3.4.96 \(\int \sqrt {x} (A+B x) (a+c x^2)^2 \, dx\)

Optimal. Leaf size=77 \[ \frac {2}{3} a^2 A x^{3/2}+\frac {2}{5} a^2 B x^{5/2}+\frac {4}{7} a A c x^{7/2}+\frac {4}{9} a B c x^{9/2}+\frac {2}{11} A c^2 x^{11/2}+\frac {2}{13} B c^2 x^{13/2} \]

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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}{3} a^2 A x^{3/2}+\frac {2}{5} a^2 B x^{5/2}+\frac {4}{7} a A c x^{7/2}+\frac {4}{9} a B c x^{9/2}+\frac {2}{11} A c^2 x^{11/2}+\frac {2}{13} B c^2 x^{13/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 60, normalized size = 0.78 \begin {gather*} \frac {2}{15} a^2 x^{3/2} (5 A+3 B x)+\frac {4}{63} a c x^{7/2} (9 A+7 B x)+\frac {2}{143} c^2 x^{11/2} (13 A+11 B x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*x^(3/2)*(5*A + 3*B*x))/15 + (4*a*c*x^(7/2)*(9*A + 7*B*x))/63 + (2*c^2*x^(11/2)*(13*A + 11*B*x))/143

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IntegrateAlgebraic [A]  time = 0.03, size = 69, normalized size = 0.90 \begin {gather*} \frac {2 \left (15015 a^2 A x^{3/2}+9009 a^2 B x^{5/2}+12870 a A c x^{7/2}+10010 a B c x^{9/2}+4095 A c^2 x^{11/2}+3465 B c^2 x^{13/2}\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15015*a^2*A*x^(3/2) + 9009*a^2*B*x^(5/2) + 12870*a*A*c*x^(7/2) + 10010*a*B*c*x^(9/2) + 4095*A*c^2*x^(11/2)
+ 3465*B*c^2*x^(13/2)))/45045

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fricas [A]  time = 0.40, size = 56, normalized size = 0.73 \begin {gather*} \frac {2}{45045} \, {\left (3465 \, B c^{2} x^{6} + 4095 \, A c^{2} x^{5} + 10010 \, B a c x^{4} + 12870 \, A a c x^{3} + 9009 \, B a^{2} x^{2} + 15015 \, A a^{2} x\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2*x^(1/2),x, algorithm="fricas")

[Out]

2/45045*(3465*B*c^2*x^6 + 4095*A*c^2*x^5 + 10010*B*a*c*x^4 + 12870*A*a*c*x^3 + 9009*B*a^2*x^2 + 15015*A*a^2*x)
*sqrt(x)

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giac [A]  time = 0.15, size = 53, normalized size = 0.69 \begin {gather*} \frac {2}{13} \, B c^{2} x^{\frac {13}{2}} + \frac {2}{11} \, A c^{2} x^{\frac {11}{2}} + \frac {4}{9} \, B a c x^{\frac {9}{2}} + \frac {4}{7} \, A a c x^{\frac {7}{2}} + \frac {2}{5} \, B a^{2} x^{\frac {5}{2}} + \frac {2}{3} \, A a^{2} x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2*x^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 54, normalized size = 0.70 \begin {gather*} \frac {2 \left (3465 B \,c^{2} x^{5}+4095 A \,c^{2} x^{4}+10010 B a c \,x^{3}+12870 A a c \,x^{2}+9009 B \,a^{2} x +15015 A \,a^{2}\right ) x^{\frac {3}{2}}}{45045} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^2*x^(1/2),x)

[Out]

2/45045*x^(3/2)*(3465*B*c^2*x^5+4095*A*c^2*x^4+10010*B*a*c*x^3+12870*A*a*c*x^2+9009*B*a^2*x+15015*A*a^2)

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maxima [A]  time = 0.48, size = 53, normalized size = 0.69 \begin {gather*} \frac {2}{13} \, B c^{2} x^{\frac {13}{2}} + \frac {2}{11} \, A c^{2} x^{\frac {11}{2}} + \frac {4}{9} \, B a c x^{\frac {9}{2}} + \frac {4}{7} \, A a c x^{\frac {7}{2}} + \frac {2}{5} \, B a^{2} x^{\frac {5}{2}} + \frac {2}{3} \, A a^{2} x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2*x^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.03, size = 53, normalized size = 0.69 \begin {gather*} \frac {2\,A\,a^2\,x^{3/2}}{3}+\frac {2\,B\,a^2\,x^{5/2}}{5}+\frac {2\,A\,c^2\,x^{11/2}}{11}+\frac {2\,B\,c^2\,x^{13/2}}{13}+\frac {4\,A\,a\,c\,x^{7/2}}{7}+\frac {4\,B\,a\,c\,x^{9/2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a + c*x^2)^2*(A + B*x),x)

[Out]

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

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sympy [A]  time = 3.08, size = 80, normalized size = 1.04 \begin {gather*} \frac {2 A a^{2} x^{\frac {3}{2}}}{3} + \frac {4 A a c x^{\frac {7}{2}}}{7} + \frac {2 A c^{2} x^{\frac {11}{2}}}{11} + \frac {2 B a^{2} x^{\frac {5}{2}}}{5} + \frac {4 B a c x^{\frac {9}{2}}}{9} + \frac {2 B c^{2} x^{\frac {13}{2}}}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**2*x**(1/2),x)

[Out]

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

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