3.170 \(\int \sqrt {x} (A+B x^2) (b x^2+c x^4)^2 \, dx\)

Optimal. Leaf size=63 \[ \frac {2}{11} A b^2 x^{11/2}+\frac {2}{19} c x^{19/2} (A c+2 b B)+\frac {2}{15} b x^{15/2} (2 A c+b B)+\frac {2}{23} B c^2 x^{23/2} \]

[Out]

2/11*A*b^2*x^(11/2)+2/15*b*(2*A*c+B*b)*x^(15/2)+2/19*c*(A*c+2*B*b)*x^(19/2)+2/23*B*c^2*x^(23/2)

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1584, 448} \[ \frac {2}{11} A b^2 x^{11/2}+\frac {2}{19} c x^{19/2} (A c+2 b B)+\frac {2}{15} b x^{15/2} (2 A c+b B)+\frac {2}{23} B c^2 x^{23/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(A + B*x^2)*(b*x^2 + c*x^4)^2,x]

[Out]

(2*A*b^2*x^(11/2))/11 + (2*b*(b*B + 2*A*c)*x^(15/2))/15 + (2*c*(2*b*B + A*c)*x^(19/2))/19 + (2*B*c^2*x^(23/2))
/23

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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \sqrt {x} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^2 \, dx &=\int x^{9/2} \left (A+B x^2\right ) \left (b+c x^2\right )^2 \, dx\\ &=\int \left (A b^2 x^{9/2}+b (b B+2 A c) x^{13/2}+c (2 b B+A c) x^{17/2}+B c^2 x^{21/2}\right ) \, dx\\ &=\frac {2}{11} A b^2 x^{11/2}+\frac {2}{15} b (b B+2 A c) x^{15/2}+\frac {2}{19} c (2 b B+A c) x^{19/2}+\frac {2}{23} B c^2 x^{23/2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.84 \[ \frac {2 x^{11/2} \left (6555 A b^2+3795 c x^4 (A c+2 b B)+4807 b x^2 (2 A c+b B)+3135 B c^2 x^6\right )}{72105} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(A + B*x^2)*(b*x^2 + c*x^4)^2,x]

[Out]

(2*x^(11/2)*(6555*A*b^2 + 4807*b*(b*B + 2*A*c)*x^2 + 3795*c*(2*b*B + A*c)*x^4 + 3135*B*c^2*x^6))/72105

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fricas [A]  time = 0.88, size = 56, normalized size = 0.89 \[ \frac {2}{72105} \, {\left (3135 \, B c^{2} x^{11} + 3795 \, {\left (2 \, B b c + A c^{2}\right )} x^{9} + 6555 \, A b^{2} x^{5} + 4807 \, {\left (B b^{2} + 2 \, A b c\right )} x^{7}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2*x^(1/2),x, algorithm="fricas")

[Out]

2/72105*(3135*B*c^2*x^11 + 3795*(2*B*b*c + A*c^2)*x^9 + 6555*A*b^2*x^5 + 4807*(B*b^2 + 2*A*b*c)*x^7)*sqrt(x)

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giac [A]  time = 0.18, size = 53, normalized size = 0.84 \[ \frac {2}{23} \, B c^{2} x^{\frac {23}{2}} + \frac {4}{19} \, B b c x^{\frac {19}{2}} + \frac {2}{19} \, A c^{2} x^{\frac {19}{2}} + \frac {2}{15} \, B b^{2} x^{\frac {15}{2}} + \frac {4}{15} \, A b c x^{\frac {15}{2}} + \frac {2}{11} \, A b^{2} x^{\frac {11}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2*x^(1/2),x, algorithm="giac")

[Out]

2/23*B*c^2*x^(23/2) + 4/19*B*b*c*x^(19/2) + 2/19*A*c^2*x^(19/2) + 2/15*B*b^2*x^(15/2) + 4/15*A*b*c*x^(15/2) +
2/11*A*b^2*x^(11/2)

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maple [A]  time = 0.05, size = 56, normalized size = 0.89 \[ \frac {2 \left (3135 B \,c^{2} x^{6}+3795 A \,c^{2} x^{4}+7590 B b c \,x^{4}+9614 A b c \,x^{2}+4807 B \,b^{2} x^{2}+6555 b^{2} A \right ) x^{\frac {11}{2}}}{72105} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^2*x^(1/2),x)

[Out]

2/72105*x^(11/2)*(3135*B*c^2*x^6+3795*A*c^2*x^4+7590*B*b*c*x^4+9614*A*b*c*x^2+4807*B*b^2*x^2+6555*A*b^2)

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maxima [A]  time = 1.38, size = 51, normalized size = 0.81 \[ \frac {2}{23} \, B c^{2} x^{\frac {23}{2}} + \frac {2}{19} \, {\left (2 \, B b c + A c^{2}\right )} x^{\frac {19}{2}} + \frac {2}{11} \, A b^{2} x^{\frac {11}{2}} + \frac {2}{15} \, {\left (B b^{2} + 2 \, A b c\right )} x^{\frac {15}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^2*x^(1/2),x, algorithm="maxima")

[Out]

2/23*B*c^2*x^(23/2) + 2/19*(2*B*b*c + A*c^2)*x^(19/2) + 2/11*A*b^2*x^(11/2) + 2/15*(B*b^2 + 2*A*b*c)*x^(15/2)

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mupad [B]  time = 0.05, size = 51, normalized size = 0.81 \[ x^{15/2}\,\left (\frac {2\,B\,b^2}{15}+\frac {4\,A\,c\,b}{15}\right )+x^{19/2}\,\left (\frac {2\,A\,c^2}{19}+\frac {4\,B\,b\,c}{19}\right )+\frac {2\,A\,b^2\,x^{11/2}}{11}+\frac {2\,B\,c^2\,x^{23/2}}{23} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(A + B*x^2)*(b*x^2 + c*x^4)^2,x)

[Out]

x^(15/2)*((2*B*b^2)/15 + (4*A*b*c)/15) + x^(19/2)*((2*A*c^2)/19 + (4*B*b*c)/19) + (2*A*b^2*x^(11/2))/11 + (2*B
*c^2*x^(23/2))/23

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sympy [A]  time = 3.73, size = 66, normalized size = 1.05 \[ \frac {2 A b^{2} x^{\frac {11}{2}}}{11} + \frac {2 B c^{2} x^{\frac {23}{2}}}{23} + \frac {2 x^{\frac {19}{2}} \left (A c^{2} + 2 B b c\right )}{19} + \frac {2 x^{\frac {15}{2}} \left (2 A b c + B b^{2}\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**2*x**(1/2),x)

[Out]

2*A*b**2*x**(11/2)/11 + 2*B*c**2*x**(23/2)/23 + 2*x**(19/2)*(A*c**2 + 2*B*b*c)/19 + 2*x**(15/2)*(2*A*b*c + B*b
**2)/15

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