3.2.35 \(\int x^{7/2} (a+b x^3)^2 (A+B x^3) \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{9} a^2 A x^{9/2}+\frac {2}{21} b x^{21/2} (2 a B+A b)+\frac {2}{15} a x^{15/2} (a B+2 A b)+\frac {2}{27} b^2 B x^{27/2} \]

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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 = {448} \begin {gather*} \frac {2}{9} a^2 A x^{9/2}+\frac {2}{21} b x^{21/2} (2 a B+A b)+\frac {2}{15} a x^{15/2} (a B+2 A b)+\frac {2}{27} b^2 B x^{27/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*(a + b*x^3)^2*(A + B*x^3),x]

[Out]

(2*a^2*A*x^(9/2))/9 + (2*a*(2*A*b + a*B)*x^(15/2))/15 + (2*b*(A*b + 2*a*B)*x^(21/2))/21 + (2*b^2*B*x^(27/2))/2
7

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin {align*} \int x^{7/2} \left (a+b x^3\right )^2 \left (A+B x^3\right ) \, dx &=\int \left (a^2 A x^{7/2}+a (2 A b+a B) x^{13/2}+b (A b+2 a B) x^{19/2}+b^2 B x^{25/2}\right ) \, dx\\ &=\frac {2}{9} a^2 A x^{9/2}+\frac {2}{15} a (2 A b+a B) x^{15/2}+\frac {2}{21} b (A b+2 a B) x^{21/2}+\frac {2}{27} b^2 B x^{27/2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{945} x^{9/2} \left (105 a^2 A+45 b x^6 (2 a B+A b)+63 a x^3 (a B+2 A b)+35 b^2 B x^9\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*(a + b*x^3)^2*(A + B*x^3),x]

[Out]

(2*x^(9/2)*(105*a^2*A + 63*a*(2*A*b + a*B)*x^3 + 45*b*(A*b + 2*a*B)*x^6 + 35*b^2*B*x^9))/945

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IntegrateAlgebraic [A]  time = 0.03, size = 59, normalized size = 0.94 \begin {gather*} \frac {2}{945} x^{9/2} \left (105 a^2 A+63 a^2 B x^3+126 a A b x^3+90 a b B x^6+45 A b^2 x^6+35 b^2 B x^9\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(7/2)*(a + b*x^3)^2*(A + B*x^3),x]

[Out]

(2*x^(9/2)*(105*a^2*A + 126*a*A*b*x^3 + 63*a^2*B*x^3 + 45*A*b^2*x^6 + 90*a*b*B*x^6 + 35*b^2*B*x^9))/945

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fricas [A]  time = 0.76, size = 56, normalized size = 0.89 \begin {gather*} \frac {2}{945} \, {\left (35 \, B b^{2} x^{13} + 45 \, {\left (2 \, B a b + A b^{2}\right )} x^{10} + 63 \, {\left (B a^{2} + 2 \, A a b\right )} x^{7} + 105 \, A a^{2} x^{4}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(b*x^3+a)^2*(B*x^3+A),x, algorithm="fricas")

[Out]

2/945*(35*B*b^2*x^13 + 45*(2*B*a*b + A*b^2)*x^10 + 63*(B*a^2 + 2*A*a*b)*x^7 + 105*A*a^2*x^4)*sqrt(x)

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giac [A]  time = 0.15, size = 53, normalized size = 0.84 \begin {gather*} \frac {2}{27} \, B b^{2} x^{\frac {27}{2}} + \frac {4}{21} \, B a b x^{\frac {21}{2}} + \frac {2}{21} \, A b^{2} x^{\frac {21}{2}} + \frac {2}{15} \, B a^{2} x^{\frac {15}{2}} + \frac {4}{15} \, A a b x^{\frac {15}{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^3+a)^2*(B*x^3+A),x, algorithm="giac")

[Out]

2/27*B*b^2*x^(27/2) + 4/21*B*a*b*x^(21/2) + 2/21*A*b^2*x^(21/2) + 2/15*B*a^2*x^(15/2) + 4/15*A*a*b*x^(15/2) +
2/9*A*a^2*x^(9/2)

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maple [A]  time = 0.05, size = 56, normalized size = 0.89 \begin {gather*} \frac {2 \left (35 b^{2} B \,x^{9}+45 A \,b^{2} x^{6}+90 B a b \,x^{6}+126 A a b \,x^{3}+63 B \,a^{2} x^{3}+105 a^{2} A \right ) x^{\frac {9}{2}}}{945} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(b*x^3+a)^2*(B*x^3+A),x)

[Out]

2/945*x^(9/2)*(35*B*b^2*x^9+45*A*b^2*x^6+90*B*a*b*x^6+126*A*a*b*x^3+63*B*a^2*x^3+105*A*a^2)

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maxima [A]  time = 0.56, size = 51, normalized size = 0.81 \begin {gather*} \frac {2}{27} \, B b^{2} x^{\frac {27}{2}} + \frac {2}{21} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {21}{2}} + \frac {2}{15} \, {\left (B a^{2} + 2 \, A a b\right )} x^{\frac {15}{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^3+a)^2*(B*x^3+A),x, algorithm="maxima")

[Out]

2/27*B*b^2*x^(27/2) + 2/21*(2*B*a*b + A*b^2)*x^(21/2) + 2/15*(B*a^2 + 2*A*a*b)*x^(15/2) + 2/9*A*a^2*x^(9/2)

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mupad [B]  time = 2.57, size = 51, normalized size = 0.81 \begin {gather*} x^{15/2}\,\left (\frac {2\,B\,a^2}{15}+\frac {4\,A\,b\,a}{15}\right )+x^{21/2}\,\left (\frac {2\,A\,b^2}{21}+\frac {4\,B\,a\,b}{21}\right )+\frac {2\,A\,a^2\,x^{9/2}}{9}+\frac {2\,B\,b^2\,x^{27/2}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(A + B*x^3)*(a + b*x^3)^2,x)

[Out]

x^(15/2)*((2*B*a^2)/15 + (4*A*a*b)/15) + x^(21/2)*((2*A*b^2)/21 + (4*B*a*b)/21) + (2*A*a^2*x^(9/2))/9 + (2*B*b
^2*x^(27/2))/27

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sympy [A]  time = 47.81, 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 {15}{2}}}{15} + \frac {2 A b^{2} x^{\frac {21}{2}}}{21} + \frac {2 B a^{2} x^{\frac {15}{2}}}{15} + \frac {4 B a b x^{\frac {21}{2}}}{21} + \frac {2 B b^{2} x^{\frac {27}{2}}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(b*x**3+a)**2*(B*x**3+A),x)

[Out]

2*A*a**2*x**(9/2)/9 + 4*A*a*b*x**(15/2)/15 + 2*A*b**2*x**(21/2)/21 + 2*B*a**2*x**(15/2)/15 + 4*B*a*b*x**(21/2)
/21 + 2*B*b**2*x**(27/2)/27

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