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

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

[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]]

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Rubi [A]  time = 0.251363, antiderivative size = 104, normalized size of antiderivative = 1., 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} \[ -\frac{A \left (b x^2+c x^4\right )^{5/2}}{5 b x^{10}}+B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )-\frac{B \left (b x^2+c x^4\right )^{3/2}}{3 x^6}-\frac{B c \sqrt{b x^2+c x^4}}{x^2} \]

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 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]

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 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 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 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^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*}

Mathematica [C]  time = 0.0469147, size = 94, normalized size = 0.9 \[ -\frac{\sqrt{x^2 \left (b+c x^2\right )} \left (3 A \left (b+c x^2\right )^2 \sqrt{\frac{c x^2}{b}+1}+5 b^2 B x^2 \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x^2}{b}\right )\right )}{15 b x^6 \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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)]))/(15*b*x^6*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.011, size = 153, normalized size = 1.5 \begin{align*} -{\frac{1}{15\,{b}^{2}{x}^{8}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( -10\,B{c}^{5/2} \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{6}+10\,B{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{4}-15\,B{c}^{5/2}\sqrt{c{x}^{2}+b}{x}^{6}b-15\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){x}^{5}{b}^{2}{c}^{2}+5\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}b+3\,A\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}b \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{c}}}} \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^9,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27494, size = 471, normalized size = 4.53 \begin{align*} \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{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^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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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^{9}}\, 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**9,x)

[Out]

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

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Giac [B]  time = 1.89374, size = 343, normalized size = 3.3 \begin{align*} -\frac{1}{2} \, B c^{\frac{3}{2}} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \,{\left (30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} B b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} A c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) - 90 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} B b^{2} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 110 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{3} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b^{2} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) - 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{4} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 20 \, B b^{5} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 3 \, A b^{4} c^{\frac{5}{2}} \mathrm{sgn}\left (x\right )\right )}}{15 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{5}} \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^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