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

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

[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])

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Rubi [A]  time = 0.21611, antiderivative size = 135, normalized size of antiderivative = 1., 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} \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (A c+4 b B)}{8 b x^5}+\frac{3 c \sqrt{b x^2+c x^4} (A c+4 b B)}{8 b x}-\frac{3 c (A c+4 b B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 \sqrt{b}}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9} \]

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 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]

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 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 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])

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*}

Mathematica [C]  time = 0.0350848, size = 63, normalized size = 0.47 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (c x^4 (A c+4 b B) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{c x^2}{b}+1\right )-5 A b^2\right )}{20 b^3 x^9} \]

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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Maple [A]  time = 0.01, size = 213, normalized size = 1.6 \begin{align*} -{\frac{1}{8\,{b}^{2}{x}^{7}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 3\,A{b}^{3/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{4}{c}^{2}-A \left ( c{x}^{2}+b \right ) ^{{\frac{3}{2}}}{x}^{4}{c}^{2}+12\,B{b}^{5/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{4}c-4\,B \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{4}bc+A \left ( c{x}^{2}+b \right ) ^{{\frac{5}{2}}}{x}^{2}c-3\,A\sqrt{c{x}^{2}+b}{x}^{4}b{c}^{2}+4\,B \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}b-12\,B\sqrt{c{x}^{2}+b}{x}^{4}{b}^{2}c+2\,A \left ( c{x}^{2}+b \right ) ^{5/2}b \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

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)*(3*A*b^(3/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*x^4*c^2-A*(c*x^2+b)^(3/2)*x^4*c^2+12
*B*b^(5/2)*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*x^4*c-4*B*(c*x^2+b)^(3/2)*x^4*b*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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}{\left (B x^{2} + A\right )}}{x^{8}}\,{d x} \end{align*}

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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Fricas [A]  time = 1.14898, size = 479, normalized size = 3.55 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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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Giac [A]  time = 1.30842, size = 196, normalized size = 1.45 \begin{align*} \frac{8 \, \sqrt{c x^{2} + b} B c^{2} \mathrm{sgn}\left (x\right ) + \frac{3 \,{\left (4 \, B b c^{2} \mathrm{sgn}\left (x\right ) + A c^{3} \mathrm{sgn}\left (x\right )\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}\left (x\right ) - 4 \, \sqrt{c x^{2} + b} B b^{2} c^{2} \mathrm{sgn}\left (x\right ) + 5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} A c^{3} \mathrm{sgn}\left (x\right ) - 3 \, \sqrt{c x^{2} + b} A b c^{3} \mathrm{sgn}\left (x\right )}{c^{2} x^{4}}}{8 \, c} \end{align*}

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