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

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

[Out]

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

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Rubi [A]  time = 0.164726, 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} \[ -\frac{3 (b c-a d) \left ((a d+b c)^2+4 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} d^{7/2}}+\frac{3 x (b c-a d)^2 (a d+3 b c)}{8 c^2 d^3 \left (c+d x^2\right )}-\frac{x (b c-a d)^3}{4 c d^3 \left (c+d x^2\right )^2}+\frac{b^3 x}{d^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3/(c + d*x^2)^3,x]

[Out]

(b^3*x)/d^3 - ((b*c - a*d)^3*x)/(4*c*d^3*(c + d*x^2)^2) + (3*(b*c - a*d)^2*(3*b*c + a*d)*x)/(8*c^2*d^3*(c + d*
x^2)) - (3*(b*c - a*d)*(4*b^2*c^2 + (b*c + a*d)^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(5/2)*d^(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 (a+b x^2\right )^3}{\left (c+d x^2\right )^3} \, dx &=\int \left (\frac{b^3}{d^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}{d^3 \left (c+d x^2\right )^3}\right ) \, dx\\ &=\frac{b^3 x}{d^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 (c+d x^2\right )^3} \, dx}{d^3}\\ &=\frac{b^3 x}{d^3}-\frac{(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac{\int \frac{-3 (b c-a d) (b c+a d)^2-12 b^2 c d (b c-a d) x^2}{\left (c+d x^2\right )^2} \, dx}{4 c d^3}\\ &=\frac{b^3 x}{d^3}-\frac{(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac{3 (b c-a d)^2 (3 b c+a d) x}{8 c^2 d^3 \left (c+d x^2\right )}-\frac{\left (3 (b c-a d) \left (4 b^2 c^2+(b c+a d)^2\right )\right ) \int \frac{1}{c+d x^2} \, dx}{8 c^2 d^3}\\ &=\frac{b^3 x}{d^3}-\frac{(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac{3 (b c-a d)^2 (3 b c+a d) x}{8 c^2 d^3 \left (c+d x^2\right )}-\frac{3 (b c-a d) \left (4 b^2 c^2+(b c+a d)^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0782675, size = 141, normalized size = 1.08 \[ -\frac{3 \left (-a^2 b c d^2-a^3 d^3-3 a b^2 c^2 d+5 b^3 c^3\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} d^{7/2}}+\frac{3 x (b c-a d)^2 (a d+3 b c)}{8 c^2 d^3 \left (c+d x^2\right )}-\frac{x (b c-a d)^3}{4 c d^3 \left (c+d x^2\right )^2}+\frac{b^3 x}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 266, normalized size = 2.1 \begin{align*}{\frac{{b}^{3}x}{{d}^{3}}}+{\frac{3\,d{x}^{3}{a}^{3}}{8\, \left ( d{x}^{2}+c \right ) ^{2}{c}^{2}}}+{\frac{3\,{x}^{3}{a}^{2}b}{8\, \left ( d{x}^{2}+c \right ) ^{2}c}}-{\frac{15\,a{b}^{2}{x}^{3}}{8\,d \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{9\,c{x}^{3}{b}^{3}}{8\,{d}^{2} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{5\,x{a}^{3}}{8\, \left ( d{x}^{2}+c \right ) ^{2}c}}-{\frac{3\,{a}^{2}bx}{8\,d \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{9\,acx{b}^{2}}{8\,{d}^{2} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{7\,{c}^{2}x{b}^{3}}{8\,{d}^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{3\,{a}^{3}}{8\,{c}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{3\,{a}^{2}b}{8\,cd}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{9\,a{b}^{2}}{8\,{d}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{15\,{b}^{3}c}{8\,{d}^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.93027, size = 1224, normalized size = 9.42 \begin{align*} \left [\frac{16 \, b^{3} c^{3} d^{3} x^{5} + 2 \,{\left (25 \, b^{3} c^{4} d^{2} - 15 \, a b^{2} c^{3} d^{3} + 3 \, a^{2} b c^{2} d^{4} + 3 \, a^{3} c d^{5}\right )} x^{3} + 3 \,{\left (5 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3} +{\left (5 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} - a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{4} + 2 \,{\left (5 \, b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} - a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} x^{2}\right )} \sqrt{-c d} \log \left (\frac{d x^{2} - 2 \, \sqrt{-c d} x - c}{d x^{2} + c}\right ) + 2 \,{\left (15 \, b^{3} c^{5} d - 9 \, a b^{2} c^{4} d^{2} - 3 \, a^{2} b c^{3} d^{3} + 5 \, a^{3} c^{2} d^{4}\right )} x}{16 \,{\left (c^{3} d^{6} x^{4} + 2 \, c^{4} d^{5} x^{2} + c^{5} d^{4}\right )}}, \frac{8 \, b^{3} c^{3} d^{3} x^{5} +{\left (25 \, b^{3} c^{4} d^{2} - 15 \, a b^{2} c^{3} d^{3} + 3 \, a^{2} b c^{2} d^{4} + 3 \, a^{3} c d^{5}\right )} x^{3} - 3 \,{\left (5 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3} +{\left (5 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} - a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{4} + 2 \,{\left (5 \, b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} - a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} x^{2}\right )} \sqrt{c d} \arctan \left (\frac{\sqrt{c d} x}{c}\right ) +{\left (15 \, b^{3} c^{5} d - 9 \, a b^{2} c^{4} d^{2} - 3 \, a^{2} b c^{3} d^{3} + 5 \, a^{3} c^{2} d^{4}\right )} x}{8 \,{\left (c^{3} d^{6} x^{4} + 2 \, c^{4} d^{5} x^{2} + c^{5} d^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.18939, size = 422, normalized size = 3.25 \begin{align*} \frac{b^{3} x}{d^{3}} - \frac{3 \sqrt{- \frac{1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right ) \log{\left (- \frac{3 c^{3} d^{3} \sqrt{- \frac{1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right )}{3 a^{3} d^{3} + 3 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - 15 b^{3} c^{3}} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right ) \log{\left (\frac{3 c^{3} d^{3} \sqrt{- \frac{1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right )}{3 a^{3} d^{3} + 3 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - 15 b^{3} c^{3}} + x \right )}}{16} + \frac{x^{3} \left (3 a^{3} d^{4} + 3 a^{2} b c d^{3} - 15 a b^{2} c^{2} d^{2} + 9 b^{3} c^{3} d\right ) + x \left (5 a^{3} c d^{3} - 3 a^{2} b c^{2} d^{2} - 9 a b^{2} c^{3} d + 7 b^{3} c^{4}\right )}{8 c^{4} d^{3} + 16 c^{3} d^{4} x^{2} + 8 c^{2} d^{5} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10929, size = 243, normalized size = 1.87 \begin{align*} \frac{b^{3} x}{d^{3}} - \frac{3 \,{\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{8 \, \sqrt{c d} c^{2} d^{3}} + \frac{9 \, b^{3} c^{3} d x^{3} - 15 \, a b^{2} c^{2} d^{2} x^{3} + 3 \, a^{2} b c d^{3} x^{3} + 3 \, a^{3} d^{4} x^{3} + 7 \, b^{3} c^{4} x - 9 \, a b^{2} c^{3} d x - 3 \, a^{2} b c^{2} d^{2} x + 5 \, a^{3} c d^{3} x}{8 \,{\left (d x^{2} + c\right )}^{2} c^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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