\(\int \frac {(a+b x^2)^2}{x^4 (c+d x^2)^{5/2}} \, dx\) [668]

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2 c^2-8 a d (b c-a d)\right ) x}{3 c^3 \left (c+d x^2\right )^{3/2}}+\frac {2 \left (b^2 c^2-8 a d (b c-a d)\right ) x}{3 c^4 \sqrt {c+d x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {473, 464, 198, 197} \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}+\frac {x \left (b^2-\frac {8 a d (b c-a d)}{c^2}\right )}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \left (b^2 c^2-8 a d (b c-a d)\right )}{3 c^4 \sqrt {c+d x^2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}} \]

[In]

Int[(a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x]

[Out]

-1/3*a^2/(c*x^3*(c + d*x^2)^(3/2)) - (2*a*(b*c - a*d))/(c^2*x*(c + d*x^2)^(3/2)) + ((b^2 - (8*a*d*(b*c - a*d))
/c^2)*x)/(3*c*(c + d*x^2)^(3/2)) + (2*(b^2*c^2 - 8*a*d*(b*c - a*d))*x)/(3*c^4*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 198

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

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 \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {6 a (b c-a d)+3 b^2 c x^2}{x^2 \left (c+d x^2\right )^{5/2}} \, dx}{3 c} \\ & = -\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}-\left (-b^2+\frac {8 a d (b c-a d)}{c^2}\right ) \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx \\ & = -\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2-\frac {8 a d (b c-a d)}{c^2}\right ) x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {\left (2 \left (b^2-\frac {8 a d (b c-a d)}{c^2}\right )\right ) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c} \\ & = -\frac {a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}-\frac {2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}}+\frac {\left (b^2-\frac {8 a d (b c-a d)}{c^2}\right ) x}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 \left (b^2-\frac {8 a d (b c-a d)}{c^2}\right ) x}{3 c^2 \sqrt {c+d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^2 c^2 x^4 \left (3 c+2 d x^2\right )-2 a b c x^2 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )+a^2 \left (-c^3+6 c^2 d x^2+24 c d^2 x^4+16 d^3 x^6\right )}{3 c^4 x^3 \left (c+d x^2\right )^{3/2}} \]

[In]

Integrate[(a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x]

[Out]

(b^2*c^2*x^4*(3*c + 2*d*x^2) - 2*a*b*c*x^2*(3*c^2 + 12*c*d*x^2 + 8*d^2*x^4) + a^2*(-c^3 + 6*c^2*d*x^2 + 24*c*d
^2*x^4 + 16*d^3*x^6))/(3*c^4*x^3*(c + d*x^2)^(3/2))

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {\left (-3 b^{2} x^{4}+6 a b \,x^{2}+a^{2}\right ) c^{3}-6 x^{2} d \left (\frac {1}{3} b^{2} x^{4}-4 a b \,x^{2}+a^{2}\right ) c^{2}-24 x^{4} \left (-\frac {2 b \,x^{2}}{3}+a \right ) d^{2} a c -16 a^{2} d^{3} x^{6}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} x^{3} c^{4}}\)	\(99\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, a \left (-8 a d \,x^{2}+6 c b \,x^{2}+a c \right )}{3 c^{4} x^{3}}+\frac {\left (a d -b c \right ) \left (8 a \,d^{2} x^{2}-2 b c d \,x^{2}+9 a c d -3 b \,c^{2}\right ) x \sqrt {d \,x^{2}+c}}{3 \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right ) c^{4}}\)	\(110\)
gosper	\(-\frac {-16 a^{2} d^{3} x^{6}+16 x^{6} d^{2} a b c -2 b^{2} c^{2} d \,x^{6}-24 a^{2} c \,d^{2} x^{4}+24 a b \,c^{2} d \,x^{4}-3 b^{2} c^{3} x^{4}-6 a^{2} c^{2} d \,x^{2}+6 a b \,c^{3} x^{2}+a^{2} c^{3}}{3 x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{4}}\)	\(116\)
trager	\(-\frac {-16 a^{2} d^{3} x^{6}+16 x^{6} d^{2} a b c -2 b^{2} c^{2} d \,x^{6}-24 a^{2} c \,d^{2} x^{4}+24 a b \,c^{2} d \,x^{4}-3 b^{2} c^{3} x^{4}-6 a^{2} c^{2} d \,x^{2}+6 a b \,c^{3} x^{2}+a^{2} c^{3}}{3 x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{4}}\)	\(116\)
default	\(b^{2} \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )+a^{2} \left (-\frac {1}{3 c \,x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {2 d \left (-\frac {1}{c x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{c}\right )}{c}\right )+2 a b \left (-\frac {1}{c x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{c}\right )\)	\(179\)

[In]

int((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*((-3*b^2*x^4+6*a*b*x^2+a^2)*c^3-6*x^2*d*(1/3*b^2*x^4-4*a*b*x^2+a^2)*c^2-24*x^4*(-2/3*b*x^2+a)*d^2*a*c-16*
a^2*d^3*x^6)/(d*x^2+c)^(3/2)/x^3/c^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, {\left (b^{2} c^{2} d - 8 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{6} - a^{2} c^{3} + 3 \, {\left (b^{2} c^{3} - 8 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} x^{4} - 6 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{4} d^{2} x^{7} + 2 \, c^{5} d x^{5} + c^{6} x^{3}\right )}} \]

