3.2.54 \(\int \frac {x^6 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [154]

Optimal. Leaf size=104 \[ -\frac {(b B-A c) x^5}{b c \sqrt {b x^2+c x^4}}-\frac {2 (4 b B-3 A c) \sqrt {b x^2+c x^4}}{3 c^3 x}+\frac {(4 b B-3 A c) x \sqrt {b x^2+c x^4}}{3 b c^2} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 104, 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 = {2062, 2041, 1602} \begin {gather*} -\frac {2 \sqrt {b x^2+c x^4} (4 b B-3 A c)}{3 c^3 x}+\frac {x \sqrt {b x^2+c x^4} (4 b B-3 A c)}{3 b c^2}-\frac {x^5 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 2041

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 2062

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 60, normalized size = 0.58 \begin {gather*} \frac {x \left (-8 b^2 B+c^2 x^2 \left (3 A+B x^2\right )+b \left (6 A c-4 B c x^2\right )\right )}{3 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(-8*b^2*B + c^2*x^2*(3*A + B*x^2) + b*(6*A*c - 4*B*c*x^2)))/(3*c^3*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.38, size = 66, normalized size = 0.63

method	result	size
gosper	\(\frac {\left (c \,x^{2}+b \right ) \left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 b B \,x^{2} c +6 A b c -8 b^{2} B \right ) x^{3}}{3 c^{3} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\)	\(66\)
default	\(\frac {\left (c \,x^{2}+b \right ) \left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 b B \,x^{2} c +6 A b c -8 b^{2} B \right ) x^{3}}{3 c^{3} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\)	\(66\)
trager	\(\frac {\left (B \,c^{2} x^{4}+3 A \,c^{2} x^{2}-4 b B \,x^{2} c +6 A b c -8 b^{2} B \right ) \sqrt {x^{4} c +b \,x^{2}}}{3 \left (c \,x^{2}+b \right ) c^{3} x}\)	\(68\)
risch	\(\frac {\left (B c \,x^{2}+3 A c -5 B b \right ) \left (c \,x^{2}+b \right ) x}{3 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b \left (A c -B b \right ) x}{c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 59, normalized size = 0.57 \begin {gather*} \frac {{\left (c x^{2} + 2 \, b\right )} A}{\sqrt {c x^{2} + b} c^{2}} + \frac {{\left (c^{2} x^{4} - 4 \, b c x^{2} - 8 \, b^{2}\right )} B}{3 \, \sqrt {c x^{2} + b} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.48, size = 68, normalized size = 0.65 \begin {gather*} \frac {{\left (B c^{2} x^{4} - 8 \, B b^{2} + 6 \, A b c - {\left (4 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3 \, {\left (c^{4} x^{3} + b c^{3} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**6*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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Giac [A]
time = 1.39, size = 107, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (4 \, B b^{2} - 3 \, A b c\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {b} c^{3}} - \frac {B b^{2} - A b c}{\sqrt {c x^{2} + b} c^{3} \mathrm {sgn}\left (x\right )} + \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}} B c^{6} - 6 \, \sqrt {c x^{2} + b} B b c^{6} + 3 \, \sqrt {c x^{2} + b} A c^{7}}{3 \, c^{9} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

2/3*(4*B*b^2 - 3*A*b*c)*sgn(x)/(sqrt(b)*c^3) - (B*b^2 - A*b*c)/(sqrt(c*x^2 + b)*c^3*sgn(x)) + 1/3*((c*x^2 + b)
^(3/2)*B*c^6 - 6*sqrt(c*x^2 + b)*B*b*c^6 + 3*sqrt(c*x^2 + b)*A*c^7)/(c^9*sgn(x))

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Mupad [B]
time = 0.28, size = 67, normalized size = 0.64 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (-8\,B\,b^2-4\,B\,b\,c\,x^2+6\,A\,b\,c+B\,c^2\,x^4+3\,A\,c^2\,x^2\right )}{3\,c^3\,x\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

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

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