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

Optimal. Leaf size=120 \[ -\frac{\sqrt{a} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b^2}+\frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{4 b^2 \sqrt{d}}+\frac{x^2 \sqrt{c+d x^4}}{4 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.154238, antiderivative size = 120, 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, 478, 523, 217, 206, 377, 205} \[ -\frac{\sqrt{a} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )}{2 b^2}+\frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x^2}{\sqrt{c+d x^4}}\right )}{4 b^2 \sqrt{d}}+\frac{x^2 \sqrt{c+d x^4}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[c + d*x^4])/(4*b) - (Sqrt[a]*Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])]
)/(2*b^2) + ((b*c - 2*a*d)*ArcTanh[(Sqrt[d]*x^2)/Sqrt[c + d*x^4]])/(4*b^2*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 478

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

Mathematica [A]  time = 0.149976, size = 114, normalized size = 0.95 \[ \frac{\frac{(b c-2 a d) \log \left (\sqrt{d} \sqrt{c+d x^4}+d x^2\right )}{\sqrt{d}}-2 \sqrt{a} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x^2 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^4}}\right )+b x^2 \sqrt{c+d x^4}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.025, size = 1066, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d*x^4+c)^(1/2)/(b*x^4+a),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.35321, size = 1577, normalized size = 13.14 \begin{align*} \left [\frac{2 \, \sqrt{d x^{4} + c} b d x^{2} -{\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 ) + \sqrt{-a b c + a^{2} d} 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 c - 2 \, a d\right )} x^{6} - a c x^{2}\right )} \sqrt{d x^{4} + c} \sqrt{-a b c + a^{2} d}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right )}{8 \, b^{2} d}, \frac{2 \, \sqrt{d x^{4} + c} b d x^{2} - 2 \,{\left (b c - 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x^{2}}{\sqrt{d x^{4} + c}}\right ) + \sqrt{-a b c + a^{2} d} 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 c - 2 \, a d\right )} x^{6} - a c x^{2}\right )} \sqrt{d x^{4} + c} \sqrt{-a b c + a^{2} d}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right )}{8 \, b^{2} d}, \frac{2 \, \sqrt{d x^{4} + c} b d x^{2} - 2 \, \sqrt{a b c - a^{2} d} d \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c} \sqrt{a b c - a^{2} d}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{6} +{\left (a b c^{2} - a^{2} c d\right )} 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}, \frac{\sqrt{d x^{4} + c} b d x^{2} -{\left (b c - 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x^{2}}{\sqrt{d x^{4} + c}}\right ) - \sqrt{a b c - a^{2} d} d \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt{d x^{4} + c} \sqrt{a b c - a^{2} d}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{6} +{\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )}}\right )}{4 \, b^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5} \sqrt{c + d x^{4}}}{a + b x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(d*x**4+c)**(1/2)/(b*x**4+a),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.58648, size = 136, normalized size = 1.13 \begin{align*} \frac{\sqrt{d x^{4} + c} b^{2} x^{2}}{384 \, d^{3}} + \frac{\sqrt{a b c - a^{2} d} \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{4}}}}{\sqrt{a b c - a^{2} d}}\right )}{2 \, b^{2}} - \frac{{\left (b^{2} c - 2 \, a b d\right )} \arctan \left (\frac{\sqrt{d + \frac{c}{x^{4}}}}{\sqrt{-d}}\right )}{384 \, \sqrt{-d} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*sqrt(d*x^4 + c)*b^2*x^2/d^3 + 1/2*sqrt(a*b*c - a^2*d)*arctan(a*sqrt(d + c/x^4)/sqrt(a*b*c - a^2*d))/b^2
- 1/384*(b^2*c - 2*a*b*d)*arctan(sqrt(d + c/x^4)/sqrt(-d))/(sqrt(-d)*d^3)