3.6.42 \(\int \frac {1}{x^7 (a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=208 \[ \frac {b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{7/2} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^4} (5 b c-2 a d)}{12 a^2 c x^6 (b c-a d)}+\frac {\sqrt {c+d x^4} \left (-4 a^2 d^2-8 a b c d+15 b^2 c^2\right )}{12 a^3 c^2 x^2 (b c-a d)}+\frac {b \sqrt {c+d x^4}}{4 a x^6 \left (a+b x^4\right ) (b c-a d)} \]

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Rubi [A]  time = 0.33, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \left (-4 a^2 d^2-8 a b c d+15 b^2 c^2\right )}{12 a^3 c^2 x^2 (b c-a d)}+\frac {b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{7/2} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^4} (5 b c-2 a d)}{12 a^2 c x^6 (b c-a d)}+\frac {b \sqrt {c+d x^4}}{4 a x^6 \left (a+b x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

-((5*b*c - 2*a*d)*Sqrt[c + d*x^4])/(12*a^2*c*(b*c - a*d)*x^6) + ((15*b^2*c^2 - 8*a*b*c*d - 4*a^2*d^2)*Sqrt[c +
d*x^4])/(12*a^3*c^2*(b*c - a*d)*x^2) + (b*Sqrt[c + d*x^4])/(4*a*(b*c - a*d)*x^6*(a + b*x^4)) + (b^2*(5*b*c -
6*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(4*a^(7/2)*(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^7 \left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \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^6 \left (a+b x^4\right )}-\frac {\operatorname {Subst}\left (\int \frac {-5 b c+2 a d-4 b d x^2}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 a (b c-a d)}\\ &=-\frac {(5 b c-2 a d) \sqrt {c+d x^4}}{12 a^2 c (b c-a d) x^6}+\frac {b \sqrt {c+d x^4}}{4 a (b c-a d) x^6 \left (a+b x^4\right )}+\frac {\operatorname {Subst}\left (\int \frac {-15 b^2 c^2+8 a b c d+4 a^2 d^2-2 b d (5 b c-2 a d) x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{12 a^2 c (b c-a d)}\\ &=-\frac {(5 b c-2 a d) \sqrt {c+d x^4}}{12 a^2 c (b c-a d) x^6}+\frac {\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt {c+d x^4}}{12 a^3 c^2 (b c-a d) x^2}+\frac {b \sqrt {c+d x^4}}{4 a (b c-a d) x^6 \left (a+b x^4\right )}-\frac {\operatorname {Subst}\left (\int -\frac {3 b^2 c^2 (5 b c-6 a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{12 a^3 c^2 (b c-a d)}\\ &=-\frac {(5 b c-2 a d) \sqrt {c+d x^4}}{12 a^2 c (b c-a d) x^6}+\frac {\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt {c+d x^4}}{12 a^3 c^2 (b c-a d) x^2}+\frac {b \sqrt {c+d x^4}}{4 a (b c-a d) x^6 \left (a+b x^4\right )}+\frac {\left (b^2 (5 b c-6 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 a^3 (b c-a d)}\\ &=-\frac {(5 b c-2 a d) \sqrt {c+d x^4}}{12 a^2 c (b c-a d) x^6}+\frac {\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt {c+d x^4}}{12 a^3 c^2 (b c-a d) x^2}+\frac {b \sqrt {c+d x^4}}{4 a (b c-a d) x^6 \left (a+b x^4\right )}+\frac {\left (b^2 (5 b c-6 a d)\right ) \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^3 (b c-a d)}\\ &=-\frac {(5 b c-2 a d) \sqrt {c+d x^4}}{12 a^2 c (b c-a d) x^6}+\frac {\left (15 b^2 c^2-8 a b c d-4 a^2 d^2\right ) \sqrt {c+d x^4}}{12 a^3 c^2 (b c-a d) x^2}+\frac {b \sqrt {c+d x^4}}{4 a (b c-a d) x^6 \left (a+b x^4\right )}+\frac {b^2 (5 b c-6 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 a^{7/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 5.87, size = 175, normalized size = 0.84 \begin {gather*} \frac {a^2 \left (c+d x^4\right ) \left (\frac {3 b^3 x^8}{\left (a+b x^4\right ) (b c-a d)}+\frac {4 x^4 (a d+3 b c)}{c^2}-\frac {2 a}{c}\right )+\frac {3 b^2 x^{12} \sqrt {\frac {d x^4}{c}+1} (5 b c-6 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}}}{12 a^5 x^6 \sqrt {c+d x^4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^7*(a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(a^2*(c + d*x^4)*((-2*a)/c + (4*(3*b*c + a*d)*x^4)/c^2 + (3*b^3*x^8)/((b*c - a*d)*(a + b*x^4))) + (3*b^2*(5*b*
c - 6*a*d)*x^12*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)))/(12*a^5*x^6*Sqrt[c + d*x^4])

