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

Optimal. Leaf size=160 \[ \frac{(b c-a d)^2 \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{9/2}}+\frac{x (b c-a d)^3 (13 a d+3 b c)}{8 a^2 b^4 \left (a+b x^2\right )}+\frac{d^3 x (4 b c-3 a d)}{b^4}+\frac{x (b c-a d)^4}{4 a b^4 \left (a+b x^2\right )^2}+\frac{d^4 x^3}{3 b^3} \]

[Out]

(d^3*(4*b*c - 3*a*d)*x)/b^4 + (d^4*x^3)/(3*b^3) + ((b*c - a*d)^4*x)/(4*a*b^4*(a + b*x^2)^2) + ((b*c - a*d)^3*(
3*b*c + 13*a*d)*x)/(8*a^2*b^4*(a + b*x^2)) + ((b*c - a*d)^2*(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(9/2))

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Rubi [A]  time = 0.197019, antiderivative size = 160, 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} \[ \frac{(b c-a d)^2 \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{9/2}}+\frac{x (b c-a d)^3 (13 a d+3 b c)}{8 a^2 b^4 \left (a+b x^2\right )}+\frac{d^3 x (4 b c-3 a d)}{b^4}+\frac{x (b c-a d)^4}{4 a b^4 \left (a+b x^2\right )^2}+\frac{d^4 x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^4/(a + b*x^2)^3,x]

[Out]

(d^3*(4*b*c - 3*a*d)*x)/b^4 + (d^4*x^3)/(3*b^3) + ((b*c - a*d)^4*x)/(4*a*b^4*(a + b*x^2)^2) + ((b*c - a*d)^3*(
3*b*c + 13*a*d)*x)/(8*a^2*b^4*(a + b*x^2)) + ((b*c - a*d)^2*(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(9/2))

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 )^4}{\left (a+b x^2\right )^3} \, dx &=\int \left (\frac{d^3 (4 b c-3 a d)}{b^4}+\frac{d^4 x^2}{b^3}+\frac{b^4 c^4-4 a^3 b c d^3+3 a^4 d^4+4 b d (b c-a d)^2 (b c+2 a d) x^2+6 b^2 d^2 (b c-a d)^2 x^4}{b^4 \left (a+b x^2\right )^3}\right ) \, dx\\ &=\frac{d^3 (4 b c-3 a d) x}{b^4}+\frac{d^4 x^3}{3 b^3}+\frac{\int \frac{b^4 c^4-4 a^3 b c d^3+3 a^4 d^4+4 b d (b c-a d)^2 (b c+2 a d) x^2+6 b^2 d^2 (b c-a d)^2 x^4}{\left (a+b x^2\right )^3} \, dx}{b^4}\\ &=\frac{d^3 (4 b c-3 a d) x}{b^4}+\frac{d^4 x^3}{3 b^3}+\frac{(b c-a d)^4 x}{4 a b^4 \left (a+b x^2\right )^2}-\frac{\int \frac{-(b c-a d)^2 \left (3 b^2 c^2+10 a b c d+11 a^2 d^2\right )-24 a b d^2 (b c-a d)^2 x^2}{\left (a+b x^2\right )^2} \, dx}{4 a b^4}\\ &=\frac{d^3 (4 b c-3 a d) x}{b^4}+\frac{d^4 x^3}{3 b^3}+\frac{(b c-a d)^4 x}{4 a b^4 \left (a+b x^2\right )^2}+\frac{(b c-a d)^3 (3 b c+13 a d) x}{8 a^2 b^4 \left (a+b x^2\right )}+\frac{\left ((b c-a d)^2 \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right )\right ) \int \frac{1}{a+b x^2} \, dx}{8 a^2 b^4}\\ &=\frac{d^3 (4 b c-3 a d) x}{b^4}+\frac{d^4 x^3}{3 b^3}+\frac{(b c-a d)^4 x}{4 a b^4 \left (a+b x^2\right )^2}+\frac{(b c-a d)^3 (3 b c+13 a d) x}{8 a^2 b^4 \left (a+b x^2\right )}+\frac{(b c-a d)^2 \left (3 b^2 c^2+10 a b c d+35 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0948017, size = 160, normalized size = 1. \[ \frac{(b c-a d)^2 \left (35 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{9/2}}+\frac{x (b c-a d)^3 (13 a d+3 b c)}{8 a^2 b^4 \left (a+b x^2\right )}+\frac{d^3 x (4 b c-3 a d)}{b^4}+\frac{x (b c-a d)^4}{4 a b^4 \left (a+b x^2\right )^2}+\frac{d^4 x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(4*b*c - 3*a*d)*x)/b^4 + (d^4*x^3)/(3*b^3) + ((b*c - a*d)^4*x)/(4*a*b^4*(a + b*x^2)^2) + ((b*c - a*d)^3*(
3*b*c + 13*a*d)*x)/(8*a^2*b^4*(a + b*x^2)) + ((b*c - a*d)^2*(3*b^2*c^2 + 10*a*b*c*d + 35*a^2*d^2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(9/2))

