3.2.13 \(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=104 \[ -\frac {A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}+B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {B c \sqrt {b x^2+c x^4}}{x^2}-\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 x^6} \]

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Rubi [A]  time = 0.25, antiderivative size = 104, 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 = {2034, 792, 662, 620, 206} \begin {gather*} -\frac {A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}+B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {B \left (b x^2+c x^4\right )^{3/2}}{3 x^6}-\frac {B c \sqrt {b x^2+c x^4}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^9,x]

[Out]

-((B*c*Sqrt[b*x^2 + c*x^4])/x^2) - (B*(b*x^2 + c*x^4)^(3/2))/(3*x^6) - (A*(b*x^2 + c*x^4)^(5/2))/(5*b*x^10) +
B*c^(3/2)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]]

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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 94, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (3 A \left (b+c x^2\right )^2 \sqrt {\frac {c x^2}{b}+1}+5 b^2 B x^2 \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x^2}{b}\right )\right )}{15 b x^6 \sqrt {\frac {c x^2}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^9,x]

[Out]

-1/15*(Sqrt[x^2*(b + c*x^2)]*(3*A*(b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] + 5*b^2*B*x^2*Hypergeometric2F1[-3/2, -3/2
, -1/2, -((c*x^2)/b)]))/(b*x^6*Sqrt[1 + (c*x^2)/b])

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IntegrateAlgebraic [A]  time = 0.45, size = 108, normalized size = 1.04 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-3 A b^2-6 A b c x^2-3 A c^2 x^4-5 b^2 B x^2-20 b B c x^4\right )}{15 b x^6}-\frac {1}{2} B c^{3/2} \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^9,x]

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*A*b^2 - 5*b^2*B*x^2 - 6*A*b*c*x^2 - 20*b*B*c*x^4 - 3*A*c^2*x^4))/(15*b*x^6) - (B*c^(3
/2)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/2

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fricas [A]  time = 0.45, size = 207, normalized size = 1.99 \begin {gather*} \left [\frac {15 \, B b c^{\frac {3}{2}} x^{6} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left ({\left (20 \, B b c + 3 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30 \, b x^{6}}, -\frac {15 \, B b \sqrt {-c} c x^{6} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left ({\left (20 \, B b c + 3 \, A c^{2}\right )} x^{4} + 3 \, A b^{2} + {\left (5 \, B b^{2} + 6 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, b x^{6}}\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^9,x, algorithm="fricas")

[Out]

[1/30*(15*B*b*c^(3/2)*x^6*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*((20*B*b*c + 3*A*c^2)*x^4 + 3*
A*b^2 + (5*B*b^2 + 6*A*b*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b*x^6), -1/15*(15*B*b*sqrt(-c)*c*x^6*arctan(sqrt(c*x^4
+ b*x^2)*sqrt(-c)/(c*x^2 + b)) + ((20*B*b*c + 3*A*c^2)*x^4 + 3*A*b^2 + (5*B*b^2 + 6*A*b*c)*x^2)*sqrt(c*x^4 + b
*x^2))/(b*x^6)]

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giac [B]  time = 1.02, size = 254, normalized size = 2.44 \begin {gather*} -\frac {1}{2} \, B c^{\frac {3}{2}} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\relax (x) + \frac {2 \, {\left (30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 110 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {5}{2}} \mathrm {sgn}\relax (x) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 20 \, B b^{5} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 3 \, A b^{4} c^{\frac {5}{2}} \mathrm {sgn}\relax (x)\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5}} \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^9,x, algorithm="giac")

[Out]

-1/2*B*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 2/15*(30*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b*c^(3
/2)*sgn(x) + 15*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*c^(5/2)*sgn(x) - 90*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^2*c^
(3/2)*sgn(x) + 110*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^3*c^(3/2)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A
*b^2*c^(5/2)*sgn(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^4*c^(3/2)*sgn(x) + 20*B*b^5*c^(3/2)*sgn(x) + 3*A*
b^4*c^(5/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5

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maple [A]  time = 0.06, size = 153, normalized size = 1.47 \begin {gather*} -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-15 B \,b^{2} c^{2} x^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 \sqrt {c \,x^{2}+b}\, B b \,c^{\frac {5}{2}} x^{6}-10 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,c^{\frac {5}{2}} x^{6}+10 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,c^{\frac {3}{2}} x^{4}+5 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \sqrt {c}\, x^{2}+3 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \sqrt {c}\right )}{15 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}\, x^{8}} \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^9,x)

[Out]

-1/15*(c*x^4+b*x^2)^(3/2)*(-10*B*(c*x^2+b)^(3/2)*c^(5/2)*x^6+10*B*(c*x^2+b)^(5/2)*c^(3/2)*x^4-15*B*(c*x^2+b)^(
1/2)*c^(5/2)*x^6*b-15*B*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*x^5*b^2*c^2+5*(c*x^2+b)^(5/2)*B*b*c^(1/2)*x^2+3*(c*x^2+b
)^(5/2)*A*b*c^(1/2))/x^8/(c*x^2+b)^(3/2)/b^2/c^(1/2)

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maxima [B]  time = 1.38, size = 177, normalized size = 1.70 \begin {gather*} \frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{4} + b x^{2}} c}{x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} b}{x^{4}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{6}}\right )} B - \frac {1}{10} \, A {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{x^{6}} + \frac {5 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{8}}\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^9,x, algorithm="maxima")

[Out]

1/6*(3*c^(3/2)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 7*sqrt(c*x^4 + b*x^2)*c/x^2 - sqrt(c*x^4 + b
*x^2)*b/x^4 - (c*x^4 + b*x^2)^(3/2)/x^6)*B - 1/10*A*(2*sqrt(c*x^4 + b*x^2)*c^2/(b*x^2) - sqrt(c*x^4 + b*x^2)*c
/x^4 - 3*sqrt(c*x^4 + b*x^2)*b/x^6 + 5*(c*x^4 + b*x^2)^(3/2)/x^8)

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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^9} \,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^9,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^9, 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^{9}}\, 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**9,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**9, x)

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