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

Optimal. Leaf size=130 \[ \frac{3 (b c-a d) \left (4 a^2 d^2+(a d+b c)^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{7/2}}+\frac{3 x (b c-a d)^2 (3 a d+b c)}{8 a^2 b^3 \left (a+b x^2\right )}+\frac{x (b c-a d)^3}{4 a b^3 \left (a+b x^2\right )^2}+\frac{d^3 x}{b^3} \]

[Out]

(d^3*x)/b^3 + ((b*c - a*d)^3*x)/(4*a*b^3*(a + b*x^2)^2) + (3*(b*c - a*d)^2*(b*c + 3*a*d)*x)/(8*a^2*b^3*(a + b*
x^2)) + (3*(b*c - a*d)*(4*a^2*d^2 + (b*c + a*d)^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(7/2))

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Rubi [A]  time = 0.165763, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {390, 1157, 385, 205} \[ \frac{3 (b c-a d) \left (4 a^2 d^2+(a d+b c)^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{7/2}}+\frac{3 x (b c-a d)^2 (3 a d+b c)}{8 a^2 b^3 \left (a+b x^2\right )}+\frac{x (b c-a d)^3}{4 a b^3 \left (a+b x^2\right )^2}+\frac{d^3 x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^3/(a + b*x^2)^3,x]

[Out]

(d^3*x)/b^3 + ((b*c - a*d)^3*x)/(4*a*b^3*(a + b*x^2)^2) + (3*(b*c - a*d)^2*(b*c + 3*a*d)*x)/(8*a^2*b^3*(a + b*
x^2)) + (3*(b*c - a*d)*(4*a^2*d^2 + (b*c + a*d)^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(7/2))

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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}{\left (a+b x^2\right )^3} \, dx &=\int \left (\frac{d^3}{b^3}+\frac{b^3 c^3-a^3 d^3+3 b d (b c-a d) (b c+a d) x^2+3 b^2 d^2 (b c-a d) x^4}{b^3 \left (a+b x^2\right )^3}\right ) \, dx\\ &=\frac{d^3 x}{b^3}+\frac{\int \frac{b^3 c^3-a^3 d^3+3 b d (b c-a d) (b c+a d) x^2+3 b^2 d^2 (b c-a d) x^4}{\left (a+b x^2\right )^3} \, dx}{b^3}\\ &=\frac{d^3 x}{b^3}+\frac{(b c-a d)^3 x}{4 a b^3 \left (a+b x^2\right )^2}-\frac{\int \frac{-3 (b c-a d) (b c+a d)^2-12 a b d^2 (b c-a d) x^2}{\left (a+b x^2\right )^2} \, dx}{4 a b^3}\\ &=\frac{d^3 x}{b^3}+\frac{(b c-a d)^3 x}{4 a b^3 \left (a+b x^2\right )^2}+\frac{3 (b c-a d)^2 (b c+3 a d) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac{\left (3 (b c-a d) \left (4 a^2 d^2+(b c+a d)^2\right )\right ) \int \frac{1}{a+b x^2} \, dx}{8 a^2 b^3}\\ &=\frac{d^3 x}{b^3}+\frac{(b c-a d)^3 x}{4 a b^3 \left (a+b x^2\right )^2}+\frac{3 (b c-a d)^2 (b c+3 a d) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac{3 (b c-a d) \left (4 a^2 d^2+(b c+a d)^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.080454, size = 139, normalized size = 1.07 \[ \frac{3 \left (3 a^2 b c d^2-5 a^3 d^3+a b^2 c^2 d+b^3 c^3\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{7/2}}+\frac{3 x (b c-a d)^2 (3 a d+b c)}{8 a^2 b^3 \left (a+b x^2\right )}+\frac{x (b c-a d)^3}{4 a b^3 \left (a+b x^2\right )^2}+\frac{d^3 x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*x)/b^3 + ((b*c - a*d)^3*x)/(4*a*b^3*(a + b*x^2)^2) + (3*(b*c - a*d)^2*(b*c + 3*a*d)*x)/(8*a^2*b^3*(a + b*
x^2)) + (3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(7/2)
)

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Maple [B]  time = 0.009, size = 266, normalized size = 2.1 \begin{align*}{\frac{{d}^{3}x}{{b}^{3}}}+{\frac{9\,a{x}^{3}{d}^{3}}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{15\,{x}^{3}c{d}^{2}}{8\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,{x}^{3}{c}^{2}d}{8\, \left ( b{x}^{2}+a \right ) ^{2}a}}+{\frac{3\,b{x}^{3}{c}^{3}}{8\, \left ( b{x}^{2}+a \right ) ^{2}{a}^{2}}}+{\frac{7\,x{a}^{2}{d}^{3}}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,acx{d}^{2}}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{3\,x{c}^{2}d}{8\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,x{c}^{3}}{8\, \left ( b{x}^{2}+a \right ) ^{2}a}}-{\frac{15\,a{d}^{3}}{8\,{b}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{9\,c{d}^{2}}{8\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,{c}^{2}d}{8\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,{c}^{3}}{8\,{a}^{2}}\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/(b*x^2+a)^3,x)

