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

Optimal. Leaf size=111 \[ \frac{4}{13} a^2 b x^{13/2} (2 a B+3 A b)+\frac{2}{11} a^3 x^{11/2} (a B+4 A b)+\frac{2}{9} a^4 A x^{9/2}+\frac{2}{17} b^3 x^{17/2} (4 a B+A b)+\frac{4}{15} a b^2 x^{15/2} (3 a B+2 A b)+\frac{2}{19} b^4 B x^{19/2} \]

[Out]

(2*a^4*A*x^(9/2))/9 + (2*a^3*(4*A*b + a*B)*x^(11/2))/11 + (4*a^2*b*(3*A*b + 2*a*B)*x^(13/2))/13 + (4*a*b^2*(2*
A*b + 3*a*B)*x^(15/2))/15 + (2*b^3*(A*b + 4*a*B)*x^(17/2))/17 + (2*b^4*B*x^(19/2))/19

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Rubi [A]  time = 0.0575564, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {27, 76} \[ \frac{4}{13} a^2 b x^{13/2} (2 a B+3 A b)+\frac{2}{11} a^3 x^{11/2} (a B+4 A b)+\frac{2}{9} a^4 A x^{9/2}+\frac{2}{17} b^3 x^{17/2} (4 a B+A b)+\frac{4}{15} a b^2 x^{15/2} (3 a B+2 A b)+\frac{2}{19} b^4 B x^{19/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*A*x^(9/2))/9 + (2*a^3*(4*A*b + a*B)*x^(11/2))/11 + (4*a^2*b*(3*A*b + 2*a*B)*x^(13/2))/13 + (4*a*b^2*(2*
A*b + 3*a*B)*x^(15/2))/15 + (2*b^3*(A*b + 4*a*B)*x^(17/2))/17 + (2*b^4*B*x^(19/2))/19

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx &=\int x^{7/2} (a+b x)^4 (A+B x) \, dx\\ &=\int \left (a^4 A x^{7/2}+a^3 (4 A b+a B) x^{9/2}+2 a^2 b (3 A b+2 a B) x^{11/2}+2 a b^2 (2 A b+3 a B) x^{13/2}+b^3 (A b+4 a B) x^{15/2}+b^4 B x^{17/2}\right ) \, dx\\ &=\frac{2}{9} a^4 A x^{9/2}+\frac{2}{11} a^3 (4 A b+a B) x^{11/2}+\frac{4}{13} a^2 b (3 A b+2 a B) x^{13/2}+\frac{4}{15} a b^2 (2 A b+3 a B) x^{15/2}+\frac{2}{17} b^3 (A b+4 a B) x^{17/2}+\frac{2}{19} b^4 B x^{19/2}\\ \end{align*}

Mathematica [A]  time = 0.0655462, size = 81, normalized size = 0.73 \[ \frac{2 \left (\frac{x^{9/2} \left (50490 a^2 b^2 x^2+39780 a^3 b x+12155 a^4+29172 a b^3 x^3+6435 b^4 x^4\right ) (19 A b-9 a B)}{109395}+B x^{9/2} (a+b x)^5\right )}{19 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*x^(9/2)*(a + b*x)^5 + ((19*A*b - 9*a*B)*x^(9/2)*(12155*a^4 + 39780*a^3*b*x + 50490*a^2*b^2*x^2 + 29172*a
*b^3*x^3 + 6435*b^4*x^4))/109395))/(19*b)

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Maple [A]  time = 0.007, size = 100, normalized size = 0.9 \begin{align*}{\frac{218790\,{b}^{4}B{x}^{5}+244530\,A{b}^{4}{x}^{4}+978120\,B{x}^{4}a{b}^{3}+1108536\,aA{b}^{3}{x}^{3}+1662804\,B{x}^{3}{a}^{2}{b}^{2}+1918620\,{a}^{2}A{b}^{2}{x}^{2}+1279080\,B{x}^{2}{a}^{3}b+1511640\,{a}^{3}Abx+377910\,{a}^{4}Bx+461890\,A{a}^{4}}{2078505}{x}^{{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/2078505*x^(9/2)*(109395*B*b^4*x^5+122265*A*b^4*x^4+489060*B*a*b^3*x^4+554268*A*a*b^3*x^3+831402*B*a^2*b^2*x^
3+959310*A*a^2*b^2*x^2+639540*B*a^3*b*x^2+755820*A*a^3*b*x+188955*B*a^4*x+230945*A*a^4)

