3.774 \(\int \frac{x^9}{(a+b x^4) (c+d x^4)} \, dx\)

Optimal. Leaf size=92 \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 b^{3/2} (b c-a d)}-\frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c}}\right )}{2 d^{3/2} (b c-a d)}+\frac{x^2}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.118431, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {465, 479, 522, 205} \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 b^{3/2} (b c-a d)}-\frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c}}\right )}{2 d^{3/2} (b c-a d)}+\frac{x^2}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^9/((a + b*x^4)*(c + d*x^4)),x]

[Out]

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

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 479

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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{x^9}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )\\ &=\frac{x^2}{2 b d}-\frac{\operatorname{Subst}\left (\int \frac{a c+(b c+a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )}{2 b d}\\ &=\frac{x^2}{2 b d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,x^2\right )}{2 b (b c-a d)}-\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{c+d x^2} \, dx,x,x^2\right )}{2 d (b c-a d)}\\ &=\frac{x^2}{2 b d}+\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 b^{3/2} (b c-a d)}-\frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c}}\right )}{2 d^{3/2} (b c-a d)}\\ \end{align*}

Mathematica [A]  time = 0.13363, size = 82, normalized size = 0.89 \[ \frac{\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{b^{3/2}}+x^2 \left (\frac{c}{d}-\frac{a}{b}\right )-\frac{c^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c}}\right )}{d^{3/2}}}{2 b c-2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/((a + b*x^4)*(c + d*x^4)),x]

[Out]

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

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Maple [A]  time = 0.009, size = 81, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,bd}}+{\frac{{c}^{2}}{2\,d \left ( ad-bc \right ) }\arctan \left ({{x}^{2}d{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{{a}^{2}}{2\,b \left ( ad-bc \right ) }\arctan \left ({b{x}^{2}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)/(d*x^4+c),x)

[Out]

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

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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(x^9/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54164, size = 848, normalized size = 9.22 \begin{align*} \left [-\frac{a d \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{4} - 2 \, b x^{2} \sqrt{-\frac{a}{b}} - a}{b x^{4} + a}\right ) + b c \sqrt{-\frac{c}{d}} \log \left (\frac{d x^{4} + 2 \, d x^{2} \sqrt{-\frac{c}{d}} - c}{d x^{4} + c}\right ) - 2 \,{\left (b c - a d\right )} x^{2}}{4 \,{\left (b^{2} c d - a b d^{2}\right )}}, \frac{2 \, a d \sqrt{\frac{a}{b}} \arctan \left (\frac{b x^{2} \sqrt{\frac{a}{b}}}{a}\right ) - b c \sqrt{-\frac{c}{d}} \log \left (\frac{d x^{4} + 2 \, d x^{2} \sqrt{-\frac{c}{d}} - c}{d x^{4} + c}\right ) + 2 \,{\left (b c - a d\right )} x^{2}}{4 \,{\left (b^{2} c d - a b d^{2}\right )}}, -\frac{2 \, b c \sqrt{\frac{c}{d}} \arctan \left (\frac{d x^{2} \sqrt{\frac{c}{d}}}{c}\right ) + a d \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{4} - 2 \, b x^{2} \sqrt{-\frac{a}{b}} - a}{b x^{4} + a}\right ) - 2 \,{\left (b c - a d\right )} x^{2}}{4 \,{\left (b^{2} c d - a b d^{2}\right )}}, \frac{a d \sqrt{\frac{a}{b}} \arctan \left (\frac{b x^{2} \sqrt{\frac{a}{b}}}{a}\right ) - b c \sqrt{\frac{c}{d}} \arctan \left (\frac{d x^{2} \sqrt{\frac{c}{d}}}{c}\right ) +{\left (b c - a d\right )} x^{2}}{2 \,{\left (b^{2} c d - a b d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 10.8265, size = 932, normalized size = 10.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)/(d*x**4+c),x)

[Out]

