\(\int \frac {(a+b x^2)^2}{x^4 (c+d x^2)^{3/2}} \, dx\) [656]

Optimal result

Integrand size = 24, antiderivative size = 97 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {a^2}{3 c x^3 \sqrt {c+d x^2}}-\frac {2 a (3 b c-2 a d)}{3 c^2 x \sqrt {c+d x^2}}+\frac {\left (3 b^2 c^2-4 a d (3 b c-2 a d)\right ) x}{3 c^3 \sqrt {c+d x^2}} \]

[Out]

-1/3*a^2/c/x^3/(d*x^2+c)^(1/2)-2/3*a*(-2*a*d+3*b*c)/c^2/x/(d*x^2+c)^(1/2)+1/3*(3*b^2*c^2-4*a*d*(-2*a*d+3*b*c))
*x/c^3/(d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {473, 464, 197} \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=\frac {x \left (8 a^2 d^2-12 a b c d+3 b^2 c^2\right )}{3 c^3 \sqrt {c+d x^2}}-\frac {a^2}{3 c x^3 \sqrt {c+d x^2}}-\frac {2 a (3 b c-2 a d)}{3 c^2 x \sqrt {c+d x^2}} \]

[In]

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

[Out]

-1/3*a^2/(c*x^3*Sqrt[c + d*x^2]) - (2*a*(3*b*c - 2*a*d))/(3*c^2*x*Sqrt[c + d*x^2]) + ((3*b^2*c^2 - 12*a*b*c*d
+ 8*a^2*d^2)*x)/(3*c^3*Sqrt[c + d*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 464

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

Rule 473

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2}{3 c x^3 \sqrt {c+d x^2}}+\frac {\int \frac {2 a (3 b c-2 a d)+3 b^2 c x^2}{x^2 \left (c+d x^2\right )^{3/2}} \, dx}{3 c} \\ & = -\frac {a^2}{3 c x^3 \sqrt {c+d x^2}}-\frac {2 a (3 b c-2 a d)}{3 c^2 x \sqrt {c+d x^2}}-\frac {1}{3} \left (-3 b^2+\frac {4 a d (3 b c-2 a d)}{c^2}\right ) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx \\ & = -\frac {a^2}{3 c x^3 \sqrt {c+d x^2}}-\frac {2 a (3 b c-2 a d)}{3 c^2 x \sqrt {c+d x^2}}+\frac {\left (3 b^2-\frac {4 a d (3 b c-2 a d)}{c^2}\right ) x}{3 c \sqrt {c+d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=\frac {3 b^2 c^2 x^4-6 a b c x^2 \left (c+2 d x^2\right )+a^2 \left (-c^2+4 c d x^2+8 d^2 x^4\right )}{3 c^3 x^3 \sqrt {c+d x^2}} \]

[In]

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

[Out]

(3*b^2*c^2*x^4 - 6*a*b*c*x^2*(c + 2*d*x^2) + a^2*(-c^2 + 4*c*d*x^2 + 8*d^2*x^4))/(3*c^3*x^3*Sqrt[c + d*x^2])

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {\left (-3 b^{2} x^{4}+6 a b \,x^{2}+a^{2}\right ) c^{2}-4 a d \,x^{2} \left (-3 b \,x^{2}+a \right ) c -8 a^{2} d^{2} x^{4}}{3 \sqrt {d \,x^{2}+c}\, x^{3} c^{3}}\)	\(69\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, a \left (-5 a d \,x^{2}+6 c b \,x^{2}+a c \right )}{3 c^{3} x^{3}}+\frac {x \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\sqrt {d \,x^{2}+c}\, c^{3}}\)	\(73\)
gosper	\(-\frac {-8 a^{2} d^{2} x^{4}+12 x^{4} a b c d -3 b^{2} c^{2} x^{4}-4 a^{2} c d \,x^{2}+6 a b \,c^{2} x^{2}+a^{2} c^{2}}{3 x^{3} \sqrt {d \,x^{2}+c}\, c^{3}}\)	\(77\)
trager	\(-\frac {-8 a^{2} d^{2} x^{4}+12 x^{4} a b c d -3 b^{2} c^{2} x^{4}-4 a^{2} c d \,x^{2}+6 a b \,c^{2} x^{2}+a^{2} c^{2}}{3 x^{3} \sqrt {d \,x^{2}+c}\, c^{3}}\)	\(77\)
default	\(\frac {b^{2} x}{c \sqrt {d \,x^{2}+c}}+a^{2} \left (-\frac {1}{3 c \,x^{3} \sqrt {d \,x^{2}+c}}-\frac {4 d \left (-\frac {1}{c x \sqrt {d \,x^{2}+c}}-\frac {2 d x}{c^{2} \sqrt {d \,x^{2}+c}}\right )}{3 c}\right )+2 a b \left (-\frac {1}{c x \sqrt {d \,x^{2}+c}}-\frac {2 d x}{c^{2} \sqrt {d \,x^{2}+c}}\right )\)	\(119\)

