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

Optimal. Leaf size=147 \[ \frac{c \left (2 a^2 d^2-9 a b c d+5 b^2 c^2\right )}{2 a^3 b x}+\frac{(b c-a d)^2 (a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2} b^{3/2}}-\frac{c^2 (5 b c-3 a d)}{6 a^2 b x^3}+\frac{\left (c+d x^2\right )^2 (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \]

[Out]

-(c^2*(5*b*c - 3*a*d))/(6*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))

________________________________________________________________________________________

Rubi [A]  time = 0.140974, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {468, 570, 205} \[ \frac{c \left (2 a^2 d^2-9 a b c d+5 b^2 c^2\right )}{2 a^3 b x}+\frac{(b c-a d)^2 (a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2} b^{3/2}}-\frac{c^2 (5 b c-3 a d)}{6 a^2 b x^3}+\frac{\left (c+d x^2\right )^2 (b c-a d)}{2 a b x^3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(5*b*c - 3*a*d))/(6*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 468

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)), I
nt[(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 570

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]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\left (c+d x^2\right )^3}{x^4 \left (a+b x^2\right )^2} \, dx &=\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]  time = 0.0622429, size = 109, normalized size = 0.74 \[ \frac{(a d-b c)^2 (a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2} b^{3/2}}-\frac{c^2 (3 a d-2 b c)}{a^3 x}-\frac{x (a d-b c)^3}{2 a^3 b \left (a+b x^2\right )}-\frac{c^3}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c^3/(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]  time = 0.011, size = 209, normalized size = 1.4 \begin{align*} -{\frac{{c}^{3}}{3\,{a}^{2}{x}^{3}}}-3\,{\frac{{c}^{2}d}{{a}^{2}x}}+2\,{\frac{{c}^{3}b}{{a}^{3}x}}-{\frac{x{d}^{3}}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{3\,cx{d}^{2}}{2\,a \left ( b{x}^{2}+a \right ) }}-{\frac{3\,bx{c}^{2}d}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{{b}^{2}{c}^{3}x}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{{d}^{3}}{2\,b}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,c{d}^{2}}{2\,a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{9\,b{c}^{2}d}{2\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{b}^{2}{c}^{3}}{2\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.68138, size = 921, normalized size = 6.27 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 2.2596, size = 321, normalized size = 2.18 \begin{align*} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]  time = 1.15098, size = 203, normalized size = 1.38 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)