3.2.39 \(\int \frac {x^6 (A+B x^2)}{\sqrt {b x^2+c x^4}} \, dx\) [139]

Optimal. Leaf size=131 \[ -\frac {8 b^2 (6 b B-7 A c) \sqrt {b x^2+c x^4}}{105 c^4 x}+\frac {4 b (6 b B-7 A c) x \sqrt {b x^2+c x^4}}{105 c^3}-\frac {(6 b B-7 A c) x^3 \sqrt {b x^2+c x^4}}{35 c^2}+\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c} \]

[Out]

-8/105*b^2*(-7*A*c+6*B*b)*(c*x^4+b*x^2)^(1/2)/c^4/x+4/105*b*(-7*A*c+6*B*b)*x*(c*x^4+b*x^2)^(1/2)/c^3-1/35*(-7*
A*c+6*B*b)*x^3*(c*x^4+b*x^2)^(1/2)/c^2+1/7*B*x^5*(c*x^4+b*x^2)^(1/2)/c

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Rubi [A]
time = 0.15, 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 = {2064, 2041, 1602} \begin {gather*} -\frac {8 b^2 \sqrt {b x^2+c x^4} (6 b B-7 A c)}{105 c^4 x}+\frac {4 b x \sqrt {b x^2+c x^4} (6 b B-7 A c)}{105 c^3}-\frac {x^3 \sqrt {b x^2+c x^4} (6 b B-7 A c)}{35 c^2}+\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b^2*(6*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(105*c^4*x) + (4*b*(6*b*B - 7*A*c)*x*Sqrt[b*x^2 + c*x^4])/(105*c^
3) - ((6*b*B - 7*A*c)*x^3*Sqrt[b*x^2 + c*x^4])/(35*c^2) + (B*x^5*Sqrt[b*x^2 + c*x^4])/(7*c)

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 2064

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 \frac {x^6 \left (A+B x^2\right )}{\sqrt {b x^2+c x^4}} \, dx &=\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}-\frac {(6 b B-7 A c) \int \frac {x^6}{\sqrt {b x^2+c x^4}} \, dx}{7 c}\\ &=-\frac {(6 b B-7 A c) x^3 \sqrt {b x^2+c x^4}}{35 c^2}+\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}+\frac {(4 b (6 b B-7 A c)) \int \frac {x^4}{\sqrt {b x^2+c x^4}} \, dx}{35 c^2}\\ &=\frac {4 b (6 b B-7 A c) x \sqrt {b x^2+c x^4}}{105 c^3}-\frac {(6 b B-7 A c) x^3 \sqrt {b x^2+c x^4}}{35 c^2}+\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}-\frac {\left (8 b^2 (6 b B-7 A c)\right ) \int \frac {x^2}{\sqrt {b x^2+c x^4}} \, dx}{105 c^3}\\ &=-\frac {8 b^2 (6 b B-7 A c) \sqrt {b x^2+c x^4}}{105 c^4 x}+\frac {4 b (6 b B-7 A c) x \sqrt {b x^2+c x^4}}{105 c^3}-\frac {(6 b B-7 A c) x^3 \sqrt {b x^2+c x^4}}{35 c^2}+\frac {B x^5 \sqrt {b x^2+c x^4}}{7 c}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 85, normalized size = 0.65 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-48 b^3 B+8 b^2 c \left (7 A+3 B x^2\right )+3 c^3 x^4 \left (7 A+5 B x^2\right )-2 b c^2 x^2 \left (14 A+9 B x^2\right )\right )}{105 c^4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-48*b^3*B + 8*b^2*c*(7*A + 3*B*x^2) + 3*c^3*x^4*(7*A + 5*B*x^2) - 2*b*c^2*x^2*(14*A +
9*B*x^2)))/(105*c^4*x)

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Maple [A]
time = 0.37, size = 89, normalized size = 0.68

