3.6.41 \(\int \frac {1}{x^3 (a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=149 \[ -\frac {b (3 b c-4 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{5/2} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^4} (3 b c-2 a d)}{4 a^2 c x^2 (b c-a d)}+\frac {b \sqrt {c+d x^4}}{4 a x^2 \left (a+b x^4\right ) (b c-a d)} \]

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Rubi [A]  time = 0.20, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {465, 472, 583, 12, 377, 205} \begin {gather*} -\frac {\sqrt {c+d x^4} (3 b c-2 a d)}{4 a^2 c x^2 (b c-a d)}-\frac {b (3 b c-4 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{5/2} (b c-a d)^{3/2}}+\frac {b \sqrt {c+d x^4}}{4 a x^2 \left (a+b x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*c - 2*a*d)*Sqrt[c + d*x^4])/(4*a^2*c*(b*c - a*d)*x^2) + (b*Sqrt[c + d*x^4])/(4*a*(b*c - a*d)*x^2*(a + b
*x^4)) - (b*(3*b*c - 4*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(4*a^(5/2)*(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 472

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 5.63, size = 155, normalized size = 1.04 \begin {gather*} \frac {a^2 \left (c+d x^4\right ) \left (\frac {b^2 x^4}{\left (a+b x^4\right ) (a d-b c)}-\frac {2}{c}\right )-\frac {b x^8 \sqrt {\frac {d x^4}{c}+1} (3 b c-4 a d) \sin ^{-1}\left (\frac {\sqrt {x^4 \left (\frac {b}{a}-\frac {d}{c}\right )}}{\sqrt {\frac {b x^4}{a}+1}}\right )}{c \left (\frac {x^4 (b c-a d)}{a c}\right )^{3/2}}}{4 a^4 x^2 \sqrt {c+d x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(c + d*x^4)*(-2/c + (b^2*x^4)/((-(b*c) + a*d)*(a + b*x^4))) - (b*(3*b*c - 4*a*d)*x^8*Sqrt[1 + (d*x^4)/c]*
ArcSin[Sqrt[(b/a - d/c)*x^4]/Sqrt[1 + (b*x^4)/a]])/(c*(((b*c - a*d)*x^4)/(a*c))^(3/2)))/(4*a^4*x^2*Sqrt[c + d*
x^4])

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IntegrateAlgebraic [A]  time = 1.37, size = 159, normalized size = 1.07 \begin {gather*} \frac {\left (4 a b d-3 b^2 c\right ) \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^{5/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^4} \left (-2 a^2 d+2 a b c-2 a b d x^4+3 b^2 c x^4\right )}{4 a^2 c x^2 \left (a+b x^4\right ) (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^4]*(2*a*b*c - 2*a^2*d + 3*b^2*c*x^4 - 2*a*b*d*x^4))/(4*a^2*c*(-(b*c) + a*d)*x^2*(a + b*x^4)) + (
(-3*b^2*c + 4*a*b*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^(5/2)*(b*c - a*d)^(3/2))

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fricas [B]  time = 0.62, size = 612, normalized size = 4.11 \begin {gather*} \left [-\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{6} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{2}\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 ) + 4 \, {\left (2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2} + {\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{4}\right )} \sqrt {d x^{4} + c}}{16 \, {\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{6} + {\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x^{2}\right )}}, -\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{6} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{2}\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 ) + 2 \, {\left (2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2} + {\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{4}\right )} \sqrt {d x^{4} + c}}{8 \, {\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{6} + {\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^6 + (3*a*b^2*c^2 - 4*a^2*b*c*d)*x^2)*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)) + 4*(2*a^2*b^2*c^2 - 4*a^3*b*c*d + 2*a^4*d^2 + (3
*a*b^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*x^4)*sqrt(d*x^4 + c))/((a^3*b^3*c^3 - 2*a^4*b^2*c^2*d + a^5*b*c*d^2)
*x^6 + (a^4*b^2*c^3 - 2*a^5*b*c^2*d + a^6*c*d^2)*x^2), -1/8*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^6 + (3*a*b^2*c^2 - 4
*a^2*b*c*d)*x^2)*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)) + 2*(2*a^2*b^2*c^2 - 4*a^3*b*c*d + 2*a^4*d^2 + (3*a*b^3*c
^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*x^4)*sqrt(d*x^4 + c))/((a^3*b^3*c^3 - 2*a^4*b^2*c^2*d + a^5*b*c*d^2)*x^6 + (
a^4*b^2*c^3 - 2*a^5*b*c^2*d + a^6*c*d^2)*x^2)]

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giac [B]  time = 1.68, size = 418, normalized size = 2.81 \begin {gather*} \frac {1}{4} \, d^{\frac {5}{2}} {\left (\frac {{\left (3 \, b^{2} c - 4 \, a b 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^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {2 \, {\left (3 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{4} b^{2} c - 4 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{4} a b d - 6 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b^{2} c^{2} + 14 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a b c d - 8 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a^{2} d^{2} + 3 \, b^{2} c^{3} - 2 \, a b c^{2} d\right )}}{{\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{6} b - 3 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{4} b c + 4 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{4} a d + 3 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b c^{2} - 4 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a c d - b c^{3}\right )} {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

1/4*d^(5/2)*((3*b^2*c - 4*a*b*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^2*b*c*d^2 - a^3*d^3)*sqrt(a*b*c*d - a^2*d^2)) + 2*(3*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^4*b^2*c -
4*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^4*a*b*d - 6*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*b^2*c^2 + 14*(sqrt(d)*x^2 - sq
rt(d*x^4 + c))^2*a*b*c*d - 8*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*a^2*d^2 + 3*b^2*c^3 - 2*a*b*c^2*d)/(((sqrt(d)*x
^2 - sqrt(d*x^4 + c))^6*b - 3*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^4*b*c + 4*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^4*a*d
+ 3*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*b*c^2 - 4*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*a*c*d - b*c^3)*(a^2*b*c*d^2
- a^3*d^3)))

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maple [B]  time = 0.27, size = 885, normalized size = 5.94 \begin {gather*} \frac {3 b \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^{2}}-\frac {3 b \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^{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}}\, a^{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}}\, a^{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}}\, b}{8 \left (a d -b c \right ) \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) a^{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}}\, b}{8 \left (a d -b c \right ) \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) a^{2}}-\frac {\sqrt {d \,x^{4}+c}}{2 a^{2} c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

-1/2/a^2/x^2*(d*x^4+c)^(1/2)/c-3/8/a^2*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))+3/8/a^2*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/a^2*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/a^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/a^2*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/a^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 \begin {gather*} \int \frac {1}{{\left (b x^{4} + a\right )}^{2} \sqrt {d x^{4} + c} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\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(1/(x^3*(a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

int(1/(x^3*(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 {1}{x^{3} \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(1/x**3/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Integral(1/(x**3*(a + b*x**4)**2*sqrt(c + d*x**4)), x)

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