3.127 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{12}} \, dx\)

Optimal. Leaf size=177 \[ \frac {c^3 (8 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{128 b^{5/2}}-\frac {c^2 \sqrt {b x^2+c x^4} (8 b B-3 A c)}{128 b^2 x^3}-\frac {\left (b x^2+c x^4\right )^{3/2} (8 b B-3 A c)}{48 b x^9}-\frac {c \sqrt {b x^2+c x^4} (8 b B-3 A c)}{64 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}} \]

[Out]

-1/48*(-3*A*c+8*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^9-1/8*A*(c*x^4+b*x^2)^(5/2)/b/x^13+1/128*c^3*(-3*A*c+8*B*b)*arcta
nh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(5/2)-1/64*c*(-3*A*c+8*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^5-1/128*c^2*(-3*A*c+8*
B*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^3

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Rubi [A]  time = 0.28, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2038, 2020, 2025, 2008, 206} \[ -\frac {c^2 \sqrt {b x^2+c x^4} (8 b B-3 A c)}{128 b^2 x^3}+\frac {c^3 (8 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{128 b^{5/2}}-\frac {c \sqrt {b x^2+c x^4} (8 b B-3 A c)}{64 b x^5}-\frac {\left (b x^2+c x^4\right )^{3/2} (8 b B-3 A c)}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12,x]

[Out]

-(c*(8*b*B - 3*A*c)*Sqrt[b*x^2 + c*x^4])/(64*b*x^5) - (c^2*(8*b*B - 3*A*c)*Sqrt[b*x^2 + c*x^4])/(128*b^2*x^3)
- ((8*b*B - 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(48*b*x^9) - (A*(b*x^2 + c*x^4)^(5/2))/(8*b*x^13) + (c^3*(8*b*B - 3*
A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(128*b^(5/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2025

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}-\frac {(-8 b B+3 A c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx}{8 b}\\ &=-\frac {(8 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}+\frac {(c (8 b B-3 A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^6} \, dx}{16 b}\\ &=-\frac {c (8 b B-3 A c) \sqrt {b x^2+c x^4}}{64 b x^5}-\frac {(8 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}+\frac {\left (c^2 (8 b B-3 A c)\right ) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{64 b}\\ &=-\frac {c (8 b B-3 A c) \sqrt {b x^2+c x^4}}{64 b x^5}-\frac {c^2 (8 b B-3 A c) \sqrt {b x^2+c x^4}}{128 b^2 x^3}-\frac {(8 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}-\frac {\left (c^3 (8 b B-3 A c)\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{128 b^2}\\ &=-\frac {c (8 b B-3 A c) \sqrt {b x^2+c x^4}}{64 b x^5}-\frac {c^2 (8 b B-3 A c) \sqrt {b x^2+c x^4}}{128 b^2 x^3}-\frac {(8 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}+\frac {\left (c^3 (8 b B-3 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{128 b^2}\\ &=-\frac {c (8 b B-3 A c) \sqrt {b x^2+c x^4}}{64 b x^5}-\frac {c^2 (8 b B-3 A c) \sqrt {b x^2+c x^4}}{128 b^2 x^3}-\frac {(8 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}+\frac {c^3 (8 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{128 b^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 66, normalized size = 0.37 \[ \frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (c^3 x^8 (8 b B-3 A c) \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {c x^2}{b}+1\right )-5 A b^4\right )}{40 b^5 x^{13}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12,x]

[Out]

((x^2*(b + c*x^2))^(5/2)*(-5*A*b^4 + c^3*(8*b*B - 3*A*c)*x^8*Hypergeometric2F1[5/2, 4, 7/2, 1 + (c*x^2)/b]))/(
40*b^5*x^13)

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fricas [A]  time = 1.00, size = 299, normalized size = 1.69 \[ \left [-\frac {3 \, {\left (8 \, B b c^{3} - 3 \, A c^{4}\right )} \sqrt {b} x^{9} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, {\left (8 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{6} + 48 \, A b^{4} + 2 \, {\left (56 \, B b^{3} c + 3 \, A b^{2} c^{2}\right )} x^{4} + 8 \, {\left (8 \, B b^{4} + 9 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, b^{3} x^{9}}, -\frac {3 \, {\left (8 \, B b c^{3} - 3 \, A c^{4}\right )} \sqrt {-b} x^{9} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (3 \, {\left (8 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{6} + 48 \, A b^{4} + 2 \, {\left (56 \, B b^{3} c + 3 \, A b^{2} c^{2}\right )} x^{4} + 8 \, {\left (8 \, B b^{4} + 9 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, b^{3} x^{9}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="fricas")

