\(\int \frac {(c+d x^2)^3}{x^4 (a+b x^2)^2} \, dx\) [288]

Optimal result

Integrand size = 22, antiderivative size = 147 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c^2 (5 b c-3 a d)}{6 a^2 b x^3}+\frac {c \left (5 b^2 c^2-9 a b c d+2 a^2 d^2\right )}{2 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x^3 \left (a+b x^2\right )}+\frac {(b c-a d)^2 (5 b c+a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {479, 584, 211} \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (b c-a d)^2 (a d+5 b c)}{2 a^{7/2} b^{3/2}}-\frac {c^2 (5 b c-3 a d)}{6 a^2 b x^3}+\frac {c \left (2 a^2 d^2-9 a b c d+5 b^2 c^2\right )}{2 a^3 b x}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \]

[In]

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

[Out]

-1/6*(c^2*(5*b*c - 3*a*d))/(a^2*b*x^3) + (c*(5*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2))/(2*a^3*b*x) + ((b*c - a*d)*(c
+ d*x^2)^2)/(2*a*b*x^3*(a + b*x^2)) + ((b*c - a*d)^2*(5*b*c + a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2)*b^
(3/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 479

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

Rule 584

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x^3 \left (a+b x^2\right )}-\frac {\int \frac {\left (c+d x^2\right ) \left (-c (5 b c-3 a d)-d (b c+a d) x^2\right )}{x^4 \left (a+b x^2\right )} \, dx}{2 a b} \\ & = \frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x^3 \left (a+b x^2\right )}-\frac {\int \left (\frac {c^2 (-5 b c+3 a d)}{a x^4}+\frac {c \left (5 b^2 c^2-9 a b c d+2 a^2 d^2\right )}{a^2 x^2}-\frac {(-b c+a d)^2 (5 b c+a d)}{a^2 \left (a+b x^2\right )}\right ) \, dx}{2 a b} \\ & = -\frac {c^2 (5 b c-3 a d)}{6 a^2 b x^3}+\frac {c \left (5 b^2 c^2-9 a b c d+2 a^2 d^2\right )}{2 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x^3 \left (a+b x^2\right )}+\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \int \frac {1}{a+b x^2} \, dx}{2 a^3 b} \\ & = -\frac {c^2 (5 b c-3 a d)}{6 a^2 b x^3}+\frac {c \left (5 b^2 c^2-9 a b c d+2 a^2 d^2\right )}{2 a^3 b x}+\frac {(b c-a d) \left (c+d x^2\right )^2}{2 a b x^3 \left (a+b x^2\right )}+\frac {(b c-a d)^2 (5 b c+a d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.74 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {c^3}{3 a^2 x^3}-\frac {c^2 (-2 b c+3 a d)}{a^3 x}-\frac {(-b c+a d)^3 x}{2 a^3 b \left (a+b x^2\right )}+\frac {(-b c+a d)^2 (5 b c+a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} b^{3/2}} \]

[In]

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

[Out]

-1/3*c^3/(a^2*x^3) - (c^2*(-2*b*c + 3*a*d))/(a^3*x) - ((-(b*c) + a*d)^3*x)/(2*a^3*b*(a + b*x^2)) + ((-(b*c) +
a*d)^2*(5*b*c + a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2)*b^(3/2))

