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

Optimal. Leaf size=96 \[ -\frac{\left (b x^2+c x^4\right )^{5/2} (4 b B-9 A c)}{63 c^2 x^3}+\frac{2 b \left (b x^2+c x^4\right )^{5/2} (4 b B-9 A c)}{315 c^3 x^5}+\frac{B \left (b x^2+c x^4\right )^{5/2}}{9 c x} \]

[Out]

(2*b*(4*b*B - 9*A*c)*(b*x^2 + c*x^4)^(5/2))/(315*c^3*x^5) - ((4*b*B - 9*A*c)*(b*x^2 + c*x^4)^(5/2))/(63*c^2*x^
3) + (B*(b*x^2 + c*x^4)^(5/2))/(9*c*x)

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Rubi [A]  time = 0.069621, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {1145, 2002, 2014} \[ -\frac{\left (b x^2+c x^4\right )^{5/2} (4 b B-9 A c)}{63 c^2 x^3}+\frac{2 b \left (b x^2+c x^4\right )^{5/2} (4 b B-9 A c)}{315 c^3 x^5}+\frac{B \left (b x^2+c x^4\right )^{5/2}}{9 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(4*b*B - 9*A*c)*(b*x^2 + c*x^4)^(5/2))/(315*c^3*x^5) - ((4*b*B - 9*A*c)*(b*x^2 + c*x^4)^(5/2))/(63*c^2*x^
3) + (B*(b*x^2 + c*x^4)^(5/2))/(9*c*x)

Rule 1145

Int[((d_) + (e_.)*(x_)^2)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*(b*x^2 + c*x^4)^(p + 1))/(c
*(4*p + 3)*x), x] - Dist[(b*e*(2*p + 1) - c*d*(4*p + 3))/(c*(4*p + 3)), Int[(b*x^2 + c*x^4)^p, x], x] /; FreeQ
[{b, c, d, e, p}, x] &&  !IntegerQ[p] && NeQ[4*p + 3, 0] && NeQ[b*e*(2*p + 1) - c*d*(4*p + 3), 0]

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac{B \left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac{(4 b B-9 A c) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{9 c}\\ &=-\frac{(4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac{B \left (b x^2+c x^4\right )^{5/2}}{9 c x}+\frac{(2 b (4 b B-9 A c)) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{63 c^2}\\ &=\frac{2 b (4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{315 c^3 x^5}-\frac{(4 b B-9 A c) \left (b x^2+c x^4\right )^{5/2}}{63 c^2 x^3}+\frac{B \left (b x^2+c x^4\right )^{5/2}}{9 c x}\\ \end{align*}

Mathematica [A]  time = 0.047036, size = 71, normalized size = 0.74 \[ \frac{x \left (b+c x^2\right )^3 \left (-2 b c \left (9 A+10 B x^2\right )+5 c^2 x^2 \left (9 A+7 B x^2\right )+8 b^2 B\right )}{315 c^3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x^2)^3*(8*b^2*B + 5*c^2*x^2*(9*A + 7*B*x^2) - 2*b*c*(9*A + 10*B*x^2)))/(315*c^3*Sqrt[x^2*(b + c*x^2)
])

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Maple [A]  time = 0.004, size = 67, normalized size = 0.7 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -35\,B{c}^{2}{x}^{4}-45\,A{x}^{2}{c}^{2}+20\,B{x}^{2}bc+18\,Abc-8\,B{b}^{2} \right ) }{315\,{c}^{3}{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(c*x^2+b)*(-35*B*c^2*x^4-45*A*c^2*x^2+20*B*b*c*x^2+18*A*b*c-8*B*b^2)*(c*x^4+b*x^2)^(3/2)/c^3/x^3

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Maxima [A]  time = 1.27695, size = 142, normalized size = 1.48 \begin{align*} \frac{{\left (5 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} + b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt{c x^{2} + b} A}{35 \, c^{2}} + \frac{{\left (35 \, c^{4} x^{8} + 50 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} - 4 \, b^{3} c x^{2} + 8 \, b^{4}\right )} \sqrt{c x^{2} + b} B}{315 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*c^3*x^6 + 8*b*c^2*x^4 + b^2*c*x^2 - 2*b^3)*sqrt(c*x^2 + b)*A/c^2 + 1/315*(35*c^4*x^8 + 50*b*c^3*x^6 +
3*b^2*c^2*x^4 - 4*b^3*c*x^2 + 8*b^4)*sqrt(c*x^2 + b)*B/c^3

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Fricas [A]  time = 1.08831, size = 228, normalized size = 2.38 \begin{align*} \frac{{\left (35 \, B c^{4} x^{8} + 5 \,{\left (10 \, B b c^{3} + 9 \, A c^{4}\right )} x^{6} + 8 \, B b^{4} - 18 \, A b^{3} c + 3 \,{\left (B b^{2} c^{2} + 24 \, A b c^{3}\right )} x^{4} -{\left (4 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{315 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*B*c^4*x^8 + 5*(10*B*b*c^3 + 9*A*c^4)*x^6 + 8*B*b^4 - 18*A*b^3*c + 3*(B*b^2*c^2 + 24*A*b*c^3)*x^4 - (
4*B*b^3*c - 9*A*b^2*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^3*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}} \left (A + B x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13837, size = 288, normalized size = 3. \begin{align*} \frac{\frac{21 \,{\left (3 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b\right )} A b \mathrm{sgn}\left (x\right )}{c} + \frac{3 \,{\left (15 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b + 35 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{2}\right )} B b \mathrm{sgn}\left (x\right )}{c^{2}} + \frac{3 \,{\left (15 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b + 35 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{2}\right )} A \mathrm{sgn}\left (x\right )}{c} + \frac{{\left (35 \,{\left (c x^{2} + b\right )}^{\frac{9}{2}} - 135 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} b + 189 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{3}\right )} B \mathrm{sgn}\left (x\right )}{c^{2}}}{315 \, c} - \frac{2 \,{\left (4 \, B b^{\frac{9}{2}} - 9 \, A b^{\frac{7}{2}} c\right )} \mathrm{sgn}\left (x\right )}{315 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(21*(3*(c*x^2 + b)^(5/2) - 5*(c*x^2 + b)^(3/2)*b)*A*b*sgn(x)/c + 3*(15*(c*x^2 + b)^(7/2) - 42*(c*x^2 + b
)^(5/2)*b + 35*(c*x^2 + b)^(3/2)*b^2)*B*b*sgn(x)/c^2 + 3*(15*(c*x^2 + b)^(7/2) - 42*(c*x^2 + b)^(5/2)*b + 35*(
c*x^2 + b)^(3/2)*b^2)*A*sgn(x)/c + (35*(c*x^2 + b)^(9/2) - 135*(c*x^2 + b)^(7/2)*b + 189*(c*x^2 + b)^(5/2)*b^2
- 105*(c*x^2 + b)^(3/2)*b^3)*B*sgn(x)/c^2)/c - 2/315*(4*B*b^(9/2) - 9*A*b^(7/2)*c)*sgn(x)/c^3