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

Optimal. Leaf size=94 \[ \frac {2 b \left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{105 c^3 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{35 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2039, 2016, 2000} \begin {gather*} -\frac {\left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{35 c^2 x}+\frac {2 b \left (b x^2+c x^4\right )^{3/2} (4 b B-7 A c)}{105 c^3 x^3}+\frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(4*b*B - 7*A*c)*(b*x^2 + c*x^4)^(3/2))/(105*c^3*x^3) - ((4*b*B - 7*A*c)*(b*x^2 + c*x^4)^(3/2))/(35*c^2*x)
+ (B*x*(b*x^2 + c*x^4)^(3/2))/(7*c)

Rule 2000

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 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 ) \sqrt {b x^2+c x^4} \, dx &=\frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {(4 b B-7 A c) \int x^2 \sqrt {b x^2+c x^4} \, dx}{7 c}\\ &=-\frac {(4 b B-7 A c) \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c}+\frac {(2 b (4 b B-7 A c)) \int \sqrt {b x^2+c x^4} \, dx}{35 c^2}\\ &=\frac {2 b (4 b B-7 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac {(4 b B-7 A c) \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {B x \left (b x^2+c x^4\right )^{3/2}}{7 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 64, normalized size = 0.68 \begin {gather*} \frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (-2 b c \left (7 A+6 B x^2\right )+3 c^2 x^2 \left (7 A+5 B x^2\right )+8 b^2 B\right )}{105 c^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + c*x^2))^(3/2)*(8*b^2*B + 3*c^2*x^2*(7*A + 5*B*x^2) - 2*b*c*(7*A + 6*B*x^2)))/(105*c^3*x^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.09, size = 63, normalized size = 0.67 \begin {gather*} \frac {\left (b x^2+c x^4\right )^{3/2} \left (-14 A b c+21 A c^2 x^2+8 b^2 B-12 b B c x^2+15 B c^2 x^4\right )}{105 c^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*x^2 + c*x^4)^(3/2)*(8*b^2*B - 14*A*b*c - 12*b*B*c*x^2 + 21*A*c^2*x^2 + 15*B*c^2*x^4))/(105*c^3*x^3)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 82, normalized size = 0.87 \begin {gather*} \frac {{\left (15 \, B c^{3} x^{6} + 3 \, {\left (B b c^{2} + 7 \, A c^{3}\right )} x^{4} + 8 \, B b^{3} - 14 \, A b^{2} c - {\left (4 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, c^{3} 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)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*B*c^3*x^6 + 3*(B*b*c^2 + 7*A*c^3)*x^4 + 8*B*b^3 - 14*A*b^2*c - (4*B*b^2*c - 7*A*b*c^2)*x^2)*sqrt(c*x
^4 + b*x^2)/(c^3*x)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 105, normalized size = 1.12 \begin {gather*} -\frac {2 \, {\left (4 \, B b^{\frac {7}{2}} - 7 \, A b^{\frac {5}{2}} c\right )} \mathrm {sgn}\relax (x)}{105 \, c^{3}} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B \mathrm {sgn}\relax (x) - 42 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b \mathrm {sgn}\relax (x) + 35 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{2} \mathrm {sgn}\relax (x) + 21 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c \mathrm {sgn}\relax (x) - 35 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b c \mathrm {sgn}\relax (x)}{105 \, c^{3}} \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)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 67, normalized size = 0.71 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-15 B \,c^{2} x^{4}-21 A \,c^{2} x^{2}+12 B b c \,x^{2}+14 A b c -8 B \,b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x} \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)^(1/2),x)

[Out]

-1/105*(c*x^2+b)*(-15*B*c^2*x^4-21*A*c^2*x^2+12*B*b*c*x^2+14*A*b*c-8*B*b^2)*(c*x^4+b*x^2)^(1/2)/c^3/x

________________________________________________________________________________________

maxima [A]  time = 1.54, size = 83, normalized size = 0.88 \begin {gather*} \frac {{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt {c x^{2} + b} A}{15 \, c^{2}} + \frac {{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{2} + b} B}{105 \, c^{3}} \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)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*c^2*x^4 + b*c*x^2 - 2*b^2)*sqrt(c*x^2 + b)*A/c^2 + 1/105*(15*c^3*x^6 + 3*b*c^2*x^4 - 4*b^2*c*x^2 + 8*b
^3)*sqrt(c*x^2 + b)*B/c^3

________________________________________________________________________________________

mupad [B]  time = 0.19, size = 83, normalized size = 0.88 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {B\,x^6}{7}+\frac {8\,B\,b^3-14\,A\,b^2\,c}{105\,c^3}+\frac {x^4\,\left (21\,A\,c^3+3\,B\,b\,c^2\right )}{105\,c^3}+\frac {b\,x^2\,\left (7\,A\,c-4\,B\,b\right )}{105\,c^2}\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)^(1/2),x)

[Out]

((b*x^2 + c*x^4)^(1/2)*((B*x^6)/7 + (8*B*b^3 - 14*A*b^2*c)/(105*c^3) + (x^4*(21*A*c^3 + 3*B*b*c^2))/(105*c^3)
+ (b*x^2*(7*A*c - 4*B*b))/(105*c^2)))/x

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt {x^{2} \left (b + c x^{2}\right )} \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)**(1/2),x)

[Out]

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

________________________________________________________________________________________