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

Optimal. Leaf size=131 \[ -\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{3465 c^4 x^5}+\frac {4 b \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{693 c^3 x^3}-\frac {\left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{99 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c} \]

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Rubi [A]  time = 0.24, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2039, 2016, 2002, 2014} \begin {gather*} -\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{3465 c^4 x^5}-\frac {\left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{99 c^2 x}+\frac {4 b \left (b x^2+c x^4\right )^{5/2} (6 b B-11 A c)}{693 c^3 x^3}+\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^2*(6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/(3465*c^4*x^5) + (4*b*(6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/
(693*c^3*x^3) - ((6*b*B - 11*A*c)*(b*x^2 + c*x^4)^(5/2))/(99*c^2*x) + (B*x*(b*x^2 + c*x^4)^(5/2))/(11*c)

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])

Rule 2016

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*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2039

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(d*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(b*(m + n + p*(j + n) + 1)), x] - Dist[(a*d*(m
+ j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)), Int[(e*x)^m*(a*x^j + b*x^(j + n))^p, x
], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[
m + n + p*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])

Rubi steps

\begin {align*} \int x^2 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {(6 b B-11 A c) \int x^2 \left (b x^2+c x^4\right )^{3/2} \, dx}{11 c}\\ &=-\frac {(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}+\frac {(4 b (6 b B-11 A c)) \int \left (b x^2+c x^4\right )^{3/2} \, dx}{99 c^2}\\ &=\frac {4 b (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{693 c^3 x^3}-\frac {(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {\left (8 b^2 (6 b B-11 A c)\right ) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{693 c^3}\\ &=-\frac {8 b^2 (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{3465 c^4 x^5}+\frac {4 b (6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{693 c^3 x^3}-\frac {(6 b B-11 A c) \left (b x^2+c x^4\right )^{5/2}}{99 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{5/2}}{11 c}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 92, normalized size = 0.70 \begin {gather*} \frac {x \left (b+c x^2\right )^3 \left (8 b^2 c \left (11 A+15 B x^2\right )-10 b c^2 x^2 \left (22 A+21 B x^2\right )+35 c^3 x^4 \left (11 A+9 B x^2\right )-48 b^3 B\right )}{3465 c^4 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x^2)^3*(-48*b^3*B + 35*c^3*x^4*(11*A + 9*B*x^2) + 8*b^2*c*(11*A + 15*B*x^2) - 10*b*c^2*x^2*(22*A + 2
1*B*x^2)))/(3465*c^4*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.56, size = 87, normalized size = 0.66 \begin {gather*} \frac {\left (b x^2+c x^4\right )^{5/2} \left (88 A b^2 c-220 A b c^2 x^2+385 A c^3 x^4-48 b^3 B+120 b^2 B c x^2-210 b B c^2 x^4+315 B c^3 x^6\right )}{3465 c^4 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*x^2 + c*x^4)^(5/2)*(-48*b^3*B + 88*A*b^2*c + 120*b^2*B*c*x^2 - 220*A*b*c^2*x^2 - 210*b*B*c^2*x^4 + 385*A*c
^3*x^4 + 315*B*c^3*x^6))/(3465*c^4*x^5)

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fricas [A]  time = 0.42, size = 131, normalized size = 1.00 \begin {gather*} \frac {{\left (315 \, B c^{5} x^{10} + 35 \, {\left (12 \, B b c^{4} + 11 \, A c^{5}\right )} x^{8} + 5 \, {\left (3 \, B b^{2} c^{3} + 110 \, A b c^{4}\right )} x^{6} - 48 \, B b^{5} + 88 \, A b^{4} c - 3 \, {\left (6 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{4} + 4 \, {\left (6 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, c^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(315*B*c^5*x^10 + 35*(12*B*b*c^4 + 11*A*c^5)*x^8 + 5*(3*B*b^2*c^3 + 110*A*b*c^4)*x^6 - 48*B*b^5 + 88*A*
b^4*c - 3*(6*B*b^3*c^2 - 11*A*b^2*c^3)*x^4 + 4*(6*B*b^4*c - 11*A*b^3*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^4*x)