[Out]

d^3*x/b^3+9/8/b^2/(b*x^2+a)^2*a*x^3*d^3-15/8/b/(b*x^2+a)^2*x^3*c*d^2+3/8/(b*x^2+a)^2/a*x^3*c^2*d+3/8*b/(b*x^2+
a)^2/a^2*x^3*c^3+7/8/b^3/(b*x^2+a)^2*x*a^2*d^3-9/8/b^2/(b*x^2+a)^2*x*a*c*d^2-3/8/b/(b*x^2+a)^2*x*c^2*d+5/8/(b*
x^2+a)^2*x/a*c^3-15/8/b^3*a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d^3+9/8/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)
)*c*d^2+3/8/b/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c^2*d+3/8/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c^3

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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/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86477, size = 1224, normalized size = 9.42 \begin{align*} \left [\frac{16 \, a^{3} b^{3} d^{3} x^{5} + 2 \,{\left (3 \, a b^{5} c^{3} + 3 \, a^{2} b^{4} c^{2} d - 15 \, a^{3} b^{3} c d^{2} + 25 \, a^{4} b^{2} d^{3}\right )} x^{3} + 3 \,{\left (a^{2} b^{3} c^{3} + a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - 5 \, a^{5} d^{3} +{\left (b^{5} c^{3} + a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - 5 \, a^{3} b^{2} d^{3}\right )} x^{4} + 2 \,{\left (a b^{4} c^{3} + a^{2} b^{3} c^{2} d + 3 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} + 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right ) + 2 \,{\left (5 \, a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d - 9 \, a^{4} b^{2} c d^{2} + 15 \, a^{5} b d^{3}\right )} x}{16 \,{\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}, \frac{8 \, a^{3} b^{3} d^{3} x^{5} +{\left (3 \, a b^{5} c^{3} + 3 \, a^{2} b^{4} c^{2} d - 15 \, a^{3} b^{3} c d^{2} + 25 \, a^{4} b^{2} d^{3}\right )} x^{3} + 3 \,{\left (a^{2} b^{3} c^{3} + a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - 5 \, a^{5} d^{3} +{\left (b^{5} c^{3} + a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - 5 \, a^{3} b^{2} d^{3}\right )} x^{4} + 2 \,{\left (a b^{4} c^{3} + a^{2} b^{3} c^{2} d + 3 \, a^{3} b^{2} c d^{2} - 5 \, a^{4} b d^{3}\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) +{\left (5 \, a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d - 9 \, a^{4} b^{2} c d^{2} + 15 \, a^{5} b d^{3}\right )} x}{8 \,{\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(16*a^3*b^3*d^3*x^5 + 2*(3*a*b^5*c^3 + 3*a^2*b^4*c^2*d - 15*a^3*b^3*c*d^2 + 25*a^4*b^2*d^3)*x^3 + 3*(a^2
*b^3*c^3 + a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - 5*a^5*d^3 + (b^5*c^3 + a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - 5*a^3*b^2*d^
3)*x^4 + 2*(a*b^4*c^3 + a^2*b^3*c^2*d + 3*a^3*b^2*c*d^2 - 5*a^4*b*d^3)*x^2)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*
b)*x - a)/(b*x^2 + a)) + 2*(5*a^2*b^4*c^3 - 3*a^3*b^3*c^2*d - 9*a^4*b^2*c*d^2 + 15*a^5*b*d^3)*x)/(a^3*b^6*x^4
+ 2*a^4*b^5*x^2 + a^5*b^4), 1/8*(8*a^3*b^3*d^3*x^5 + (3*a*b^5*c^3 + 3*a^2*b^4*c^2*d - 15*a^3*b^3*c*d^2 + 25*a^
4*b^2*d^3)*x^3 + 3*(a^2*b^3*c^3 + a^3*b^2*c^2*d + 3*a^4*b*c*d^2 - 5*a^5*d^3 + (b^5*c^3 + a*b^4*c^2*d + 3*a^2*b
^3*c*d^2 - 5*a^3*b^2*d^3)*x^4 + 2*(a*b^4*c^3 + a^2*b^3*c^2*d + 3*a^3*b^2*c*d^2 - 5*a^4*b*d^3)*x^2)*sqrt(a*b)*a
rctan(sqrt(a*b)*x/a) + (5*a^2*b^4*c^3 - 3*a^3*b^3*c^2*d - 9*a^4*b^2*c*d^2 + 15*a^5*b*d^3)*x)/(a^3*b^6*x^4 + 2*
a^4*b^5*x^2 + a^5*b^4)]