-sqrt(-a**3/b**3)*log(x**2 + (-a**4*d**4*sqrt(-a**3/b**3)/(a*d - b*c) - a**3*b**3*d**6*(-a**3/b**3)**(3/2)/(a*
d - b*c)**3 + a**2*b**4*c*d**5*(-a**3/b**3)**(3/2)/(a*d - b*c)**3 + a*b**5*c**2*d**4*(-a**3/b**3)**(3/2)/(a*d
- b*c)**3 - b**6*c**3*d**3*(-a**3/b**3)**(3/2)/(a*d - b*c)**3 - b**4*c**4*sqrt(-a**3/b**3)/(a*d - b*c))/(a**3*
c*d**2 + a**2*b*c**2*d + a*b**2*c**3))/(4*(a*d - b*c)) + sqrt(-a**3/b**3)*log(x**2 + (a**4*d**4*sqrt(-a**3/b**
3)/(a*d - b*c) + a**3*b**3*d**6*(-a**3/b**3)**(3/2)/(a*d - b*c)**3 - a**2*b**4*c*d**5*(-a**3/b**3)**(3/2)/(a*d
- b*c)**3 - a*b**5*c**2*d**4*(-a**3/b**3)**(3/2)/(a*d - b*c)**3 + b**6*c**3*d**3*(-a**3/b**3)**(3/2)/(a*d - b
*c)**3 + b**4*c**4*sqrt(-a**3/b**3)/(a*d - b*c))/(a**3*c*d**2 + a**2*b*c**2*d + a*b**2*c**3))/(4*(a*d - b*c))
- sqrt(-c**3/d**3)*log(x**2 + (-a**4*d**4*sqrt(-c**3/d**3)/(a*d - b*c) - a**3*b**3*d**6*(-c**3/d**3)**(3/2)/(a
*d - b*c)**3 + a**2*b**4*c*d**5*(-c**3/d**3)**(3/2)/(a*d - b*c)**3 + a*b**5*c**2*d**4*(-c**3/d**3)**(3/2)/(a*d
- b*c)**3 - b**6*c**3*d**3*(-c**3/d**3)**(3/2)/(a*d - b*c)**3 - b**4*c**4*sqrt(-c**3/d**3)/(a*d - b*c))/(a**3
*c*d**2 + a**2*b*c**2*d + a*b**2*c**3))/(4*(a*d - b*c)) + sqrt(-c**3/d**3)*log(x**2 + (a**4*d**4*sqrt(-c**3/d*
*3)/(a*d - b*c) + a**3*b**3*d**6*(-c**3/d**3)**(3/2)/(a*d - b*c)**3 - a**2*b**4*c*d**5*(-c**3/d**3)**(3/2)/(a*
d - b*c)**3 - a*b**5*c**2*d**4*(-c**3/d**3)**(3/2)/(a*d - b*c)**3 + b**6*c**3*d**3*(-c**3/d**3)**(3/2)/(a*d -
b*c)**3 + b**4*c**4*sqrt(-c**3/d**3)/(a*d - b*c))/(a**3*c*d**2 + a**2*b*c**2*d + a*b**2*c**3))/(4*(a*d - b*c))
+ x**2/(2*b*d)

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Giac [B]  time = 1.19765, size = 482, normalized size = 5.24 \begin{align*} \frac{{\left (\sqrt{a b} b c d^{2} x^{4}{\left | b \right |} + \sqrt{a b} a d^{3} x^{4}{\left | b \right |} + \sqrt{a b} a c d^{2}{\left | b \right |}\right )} \arctan \left (\frac{4 \, \sqrt{\frac{1}{2}} x^{2}}{\sqrt{\frac{4 \, b^{2} c d + 4 \, a b d^{2} + \sqrt{-64 \, a b^{3} c d^{3} + 16 \,{\left (b^{2} c d + a b d^{2}\right )}^{2}}}{b^{2} d^{2}}}}\right )}{b^{2} c d{\left | -b^{2} c d + a b d^{2} \right |} + a b d^{2}{\left | -b^{2} c d + a b d^{2} \right |} +{\left (b^{2} c d - a b d^{2}\right )}^{2}} - \frac{{\left (\sqrt{c d} b^{3} c x^{4}{\left | d \right |} + \sqrt{c d} a b^{2} d x^{4}{\left | d \right |} + \sqrt{c d} a b^{2} c{\left | d \right |}\right )} \arctan \left (\frac{4 \, \sqrt{\frac{1}{2}} x^{2}}{\sqrt{\frac{4 \, b^{2} c d + 4 \, a b d^{2} - \sqrt{-64 \, a b^{3} c d^{3} + 16 \,{\left (b^{2} c d + a b d^{2}\right )}^{2}}}{b^{2} d^{2}}}}\right )}{b^{2} c d{\left | -b^{2} c d + a b d^{2} \right |} + a b d^{2}{\left | -b^{2} c d + a b d^{2} \right |} -{\left (b^{2} c d - a b d^{2}\right )}^{2}} + \frac{x^{2}}{2 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")

[Out]

(sqrt(a*b)*b*c*d^2*x^4*abs(b) + sqrt(a*b)*a*d^3*x^4*abs(b) + sqrt(a*b)*a*c*d^2*abs(b))*arctan(4*sqrt(1/2)*x^2/
sqrt((4*b^2*c*d + 4*a*b*d^2 + sqrt(-64*a*b^3*c*d^3 + 16*(b^2*c*d + a*b*d^2)^2))/(b^2*d^2)))/(b^2*c*d*abs(-b^2*
c*d + a*b*d^2) + a*b*d^2*abs(-b^2*c*d + a*b*d^2) + (b^2*c*d - a*b*d^2)^2) - (sqrt(c*d)*b^3*c*x^4*abs(d) + sqrt
(c*d)*a*b^2*d*x^4*abs(d) + sqrt(c*d)*a*b^2*c*abs(d))*arctan(4*sqrt(1/2)*x^2/sqrt((4*b^2*c*d + 4*a*b*d^2 - sqrt
(-64*a*b^3*c*d^3 + 16*(b^2*c*d + a*b*d^2)^2))/(b^2*d^2)))/(b^2*c*d*abs(-b^2*c*d + a*b*d^2) + a*b*d^2*abs(-b^2*
c*d + a*b*d^2) - (b^2*c*d - a*b*d^2)^2) + 1/2*x^2/(b*d)