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {c^{3}}{3 a^{2} x^{3}}-\frac {c^{2} \left (3 a d -2 b c \right )}{a^{3} x}+\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{2 b \left (b \,x^{2}+a \right )}+\frac {\left (a^{3} d^{3}+3 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 b \sqrt {a b}}}{a^{3}}\)	\(144\)
risch	\(\frac {-\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+9 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) x^{4}}{2 a^{3} b}-\frac {c^{2} \left (9 a d -5 b c \right ) x^{2}}{3 a^{2}}-\frac {c^{3}}{3 a}}{x^{3} \left (b \,x^{2}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2} b^{3}+a^{6} d^{6}+6 a^{5} b c \,d^{5}-9 a^{4} b^{2} c^{2} d^{4}-44 a^{3} b^{3} c^{3} d^{3}+111 a^{2} b^{4} c^{4} d^{2}-90 a \,b^{5} c^{5} d +25 b^{6} c^{6}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7} b^{3}+2 a^{6} d^{6}+12 a^{5} b c \,d^{5}-18 a^{4} b^{2} c^{2} d^{4}-88 a^{3} b^{3} c^{3} d^{3}+222 a^{2} b^{4} c^{4} d^{2}-180 a \,b^{5} c^{5} d +50 b^{6} c^{6}\right ) x +\left (-a^{7} d^{3} b -3 b^{2} c \,d^{2} a^{6}+9 b^{3} c^{2} d \,a^{5}-5 b^{4} c^{3} a^{4}\right ) \textit {\_R} \right )\right )}{4}\)	\(328\)

[In]

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

[Out]

-1/3*c^3/a^2/x^3-c^2*(3*a*d-2*b*c)/a^3/x+1/a^3*(-1/2*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/b*x/(b*x^2+
a)+1/2*(a^3*d^3+3*a^2*b*c*d^2-9*a*b^2*c^2*d+5*b^3*c^3)/b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 458, normalized size of antiderivative = 3.12 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=\left [-\frac {4 \, a^{3} b^{2} c^{3} - 6 \, {\left (5 \, a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 3 \, a^{3} b^{2} c d^{2} - a^{4} b d^{3}\right )} x^{4} - 4 \, {\left (5 \, a^{2} b^{3} c^{3} - 9 \, a^{3} b^{2} c^{2} d\right )} x^{2} + 3 \, {\left ({\left (5 \, b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} + a^{3} b d^{3}\right )} x^{5} + {\left (5 \, a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} + a^{4} d^{3}\right )} x^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{12 \, {\left (a^{4} b^{3} x^{5} + a^{5} b^{2} x^{3}\right )}}, -\frac {2 \, a^{3} b^{2} c^{3} - 3 \, {\left (5 \, a b^{4} c^{3} - 9 \, a^{2} b^{3} c^{2} d + 3 \, a^{3} b^{2} c d^{2} - a^{4} b d^{3}\right )} x^{4} - 2 \, {\left (5 \, a^{2} b^{3} c^{3} - 9 \, a^{3} b^{2} c^{2} d\right )} x^{2} - 3 \, {\left ({\left (5 \, b^{4} c^{3} - 9 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} + a^{3} b d^{3}\right )} x^{5} + {\left (5 \, a b^{3} c^{3} - 9 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} + a^{4} d^{3}\right )} x^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{6 \, {\left (a^{4} b^{3} x^{5} + a^{5} b^{2} x^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/12*(4*a^3*b^2*c^3 - 6*(5*a*b^4*c^3 - 9*a^2*b^3*c^2*d + 3*a^3*b^2*c*d^2 - a^4*b*d^3)*x^4 - 4*(5*a^2*b^3*c^3
- 9*a^3*b^2*c^2*d)*x^2 + 3*((5*b^4*c^3 - 9*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 + a^3*b*d^3)*x^5 + (5*a*b^3*c^3 - 9*
a^2*b^2*c^2*d + 3*a^3*b*c*d^2 + a^4*d^3)*x^3)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^4*b
^3*x^5 + a^5*b^2*x^3), -1/6*(2*a^3*b^2*c^3 - 3*(5*a*b^4*c^3 - 9*a^2*b^3*c^2*d + 3*a^3*b^2*c*d^2 - a^4*b*d^3)*x
^4 - 2*(5*a^2*b^3*c^3 - 9*a^3*b^2*c^2*d)*x^2 - 3*((5*b^4*c^3 - 9*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 + a^3*b*d^3)*x^
5 + (5*a*b^3*c^3 - 9*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 + a^4*d^3)*x^3)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^4*b^3*x
^5 + a^5*b^2*x^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (133) = 266\).

