3.4.95 \(\int x^{3/2} (A+B x) (a+c x^2)^2 \, dx\)

Optimal. Leaf size=77 \[ \frac {2}{5} a^2 A x^{5/2}+\frac {2}{7} a^2 B x^{7/2}+\frac {4}{9} a A c x^{9/2}+\frac {4}{11} a B c x^{11/2}+\frac {2}{13} A c^2 x^{13/2}+\frac {2}{15} B c^2 x^{15/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {766} \begin {gather*} \frac {2}{5} a^2 A x^{5/2}+\frac {2}{7} a^2 B x^{7/2}+\frac {4}{9} a A c x^{9/2}+\frac {4}{11} a B c x^{11/2}+\frac {2}{13} A c^2 x^{13/2}+\frac {2}{15} B c^2 x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*A*x^(5/2))/5 + (2*a^2*B*x^(7/2))/7 + (4*a*A*c*x^(9/2))/9 + (4*a*B*c*x^(11/2))/11 + (2*A*c^2*x^(13/2))/1
3 + (2*B*c^2*x^(15/2))/15

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^{3/2} (A+B x) \left (a+c x^2\right )^2 \, dx &=\int \left (a^2 A x^{3/2}+a^2 B x^{5/2}+2 a A c x^{7/2}+2 a B c x^{9/2}+A c^2 x^{11/2}+B c^2 x^{13/2}\right ) \, dx\\ &=\frac {2}{5} a^2 A x^{5/2}+\frac {2}{7} a^2 B x^{7/2}+\frac {4}{9} a A c x^{9/2}+\frac {4}{11} a B c x^{11/2}+\frac {2}{13} A c^2 x^{13/2}+\frac {2}{15} B c^2 x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 54, normalized size = 0.70 \begin {gather*} \frac {2 x^{5/2} \left (1287 a^2 (7 A+5 B x)+910 a c x^2 (11 A+9 B x)+231 c^2 x^4 (15 A+13 B x)\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(1287*a^2*(7*A + 5*B*x) + 910*a*c*x^2*(11*A + 9*B*x) + 231*c^2*x^4*(15*A + 13*B*x)))/45045

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 0.90 \begin {gather*} \frac {2 \left (9009 a^2 A x^{5/2}+6435 a^2 B x^{7/2}+10010 a A c x^{9/2}+8190 a B c x^{11/2}+3465 A c^2 x^{13/2}+3003 B c^2 x^{15/2}\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9009*a^2*A*x^(5/2) + 6435*a^2*B*x^(7/2) + 10010*a*A*c*x^(9/2) + 8190*a*B*c*x^(11/2) + 3465*A*c^2*x^(13/2)
+ 3003*B*c^2*x^(15/2)))/45045

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fricas [A]  time = 0.40, size = 58, normalized size = 0.75 \begin {gather*} \frac {2}{45045} \, {\left (3003 \, B c^{2} x^{7} + 3465 \, A c^{2} x^{6} + 8190 \, B a c x^{5} + 10010 \, A a c x^{4} + 6435 \, B a^{2} x^{3} + 9009 \, A a^{2} x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3003*B*c^2*x^7 + 3465*A*c^2*x^6 + 8190*B*a*c*x^5 + 10010*A*a*c*x^4 + 6435*B*a^2*x^3 + 9009*A*a^2*x^2)
*sqrt(x)

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giac [A]  time = 0.15, size = 53, normalized size = 0.69 \begin {gather*} \frac {2}{15} \, B c^{2} x^{\frac {15}{2}} + \frac {2}{13} \, A c^{2} x^{\frac {13}{2}} + \frac {4}{11} \, B a c x^{\frac {11}{2}} + \frac {4}{9} \, A a c x^{\frac {9}{2}} + \frac {2}{7} \, B a^{2} x^{\frac {7}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)*(c*x^2+a)^2,x, algorithm="giac")

[Out]

2/15*B*c^2*x^(15/2) + 2/13*A*c^2*x^(13/2) + 4/11*B*a*c*x^(11/2) + 4/9*A*a*c*x^(9/2) + 2/7*B*a^2*x^(7/2) + 2/5*
A*a^2*x^(5/2)

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maple [A]  time = 0.05, size = 54, normalized size = 0.70 \begin {gather*} \frac {2 \left (3003 B \,c^{2} x^{5}+3465 A \,c^{2} x^{4}+8190 B a c \,x^{3}+10010 A a c \,x^{2}+6435 B \,a^{2} x +9009 A \,a^{2}\right ) x^{\frac {5}{2}}}{45045} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*x^(5/2)*(3003*B*c^2*x^5+3465*A*c^2*x^4+8190*B*a*c*x^3+10010*A*a*c*x^2+6435*B*a^2*x+9009*A*a^2)

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maxima [A]  time = 0.51, size = 53, normalized size = 0.69 \begin {gather*} \frac {2}{15} \, B c^{2} x^{\frac {15}{2}} + \frac {2}{13} \, A c^{2} x^{\frac {13}{2}} + \frac {4}{11} \, B a c x^{\frac {11}{2}} + \frac {4}{9} \, A a c x^{\frac {9}{2}} + \frac {2}{7} \, B a^{2} x^{\frac {7}{2}} + \frac {2}{5} \, A a^{2} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x+A)*(c*x^2+a)^2,x, algorithm="maxima")

[Out]

2/15*B*c^2*x^(15/2) + 2/13*A*c^2*x^(13/2) + 4/11*B*a*c*x^(11/2) + 4/9*A*a*c*x^(9/2) + 2/7*B*a^2*x^(7/2) + 2/5*
A*a^2*x^(5/2)

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mupad [B]  time = 0.03, size = 53, normalized size = 0.69 \begin {gather*} \frac {2\,A\,a^2\,x^{5/2}}{5}+\frac {2\,B\,a^2\,x^{7/2}}{7}+\frac {2\,A\,c^2\,x^{13/2}}{13}+\frac {2\,B\,c^2\,x^{15/2}}{15}+\frac {4\,A\,a\,c\,x^{9/2}}{9}+\frac {4\,B\,a\,c\,x^{11/2}}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*a^2*x^(5/2))/5 + (2*B*a^2*x^(7/2))/7 + (2*A*c^2*x^(13/2))/13 + (2*B*c^2*x^(15/2))/15 + (4*A*a*c*x^(9/2))/
9 + (4*B*a*c*x^(11/2))/11

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sympy [A]  time = 4.26, size = 80, normalized size = 1.04 \begin {gather*} \frac {2 A a^{2} x^{\frac {5}{2}}}{5} + \frac {4 A a c x^{\frac {9}{2}}}{9} + \frac {2 A c^{2} x^{\frac {13}{2}}}{13} + \frac {2 B a^{2} x^{\frac {7}{2}}}{7} + \frac {4 B a c x^{\frac {11}{2}}}{11} + \frac {2 B c^{2} x^{\frac {15}{2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*x**(5/2)/5 + 4*A*a*c*x**(9/2)/9 + 2*A*c**2*x**(13/2)/13 + 2*B*a**2*x**(7/2)/7 + 4*B*a*c*x**(11/2)/11
+ 2*B*c**2*x**(15/2)/15

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