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Maple [B]  time = 0.01, size = 367, normalized size = 2.3 \begin{align*}{\frac{{d}^{4}{x}^{3}}{3\,{b}^{3}}}-3\,{\frac{a{d}^{4}x}{{b}^{4}}}+4\,{\frac{{d}^{3}xc}{{b}^{3}}}-{\frac{13\,{a}^{2}{x}^{3}{d}^{4}}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{9\,a{x}^{3}c{d}^{3}}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{15\,{x}^{3}{c}^{2}{d}^{2}}{4\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{{x}^{3}{c}^{3}d}{2\, \left ( b{x}^{2}+a \right ) ^{2}a}}+{\frac{3\,b{x}^{3}{c}^{4}}{8\, \left ( b{x}^{2}+a \right ) ^{2}{a}^{2}}}-{\frac{11\,x{a}^{3}{d}^{4}}{8\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{a}^{2}cx{d}^{3}}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,a{c}^{2}x{d}^{2}}{4\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{x{c}^{3}d}{2\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,x{c}^{4}}{8\, \left ( b{x}^{2}+a \right ) ^{2}a}}+{\frac{35\,{a}^{2}{d}^{4}}{8\,{b}^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{15\,ac{d}^{3}}{2\,{b}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{9\,{c}^{2}{d}^{2}}{4\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{c}^{3}d}{2\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,{c}^{4}}{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)^4/(b*x^2+a)^3,x)

[Out]

1/3*d^4*x^3/b^3-3*d^4/b^4*a*x+4*d^3/b^3*x*c-13/8/b^3/(b*x^2+a)^2*a^2*x^3*d^4+9/2/b^2/(b*x^2+a)^2*a*x^3*c*d^3-1
5/4/b/(b*x^2+a)^2*x^3*c^2*d^2+1/2/(b*x^2+a)^2/a*x^3*c^3*d+3/8*b/(b*x^2+a)^2/a^2*x^3*c^4-11/8/b^4/(b*x^2+a)^2*x
*a^3*d^4+7/2/b^3/(b*x^2+a)^2*x*a^2*c*d^3-9/4/b^2/(b*x^2+a)^2*x*a*c^2*d^2-1/2/b/(b*x^2+a)^2*x*c^3*d+5/8/(b*x^2+
a)^2*x/a*c^4+35/8/b^4*a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d^4-15/2/b^3*a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2
))*c*d^3+9/4/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c^2*d^2+1/2/b/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c^3*d
+3/8/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c^4

