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

Optimal. Leaf size=131 \[ -\frac {8 b^2 \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{315 c^4 x^3}+\frac {4 b \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{105 c^3 x}-\frac {x \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{21 c^2}+\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c} \]

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Rubi [A]  time = 0.22, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {8 b^2 \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{315 c^4 x^3}-\frac {x \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{21 c^2}+\frac {4 b \left (b x^2+c x^4\right )^{3/2} (2 b B-3 A c)}{105 c^3 x}+\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^2*(2*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(315*c^4*x^3) + (4*b*(2*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(10
5*c^3*x) - ((2*b*B - 3*A*c)*x*(b*x^2 + c*x^4)^(3/2))/(21*c^2) + (B*x^3*(b*x^2 + c*x^4)^(3/2))/(9*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^4 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac {(6 b B-9 A c) \int x^4 \sqrt {b x^2+c x^4} \, dx}{9 c}\\ &=-\frac {(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}+\frac {(4 b (2 b B-3 A c)) \int x^2 \sqrt {b x^2+c x^4} \, dx}{21 c^2}\\ &=\frac {4 b (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x}-\frac {(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}-\frac {\left (8 b^2 (2 b B-3 A c)\right ) \int \sqrt {b x^2+c x^4} \, dx}{105 c^3}\\ &=-\frac {8 b^2 (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{315 c^4 x^3}+\frac {4 b (2 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x}-\frac {(2 b B-3 A c) x \left (b x^2+c x^4\right )^{3/2}}{21 c^2}+\frac {B x^3 \left (b x^2+c x^4\right )^{3/2}}{9 c}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.63 \begin {gather*} \frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (24 b^2 c \left (A+B x^2\right )-6 b c^2 x^2 \left (6 A+5 B x^2\right )+5 c^3 x^4 \left (9 A+7 B x^2\right )-16 b^3 B\right )}{315 c^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + c*x^2))^(3/2)*(-16*b^3*B + 24*b^2*c*(A + B*x^2) - 6*b*c^2*x^2*(6*A + 5*B*x^2) + 5*c^3*x^4*(9*A + 7*
B*x^2)))/(315*c^4*x^3)

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IntegrateAlgebraic [A]  time = 0.11, size = 87, normalized size = 0.66 \begin {gather*} \frac {\left (b x^2+c x^4\right )^{3/2} \left (24 A b^2 c-36 A b c^2 x^2+45 A c^3 x^4-16 b^3 B+24 b^2 B c x^2-30 b B c^2 x^4+35 B c^3 x^6\right )}{315 c^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*x^2 + c*x^4)^(3/2)*(-16*b^3*B + 24*A*b^2*c + 24*b^2*B*c*x^2 - 36*A*b*c^2*x^2 - 30*b*B*c^2*x^4 + 45*A*c^3*x
^4 + 35*B*c^3*x^6))/(315*c^4*x^3)

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fricas [A]  time = 0.41, size = 106, normalized size = 0.81 \begin {gather*} \frac {{\left (35 \, B c^{4} x^{8} + 5 \, {\left (B b c^{3} + 9 \, A c^{4}\right )} x^{6} - 16 \, B b^{4} + 24 \, A b^{3} c - 3 \, {\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{4} + 4 \, {\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, c^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/315*(35*B*c^4*x^8 + 5*(B*b*c^3 + 9*A*c^4)*x^6 - 16*B*b^4 + 24*A*b^3*c - 3*(2*B*b^2*c^2 - 3*A*b*c^3)*x^4 + 4*
(2*B*b^3*c - 3*A*b^2*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^4*x)

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giac [A]  time = 0.17, size = 140, normalized size = 1.07 \begin {gather*} \frac {8 \, {\left (2 \, B b^{\frac {9}{2}} - 3 \, A b^{\frac {7}{2}} c\right )} \mathrm {sgn}\relax (x)}{315 \, c^{4}} + \frac {35 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B \mathrm {sgn}\relax (x) - 135 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b \mathrm {sgn}\relax (x) + 189 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{2} \mathrm {sgn}\relax (x) - 105 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{3} \mathrm {sgn}\relax (x) + 45 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A c \mathrm {sgn}\relax (x) - 126 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b c \mathrm {sgn}\relax (x) + 105 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{2} c \mathrm {sgn}\relax (x)}{315 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

8/315*(2*B*b^(9/2) - 3*A*b^(7/2)*c)*sgn(x)/c^4 + 1/315*(35*(c*x^2 + b)^(9/2)*B*sgn(x) - 135*(c*x^2 + b)^(7/2)*
B*b*sgn(x) + 189*(c*x^2 + b)^(5/2)*B*b^2*sgn(x) - 105*(c*x^2 + b)^(3/2)*B*b^3*sgn(x) + 45*(c*x^2 + b)^(7/2)*A*
c*sgn(x) - 126*(c*x^2 + b)^(5/2)*A*b*c*sgn(x) + 105*(c*x^2 + b)^(3/2)*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 (35 B \,c^{3} x^{6}+45 A \,c^{3} x^{4}-30 B b \,c^{2} x^{4}-36 A b \,c^{2} x^{2}+24 B \,b^{2} c \,x^{2}+24 A \,b^{2} c -16 B \,b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 c^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x)

[Out]

1/315*(c*x^2+b)*(35*B*c^3*x^6+45*A*c^3*x^4-30*B*b*c^2*x^4-36*A*b*c^2*x^2+24*B*b^2*c*x^2+24*A*b^2*c-16*B*b^3)*(
c*x^4+b*x^2)^(1/2)/c^4/x

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maxima [A]  time = 1.42, size = 106, normalized size = 0.81 \begin {gather*} \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} A}{105 \, c^{3}} + \frac {{\left (35 \, c^{4} x^{8} + 5 \, b c^{3} x^{6} - 6 \, b^{2} c^{2} x^{4} + 8 \, b^{3} c x^{2} - 16 \, b^{4}\right )} \sqrt {c x^{2} + b} B}{315 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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)*A/c^3 + 1/315*(35*c^4*x^8 + 5*b*c^3*x^6
- 6*b^2*c^2*x^4 + 8*b^3*c*x^2 - 16*b^4)*sqrt(c*x^2 + b)*B/c^4

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mupad [B]  time = 0.25, size = 103, normalized size = 0.79 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {B\,x^8}{9}-\frac {16\,B\,b^4-24\,A\,b^3\,c}{315\,c^4}+\frac {x^6\,\left (45\,A\,c^4+5\,B\,b\,c^3\right )}{315\,c^4}-\frac {4\,b^2\,x^2\,\left (3\,A\,c-2\,B\,b\right )}{315\,c^3}+\frac {b\,x^4\,\left (3\,A\,c-2\,B\,b\right )}{105\,c^2}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)

[Out]

((b*x^2 + c*x^4)^(1/2)*((B*x^8)/9 - (16*B*b^4 - 24*A*b^3*c)/(315*c^4) + (x^6*(45*A*c^4 + 5*B*b*c^3))/(315*c^4)
- (4*b^2*x^2*(3*A*c - 2*B*b))/(315*c^3) + (b*x^4*(3*A*c - 2*B*b))/(105*c^2)))/x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \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**4*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**4*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)

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