3.112 \(\int \frac{(A+B x^2) (b x^2+c x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=136 \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (2 A c+3 b B)}{3 b x^4}+\frac{c \sqrt{b x^2+c x^4} (2 A c+3 b B)}{2 b}+\frac{1}{2} \sqrt{c} (2 A c+3 b B) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8} \]

[Out]

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

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Rubi [A]  time = 0.273528, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2034, 792, 662, 664, 620, 206} \[ -\frac{\left (b x^2+c x^4\right )^{3/2} (2 A c+3 b B)}{3 b x^4}+\frac{c \sqrt{b x^2+c x^4} (2 A c+3 b B)}{2 b}+\frac{1}{2} \sqrt{c} (2 A c+3 b B) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^7,x]

[Out]

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

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 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}+\frac{\left (-4 (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{\left (b x+c x^2\right )^{3/2}}{x^3} \, dx,x,x^2\right )}{3 b}\\ &=-\frac{(3 b B+2 A c) \left (b x^2+c x^4\right )^{3/2}}{3 b x^4}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}+\frac{\left (c \left (-4 (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x} \, dx,x,x^2\right )}{b}\\ &=\frac{c (3 b B+2 A c) \sqrt{b x^2+c x^4}}{2 b}-\frac{(3 b B+2 A c) \left (b x^2+c x^4\right )^{3/2}}{3 b x^4}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}+\frac{1}{4} (c (3 b B+2 A c)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{c (3 b B+2 A c) \sqrt{b x^2+c x^4}}{2 b}-\frac{(3 b B+2 A c) \left (b x^2+c x^4\right )^{3/2}}{3 b x^4}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}+\frac{1}{2} (c (3 b B+2 A c)) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )\\ &=\frac{c (3 b B+2 A c) \sqrt{b x^2+c x^4}}{2 b}-\frac{(3 b B+2 A c) \left (b x^2+c x^4\right )^{3/2}}{3 b x^4}-\frac{A \left (b x^2+c x^4\right )^{5/2}}{3 b x^8}+\frac{1}{2} \sqrt{c} (3 b B+2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0505867, size = 98, normalized size = 0.72 \[ -\frac{\sqrt{x^2 \left (b+c x^2\right )} \left (b x^2 (2 A c+3 b B) \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{c x^2}{b}\right )+A \left (b+c x^2\right )^2 \sqrt{\frac{c x^2}{b}+1}\right )}{3 b x^4 \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^7,x]

[Out]

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

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Maple [A]  time = 0.01, size = 219, normalized size = 1.6 \begin{align*}{\frac{1}{6\,{b}^{2}{x}^{6}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 4\,A{c}^{5/2} \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{4}+6\,A{c}^{5/2}\sqrt{c{x}^{2}+b}{x}^{4}b+6\,B{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{3/2}{x}^{4}b-4\,A{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}+9\,B{c}^{3/2}\sqrt{c{x}^{2}+b}{x}^{4}{b}^{2}-6\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{2}b+6\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){x}^{3}{b}^{2}{c}^{2}+9\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){x}^{3}{b}^{3}c-2\,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^7,x)

[Out]

1/6*(c*x^4+b*x^2)^(3/2)*(4*A*c^(5/2)*(c*x^2+b)^(3/2)*x^4+6*A*c^(5/2)*(c*x^2+b)^(1/2)*x^4*b+6*B*c^(3/2)*(c*x^2+
b)^(3/2)*x^4*b-4*A*c^(3/2)*(c*x^2+b)^(5/2)*x^2+9*B*c^(3/2)*(c*x^2+b)^(1/2)*x^4*b^2-6*B*c^(1/2)*(c*x^2+b)^(5/2)
*x^2*b+6*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*x^3*b^2*c^2+9*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*x^3*b^3*c-2*A*c^(1/2)*(
c*x^2+b)^(5/2)*b)/x^6/(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^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27299, size = 435, normalized size = 3.2 \begin{align*} \left [\frac{3 \,{\left (3 \, B b + 2 \, A c\right )} \sqrt{c} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) + 2 \,{\left (3 \, B c x^{4} - 2 \,{\left (3 \, B b + 4 \, A c\right )} x^{2} - 2 \, A b\right )} \sqrt{c x^{4} + b x^{2}}}{12 \, x^{4}}, -\frac{3 \,{\left (3 \, B b + 2 \, A c\right )} \sqrt{-c} x^{4} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) -{\left (3 \, B c x^{4} - 2 \,{\left (3 \, B b + 4 \, A c\right )} x^{2} - 2 \, A b\right )} \sqrt{c x^{4} + b x^{2}}}{6 \, x^{4}}\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^7,x, algorithm="fricas")

[Out]

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

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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^{7}}\, 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**7,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**7, x)

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Giac [A]  time = 1.45597, size = 304, normalized size = 2.24 \begin{align*} \frac{1}{2} \, \sqrt{c x^{2} + b} B c x \mathrm{sgn}\left (x\right ) - \frac{1}{4} \,{\left (3 \, B b \sqrt{c} \mathrm{sgn}\left (x\right ) + 2 \, A c^{\frac{3}{2}} \mathrm{sgn}\left (x\right )\right )} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2}\right ) + \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} B b^{2} \sqrt{c} \mathrm{sgn}\left (x\right ) + 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} A b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} B b^{3} \sqrt{c} \mathrm{sgn}\left (x\right ) - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} A b^{2} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 3 \, B b^{4} \sqrt{c} \mathrm{sgn}\left (x\right ) + 4 \, A b^{3} c^{\frac{3}{2}} \mathrm{sgn}\left (x\right )\right )}}{3 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{3}} \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^7,x, algorithm="giac")

[Out]

1/2*sqrt(c*x^2 + b)*B*c*x*sgn(x) - 1/4*(3*B*b*sqrt(c)*sgn(x) + 2*A*c^(3/2)*sgn(x))*log((sqrt(c)*x - sqrt(c*x^2
+ b))^2) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^2*sqrt(c)*sgn(x) + 6*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*
b*c^(3/2)*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^3*sqrt(c)*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + b))^2*
A*b^2*c^(3/2)*sgn(x) + 3*B*b^4*sqrt(c)*sgn(x) + 4*A*b^3*c^(3/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^
3