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.7646, size = 1689, normalized size = 10.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(16*a^3*b^4*d^4*x^7 + 16*(12*a^3*b^4*c*d^3 - 7*a^4*b^3*d^4)*x^5 + 2*(9*a*b^6*c^4 + 12*a^2*b^5*c^3*d - 90
*a^3*b^4*c^2*d^2 + 300*a^4*b^3*c*d^3 - 175*a^5*b^2*d^4)*x^3 - 3*(3*a^2*b^4*c^4 + 4*a^3*b^3*c^3*d + 18*a^4*b^2*
c^2*d^2 - 60*a^5*b*c*d^3 + 35*a^6*d^4 + (3*b^6*c^4 + 4*a*b^5*c^3*d + 18*a^2*b^4*c^2*d^2 - 60*a^3*b^3*c*d^3 + 3
5*a^4*b^2*d^4)*x^4 + 2*(3*a*b^5*c^4 + 4*a^2*b^4*c^3*d + 18*a^3*b^3*c^2*d^2 - 60*a^4*b^2*c*d^3 + 35*a^5*b*d^4)*
x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 6*(5*a^2*b^5*c^4 - 4*a^3*b^4*c^3*d - 18*a^4*b^
3*c^2*d^2 + 60*a^5*b^2*c*d^3 - 35*a^6*b*d^4)*x)/(a^3*b^7*x^4 + 2*a^4*b^6*x^2 + a^5*b^5), 1/24*(8*a^3*b^4*d^4*x
^7 + 8*(12*a^3*b^4*c*d^3 - 7*a^4*b^3*d^4)*x^5 + (9*a*b^6*c^4 + 12*a^2*b^5*c^3*d - 90*a^3*b^4*c^2*d^2 + 300*a^4
*b^3*c*d^3 - 175*a^5*b^2*d^4)*x^3 + 3*(3*a^2*b^4*c^4 + 4*a^3*b^3*c^3*d + 18*a^4*b^2*c^2*d^2 - 60*a^5*b*c*d^3 +
35*a^6*d^4 + (3*b^6*c^4 + 4*a*b^5*c^3*d + 18*a^2*b^4*c^2*d^2 - 60*a^3*b^3*c*d^3 + 35*a^4*b^2*d^4)*x^4 + 2*(3*
a*b^5*c^4 + 4*a^2*b^4*c^3*d + 18*a^3*b^3*c^2*d^2 - 60*a^4*b^2*c*d^3 + 35*a^5*b*d^4)*x^2)*sqrt(a*b)*arctan(sqrt
(a*b)*x/a) + 3*(5*a^2*b^5*c^4 - 4*a^3*b^4*c^3*d - 18*a^4*b^3*c^2*d^2 + 60*a^5*b^2*c*d^3 - 35*a^6*b*d^4)*x)/(a^
3*b^7*x^4 + 2*a^4*b^6*x^2 + a^5*b^5)]

