3.2.25 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=135 \[ \frac {3 c \sqrt {b x^2+c x^4} (A c+4 b B)}{8 b x}-\frac {3 c (A c+4 b B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}}-\frac {\left (b x^2+c x^4\right )^{3/2} (A c+4 b B)}{8 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9} \]

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Rubi [A]  time = 0.22, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2038, 2020, 2021, 2008, 206} \begin {gather*} -\frac {\left (b x^2+c x^4\right )^{3/2} (A c+4 b B)}{8 b x^5}+\frac {3 c \sqrt {b x^2+c x^4} (A c+4 b B)}{8 b x}-\frac {3 c (A c+4 b B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8,x]

[Out]

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

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 2021

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 + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*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^8} \, dx &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}-\frac {(-4 b B-A c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx}{4 b}\\ &=-\frac {(4 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{8 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}+\frac {(3 c (4 b B+A c)) \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx}{8 b}\\ &=\frac {3 c (4 b B+A c) \sqrt {b x^2+c x^4}}{8 b x}-\frac {(4 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{8 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}+\frac {1}{8} (3 c (4 b B+A c)) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {3 c (4 b B+A c) \sqrt {b x^2+c x^4}}{8 b x}-\frac {(4 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{8 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}-\frac {1}{8} (3 c (4 b B+A c)) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )\\ &=\frac {3 c (4 b B+A c) \sqrt {b x^2+c x^4}}{8 b x}-\frac {(4 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{8 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}-\frac {3 c (4 b B+A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 63, normalized size = 0.47 \begin {gather*} \frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (c x^4 (A c+4 b B) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {c x^2}{b}+1\right )-5 A b^2\right )}{20 b^3 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8,x]

[Out]

((x^2*(b + c*x^2))^(5/2)*(-5*A*b^2 + c*(4*b*B + A*c)*x^4*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/b]))/(20*b
^3*x^9)

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IntegrateAlgebraic [A]  time = 0.88, size = 92, normalized size = 0.68 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-2 A b-5 A c x^2-4 b B x^2+8 B c x^4\right )}{8 x^5}-\frac {3 \left (A c^2+4 b B c\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-2*A*b - 4*b*B*x^2 - 5*A*c*x^2 + 8*B*c*x^4))/(8*x^5) - (3*(4*b*B*c + A*c^2)*ArcTanh[(Sqr
t[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*Sqrt[b])

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fricas [A]  time = 0.45, size = 217, normalized size = 1.61 \begin {gather*} \left [\frac {3 \, {\left (4 \, B b c + A c^{2}\right )} \sqrt {b} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (8 \, B b c x^{4} - 2 \, A b^{2} - {\left (4 \, B b^{2} + 5 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, b x^{5}}, \frac {3 \, {\left (4 \, B b c + A c^{2}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (8 \, B b c x^{4} - 2 \, A b^{2} - {\left (4 \, B b^{2} + 5 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, b x^{5}}\right ] \end {gather*}

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^8,x, algorithm="fricas")

[Out]

[1/16*(3*(4*B*b*c + A*c^2)*sqrt(b)*x^5*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(8*B*b*c*
x^4 - 2*A*b^2 - (4*B*b^2 + 5*A*b*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b*x^5), 1/8*(3*(4*B*b*c + A*c^2)*sqrt(-b)*x^5*a
rctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (8*B*b*c*x^4 - 2*A*b^2 - (4*B*b^2 + 5*A*b*c)*x^2)*sqrt(c*x
^4 + b*x^2))/(b*x^5)]

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giac [A]  time = 0.26, size = 145, normalized size = 1.07 \begin {gather*} \frac {8 \, \sqrt {c x^{2} + b} B c^{2} \mathrm {sgn}\relax (x) + \frac {3 \, {\left (4 \, B b c^{2} \mathrm {sgn}\relax (x) + A c^{3} \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {4 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b c^{2} \mathrm {sgn}\relax (x) - 4 \, \sqrt {c x^{2} + b} B b^{2} c^{2} \mathrm {sgn}\relax (x) + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{3} \mathrm {sgn}\relax (x) - 3 \, \sqrt {c x^{2} + b} A b c^{3} \mathrm {sgn}\relax (x)}{c^{2} x^{4}}}{8 \, c} \end {gather*}

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^8,x, algorithm="giac")

[Out]

1/8*(8*sqrt(c*x^2 + b)*B*c^2*sgn(x) + 3*(4*B*b*c^2*sgn(x) + A*c^3*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/sqr
t(-b) - (4*(c*x^2 + b)^(3/2)*B*b*c^2*sgn(x) - 4*sqrt(c*x^2 + b)*B*b^2*c^2*sgn(x) + 5*(c*x^2 + b)^(3/2)*A*c^3*s
gn(x) - 3*sqrt(c*x^2 + b)*A*b*c^3*sgn(x))/(c^2*x^4))/c

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maple [A]  time = 0.06, size = 213, normalized size = 1.58 \begin {gather*} -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 A \,b^{\frac {3}{2}} c^{2} x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+12 B \,b^{\frac {5}{2}} c \,x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {c \,x^{2}+b}\, A b \,c^{2} x^{4}-12 \sqrt {c \,x^{2}+b}\, B \,b^{2} c \,x^{4}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{2} x^{4}-4 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b c \,x^{4}+\left (c \,x^{2}+b \right )^{\frac {5}{2}} A c \,x^{2}+4 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \,x^{2}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \right )}{8 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} x^{7}} \end {gather*}

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^8,x)

[Out]

-1/8*(c*x^4+b*x^2)^(3/2)*(-A*(c*x^2+b)^(3/2)*x^4*c^2+3*A*b^(3/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^4*c^2-4
*B*(c*x^2+b)^(3/2)*x^4*b*c+12*B*b^(5/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^4*c+A*(c*x^2+b)^(5/2)*x^2*c-3*A*
(c*x^2+b)^(1/2)*x^4*b*c^2+4*B*(c*x^2+b)^(5/2)*x^2*b-12*B*(c*x^2+b)^(1/2)*x^4*b^2*c+2*A*(c*x^2+b)^(5/2)*b)/x^7/
(c*x^2+b)^(3/2)/b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{8}}\,{d x} \end {gather*}

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^8,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^8, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^8} \,d x \end {gather*}

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^8,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{8}}\, dx \end {gather*}

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**8,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**8, x)

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