3.164 \(\int \frac {(A+B x^2) (b x^2+c x^4)}{x^{3/2}} \, dx\)

Optimal. Leaf size=39 \[ \frac {2}{7} x^{7/2} (A c+b B)+\frac {2}{3} A b x^{3/2}+\frac {2}{11} B c x^{11/2} \]

[Out]

2/3*A*b*x^(3/2)+2/7*(A*c+B*b)*x^(7/2)+2/11*B*c*x^(11/2)

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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} \[ \frac {2}{7} x^{7/2} (A c+b B)+\frac {2}{3} A b x^{3/2}+\frac {2}{11} B c x^{11/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4))/x^(3/2),x]

[Out]

(2*A*b*x^(3/2))/3 + (2*(b*B + A*c)*x^(7/2))/7 + (2*B*c*x^(11/2))/11

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

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.85 \[ \frac {2}{231} x^{3/2} \left (33 x^2 (A c+b B)+77 A b+21 B c x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4))/x^(3/2),x]

[Out]

(2*x^(3/2)*(77*A*b + 33*(b*B + A*c)*x^2 + 21*B*c*x^4))/231

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fricas [A]  time = 0.93, size = 30, normalized size = 0.77 \[ \frac {2}{231} \, {\left (21 \, B c x^{5} + 33 \, {\left (B b + A c\right )} x^{3} + 77 \, A b x\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^(3/2),x, algorithm="fricas")

[Out]

2/231*(21*B*c*x^5 + 33*(B*b + A*c)*x^3 + 77*A*b*x)*sqrt(x)

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giac [A]  time = 0.16, size = 29, normalized size = 0.74 \[ \frac {2}{11} \, B c x^{\frac {11}{2}} + \frac {2}{7} \, B b x^{\frac {7}{2}} + \frac {2}{7} \, A c x^{\frac {7}{2}} + \frac {2}{3} \, A b x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^(3/2),x, algorithm="giac")

[Out]

2/11*B*c*x^(11/2) + 2/7*B*b*x^(7/2) + 2/7*A*c*x^(7/2) + 2/3*A*b*x^(3/2)

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maple [A]  time = 0.05, size = 32, normalized size = 0.82 \[ \frac {2 \left (21 B c \,x^{4}+33 A c \,x^{2}+33 B b \,x^{2}+77 A b \right ) x^{\frac {3}{2}}}{231} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)/x^(3/2),x)

[Out]

2/231*x^(3/2)*(21*B*c*x^4+33*A*c*x^2+33*B*b*x^2+77*A*b)

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maxima [A]  time = 1.38, size = 27, normalized size = 0.69 \[ \frac {2}{11} \, B c x^{\frac {11}{2}} + \frac {2}{7} \, {\left (B b + A c\right )} x^{\frac {7}{2}} + \frac {2}{3} \, A b x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^(3/2),x, algorithm="maxima")

[Out]

2/11*B*c*x^(11/2) + 2/7*(B*b + A*c)*x^(7/2) + 2/3*A*b*x^(3/2)

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mupad [B]  time = 0.04, size = 31, normalized size = 0.79 \[ \frac {2\,x^{3/2}\,\left (77\,A\,b+33\,A\,c\,x^2+33\,B\,b\,x^2+21\,B\,c\,x^4\right )}{231} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4))/x^(3/2),x)

[Out]

(2*x^(3/2)*(77*A*b + 33*A*c*x^2 + 33*B*b*x^2 + 21*B*c*x^4))/231

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sympy [A]  time = 2.26, size = 46, normalized size = 1.18 \[ \frac {2 A b x^{\frac {3}{2}}}{3} + \frac {2 A c x^{\frac {7}{2}}}{7} + \frac {2 B b x^{\frac {7}{2}}}{7} + \frac {2 B c x^{\frac {11}{2}}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)/x**(3/2),x)

[Out]

2*A*b*x**(3/2)/3 + 2*A*c*x**(7/2)/7 + 2*B*b*x**(7/2)/7 + 2*B*c*x**(11/2)/11

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