3.9.32 \(\int \frac {1}{x^3 (a+b x^4)^2 \sqrt {c+d x^4}} \, dx\) [832]

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

[Out]

-1/4*b*(-4*a*d+3*b*c)*arctan(x^2*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^4+c)^(1/2))/a^(5/2)/(-a*d+b*c)^(3/2)-1/4*(-2*a*
d+3*b*c)*(d*x^4+c)^(1/2)/a^2/c/(-a*d+b*c)/x^2+1/4*b*(d*x^4+c)^(1/2)/a/(-a*d+b*c)/x^2/(b*x^4+a)

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Rubi [A]
time = 0.13, 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 = {476, 483, 597, 12, 385, 211} \begin {gather*} -\frac {b (3 b c-4 a d) \text {ArcTan}\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((3*b*c - 2*a*d)*Sqrt[c + d*x^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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 385

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 476

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_.)*((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 597

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} \text {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 {\text {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 {\text {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)) \text {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)) \text {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 = 0.99, size = 157, normalized size = 1.05 \begin {gather*} \frac {\sqrt {c+d x^4} \left (2 a b c-2 a^2 d+3 b^2 c x^4-2 a b d x^4\right )}{4 a^2 c (-b c+a d) x^2 \left (a+b x^4\right )}-\frac {b (3 b c-4 a d) \tan ^{-1}\left (\frac {a \sqrt {d}+b \sqrt {d} x^4+b x^2 \sqrt {c+d x^4}}{\sqrt {a} \sqrt {b c-a d}}\right )}{4 a^{5/2} (b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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)) - (
b*(3*b*c - 4*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^
(5/2)*(b*c - a*d)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1220\) vs. \(2(129)=258\).
time = 0.40, size = 1221, normalized size = 8.19 Too large to display

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,method=_RETURNVERBOSE)

[Out]

-1/2/a^2/x^2*(d*x^4+c)^(1/2)/c-b/a^2*(-1/4/(-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^2-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a
*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2-1/b*(-a*b)^(1/2)))+1/4/(-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+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2
)/b*(x^2+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2+1/b*(-a*b)^(1/2))))-b/a*(-1/8/a/(a*d-b*c)/(x^2-1/b*(-a*b)^
(1/2))*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/8/b/a*d*(-a*
b)^(1/2)/(a*d-b*c)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))+2*(-(a*d-
b*c)/b)^(1/2)*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2-1
/b*(-a*b)^(1/2)))-1/8/a/(-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^2-1/b*(-a*b
)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x^2-1/b*(-a*b)^(1/2))-(a*d-b*c
)/b)^(1/2))/(x^2-1/b*(-a*b)^(1/2)))+1/8/a/(-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+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*
b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2+1/b*(-a*b)^(1/2)))-1/8/a/(a*d-b*c)/(x^2+1/b*(-a*b)^(1/2))*((x^2+1/b*(-a*b)^
(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/8/b/a*d*(-a*b)^(1/2)/(a*d-b*c)/(-(a*
d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x^2+1/b*
(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x^2+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x^2+1/b*(-a*b)^(1/2))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (129) = 258\).
time = 2.76, 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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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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (129) = 258\).
time = 3.17, 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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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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