[Out]

[-1/768*(3*(8*B*b*c^3 - 3*A*c^4)*sqrt(b)*x^9*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*
(8*B*b^2*c^2 - 3*A*b*c^3)*x^6 + 48*A*b^4 + 2*(56*B*b^3*c + 3*A*b^2*c^2)*x^4 + 8*(8*B*b^4 + 9*A*b^3*c)*x^2)*sqr
t(c*x^4 + b*x^2))/(b^3*x^9), -1/384*(3*(8*B*b*c^3 - 3*A*c^4)*sqrt(-b)*x^9*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/
(c*x^3 + b*x)) + (3*(8*B*b^2*c^2 - 3*A*b*c^3)*x^6 + 48*A*b^4 + 2*(56*B*b^3*c + 3*A*b^2*c^2)*x^4 + 8*(8*B*b^4 +
9*A*b^3*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^3*x^9)]

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giac [A]  time = 0.32, size = 214, normalized size = 1.21 \[ -\frac {\frac {3 \, {\left (8 \, B b c^{4} \mathrm {sgn}\relax (x) - 3 \, A c^{5} \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {24 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b c^{4} \mathrm {sgn}\relax (x) + 40 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{2} c^{4} \mathrm {sgn}\relax (x) - 88 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{3} c^{4} \mathrm {sgn}\relax (x) + 24 \, \sqrt {c x^{2} + b} B b^{4} c^{4} \mathrm {sgn}\relax (x) - 9 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A c^{5} \mathrm {sgn}\relax (x) + 33 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b c^{5} \mathrm {sgn}\relax (x) + 33 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{2} c^{5} \mathrm {sgn}\relax (x) - 9 \, \sqrt {c x^{2} + b} A b^{3} c^{5} \mathrm {sgn}\relax (x)}{b^{2} c^{4} x^{8}}}{384 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="giac")

[Out]

-1/384*(3*(8*B*b*c^4*sgn(x) - 3*A*c^5*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2) + (24*(c*x^2 + b
)^(7/2)*B*b*c^4*sgn(x) + 40*(c*x^2 + b)^(5/2)*B*b^2*c^4*sgn(x) - 88*(c*x^2 + b)^(3/2)*B*b^3*c^4*sgn(x) + 24*sq
rt(c*x^2 + b)*B*b^4*c^4*sgn(x) - 9*(c*x^2 + b)^(7/2)*A*c^5*sgn(x) + 33*(c*x^2 + b)^(5/2)*A*b*c^5*sgn(x) + 33*(
c*x^2 + b)^(3/2)*A*b^2*c^5*sgn(x) - 9*sqrt(c*x^2 + b)*A*b^3*c^5*sgn(x))/(b^2*c^4*x^8))/c

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maple [A]  time = 0.08, size = 302, normalized size = 1.71 \[ -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (9 A \,b^{\frac {3}{2}} c^{4} x^{8} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-24 B \,b^{\frac {5}{2}} c^{3} x^{8} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-9 \sqrt {c \,x^{2}+b}\, A b \,c^{4} x^{8}+24 \sqrt {c \,x^{2}+b}\, B \,b^{2} c^{3} x^{8}-3 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{4} x^{8}+8 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,c^{3} x^{8}+3 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{3} x^{6}-8 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \,c^{2} x^{6}+6 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \,c^{2} x^{4}-16 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{2} c \,x^{4}-24 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{2} c \,x^{2}+64 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{3} x^{2}+48 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{3}\right )}{384 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{4} x^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x)

[Out]

-1/384*(c*x^4+b*x^2)^(3/2)*(-3*A*(c*x^2+b)^(3/2)*x^8*c^4+9*A*b^(3/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^8*c
^4+8*B*(c*x^2+b)^(3/2)*x^8*b*c^3-24*B*b^(5/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^8*c^3+3*A*(c*x^2+b)^(5/2)*
x^6*c^3-9*A*(c*x^2+b)^(1/2)*x^8*b*c^4-8*B*(c*x^2+b)^(5/2)*x^6*b*c^2+24*B*(c*x^2+b)^(1/2)*x^8*b^2*c^3+6*A*(c*x^
2+b)^(5/2)*x^4*b*c^2-16*B*(c*x^2+b)^(5/2)*x^4*b^2*c-24*A*(c*x^2+b)^(5/2)*x^2*b^2*c+64*B*(c*x^2+b)^(5/2)*x^2*b^
3+48*A*(c*x^2+b)^(5/2)*b^3)/x^11/(c*x^2+b)^(3/2)/b^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{12}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^12, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{12}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{12}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**12,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**12, x)

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