\left (\frac {2\,A\,b^2}{13}+\frac {4\,B\,a\,b}{13}\right )+\frac {2\,A\,a^2\,x^{9/2}}{9}+\frac {2\,B\,b^2\,x^{15/2}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(A + B*x)*(a + b*x)^2,x)

[Out]

x^(11/2)*((2*B*a^2)/11 + (4*A*a*b)/11) + x^(13/2)*((2*A*b^2)/13 + (4*B*a*b)/13) + (2*A*a^2*x^(9/2))/9 + (2*B*b
^2*x^(15/2))/15

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sympy [A]  time = 8.52, size = 80, normalized size = 1.27 \begin {gather*} \frac {2 A a^{2} x^{\frac {9}{2}}}{9} + \frac {4 A a b x^{\frac {11}{2}}}{11} + \frac {2 A b^{2} x^{\frac {13}{2}}}{13} + \frac {2 B a^{2} x^{\frac {11}{2}}}{11} + \frac {4 B a b x^{\frac {13}{2}}}{13} + \frac {2 B b^{2} x^{\frac {15}{2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(9/2)/9 + 4*A*a*b*x**(11/2)/11 + 2*A*b**2*x**(13/2)/13 + 2*B*a**2*x**(11/2)/11 + 4*B*a*b*x**(13/2)
/13 + 2*B*b**2*x**(15/2)/15

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