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Sympy [B]  time = 3.71684, size = 513, normalized size = 3.21 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b^{9}}} \left (a d - b c\right )^{2} \left (35 a^{2} d^{2} + 10 a b c d + 3 b^{2} c^{2}\right ) \log{\left (- \frac{a^{3} b^{4} \sqrt{- \frac{1}{a^{5} b^{9}}} \left (a d - b c\right )^{2} \left (35 a^{2} d^{2} + 10 a b c d + 3 b^{2} c^{2}\right )}{35 a^{4} d^{4} - 60 a^{3} b c d^{3} + 18 a^{2} b^{2} c^{2} d^{2} + 4 a b^{3} c^{3} d + 3 b^{4} c^{4}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{5} b^{9}}} \left (a d - b c\right )^{2} \left (35 a^{2} d^{2} + 10 a b c d + 3 b^{2} c^{2}\right ) \log{\left (\frac{a^{3} b^{4} \sqrt{- \frac{1}{a^{5} b^{9}}} \left (a d - b c\right )^{2} \left (35 a^{2} d^{2} + 10 a b c d + 3 b^{2} c^{2}\right )}{35 a^{4} d^{4} - 60 a^{3} b c d^{3} + 18 a^{2} b^{2} c^{2} d^{2} + 4 a b^{3} c^{3} d + 3 b^{4} c^{4}} + x \right )}}{16} - \frac{x^{3} \left (13 a^{4} b d^{4} - 36 a^{3} b^{2} c d^{3} + 30 a^{2} b^{3} c^{2} d^{2} - 4 a b^{4} c^{3} d - 3 b^{5} c^{4}\right ) + x \left (11 a^{5} d^{4} - 28 a^{4} b c d^{3} + 18 a^{3} b^{2} c^{2} d^{2} + 4 a^{2} b^{3} c^{3} d - 5 a b^{4} c^{4}\right )}{8 a^{4} b^{4} + 16 a^{3} b^{5} x^{2} + 8 a^{2} b^{6} x^{4}} + \frac{d^{4} x^{3}}{3 b^{3}} - \frac{x \left (3 a d^{4} - 4 b c d^{3}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**5*b**9))*(a*d - b*c)**2*(35*a**2*d**2 + 10*a*b*c*d + 3*b**2*c**2)*log(-a**3*b**4*sqrt(-1/(a**5*b*
*9))*(a*d - b*c)**2*(35*a**2*d**2 + 10*a*b*c*d + 3*b**2*c**2)/(35*a**4*d**4 - 60*a**3*b*c*d**3 + 18*a**2*b**2*
c**2*d**2 + 4*a*b**3*c**3*d + 3*b**4*c**4) + x)/16 + sqrt(-1/(a**5*b**9))*(a*d - b*c)**2*(35*a**2*d**2 + 10*a*
b*c*d + 3*b**2*c**2)*log(a**3*b**4*sqrt(-1/(a**5*b**9))*(a*d - b*c)**2*(35*a**2*d**2 + 10*a*b*c*d + 3*b**2*c**
2)/(35*a**4*d**4 - 60*a**3*b*c*d**3 + 18*a**2*b**2*c**2*d**2 + 4*a*b**3*c**3*d + 3*b**4*c**4) + x)/16 - (x**3*
(13*a**4*b*d**4 - 36*a**3*b**2*c*d**3 + 30*a**2*b**3*c**2*d**2 - 4*a*b**4*c**3*d - 3*b**5*c**4) + x*(11*a**5*d
**4 - 28*a**4*b*c*d**3 + 18*a**3*b**2*c**2*d**2 + 4*a**2*b**3*c**3*d - 5*a*b**4*c**4))/(8*a**4*b**4 + 16*a**3*
b**5*x**2 + 8*a**2*b**6*x**4) + d**4*x**3/(3*b**3) - x*(3*a*d**4 - 4*b*c*d**3)/b**4

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Giac [A]  time = 1.10662, size = 343, normalized size = 2.14 \begin{align*} \frac{{\left (3 \, b^{4} c^{4} + 4 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} - 60 \, a^{3} b c d^{3} + 35 \, a^{4} d^{4}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2} b^{4}} + \frac{3 \, b^{5} c^{4} x^{3} + 4 \, a b^{4} c^{3} d x^{3} - 30 \, a^{2} b^{3} c^{2} d^{2} x^{3} + 36 \, a^{3} b^{2} c d^{3} x^{3} - 13 \, a^{4} b d^{4} x^{3} + 5 \, a b^{4} c^{4} x - 4 \, a^{2} b^{3} c^{3} d x - 18 \, a^{3} b^{2} c^{2} d^{2} x + 28 \, a^{4} b c d^{3} x - 11 \, a^{5} d^{4} x}{8 \,{\left (b x^{2} + a\right )}^{2} a^{2} b^{4}} + \frac{b^{6} d^{4} x^{3} + 12 \, b^{6} c d^{3} x - 9 \, a b^{5} d^{4} x}{3 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b^4*c^4 + 4*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 - 60*a^3*b*c*d^3 + 35*a^4*d^4)*arctan(b*x/sqrt(a*b))/(sqrt
(a*b)*a^2*b^4) + 1/8*(3*b^5*c^4*x^3 + 4*a*b^4*c^3*d*x^3 - 30*a^2*b^3*c^2*d^2*x^3 + 36*a^3*b^2*c*d^3*x^3 - 13*a
^4*b*d^4*x^3 + 5*a*b^4*c^4*x - 4*a^2*b^3*c^3*d*x - 18*a^3*b^2*c^2*d^2*x + 28*a^4*b*c*d^3*x - 11*a^5*d^4*x)/((b
*x^2 + a)^2*a^2*b^4) + 1/3*(b^6*d^4*x^3 + 12*b^6*c*d^3*x - 9*a*b^5*d^4*x)/b^9