Time = 1.11 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.18 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} b^{3}}} \left (a d - b c\right )^{2} \left (a d + 5 b c\right ) \log {\left (- \frac {a^{4} b \sqrt {- \frac {1}{a^{7} b^{3}}} \left (a d - b c\right )^{2} \left (a d + 5 b c\right )}{a^{3} d^{3} + 3 a^{2} b c d^{2} - 9 a b^{2} c^{2} d + 5 b^{3} c^{3}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{7} b^{3}}} \left (a d - b c\right )^{2} \left (a d + 5 b c\right ) \log {\left (\frac {a^{4} b \sqrt {- \frac {1}{a^{7} b^{3}}} \left (a d - b c\right )^{2} \left (a d + 5 b c\right )}{a^{3} d^{3} + 3 a^{2} b c d^{2} - 9 a b^{2} c^{2} d + 5 b^{3} c^{3}} + x \right )}}{4} + \frac {- 2 a^{2} b c^{3} + x^{4} \left (- 3 a^{3} d^{3} + 9 a^{2} b c d^{2} - 27 a b^{2} c^{2} d + 15 b^{3} c^{3}\right ) + x^{2} \left (- 18 a^{2} b c^{2} d + 10 a b^{2} c^{3}\right )}{6 a^{4} b x^{3} + 6 a^{3} b^{2} x^{5}} \]

[In]

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

[Out]

-sqrt(-1/(a**7*b**3))*(a*d - b*c)**2*(a*d + 5*b*c)*log(-a**4*b*sqrt(-1/(a**7*b**3))*(a*d - b*c)**2*(a*d + 5*b*
c)/(a**3*d**3 + 3*a**2*b*c*d**2 - 9*a*b**2*c**2*d + 5*b**3*c**3) + x)/4 + sqrt(-1/(a**7*b**3))*(a*d - b*c)**2*
(a*d + 5*b*c)*log(a**4*b*sqrt(-1/(a**7*b**3))*(a*d - b*c)**2*(a*d + 5*b*c)/(a**3*d**3 + 3*a**2*b*c*d**2 - 9*a*
b**2*c**2*d + 5*b**3*c**3) + x)/4 + (-2*a**2*b*c**3 + x**4*(-3*a**3*d**3 + 9*a**2*b*c*d**2 - 27*a*b**2*c**2*d
+ 15*b**3*c**3) + x**2*(-18*a**2*b*c**2*d + 10*a*b**2*c**3))/(6*a**4*b*x**3 + 6*a**3*b**2*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.08 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=-\frac {2 \, a^{2} b c^{3} - 3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{4} - 2 \, {\left (5 \, a b^{2} c^{3} - 9 \, a^{2} b c^{2} d\right )} x^{2}}{6 \, {\left (a^{3} b^{2} x^{5} + a^{4} b x^{3}\right )}} + \frac {{\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {{\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} b} + \frac {b^{3} c^{3} x - 3 \, a b^{2} c^{2} d x + 3 \, a^{2} b c d^{2} x - a^{3} d^{3} x}{2 \, {\left (b x^{2} + a\right )} a^{3} b} + \frac {6 \, b c^{3} x^{2} - 9 \, a c^{2} d x^{2} - a c^{3}}{3 \, a^{3} x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.12 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+5\,b\,c\right )}{\sqrt {a}\,\left (a^3\,d^3+3\,a^2\,b\,c\,d^2-9\,a\,b^2\,c^2\,d+5\,b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (a\,d+5\,b\,c\right )}{2\,a^{7/2}\,b^{3/2}}-\frac {\frac {c^3}{3\,a}+\frac {x^4\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+9\,a\,b^2\,c^2\,d-5\,b^3\,c^3\right )}{2\,a^3\,b}+\frac {c^2\,x^2\,\left (9\,a\,d-5\,b\,c\right )}{3\,a^2}}{b\,x^5+a\,x^3} \]

[In]

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

[Out]

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