[In]

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

[Out]

-1/3/(d*x^2+c)^(1/2)*((-3*b^2*x^4+6*a*b*x^2+a^2)*c^2-4*a*d*x^2*(-3*b*x^2+a)*c-8*a^2*d^2*x^4)/x^3/c^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (3 \, b^{2} c^{2} - 12 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} - a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{3} d x^{5} + c^{4} x^{3}\right )}} \]

[In]

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

[Out]

1/3*((3*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)*x^4 - a^2*c^2 - 2*(3*a*b*c^2 - 2*a^2*c*d)*x^2)*sqrt(d*x^2 + c)/(c^3*
d*x^5 + c^4*x^3)

Sympy [F]

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

[In]

integrate((b*x**2+a)**2/x**4/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/(x**4*(c + d*x**2)**(3/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x}{\sqrt {d x^{2} + c} c} - \frac {4 \, a b d x}{\sqrt {d x^{2} + c} c^{2}} + \frac {8 \, a^{2} d^{2} x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {2 \, a b}{\sqrt {d x^{2} + c} c x} + \frac {4 \, a^{2} d}{3 \, \sqrt {d x^{2} + c} c^{2} x} - \frac {a^{2}}{3 \, \sqrt {d x^{2} + c} c x^{3}} \]

[In]

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

[Out]

b^2*x/(sqrt(d*x^2 + c)*c) - 4*a*b*d*x/(sqrt(d*x^2 + c)*c^2) + 8/3*a^2*d^2*x/(sqrt(d*x^2 + c)*c^3) - 2*a*b/(sqr
t(d*x^2 + c)*c*x) + 4/3*a^2*d/(sqrt(d*x^2 + c)*c^2*x) - 1/3*a^2/(sqrt(d*x^2 + c)*c*x^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (85) = 170\).

Time = 0.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.05 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{\sqrt {d x^{2} + c} c^{3}} + \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{2} \sqrt {d} + 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c d^{\frac {3}{2}} + 6 \, a b c^{3} \sqrt {d} - 5 \, a^{2} c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} c^{2}} \]

[In]

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

[Out]

(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x/(sqrt(d*x^2 + c)*c^3) + 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c*sqrt(d)
- 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*d^(3/2) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^2*sqrt(d) + 12*(sq
rt(d)*x - sqrt(d*x^2 + c))^2*a^2*c*d^(3/2) + 6*a*b*c^3*sqrt(d) - 5*a^2*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2
+ c))^2 - c)^3*c^2)

Mupad [B] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {a^2\,c^2-4\,a^2\,c\,d\,x^2-8\,a^2\,d^2\,x^4+6\,a\,b\,c^2\,x^2+12\,a\,b\,c\,d\,x^4-3\,b^2\,c^2\,x^4}{3\,c^3\,x^3\,\sqrt {d\,x^2+c}} \]

[In]

int((a + b*x^2)^2/(x^4*(c + d*x^2)^(3/2)),x)

[Out]

-(a^2*c^2 - 8*a^2*d^2*x^4 - 3*b^2*c^2*x^4 + 6*a*b*c^2*x^2 - 4*a^2*c*d*x^2 + 12*a*b*c*d*x^4)/(3*c^3*x^3*(c + d*
x^2)^(1/2))