3.2.24 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=133 \[ \frac {\sqrt {b x^2+c x^4} (3 A c+2 b B)}{2 x}-\frac {1}{2} \sqrt {b} (3 A c+2 b B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )+\frac {\left (b x^2+c x^4\right )^{3/2} (3 A c+2 b B)}{6 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7} \]

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Rubi [A]  time = 0.22, antiderivative size = 133, 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, 2021, 2008, 206} \begin {gather*} \frac {\left (b x^2+c x^4\right )^{3/2} (3 A c+2 b B)}{6 b x^3}+\frac {\sqrt {b x^2+c x^4} (3 A c+2 b B)}{2 x}-\frac {1}{2} \sqrt {b} (3 A c+2 b B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6,x]

[Out]

((2*b*B + 3*A*c)*Sqrt[b*x^2 + c*x^4])/(2*x) + ((2*b*B + 3*A*c)*(b*x^2 + c*x^4)^(3/2))/(6*b*x^3) - (A*(b*x^2 +
c*x^4)^(5/2))/(2*b*x^7) - (Sqrt[b]*(2*b*B + 3*A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/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 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^6} \, dx &=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}-\frac {(-2 b B-3 A c) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx}{2 b}\\ &=\frac {(2 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}-\frac {1}{2} (-2 b B-3 A c) \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx\\ &=\frac {(2 b B+3 A c) \sqrt {b x^2+c x^4}}{2 x}+\frac {(2 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}+\frac {1}{2} (b (2 b B+3 A c)) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {(2 b B+3 A c) \sqrt {b x^2+c x^4}}{2 x}+\frac {(2 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}-\frac {1}{2} (b (2 b B+3 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 {(2 b B+3 A c) \sqrt {b x^2+c x^4}}{2 x}+\frac {(2 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}-\frac {1}{2} \sqrt {b} (2 b B+3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 109, normalized size = 0.82 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b+c x^2} \left (-3 A b+6 A c x^2+8 b B x^2+2 B c x^4\right )-3 \sqrt {b} x^2 (3 A c+2 b B) \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{6 x^3 \sqrt {b+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.80, size = 94, normalized size = 0.71 \begin {gather*} \frac {1}{2} \left (-3 A \sqrt {b} c-2 b^{3/2} B\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )+\frac {\sqrt {b x^2+c x^4} \left (-3 A b+6 A c x^2+8 b B x^2+2 B c x^4\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6,x]

[Out]

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

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fricas [A]  time = 0.44, size = 195, normalized size = 1.47 \begin {gather*} \left [\frac {3 \, {\left (2 \, B b + 3 \, A c\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (2 \, B c x^{4} + 2 \, {\left (4 \, B b + 3 \, A c\right )} x^{2} - 3 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{12 \, x^{3}}, \frac {3 \, {\left (2 \, B b + 3 \, A c\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (2 \, B c x^{4} + 2 \, {\left (4 \, B b + 3 \, A c\right )} x^{2} - 3 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{6 \, x^{3}}\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^6,x, algorithm="fricas")

[Out]

[1/12*(3*(2*B*b + 3*A*c)*sqrt(b)*x^3*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(2*B*c*x^4
+ 2*(4*B*b + 3*A*c)*x^2 - 3*A*b)*sqrt(c*x^4 + b*x^2))/x^3, 1/6*(3*(2*B*b + 3*A*c)*sqrt(-b)*x^3*arctan(sqrt(c*x
^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (2*B*c*x^4 + 2*(4*B*b + 3*A*c)*x^2 - 3*A*b)*sqrt(c*x^4 + b*x^2))/x^3]

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giac [A]  time = 0.28, size = 115, normalized size = 0.86 \begin {gather*} \frac {2 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B c \mathrm {sgn}\relax (x) + 6 \, \sqrt {c x^{2} + b} B b c \mathrm {sgn}\relax (x) + 6 \, \sqrt {c x^{2} + b} A c^{2} \mathrm {sgn}\relax (x) - \frac {3 \, \sqrt {c x^{2} + b} A b c \mathrm {sgn}\relax (x)}{x^{2}} + \frac {3 \, {\left (2 \, B b^{2} c \mathrm {sgn}\relax (x) + 3 \, A b c^{2} \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}}}{6 \, 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^6,x, algorithm="giac")

[Out]

1/6*(2*(c*x^2 + b)^(3/2)*B*c*sgn(x) + 6*sqrt(c*x^2 + b)*B*b*c*sgn(x) + 6*sqrt(c*x^2 + b)*A*c^2*sgn(x) - 3*sqrt
(c*x^2 + b)*A*b*c*sgn(x)/x^2 + 3*(2*B*b^2*c*sgn(x) + 3*A*b*c^2*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b))/sqrt(-
b))/c

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maple [A]  time = 0.06, size = 172, normalized size = 1.29 \begin {gather*} -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (9 A \,b^{\frac {3}{2}} c \,x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+6 B \,b^{\frac {5}{2}} x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-9 \sqrt {c \,x^{2}+b}\, A b c \,x^{2}-6 \sqrt {c \,x^{2}+b}\, B \,b^{2} x^{2}-3 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A c \,x^{2}-2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,x^{2}+3 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \right )}{6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{5}} \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^6,x)

[Out]

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

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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^{6}}\,{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^6,x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^6, 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^6} \,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^6,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6, 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^{6}}\, 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**6,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**6, x)

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