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Maxima [A]  time = 1.00005, size = 134, normalized size = 1.21 \begin{align*} \frac{2}{19} \, B b^{4} x^{\frac{19}{2}} + \frac{2}{9} \, A a^{4} x^{\frac{9}{2}} + \frac{2}{17} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac{17}{2}} + \frac{4}{15} \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac{15}{2}} + \frac{4}{13} \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{\frac{13}{2}} + \frac{2}{11} \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x^{\frac{11}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*B*b^4*x^(19/2) + 2/9*A*a^4*x^(9/2) + 2/17*(4*B*a*b^3 + A*b^4)*x^(17/2) + 4/15*(3*B*a^2*b^2 + 2*A*a*b^3)*x
^(15/2) + 4/13*(2*B*a^3*b + 3*A*a^2*b^2)*x^(13/2) + 2/11*(B*a^4 + 4*A*a^3*b)*x^(11/2)

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Fricas [A]  time = 1.47288, size = 271, normalized size = 2.44 \begin{align*} \frac{2}{2078505} \,{\left (109395 \, B b^{4} x^{9} + 230945 \, A a^{4} x^{4} + 122265 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{8} + 277134 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{7} + 319770 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{6} + 188955 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x^{5}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/2078505*(109395*B*b^4*x^9 + 230945*A*a^4*x^4 + 122265*(4*B*a*b^3 + A*b^4)*x^8 + 277134*(3*B*a^2*b^2 + 2*A*a*
b^3)*x^7 + 319770*(2*B*a^3*b + 3*A*a^2*b^2)*x^6 + 188955*(B*a^4 + 4*A*a^3*b)*x^5)*sqrt(x)

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Sympy [A]  time = 17.3656, size = 148, normalized size = 1.33 \begin{align*} \frac{2 A a^{4} x^{\frac{9}{2}}}{9} + \frac{8 A a^{3} b x^{\frac{11}{2}}}{11} + \frac{12 A a^{2} b^{2} x^{\frac{13}{2}}}{13} + \frac{8 A a b^{3} x^{\frac{15}{2}}}{15} + \frac{2 A b^{4} x^{\frac{17}{2}}}{17} + \frac{2 B a^{4} x^{\frac{11}{2}}}{11} + \frac{8 B a^{3} b x^{\frac{13}{2}}}{13} + \frac{4 B a^{2} b^{2} x^{\frac{15}{2}}}{5} + \frac{8 B a b^{3} x^{\frac{17}{2}}}{17} + \frac{2 B b^{4} x^{\frac{19}{2}}}{19} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**4*x**(9/2)/9 + 8*A*a**3*b*x**(11/2)/11 + 12*A*a**2*b**2*x**(13/2)/13 + 8*A*a*b**3*x**(15/2)/15 + 2*A*b*
*4*x**(17/2)/17 + 2*B*a**4*x**(11/2)/11 + 8*B*a**3*b*x**(13/2)/13 + 4*B*a**2*b**2*x**(15/2)/5 + 8*B*a*b**3*x**
(17/2)/17 + 2*B*b**4*x**(19/2)/19

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Giac [A]  time = 1.23481, size = 136, normalized size = 1.23 \begin{align*} \frac{2}{19} \, B b^{4} x^{\frac{19}{2}} + \frac{8}{17} \, B a b^{3} x^{\frac{17}{2}} + \frac{2}{17} \, A b^{4} x^{\frac{17}{2}} + \frac{4}{5} \, B a^{2} b^{2} x^{\frac{15}{2}} + \frac{8}{15} \, A a b^{3} x^{\frac{15}{2}} + \frac{8}{13} \, B a^{3} b x^{\frac{13}{2}} + \frac{12}{13} \, A a^{2} b^{2} x^{\frac{13}{2}} + \frac{2}{11} \, B a^{4} x^{\frac{11}{2}} + \frac{8}{11} \, A a^{3} b x^{\frac{11}{2}} + \frac{2}{9} \, A a^{4} x^{\frac{9}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/19*B*b^4*x^(19/2) + 8/17*B*a*b^3*x^(17/2) + 2/17*A*b^4*x^(17/2) + 4/5*B*a^2*b^2*x^(15/2) + 8/15*A*a*b^3*x^(1
5/2) + 8/13*B*a^3*b*x^(13/2) + 12/13*A*a^2*b^2*x^(13/2) + 2/11*B*a^4*x^(11/2) + 8/11*A*a^3*b*x^(11/2) + 2/9*A*
a^4*x^(9/2)