method	result	size
trager	\(\frac {\left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 B \,b^{2} c \,x^{2}+56 A \,b^{2} c -48 B \,b^{3}\right ) \sqrt {x^{4} c +b \,x^{2}}}{105 c^{4} x}\)	\(84\)
gosper	\(\frac {\left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 B \,b^{2} c \,x^{2}+56 A \,b^{2} c -48 B \,b^{3}\right ) x}{105 c^{4} \sqrt {x^{4} c +b \,x^{2}}}\)	\(89\)
default	\(\frac {\left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 B \,b^{2} c \,x^{2}+56 A \,b^{2} c -48 B \,b^{3}\right ) x}{105 c^{4} \sqrt {x^{4} c +b \,x^{2}}}\)	\(89\)
risch	\(\frac {x \left (c \,x^{2}+b \right ) \left (15 B \,c^{3} x^{6}+21 A \,c^{3} x^{4}-18 B b \,c^{2} x^{4}-28 A b \,c^{2} x^{2}+24 B \,b^{2} c \,x^{2}+56 A \,b^{2} c -48 B \,b^{3}\right )}{105 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{4}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/105*(c*x^2+b)*(15*B*c^3*x^6+21*A*c^3*x^4-18*B*b*c^2*x^4-28*A*b*c^2*x^2+24*B*b^2*c*x^2+56*A*b^2*c-48*B*b^3)*x
/c^4/(c*x^4+b*x^2)^(1/2)

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Maxima [A]
time = 0.28, size = 106, normalized size = 0.81 \begin {gather*} \frac {{\left (3 \, c^{3} x^{6} - b c^{2} x^{4} + 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} A}{15 \, \sqrt {c x^{2} + b} c^{3}} + \frac {{\left (5 \, c^{4} x^{8} - b c^{3} x^{6} + 2 \, b^{2} c^{2} x^{4} - 8 \, b^{3} c x^{2} - 16 \, b^{4}\right )} B}{35 \, \sqrt {c x^{2} + b} c^{4}} \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)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*c^3*x^6 - b*c^2*x^4 + 4*b^2*c*x^2 + 8*b^3)*A/(sqrt(c*x^2 + b)*c^3) + 1/35*(5*c^4*x^8 - b*c^3*x^6 + 2*b
^2*c^2*x^4 - 8*b^3*c*x^2 - 16*b^4)*B/(sqrt(c*x^2 + b)*c^4)

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Fricas [A]
time = 1.02, size = 83, normalized size = 0.63 \begin {gather*} \frac {{\left (15 \, B c^{3} x^{6} - 3 \, {\left (6 \, B b c^{2} - 7 \, A c^{3}\right )} x^{4} - 48 \, B b^{3} + 56 \, A b^{2} c + 4 \, {\left (6 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, c^{4} x} \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)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*B*c^3*x^6 - 3*(6*B*b*c^2 - 7*A*c^3)*x^4 - 48*B*b^3 + 56*A*b^2*c + 4*(6*B*b^2*c - 7*A*b*c^2)*x^2)*sqr
t(c*x^4 + b*x^2)/(c^4*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 )}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, 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)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.47, size = 130, normalized size = 0.99 \begin {gather*} \frac {8 \, {\left (6 \, B b^{\frac {7}{2}} - 7 \, A b^{\frac {5}{2}} c\right )} \mathrm {sgn}\left (x\right )}{105 \, c^{4}} - \frac {{\left (B b^{3} - A b^{2} c\right )} \sqrt {c x^{2} + b}}{c^{4} \mathrm {sgn}\left (x\right )} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B - 63 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b + 105 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{2} + 21 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c - 70 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b c}{105 \, c^{4} \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)^(1/2),x, algorithm="giac")

[Out]

8/105*(6*B*b^(7/2) - 7*A*b^(5/2)*c)*sgn(x)/c^4 - (B*b^3 - A*b^2*c)*sqrt(c*x^2 + b)/(c^4*sgn(x)) + 1/105*(15*(c
*x^2 + b)^(7/2)*B - 63*(c*x^2 + b)^(5/2)*B*b + 105*(c*x^2 + b)^(3/2)*B*b^2 + 21*(c*x^2 + b)^(5/2)*A*c - 70*(c*
x^2 + b)^(3/2)*A*b*c)/(c^4*sgn(x))

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Mupad [B]
time = 0.26, size = 87, normalized size = 0.66 \begin {gather*} -\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {48\,B\,b^3-56\,A\,b^2\,c}{105\,c^4}-\frac {B\,x^6}{7\,c}-\frac {x^4\,\left (21\,A\,c^3-18\,B\,b\,c^2\right )}{105\,c^4}+\frac {4\,b\,x^2\,\left (7\,A\,c-6\,B\,b\right )}{105\,c^3}\right )}{x} \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)^(1/2),x)

[Out]

-((b*x^2 + c*x^4)^(1/2)*((48*B*b^3 - 56*A*b^2*c)/(105*c^4) - (B*x^6)/(7*c) - (x^4*(21*A*c^3 - 18*B*b*c^2))/(10
5*c^4) + (4*b*x^2*(7*A*c - 6*B*b))/(105*c^3)))/x

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