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

Optimal. Leaf size=104 \[ \frac {(b c-2 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{3/2} (b c-a d)^{3/2}}+\frac {b x^2 \sqrt {c+d x^4}}{4 a \left (a+b x^4\right ) (b c-a d)} \]

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Rubi [A]  time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {465, 382, 377, 205} \begin {gather*} \frac {(b c-2 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{3/2} (b c-a d)^{3/2}}+\frac {b x^2 \sqrt {c+d x^4}}{4 a \left (a+b x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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]

Rubi steps

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

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Mathematica [C]  time = 6.01, size = 407, normalized size = 3.91 \begin {gather*} \frac {x^2 \sqrt {c+d x^4} \left (-30 d x^4 \sqrt {\frac {a x^4 \left (c+d x^4\right ) (b c-a d)}{c^2 \left (a+b x^4\right )^2}}-45 c \sqrt {\frac {a x^4 \left (c+d x^4\right ) (b c-a d)}{c^2 \left (a+b x^4\right )^2}}+16 d x^4 \left (\frac {x^4 (b c-a d)}{c \left (a+b x^4\right )}\right )^{5/2} \sqrt {\frac {a \left (c+d x^4\right )}{c \left (a+b x^4\right )}} \, _2F_1\left (2,3;\frac {7}{2};\frac {(b c-a d) x^4}{c \left (b x^4+a\right )}\right )+16 c \left (\frac {x^4 (b c-a d)}{c \left (a+b x^4\right )}\right )^{5/2} \sqrt {\frac {a \left (c+d x^4\right )}{c \left (a+b x^4\right )}} \, _2F_1\left (2,3;\frac {7}{2};\frac {(b c-a d) x^4}{c \left (b x^4+a\right )}\right )+30 d x^4 \sin ^{-1}\left (\sqrt {\frac {x^4 (b c-a d)}{c \left (a+b x^4\right )}}\right )+45 c \sin ^{-1}\left (\sqrt {\frac {x^4 (b c-a d)}{c \left (a+b x^4\right )}}\right )\right )}{60 c^2 \left (a+b x^4\right )^2 \left (\frac {x^4 (b c-a d)}{c \left (a+b x^4\right )}\right )^{3/2} \sqrt {\frac {a \left (c+d x^4\right )}{c \left (a+b x^4\right )}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^2*Sqrt[c + d*x^4]*(-45*c*Sqrt[(a*(b*c - a*d)*x^4*(c + d*x^4))/(c^2*(a + b*x^4)^2)] - 30*d*x^4*Sqrt[(a*(b*c
- a*d)*x^4*(c + d*x^4))/(c^2*(a + b*x^4)^2)] + 45*c*ArcSin[Sqrt[((b*c - a*d)*x^4)/(c*(a + b*x^4))]] + 30*d*x^4
*ArcSin[Sqrt[((b*c - a*d)*x^4)/(c*(a + b*x^4))]] + 16*c*(((b*c - a*d)*x^4)/(c*(a + b*x^4)))^(5/2)*Sqrt[(a*(c +
d*x^4))/(c*(a + b*x^4))]*Hypergeometric2F1[2, 3, 7/2, ((b*c - a*d)*x^4)/(c*(a + b*x^4))] + 16*d*x^4*(((b*c -
a*d)*x^4)/(c*(a + b*x^4)))^(5/2)*Sqrt[(a*(c + d*x^4))/(c*(a + b*x^4))]*Hypergeometric2F1[2, 3, 7/2, ((b*c - a*
d)*x^4)/(c*(a + b*x^4))]))/(60*c^2*(((b*c - a*d)*x^4)/(c*(a + b*x^4)))^(3/2)*(a + b*x^4)^2*Sqrt[(a*(c + d*x^4)
)/(c*(a + b*x^4))])

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IntegrateAlgebraic [A]  time = 0.64, size = 124, normalized size = 1.19 \begin {gather*} \frac {(b c-2 a d) \tan ^{-1}\left (\frac {a \sqrt {d}+b x^2 \sqrt {c+d x^4}+b \sqrt {d} x^4}{\sqrt {a} \sqrt {b c-a d}}\right )}{4 a^{3/2} (b c-a d)^{3/2}}-\frac {b x^2 \sqrt {c+d x^4}}{4 a \left (a+b x^4\right ) (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*x^2*Sqrt[c + d*x^4])/(a*(-(b*c) + a*d)*(a + b*x^4)) + ((b*c - 2*a*d)*ArcTan[(a*Sqrt[d] + b*Sqrt[d]*x^4
+ b*x^2*Sqrt[c + d*x^4])/(Sqrt[a]*Sqrt[b*c - a*d])])/(4*a^(3/2)*(b*c - a*d)^(3/2))

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fricas [B]  time = 0.60, size = 467, normalized size = 4.49 \begin {gather*} \left [\frac {4 \, \sqrt {d x^{4} + c} {\left (a b^{2} c - a^{2} b d\right )} x^{2} - {\left ({\left (b^{2} c - 2 \, a b d\right )} x^{4} + a b c - 2 \, a^{2} d\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^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2} + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{4}\right )}}, \frac {2 \, \sqrt {d x^{4} + c} {\left (a b^{2} c - a^{2} b d\right )} x^{2} + {\left ({\left (b^{2} c - 2 \, a b d\right )} x^{4} + a b c - 2 \, a^{2} d\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^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2} + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(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^2*c - a^2*b*d)*x^2 - ((b^2*c - 2*a*b*d)*x^4 + a*b*c - 2*a^2*d)*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^3*b^2*c^2 - 2*a^4*b*c*d +
a^5*d^2 + (a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*x^4), 1/8*(2*sqrt(d*x^4 + c)*(a*b^2*c - a^2*b*d)*x^2 + ((b
^2*c - 2*a*b*d)*x^4 + a*b*c - 2*a^2*d)*sqrt(a*b*c - a^2*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)))/(a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d
^2 + (a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*x^4)]

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giac [B]  time = 0.38, size = 237, normalized size = 2.28 \begin {gather*} -\frac {1}{4} \, d^{\frac {3}{2}} {\left (\frac {{\left (b c - 2 \, a d\right )} \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 )}{{\left (a b c d - a^{2} d^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b c - 2 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a d - b c^{2}\right )}}{{\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 (a b c d - a^{2} d^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.19, size = 867, normalized size = 8.34 \begin {gather*} \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}}\, a 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}}\, a 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}}\, a}+\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}}\, a}-\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 ) a}-\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 ) a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/a/(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/a*(-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/a/(-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/a/(-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/a/(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/a*(-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 \begin {gather*} \int \frac {x}{{\left (b x^{4} + a\right )}^{2} \sqrt {d x^{4} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/((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 \begin {gather*} \int \frac {x}{{\left (b\,x^4+a\right )}^2\,\sqrt {d\,x^4+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/((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 \begin {gather*} \int \frac {x}{\left (a + b x^{4}\right )^{2} \sqrt {c + d x^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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