[In]

integrate((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*(b^2*c^2*d - 8*a*b*c*d^2 + 8*a^2*d^3)*x^6 - a^2*c^3 + 3*(b^2*c^3 - 8*a*b*c^2*d + 8*a^2*c*d^2)*x^4 - 6*(
a*b*c^3 - a^2*c^2*d)*x^2)*sqrt(d*x^2 + c)/(c^4*d^2*x^7 + 2*c^5*d*x^5 + c^6*x^3)

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{x^{4} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((b*x**2+a)**2/x**4/(d*x**2+c)**(5/2),x)

[Out]

Integral((a + b*x**2)**2/(x**4*(c + d*x**2)**(5/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=\frac {2 \, b^{2} x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {b^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {16 \, a b d x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {8 \, a b d x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {16 \, a^{2} d^{2} x}{3 \, \sqrt {d x^{2} + c} c^{4}} + \frac {8 \, a^{2} d^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} - \frac {2 \, a b}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x} + \frac {2 \, a^{2} d}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x} - \frac {a^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{3}} \]

[In]

integrate((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

2/3*b^2*x/(sqrt(d*x^2 + c)*c^2) + 1/3*b^2*x/((d*x^2 + c)^(3/2)*c) - 16/3*a*b*d*x/(sqrt(d*x^2 + c)*c^3) - 8/3*a
*b*d*x/((d*x^2 + c)^(3/2)*c^2) + 16/3*a^2*d^2*x/(sqrt(d*x^2 + c)*c^4) + 8/3*a^2*d^2*x/((d*x^2 + c)^(3/2)*c^3)
- 2*a*b/((d*x^2 + c)^(3/2)*c*x) + 2*a^2*d/((d*x^2 + c)^(3/2)*c^2*x) - 1/3*a^2/((d*x^2 + c)^(3/2)*c*x^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (117) = 234\).

Time = 0.30 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=\frac {x {\left (\frac {2 \, {\left (b^{2} c^{5} d^{2} - 5 \, a b c^{4} d^{3} + 4 \, a^{2} c^{3} d^{4}\right )} x^{2}}{c^{7} d} + \frac {3 \, {\left (b^{2} c^{6} d - 4 \, a b c^{5} d^{2} + 3 \, a^{2} c^{4} d^{3}\right )}}{c^{7} d}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {4 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{2} \sqrt {d} + 9 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c d^{\frac {3}{2}} + 3 \, a b c^{3} \sqrt {d} - 4 \, a^{2} c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} c^{3}} \]

[In]

integrate((b*x^2+a)^2/x^4/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*x*(2*(b^2*c^5*d^2 - 5*a*b*c^4*d^3 + 4*a^2*c^3*d^4)*x^2/(c^7*d) + 3*(b^2*c^6*d - 4*a*b*c^5*d^2 + 3*a^2*c^4*
d^3)/(c^7*d))/(d*x^2 + c)^(3/2) + 4/3*(3*(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) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^2*sqrt(d) + 9*(sqrt(d)*x - sqrt(d*x^2 + c))
^2*a^2*c*d^(3/2) + 3*a*b*c^3*sqrt(d) - 4*a^2*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*c^3)

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a+b x^2\right )^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^2\,c^4\,x^2-a^2\,c^3\,d-16\,a^2\,d\,{\left (d\,x^2+c\right )}^3+2\,a\,b\,c^4+b^2\,c^3\,x^2\,\left (d\,x^2+c\right )+16\,a\,b\,c\,{\left (d\,x^2+c\right )}^3+6\,a\,b\,c^3\,\left (d\,x^2+c\right )-2\,b^2\,c^2\,x^2\,{\left (d\,x^2+c\right )}^2-24\,a\,b\,c^2\,{\left (d\,x^2+c\right )}^2+24\,a^2\,c\,d\,{\left (d\,x^2+c\right )}^2-6\,a^2\,c^2\,d\,\left (d\,x^2+c\right )}{{\left (d\,x^2+c\right )}^{3/2}\,\left (3\,c^5\,x-3\,c^4\,x\,\left (d\,x^2+c\right )\right )} \]

[In]

int((a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x)

[Out]

(b^2*c^4*x^2 - a^2*c^3*d - 16*a^2*d*(c + d*x^2)^3 + 2*a*b*c^4 + b^2*c^3*x^2*(c + d*x^2) + 16*a*b*c*(c + d*x^2)
^3 + 6*a*b*c^3*(c + d*x^2) - 2*b^2*c^2*x^2*(c + d*x^2)^2 - 24*a*b*c^2*(c + d*x^2)^2 + 24*a^2*c*d*(c + d*x^2)^2
- 6*a^2*c^2*d*(c + d*x^2))/((c + d*x^2)^(3/2)*(3*c^5*x - 3*c^4*x*(c + d*x^2)))