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Sympy [B]  time = 2.18924, size = 422, normalized size = 3.25 \begin{align*} \frac{3 \sqrt{- \frac{1}{a^{5} b^{7}}} \left (a d - b c\right ) \left (5 a^{2} d^{2} + 2 a b c d + b^{2} c^{2}\right ) \log{\left (- \frac{3 a^{3} b^{3} \sqrt{- \frac{1}{a^{5} b^{7}}} \left (a d - b c\right ) \left (5 a^{2} d^{2} + 2 a b c d + b^{2} c^{2}\right )}{15 a^{3} d^{3} - 9 a^{2} b c d^{2} - 3 a b^{2} c^{2} d - 3 b^{3} c^{3}} + x \right )}}{16} - \frac{3 \sqrt{- \frac{1}{a^{5} b^{7}}} \left (a d - b c\right ) \left (5 a^{2} d^{2} + 2 a b c d + b^{2} c^{2}\right ) \log{\left (\frac{3 a^{3} b^{3} \sqrt{- \frac{1}{a^{5} b^{7}}} \left (a d - b c\right ) \left (5 a^{2} d^{2} + 2 a b c d + b^{2} c^{2}\right )}{15 a^{3} d^{3} - 9 a^{2} b c d^{2} - 3 a b^{2} c^{2} d - 3 b^{3} c^{3}} + x \right )}}{16} + \frac{x^{3} \left (9 a^{3} b d^{3} - 15 a^{2} b^{2} c d^{2} + 3 a b^{3} c^{2} d + 3 b^{4} c^{3}\right ) + x \left (7 a^{4} d^{3} - 9 a^{3} b c d^{2} - 3 a^{2} b^{2} c^{2} d + 5 a b^{3} c^{3}\right )}{8 a^{4} b^{3} + 16 a^{3} b^{4} x^{2} + 8 a^{2} b^{5} x^{4}} + \frac{d^{3} x}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(-1/(a**5*b**7))*(a*d - b*c)*(5*a**2*d**2 + 2*a*b*c*d + b**2*c**2)*log(-3*a**3*b**3*sqrt(-1/(a**5*b**7))
*(a*d - b*c)*(5*a**2*d**2 + 2*a*b*c*d + b**2*c**2)/(15*a**3*d**3 - 9*a**2*b*c*d**2 - 3*a*b**2*c**2*d - 3*b**3*
c**3) + x)/16 - 3*sqrt(-1/(a**5*b**7))*(a*d - b*c)*(5*a**2*d**2 + 2*a*b*c*d + b**2*c**2)*log(3*a**3*b**3*sqrt(
-1/(a**5*b**7))*(a*d - b*c)*(5*a**2*d**2 + 2*a*b*c*d + b**2*c**2)/(15*a**3*d**3 - 9*a**2*b*c*d**2 - 3*a*b**2*c
**2*d - 3*b**3*c**3) + x)/16 + (x**3*(9*a**3*b*d**3 - 15*a**2*b**2*c*d**2 + 3*a*b**3*c**2*d + 3*b**4*c**3) + x
*(7*a**4*d**3 - 9*a**3*b*c*d**2 - 3*a**2*b**2*c**2*d + 5*a*b**3*c**3))/(8*a**4*b**3 + 16*a**3*b**4*x**2 + 8*a*
*2*b**5*x**4) + d**3*x/b**3

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Giac [A]  time = 1.10709, size = 240, normalized size = 1.85 \begin{align*} \frac{d^{3} x}{b^{3}} + \frac{3 \,{\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2} b^{3}} + \frac{3 \, b^{4} c^{3} x^{3} + 3 \, a b^{3} c^{2} d x^{3} - 15 \, a^{2} b^{2} c d^{2} x^{3} + 9 \, a^{3} b d^{3} x^{3} + 5 \, a b^{3} c^{3} x - 3 \, a^{2} b^{2} c^{2} d x - 9 \, a^{3} b c d^{2} x + 7 \, a^{4} d^{3} x}{8 \,{\left (b x^{2} + a\right )}^{2} a^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^3*x/b^3 + 3/8*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b^3)
+ 1/8*(3*b^4*c^3*x^3 + 3*a*b^3*c^2*d*x^3 - 15*a^2*b^2*c*d^2*x^3 + 9*a^3*b*d^3*x^3 + 5*a*b^3*c^3*x - 3*a^2*b^2*
c^2*d*x - 9*a^3*b*c*d^2*x + 7*a^4*d^3*x)/((b*x^2 + a)^2*a^2*b^3)