\(\int \frac {A+B x^2}{x^3 (b x^2+c x^4)^{3/2}} \, dx\) [151]

Optimal result

Integrand size = 26, antiderivative size = 101 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}+\frac {4 c (5 b B-6 A c) \left (b+2 c x^2\right )}{15 b^4 \sqrt {b x^2+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2059, 806, 672, 627} \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {4 c \left (b+2 c x^2\right ) (5 b B-6 A c)}{15 b^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}} \]

[In]

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

[Out]

-1/5*A/(b*x^4*Sqrt[b*x^2 + c*x^4]) - (5*b*B - 6*A*c)/(15*b^2*x^2*Sqrt[b*x^2 + c*x^4]) + (4*c*(5*b*B - 6*A*c)*(
b + 2*c*x^2))/(15*b^4*Sqrt[b*x^2 + c*x^4])

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 672

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

Rule 806

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 2059

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {A+B x}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}+\frac {\left (\frac {1}{2} (b B-2 A c)-2 (-b B+A c)\right ) \text {Subst}\left (\int \frac {1}{x \left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{5 b} \\ & = -\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {(2 c (5 b B-6 A c)) \text {Subst}\left (\int \frac {1}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )}{15 b^2} \\ & = -\frac {A}{5 b x^4 \sqrt {b x^2+c x^4}}-\frac {5 b B-6 A c}{15 b^2 x^2 \sqrt {b x^2+c x^4}}+\frac {4 c (5 b B-6 A c) \left (b+2 c x^2\right )}{15 b^4 \sqrt {b x^2+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-5 b B x^2 \left (b^2-4 b c x^2-8 c^2 x^4\right )-3 A \left (b^3-2 b^2 c x^2+8 b c^2 x^4+16 c^3 x^6\right )}{15 b^4 x^4 \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

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

[Out]

(-5*b*B*x^2*(b^2 - 4*b*c*x^2 - 8*c^2*x^4) - 3*A*(b^3 - 2*b^2*c*x^2 + 8*b*c^2*x^4 + 16*c^3*x^6))/(15*b^4*x^4*Sq
rt[x^2*(b + c*x^2)])

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {\left (-5 x^{2} B -3 A \right ) b^{3}+6 x^{2} \left (\frac {10 x^{2} B}{3}+A \right ) c \,b^{2}-24 x^{4} c^{2} \left (-\frac {5 x^{2} B}{3}+A \right ) b -48 A \,c^{3} x^{6}}{15 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{4} b^{4}}\)	\(80\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (48 A \,c^{3} x^{6}-40 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-20 x^{4} B \,b^{2} c -6 A \,b^{2} c \,x^{2}+5 b^{3} B \,x^{2}+3 b^{3} A \right )}{15 x^{2} b^{4} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\)	\(94\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (48 A \,c^{3} x^{6}-40 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-20 x^{4} B \,b^{2} c -6 A \,b^{2} c \,x^{2}+5 b^{3} B \,x^{2}+3 b^{3} A \right )}{15 x^{2} b^{4} \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}\)	\(94\)
trager	\(-\frac {\left (48 A \,c^{3} x^{6}-40 x^{6} B b \,c^{2}+24 A b \,c^{2} x^{4}-20 x^{4} B \,b^{2} c -6 A \,b^{2} c \,x^{2}+5 b^{3} B \,x^{2}+3 b^{3} A \right ) \sqrt {x^{4} c +b \,x^{2}}}{15 \left (c \,x^{2}+b \right ) b^{4} x^{6}}\)	\(96\)
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (33 A \,c^{2} x^{4}-25 x^{4} B b c -9 A b c \,x^{2}+5 b^{2} B \,x^{2}+3 b^{2} A \right )}{15 b^{4} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {x^{2} c^{2} \left (A c -B b \right )}{b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(103\)

[In]

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

[Out]

1/15*((-5*B*x^2-3*A)*b^3+6*x^2*(10/3*x^2*B+A)*c*b^2-24*x^4*c^2*(-5/3*x^2*B+A)*b-48*A*c^3*x^6)/(x^2*(c*x^2+b))^
(1/2)/x^4/b^4

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (8 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x^{6} + 4 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} x^{4} - 3 \, A b^{3} - {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \]

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {A + B x^{2}}{x^{3} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((B*x**2+A)/x**3/(c*x**4+b*x**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{3} \, B {\left (\frac {8 \, c^{2} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {4 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2}} - \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{2}}\right )} - \frac {1}{5} \, A {\left (\frac {16 \, c^{3} x^{2}}{\sqrt {c x^{4} + b x^{2}} b^{4}} + \frac {8 \, c^{2}}{\sqrt {c x^{4} + b x^{2}} b^{3}} - \frac {2 \, c}{\sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} + \frac {1}{\sqrt {c x^{4} + b x^{2}} b x^{4}}\right )} \]

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/3*B*(8*c^2*x^2/(sqrt(c*x^4 + b*x^2)*b^3) + 4*c/(sqrt(c*x^4 + b*x^2)*b^2) - 1/(sqrt(c*x^4 + b*x^2)*b*x^2)) -
1/5*A*(16*c^3*x^2/(sqrt(c*x^4 + b*x^2)*b^4) + 8*c^2/(sqrt(c*x^4 + b*x^2)*b^3) - 2*c/(sqrt(c*x^4 + b*x^2)*b^2*x
^2) + 1/(sqrt(c*x^4 + b*x^2)*b*x^4))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (89) = 178\).

Time = 0.92 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.99 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (B b c^{2} - A c^{3}\right )} x}{\sqrt {c x^{2} + b} b^{4} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {3}{2}} - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {5}{2}} - 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {3}{2}} + 90 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b c^{\frac {5}{2}} + 160 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {3}{2}} - 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {5}{2}} - 110 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {3}{2}} + 150 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{3} c^{\frac {5}{2}} + 25 \, B b^{5} c^{\frac {3}{2}} - 33 \, A b^{4} c^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5} b^{3} \mathrm {sgn}\left (x\right )} \]

[In]

integrate((B*x^2+A)/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

(B*b*c^2 - A*c^3)*x/(sqrt(c*x^2 + b)*b^4*sgn(x)) - 2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b*c^(3/2) - 15*(
sqrt(c)*x - sqrt(c*x^2 + b))^8*A*c^(5/2) - 90*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^2*c^(3/2) + 90*(sqrt(c)*x -
sqrt(c*x^2 + b))^6*A*b*c^(5/2) + 160*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^3*c^(3/2) - 240*(sqrt(c)*x - sqrt(c*x
^2 + b))^4*A*b^2*c^(5/2) - 110*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^4*c^(3/2) + 150*(sqrt(c)*x - sqrt(c*x^2 + b
))^2*A*b^3*c^(5/2) + 25*B*b^5*c^(3/2) - 33*A*b^4*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5*b^3*sgn(x))

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (5\,B\,b^3\,x^2+3\,A\,b^3-20\,B\,b^2\,c\,x^4-6\,A\,b^2\,c\,x^2-40\,B\,b\,c^2\,x^6+24\,A\,b\,c^2\,x^4+48\,A\,c^3\,x^6\right )}{15\,b^4\,x^6\,\left (c\,x^2+b\right )} \]

[In]

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

[Out]

-((b*x^2 + c*x^4)^(1/2)*(3*A*b^3 + 5*B*b^3*x^2 + 48*A*c^3*x^6 - 6*A*b^2*c*x^2 + 24*A*b*c^2*x^4 - 20*B*b^2*c*x^
4 - 40*B*b*c^2*x^6))/(15*b^4*x^6*(b + c*x^2))