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

Optimal. Leaf size=61 \[ \frac {B \left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {\left (b x^2+c x^4\right )^{3/2} (2 b B-5 A c)}{15 c^2 x^3} \]

[Out]

-1/15*(-5*A*c+2*B*b)*(c*x^4+b*x^2)^(3/2)/c^2/x^3+1/5*B*(c*x^4+b*x^2)^(3/2)/c/x

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Rubi [A]  time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1145, 2000} \[ \frac {B \left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {\left (b x^2+c x^4\right )^{3/2} (2 b B-5 A c)}{15 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*b*B - 5*A*c)*(b*x^2 + c*x^4)^(3/2))/(15*c^2*x^3) + (B*(b*x^2 + c*x^4)^(3/2))/(5*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 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]

Rubi steps

\begin {align*} \int \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {B \left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {(2 b B-5 A c) \int \sqrt {b x^2+c x^4} \, dx}{5 c}\\ &=-\frac {(2 b B-5 A c) \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{5 c x}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.67 \[ \frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (5 A c-2 b B+3 B c x^2\right )}{15 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + c*x^2))^(3/2)*(-2*b*B + 5*A*c + 3*B*c*x^2))/(15*c^2*x^3)

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fricas [A]  time = 0.85, size = 57, normalized size = 0.93 \[ \frac {{\left (3 \, B c^{2} x^{4} - 2 \, B b^{2} + 5 \, A b c + {\left (B b c + 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*B*c^2*x^4 - 2*B*b^2 + 5*A*b*c + (B*b*c + 5*A*c^2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^2*x)

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giac [A]  time = 0.16, size = 72, normalized size = 1.18 \[ \frac {{\left (2 \, B b^{\frac {5}{2}} - 5 \, A b^{\frac {3}{2}} c\right )} \mathrm {sgn}\relax (x)}{15 \, c^{2}} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B \mathrm {sgn}\relax (x) - 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b \mathrm {sgn}\relax (x) + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c \mathrm {sgn}\relax (x)}{15 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(2*B*b^(5/2) - 5*A*b^(3/2)*c)*sgn(x)/c^2 + 1/15*(3*(c*x^2 + b)^(5/2)*B*sgn(x) - 5*(c*x^2 + b)^(3/2)*B*b*s
gn(x) + 5*(c*x^2 + b)^(3/2)*A*c*sgn(x))/c^2

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maple [A]  time = 0.05, size = 45, normalized size = 0.74 \[ \frac {\left (c \,x^{2}+b \right ) \left (3 B c \,x^{2}+5 A c -2 b B \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(c*x^2+b)*(3*B*c*x^2+5*A*c-2*B*b)*(c*x^4+b*x^2)^(1/2)/c^2/x

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maxima [A]  time = 1.46, size = 51, normalized size = 0.84 \[ \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}} A}{3 \, c} + \frac {{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt {c x^{2} + b} B}{15 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.15, size = 60, normalized size = 0.98 \[ \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {B\,x^4}{5}-\frac {2\,B\,b^2-5\,A\,b\,c}{15\,c^2}+\frac {x^2\,\left (5\,A\,c^2+B\,b\,c\right )}{15\,c^2}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x^2 + c*x^4)^(1/2)*((B*x^4)/5 - (2*B*b^2 - 5*A*b*c)/(15*c^2) + (x^2*(5*A*c^2 + B*b*c))/(15*c^2)))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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