3.830 \(\int \frac {x^5}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=93 \[ \frac {c \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 \sqrt {a} (b c-a d)^{3/2}}-\frac {x^2 \sqrt {c+d x^4}}{4 \left (a+b x^4\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {465, 471, 12, 377, 205} \[ \frac {c \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 \sqrt {a} (b c-a d)^{3/2}}-\frac {x^2 \sqrt {c+d x^4}}{4 \left (a+b x^4\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^5/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

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 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 471

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 + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rubi steps

\begin {align*} \int \frac {x^5}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^2 \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}+\frac {\operatorname {Subst}\left (\int \frac {c}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 (b c-a d)}\\ &=-\frac {x^2 \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}+\frac {c \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 (b c-a d)}\\ &=-\frac {x^2 \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}+\frac {c \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x^2}{\sqrt {c+d x^4}}\right )}{4 (b c-a d)}\\ &=-\frac {x^2 \sqrt {c+d x^4}}{4 (b c-a d) \left (a+b x^4\right )}+\frac {c \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 \sqrt {a} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 124, normalized size = 1.33 \[ \frac {\sqrt {c+d x^4} \left (-\frac {x^4 (b c-a d)}{a+b x^4}-\frac {c \sqrt {x^4 \left (\frac {d}{c}-\frac {b}{a}\right )} \tanh ^{-1}\left (\frac {\sqrt {x^4 \left (\frac {d}{c}-\frac {b}{a}\right )}}{\sqrt {\frac {d x^4}{c}+1}}\right )}{\sqrt {\frac {d x^4}{c}+1}}\right )}{4 x^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(Sqrt[c + d*x^4]*(-(((b*c - a*d)*x^4)/(a + b*x^4)) - (c*Sqrt[(-(b/a) + d/c)*x^4]*ArcTanh[Sqrt[(-(b/a) + d/c)*x
^4]/Sqrt[1 + (d*x^4)/c]])/Sqrt[1 + (d*x^4)/c]))/(4*(b*c - a*d)^2*x^2)

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fricas [B]  time = 1.03, size = 426, normalized size = 4.58 \[ \left [-\frac {4 \, \sqrt {d x^{4} + c} {\left (a b c - a^{2} d\right )} x^{2} - {\left (b c x^{4} + a c\right )} \sqrt {-a b c + a^{2} d} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{8} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{4} + a^{2} c^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{6} - a c x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right )}{16 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{4}\right )}}, -\frac {2 \, \sqrt {d x^{4} + c} {\left (a b c - a^{2} d\right )} x^{2} - {\left (b c x^{4} + a c\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{6} + {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )}}\right )}{8 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(4*sqrt(d*x^4 + c)*(a*b*c - a^2*d)*x^2 - (b*c*x^4 + a*c)*sqrt(-a*b*c + a^2*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)))/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (a*b^3*c^2 - 2*a^2*
b^2*c*d + a^3*b*d^2)*x^4), -1/8*(2*sqrt(d*x^4 + c)*(a*b*c - a^2*d)*x^2 - (b*c*x^4 + a*c)*sqrt(a*b*c - a^2*d)*a
rctan(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)))/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^4)]

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giac [B]  time = 1.63, size = 244, normalized size = 2.62 \[ \frac {c \sqrt {d} \arctan \left (-\frac {{\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{4 \, \sqrt {a b c d - a^{2} d^{2}} {\left (b c - a d\right )}} + \frac {{\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a d^{\frac {3}{2}} - b c^{2} \sqrt {d}}{2 \, {\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{2} c - a b d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.24, size = 861, normalized size = 9.26 \[ -\frac {\ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{8 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}+\frac {\ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{8 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b}-\frac {\sqrt {-a b}\, d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{8 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {\sqrt {-a b}\, d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{8 \left (a d -b c \right ) \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {\sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{8 \left (a d -b c \right ) \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) b}+\frac {\sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{8 \left (a d -b c \right ) \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x^2+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*
c)/b)^(1/2)*((x^2+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x^2+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x^2+(-a*b)^
(1/2)/b))-1/8/b/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x^2-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*
(-(a*d-b*c)/b)^(1/2)*((x^2-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x^2-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x^
2-(-a*b)^(1/2)/b))+1/8/b/(a*d-b*c)/(x^2-(-a*b)^(1/2)/b)*((x^2-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x^2-(-a*b)^(
1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)-1/8/b^2*(-a*b)^(1/2)*d/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)*(x^2-(
-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x^2-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x^2-(-a*b)^(
1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x^2-(-a*b)^(1/2)/b))+1/8/b/(a*d-b*c)/(x^2+(-a*b)^(1/2)/b)*((x^2+(-a*b)^(1/2)/
b)^2*d-2*(-a*b)^(1/2)*(x^2+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)+1/8/b^2*(-a*b)^(1/2)*d/(a*d-b*c)/(-(a*d-b*c)
/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x^2+(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x^2+(-a*b)^(1/2)
/b)^2*d-2*(-a*b)^(1/2)*(x^2+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x^2+(-a*b)^(1/2)/b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{{\left (b x^{4} + a\right )}^{2} \sqrt {d x^{4} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5}{{\left (b\,x^4+a\right )}^2\,\sqrt {d\,x^4+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\left (a + b x^{4}\right )^{2} \sqrt {c + d x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5/((a + b*x**4)**2*sqrt(c + d*x**4)), x)

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