3.160 \(\int x^{5/2} (A+B x^2) (b x^2+c x^4) \, dx\)

Optimal. Leaf size=39 \[ \frac {2}{15} x^{15/2} (A c+b B)+\frac {2}{11} A b x^{11/2}+\frac {2}{19} B c x^{19/2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1584, 448} \[ \frac {2}{15} x^{15/2} (A c+b B)+\frac {2}{11} A b x^{11/2}+\frac {2}{19} B c x^{19/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4),x]

[Out]

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

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 x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right ) \, dx &=\int x^{9/2} \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\int \left (A b x^{9/2}+(b B+A c) x^{13/2}+B c x^{17/2}\right ) \, dx\\ &=\frac {2}{11} A b x^{11/2}+\frac {2}{15} (b B+A c) x^{15/2}+\frac {2}{19} B c x^{19/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.85 \[ \frac {2 x^{11/2} \left (209 x^2 (A c+b B)+285 A b+165 B c x^4\right )}{3135} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4),x]

[Out]

(2*x^(11/2)*(285*A*b + 209*(b*B + A*c)*x^2 + 165*B*c*x^4))/3135

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fricas [A]  time = 0.97, size = 32, normalized size = 0.82 \[ \frac {2}{3135} \, {\left (165 \, B c x^{9} + 209 \, {\left (B b + A c\right )} x^{7} + 285 \, A b x^{5}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

2/3135*(165*B*c*x^9 + 209*(B*b + A*c)*x^7 + 285*A*b*x^5)*sqrt(x)

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giac [A]  time = 0.15, size = 29, normalized size = 0.74 \[ \frac {2}{19} \, B c x^{\frac {19}{2}} + \frac {2}{15} \, B b x^{\frac {15}{2}} + \frac {2}{15} \, A c x^{\frac {15}{2}} + \frac {2}{11} \, A b x^{\frac {11}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 32, normalized size = 0.82 \[ \frac {2 \left (165 B c \,x^{4}+209 A c \,x^{2}+209 B b \,x^{2}+285 A b \right ) x^{\frac {11}{2}}}{3135} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2),x)

[Out]

2/3135*x^(11/2)*(165*B*c*x^4+209*A*c*x^2+209*B*b*x^2+285*A*b)

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maxima [A]  time = 1.35, size = 27, normalized size = 0.69 \[ \frac {2}{19} \, B c x^{\frac {19}{2}} + \frac {2}{15} \, {\left (B b + A c\right )} x^{\frac {15}{2}} + \frac {2}{11} \, A b x^{\frac {11}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.11, size = 31, normalized size = 0.79 \[ \frac {2\,x^{11/2}\,\left (285\,A\,b+209\,A\,c\,x^2+209\,B\,b\,x^2+165\,B\,c\,x^4\right )}{3135} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4),x)

[Out]

(2*x^(11/2)*(285*A*b + 209*A*c*x^2 + 209*B*b*x^2 + 165*B*c*x^4))/3135

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sympy [A]  time = 11.25, size = 46, normalized size = 1.18 \[ \frac {2 A b x^{\frac {11}{2}}}{11} + \frac {2 A c x^{\frac {15}{2}}}{15} + \frac {2 B b x^{\frac {15}{2}}}{15} + \frac {2 B c x^{\frac {19}{2}}}{19} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x**2+A)*(c*x**4+b*x**2),x)

[Out]

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

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