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giac [A]  time = 0.21, size = 140, normalized size = 1.07 \begin {gather*} \frac {8 \, {\left (6 \, B b^{\frac {11}{2}} - 11 \, A b^{\frac {9}{2}} c\right )} \mathrm {sgn}\relax (x)}{3465 \, c^{4}} + \frac {315 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} B \mathrm {sgn}\relax (x) - 1155 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B b \mathrm {sgn}\relax (x) + 1485 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b^{2} \mathrm {sgn}\relax (x) - 693 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{3} \mathrm {sgn}\relax (x) + 385 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} A c \mathrm {sgn}\relax (x) - 990 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A b c \mathrm {sgn}\relax (x) + 693 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b^{2} c \mathrm {sgn}\relax (x)}{3465 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/3465*(6*B*b^(11/2) - 11*A*b^(9/2)*c)*sgn(x)/c^4 + 1/3465*(315*(c*x^2 + b)^(11/2)*B*sgn(x) - 1155*(c*x^2 + b)
^(9/2)*B*b*sgn(x) + 1485*(c*x^2 + b)^(7/2)*B*b^2*sgn(x) - 693*(c*x^2 + b)^(5/2)*B*b^3*sgn(x) + 385*(c*x^2 + b)
^(9/2)*A*c*sgn(x) - 990*(c*x^2 + b)^(7/2)*A*b*c*sgn(x) + 693*(c*x^2 + b)^(5/2)*A*b^2*c*sgn(x))/c^4

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maple [A]  time = 0.05, size = 91, normalized size = 0.69 \begin {gather*} \frac {\left (c \,x^{2}+b \right ) \left (315 B \,c^{3} x^{6}+385 A \,c^{3} x^{4}-210 B b \,c^{2} x^{4}-220 A b \,c^{2} x^{2}+120 B \,b^{2} c \,x^{2}+88 A \,b^{2} c -48 B \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 c^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(c*x^2+b)*(315*B*c^3*x^6+385*A*c^3*x^4-210*B*b*c^2*x^4-220*A*b*c^2*x^2+120*B*b^2*c*x^2+88*A*b^2*c-48*B*
b^3)*(c*x^4+b*x^2)^(3/2)/c^4/x^3

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maxima [A]  time = 1.53, size = 128, normalized size = 0.98 \begin {gather*} \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} A}{315 \, c^{3}} + \frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{2} + b} B}{1155 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*A/c^3 + 1/1155*(105*c^
5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b^4*c*x^2 - 16*b^5)*sqrt(c*x^2 + b)*B/c^4

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mupad [B]  time = 0.29, size = 124, normalized size = 0.95 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {x^8\,\left (385\,A\,c^5+420\,B\,b\,c^4\right )}{3465\,c^4}-\frac {48\,B\,b^5-88\,A\,b^4\,c}{3465\,c^4}+\frac {B\,c\,x^{10}}{11}+\frac {b^2\,x^4\,\left (11\,A\,c-6\,B\,b\right )}{1155\,c^2}-\frac {4\,b^3\,x^2\,\left (11\,A\,c-6\,B\,b\right )}{3465\,c^3}+\frac {b\,x^6\,\left (110\,A\,c+3\,B\,b\right )}{693\,c}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x^2 + c*x^4)^(1/2)*((x^8*(385*A*c^5 + 420*B*b*c^4))/(3465*c^4) - (48*B*b^5 - 88*A*b^4*c)/(3465*c^4) + (B*c
*x^10)/11 + (b^2*x^4*(11*A*c - 6*B*b))/(1155*c^2) - (4*b^3*x^2*(11*A*c - 6*B*b))/(3465*c^3) + (b*x^6*(110*A*c
+ 3*B*b))/(693*c)))/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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