3.126 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{10}} \, dx\)

Optimal. Leaf size=140 \[ -\frac {c^2 (6 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}}-\frac {\left (b x^2+c x^4\right )^{3/2} (6 b B-A c)}{24 b x^7}-\frac {c \sqrt {b x^2+c x^4} (6 b B-A c)}{16 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}} \]

[Out]

-1/24*(-A*c+6*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^7-1/6*A*(c*x^4+b*x^2)^(5/2)/b/x^11-1/16*c^2*(-A*c+6*B*b)*arctanh(x*
b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(3/2)-1/16*c*(-A*c+6*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^3

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Rubi [A]  time = 0.22, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2038, 2020, 2008, 206} \[ -\frac {c^2 (6 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}}-\frac {\left (b x^2+c x^4\right )^{3/2} (6 b B-A c)}{24 b x^7}-\frac {c \sqrt {b x^2+c x^4} (6 b B-A c)}{16 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10,x]

[Out]

-(c*(6*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(16*b*x^3) - ((6*b*B - A*c)*(b*x^2 + c*x^4)^(3/2))/(24*b*x^7) - (A*(b*x
^2 + c*x^4)^(5/2))/(6*b*x^11) - (c^2*(6*b*B - A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(16*b^(3/2))

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 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^{10}} \, dx &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}-\frac {(-6 b B+A c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx}{6 b}\\ &=-\frac {(6 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{24 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}+\frac {(c (6 b B-A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^4} \, dx}{8 b}\\ &=-\frac {c (6 b B-A c) \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {(6 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{24 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}+\frac {\left (c^2 (6 b B-A c)\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{16 b}\\ &=-\frac {c (6 b B-A c) \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {(6 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{24 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}-\frac {\left (c^2 (6 b B-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 )}{16 b}\\ &=-\frac {c (6 b B-A c) \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {(6 b B-A c) \left (b x^2+c x^4\right )^{3/2}}{24 b x^7}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{6 b x^{11}}-\frac {c^2 (6 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 121, normalized size = 0.86 \[ -\frac {\left (b+c x^2\right ) \left (A \left (8 b^2+14 b c x^2+3 c^2 x^4\right )+6 b B x^2 \left (2 b+5 c x^2\right )\right )+3 c^2 x^6 \sqrt {\frac {c x^2}{b}+1} (6 b B-A c) \tanh ^{-1}\left (\sqrt {\frac {c x^2}{b}+1}\right )}{48 b x^5 \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10,x]

[Out]

-1/48*((b + c*x^2)*(6*b*B*x^2*(2*b + 5*c*x^2) + A*(8*b^2 + 14*b*c*x^2 + 3*c^2*x^4)) + 3*c^2*(6*b*B - A*c)*x^6*
Sqrt[1 + (c*x^2)/b]*ArcTanh[Sqrt[1 + (c*x^2)/b]])/(b*x^5*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 0.94, size = 250, normalized size = 1.79 \[ \left [-\frac {3 \, {\left (6 \, B b c^{2} - A c^{3}\right )} \sqrt {b} x^{7} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, {\left (10 \, B b^{2} c + A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 2 \, {\left (6 \, B b^{3} + 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, b^{2} x^{7}}, \frac {3 \, {\left (6 \, B b c^{2} - A c^{3}\right )} \sqrt {-b} x^{7} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) - {\left (3 \, {\left (10 \, B b^{2} c + A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 2 \, {\left (6 \, B b^{3} + 7 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, b^{2} x^{7}}\right ] \]

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^10,x, algorithm="fricas")

[Out]

[-1/96*(3*(6*B*b*c^2 - A*c^3)*sqrt(b)*x^7*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*(10
*B*b^2*c + A*b*c^2)*x^4 + 8*A*b^3 + 2*(6*B*b^3 + 7*A*b^2*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^2*x^7), 1/48*(3*(6*B*
b*c^2 - A*c^3)*sqrt(-b)*x^7*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) - (3*(10*B*b^2*c + A*b*c^2)*x^4
+ 8*A*b^3 + 2*(6*B*b^3 + 7*A*b^2*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^2*x^7)]

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giac [A]  time = 0.27, size = 175, normalized size = 1.25 \[ \frac {\frac {3 \, {\left (6 \, B b c^{3} \mathrm {sgn}\relax (x) - A c^{4} \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {30 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b c^{3} \mathrm {sgn}\relax (x) - 48 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{2} c^{3} \mathrm {sgn}\relax (x) + 18 \, \sqrt {c x^{2} + b} B b^{3} c^{3} \mathrm {sgn}\relax (x) + 3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c^{4} \mathrm {sgn}\relax (x) + 8 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b c^{4} \mathrm {sgn}\relax (x) - 3 \, \sqrt {c x^{2} + b} A b^{2} c^{4} \mathrm {sgn}\relax (x)}{b c^{3} x^{6}}}{48 \, c} \]

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^10,x, algorithm="giac")

[Out]

1/48*(3*(6*B*b*c^3*sgn(x) - A*c^4*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b) - (30*(c*x^2 + b)^(5/2
)*B*b*c^3*sgn(x) - 48*(c*x^2 + b)^(3/2)*B*b^2*c^3*sgn(x) + 18*sqrt(c*x^2 + b)*B*b^3*c^3*sgn(x) + 3*(c*x^2 + b)
^(5/2)*A*c^4*sgn(x) + 8*(c*x^2 + b)^(3/2)*A*b*c^4*sgn(x) - 3*sqrt(c*x^2 + b)*A*b^2*c^4*sgn(x))/(b*c^3*x^6))/c

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maple [B]  time = 0.08, size = 259, normalized size = 1.85 \[ \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 A \,b^{\frac {3}{2}} c^{3} x^{6} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-18 B \,b^{\frac {5}{2}} c^{2} x^{6} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {c \,x^{2}+b}\, A b \,c^{3} x^{6}+18 \sqrt {c \,x^{2}+b}\, B \,b^{2} c^{2} x^{6}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{3} x^{6}+6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,c^{2} x^{6}+\left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{2} x^{4}-6 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b c \,x^{4}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b c \,x^{2}-12 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{2} x^{2}-8 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{2}\right )}{48 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3} x^{9}} \]

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^10,x)

[Out]

1/48*(c*x^4+b*x^2)^(3/2)*(-A*(c*x^2+b)^(3/2)*x^6*c^3+3*A*b^(3/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^6*c^3+6
*B*(c*x^2+b)^(3/2)*x^6*b*c^2-18*B*b^(5/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^6*c^2+A*(c*x^2+b)^(5/2)*x^4*c^
2-3*A*(c*x^2+b)^(1/2)*x^6*b*c^3-6*B*(c*x^2+b)^(5/2)*x^4*b*c+18*B*(c*x^2+b)^(1/2)*x^6*b^2*c^2+2*A*(c*x^2+b)^(5/
2)*x^2*b*c-12*B*(c*x^2+b)^(5/2)*x^2*b^2-8*A*(c*x^2+b)^(5/2)*b^2)/x^9/(c*x^2+b)^(3/2)/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{10}}\,{d x} \]

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^10,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^10, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{10}} \,d x \]

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^10,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^10, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{10}}\, dx \]

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**10,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**10, x)

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