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

Optimal. Leaf size=123 \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b^2 \sqrt{b c-a d}}-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{4 b^2 d^{3/2}}+\frac{x^2 \sqrt{c+d x^4}}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.149109, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {465, 479, 523, 217, 206, 377, 205} \[ \frac{a^{3/2} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b^2 \sqrt{b c-a d}}-\frac{(2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{4 b^2 d^{3/2}}+\frac{x^2 \sqrt{c+d x^4}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 523

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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

Mathematica [A]  time = 0.129691, size = 118, normalized size = 0.96 \[ \frac{\frac{2 a^{3/2} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{\sqrt{b c-a d}}-\frac{(2 a d+b c) \log \left (\sqrt{d} \sqrt{c+d x^4}+d x^2\right )}{d^{3/2}}+\frac{b x^2 \sqrt{c+d x^4}}{d}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.027, size = 408, normalized size = 3.3 \begin{align*}{\frac{{x}^{2}}{4\,bd}\sqrt{d{x}^{4}+c}}-{\frac{c}{4\,b}\ln \left ({x}^{2}\sqrt{d}+\sqrt{d{x}^{4}+c} \right ){d}^{-{\frac{3}{2}}}}-{\frac{a}{2\,{b}^{2}}\ln \left ({x}^{2}\sqrt{d}+\sqrt{d{x}^{4}+c} \right ){\frac{1}{\sqrt{d}}}}-{\frac{{a}^{2}}{4\,{b}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ({x}^{2}-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ({x}^{2}-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}+{\frac{{a}^{2}}{4\,{b}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ({x}^{2}+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ({x}^{2}+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ({x}^{2}+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)/(d*x^4+c)^(1/2),x)

[Out]

1/4*x^2*(d*x^4+c)^(1/2)/b/d-1/4/b*c/d^(3/2)*ln(x^2*d^(1/2)+(d*x^4+c)^(1/2))-1/2/b^2*a*ln(x^2*d^(1/2)+(d*x^4+c)
^(1/2))/d^(1/2)-1/4*a^2/b^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x^2-(-a*b
)^(1/2)/b)+2*(-(a*d-b*c)/b)^(1/2)*((x^2-(-a*b)^(1/2)/b)^2*d+2*d*(-a*b)^(1/2)/b*(x^2-(-a*b)^(1/2)/b)-(a*d-b*c)/
b)^(1/2))/(x^2-(-a*b)^(1/2)/b))+1/4*a^2/b^2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1
/2)/b*(x^2+(-a*b)^(1/2)/b)+2*(-(a*d-b*c)/b)^(1/2)*((x^2+(-a*b)^(1/2)/b)^2*d-2*d*(-a*b)^(1/2)/b*(x^2+(-a*b)^(1/
2)/b)-(a*d-b*c)/b)^(1/2))/(x^2+(-a*b)^(1/2)/b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9}}{{\left (b x^{4} + a\right )} \sqrt{d x^{4} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.4851, size = 1623, normalized size = 13.2 \begin{align*} \left [\frac{2 \, \sqrt{d x^{4} + c} b d x^{2} + a d^{2} \sqrt{-\frac{a}{b c - a d}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} + 4 \,{\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{6} -{\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{4} + c} \sqrt{-\frac{a}{b c - a d}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) +{\left (b c + 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{4} + 2 \, \sqrt{d x^{4} + c} \sqrt{d} x^{2} - c\right )}{8 \, b^{2} d^{2}}, \frac{2 \, \sqrt{d x^{4} + c} b d x^{2} + a d^{2} \sqrt{-\frac{a}{b c - a d}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} + 4 \,{\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{6} -{\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{4} + c} \sqrt{-\frac{a}{b c - a d}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right ) + 2 \,{\left (b c + 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x^{2}}{\sqrt{d x^{4} + c}}\right )}{8 \, b^{2} d^{2}}, \frac{2 \, \sqrt{d x^{4} + c} b d x^{2} - 2 \, a d^{2} \sqrt{\frac{a}{b c - a d}} \arctan \left (-\frac{{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c} \sqrt{\frac{a}{b c - a d}}}{2 \,{\left (a d x^{6} + a c x^{2}\right )}}\right ) +{\left (b c + 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{4} + 2 \, \sqrt{d x^{4} + c} \sqrt{d} x^{2} - c\right )}{8 \, b^{2} d^{2}}, \frac{\sqrt{d x^{4} + c} b d x^{2} - a d^{2} \sqrt{\frac{a}{b c - a d}} \arctan \left (-\frac{{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c} \sqrt{\frac{a}{b c - a d}}}{2 \,{\left (a d x^{6} + a c x^{2}\right )}}\right ) +{\left (b c + 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x^{2}}{\sqrt{d x^{4} + c}}\right )}{4 \, b^{2} d^{2}}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(2*sqrt(d*x^4 + c)*b*d*x^2 + a*d^2*sqrt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3
*a*b*c^2 - 4*a^2*c*d)*x^4 + a^2*c^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^6 - (a*b*c^2 - a^2*c*d)*x^2)*sqrt
(d*x^4 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^8 + 2*a*b*x^4 + a^2)) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^4 + 2*sqrt(d
*x^4 + c)*sqrt(d)*x^2 - c))/(b^2*d^2), 1/8*(2*sqrt(d*x^4 + c)*b*d*x^2 + a*d^2*sqrt(-a/(b*c - a*d))*log(((b^2*c
^2 - 8*a*b*c*d + 8*a^2*d^2)*x^8 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^4 + a^2*c^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^
2)*x^6 - (a*b*c^2 - a^2*c*d)*x^2)*sqrt(d*x^4 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^8 + 2*a*b*x^4 + a^2)) + 2*(b*c
+ 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c)))/(b^2*d^2), 1/8*(2*sqrt(d*x^4 + c)*b*d*x^2 - 2*a*d^2*sq
rt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a/(b*c - a*d))/(a*d*x^6 + a*c*x^2
)) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^4 + 2*sqrt(d*x^4 + c)*sqrt(d)*x^2 - c))/(b^2*d^2), 1/4*(sqrt(d*x^4 + c)*
b*d*x^2 - a*d^2*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a/(b*c - a*d))/
(a*d*x^6 + a*c*x^2)) + (b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c)))/(b^2*d^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{9}}{\left (a + b x^{4}\right ) \sqrt{c + d x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.33182, size = 140, normalized size = 1.14 \begin{align*} \frac{\sqrt{d x^{4} + c} x^{2}}{4 \, b d} - \frac{a^{2} \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{4}}}}{\sqrt{a b c - a^{2} d}}\right )}{2 \, \sqrt{a b c - a^{2} d} b^{2}} + \frac{{\left (b c + 2 \, a d\right )} \arctan \left (\frac{\sqrt{d + \frac{c}{x^{4}}}}{\sqrt{-d}}\right )}{4 \, b^{2} \sqrt{-d} 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)^(1/2),x, algorithm="giac")

[Out]

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