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]

-1/3*(2*A*c+3*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^4-1/3*A*(c*x^4+b*x^2)^(5/2)/b/x^8+1/2*(2*A*c+3*B*b)*arctanh(x^2*c^(
1/2)/(c*x^4+b*x^2)^(1/2))*c^(1/2)+1/2*c*(2*A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/b

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Rubi [A]  time = 0.27, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 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 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 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^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*}

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Mathematica [C]  time = 0.05, 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]

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

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fricas [A]  time = 1.10, size = 189, normalized size = 1.39 \[ \left [\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {c} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (3 \, B c x^{4} - 2 \, {\left (3 \, B b + 4 \, A c\right )} x^{2} - 2 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{12 \, x^{4}}, -\frac {3 \, {\left (3 \, B b + 2 \, A c\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (3 \, B c x^{4} - 2 \, {\left (3 \, B b + 4 \, A c\right )} x^{2} - 2 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{6 \, x^{4}}\right ] \]

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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giac [A]  time = 0.53, size = 225, normalized size = 1.65 \[ \frac {1}{2} \, \sqrt {c x^{2} + b} B c x \mathrm {sgn}\relax (x) - \frac {1}{4} \, {\left (3 \, B b \sqrt {c} \mathrm {sgn}\relax (x) + 2 \, A c^{\frac {3}{2}} \mathrm {sgn}\relax (x)\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{2} \sqrt {c} \mathrm {sgn}\relax (x) + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b c^{\frac {3}{2}} \mathrm {sgn}\relax (x) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{3} \sqrt {c} \mathrm {sgn}\relax (x) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{2} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 3 \, B b^{4} \sqrt {c} \mathrm {sgn}\relax (x) + 4 \, A b^{3} c^{\frac {3}{2}} \mathrm {sgn}\relax (x)\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3}} \]

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

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maple [A]  time = 0.06, size = 219, normalized size = 1.61 \[ -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-6 A \,b^{2} c^{2} x^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-9 B \,b^{3} c \,x^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-6 \sqrt {c \,x^{2}+b}\, A b \,c^{\frac {5}{2}} x^{4}-9 \sqrt {c \,x^{2}+b}\, B \,b^{2} c^{\frac {3}{2}} x^{4}-4 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{\frac {5}{2}} x^{4}-6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,c^{\frac {3}{2}} x^{4}+4 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{\frac {3}{2}} x^{2}+6 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \sqrt {c}\, x^{2}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \sqrt {c}\right )}{6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}\, x^{6}} \]

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

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maxima [A]  time = 1.50, size = 167, normalized size = 1.23 \[ \frac {1}{6} \, {\left (3 \, c^{\frac {3}{2}} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{4} + b x^{2}} c}{x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} b}{x^{4}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{6}}\right )} A + \frac {1}{4} \, {\left (3 \, b \sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b}{x^{2}} + \frac {2 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{4}}\right )} B \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^7} \,d x \]

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^7,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{7}}\, dx \]

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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