3.811 \(\int \frac{x^5}{(a+b x^4) \sqrt{c+d x^4}} \, dx\)

Optimal. Leaf size=91 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{2 b \sqrt{d}}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b \sqrt{b c-a d}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0874657, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {465, 483, 217, 206, 377, 205} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{2 b \sqrt{d}}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b \sqrt{b c-a d}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

-(Sqrt[a]*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(2*b*Sqrt[b*c - a*d]) + ArcTanh[(Sqrt[d]*x^
2)/Sqrt[c + d*x^4]]/(2*b*Sqrt[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 483

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Dist[e^n/b, Int[
(e*x)^(m - n)*(c + d*x^n)^q, x], x] - Dist[(a*e^n)/b, Int[((e*x)^(m - n)*(c + d*x^n)^q)/(a + b*x^n), x], x] /;
FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1] && IntBinomialQ[a, b
, c, d, e, m, n, -1, q, 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^5}{\left (a+b x^4\right ) \sqrt{c+d x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,x^2\right )}{2 b}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^2\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x^2}{\sqrt{c+d x^4}}\right )}{2 b}-\frac{a \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}\\ &=-\frac{\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^2}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b \sqrt{b c-a d}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{2 b \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.0571033, size = 90, normalized size = 0.99 \[ \frac{\frac{\log \left (\sqrt{d} \sqrt{c+d x^4}+d x^2\right )}{\sqrt{d}}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{\sqrt{b c-a d}}}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((a + b*x^4)*Sqrt[c + d*x^4]),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.01, size = 356, normalized size = 3.9 \begin{align*}{\frac{1}{2\,b}\ln \left ({x}^{2}\sqrt{d}+\sqrt{d{x}^{4}+c} \right ){\frac{1}{\sqrt{d}}}}+{\frac{a}{4\,b}\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}{4\,b}\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^5/(b*x^4+a)/(d*x^4+c)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{{\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^5/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^5/((b*x^4 + a)*sqrt(d*x^4 + c)), x)

________________________________________________________________________________________

Fricas [A]  time = 2.22666, size = 1358, normalized size = 14.92 \begin{align*} \left [\frac{d \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 \, \sqrt{d} \log \left (-2 \, d x^{4} - 2 \, \sqrt{d x^{4} + c} \sqrt{d} x^{2} - c\right )}{8 \, b d}, \frac{d \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 ) - 4 \, \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x^{2}}{\sqrt{d x^{4} + c}}\right )}{8 \, b d}, \frac{d \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 ) + \sqrt{d} \log \left (-2 \, d x^{4} - 2 \, \sqrt{d x^{4} + c} \sqrt{d} x^{2} - c\right )}{4 \, b d}, \frac{d \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 ) - 2 \, \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x^{2}}{\sqrt{d x^{4} + c}}\right )}{4 \, b d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\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**5/(b*x**4+a)/(d*x**4+c)**(1/2),x)

[Out]

Integral(x**5/((a + b*x**4)*sqrt(c + d*x**4)), x)

________________________________________________________________________________________

Giac [A]  time = 1.13233, size = 107, normalized size = 1.18 \begin{align*} \frac{1}{2} \, c{\left (\frac{a \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{4}}}}{\sqrt{a b c - a^{2} d}}\right )}{\sqrt{a b c - a^{2} d} b c} - \frac{\arctan \left (\frac{\sqrt{d + \frac{c}{x^{4}}}}{\sqrt{-d}}\right )}{b c \sqrt{-d}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(b*x^4+a)/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

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