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IntegrateAlgebraic [A]  time = 2.28, size = 217, normalized size = 1.04 \begin {gather*} \frac {\left (5 b^3 c-6 a b^2 d\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^{7/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^4} \left (-2 a^3 c d+4 a^3 d^2 x^4+2 a^2 b c^2+6 a^2 b c d x^4+4 a^2 b d^2 x^8-10 a b^2 c^2 x^4+8 a b^2 c d x^8-15 b^3 c^2 x^8\right )}{12 a^3 c^2 x^6 \left (a+b x^4\right ) (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^7*(a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(Sqrt[c + d*x^4]*(2*a^2*b*c^2 - 2*a^3*c*d - 10*a*b^2*c^2*x^4 + 6*a^2*b*c*d*x^4 + 4*a^3*d^2*x^4 - 15*b^3*c^2*x^
8 + 8*a*b^2*c*d*x^8 + 4*a^2*b*d^2*x^8))/(12*a^3*c^2*(-(b*c) + a*d)*x^6*(a + b*x^4)) + ((5*b^3*c - 6*a*b^2*d)*A
rcTan[(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^(7/2)*(b*c - a*d)^(
3/2))

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fricas [A]  time = 0.78, size = 760, normalized size = 3.65 \begin {gather*} \left [-\frac {3 \, {\left ({\left (5 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d\right )} x^{10} + {\left (5 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d\right )} x^{6}\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 ({\left (15 \, a b^{4} c^{3} - 23 \, a^{2} b^{3} c^{2} d + 4 \, a^{3} b^{2} c d^{2} + 4 \, a^{4} b d^{3}\right )} x^{8} - 2 \, a^{3} b^{2} c^{3} + 4 \, a^{4} b c^{2} d - 2 \, a^{5} c d^{2} + 2 \, {\left (5 \, a^{2} b^{3} c^{3} - 8 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{4}\right )} \sqrt {d x^{4} + c}}{48 \, {\left ({\left (a^{4} b^{3} c^{4} - 2 \, a^{5} b^{2} c^{3} d + a^{6} b c^{2} d^{2}\right )} x^{10} + {\left (a^{5} b^{2} c^{4} - 2 \, a^{6} b c^{3} d + a^{7} c^{2} d^{2}\right )} x^{6}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d\right )} x^{10} + {\left (5 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d\right )} x^{6}\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 ({\left (15 \, a b^{4} c^{3} - 23 \, a^{2} b^{3} c^{2} d + 4 \, a^{3} b^{2} c d^{2} + 4 \, a^{4} b d^{3}\right )} x^{8} - 2 \, a^{3} b^{2} c^{3} + 4 \, a^{4} b c^{2} d - 2 \, a^{5} c d^{2} + 2 \, {\left (5 \, a^{2} b^{3} c^{3} - 8 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{4}\right )} \sqrt {d x^{4} + c}}{24 \, {\left ({\left (a^{4} b^{3} c^{4} - 2 \, a^{5} b^{2} c^{3} d + a^{6} b c^{2} d^{2}\right )} x^{10} + {\left (a^{5} b^{2} c^{4} - 2 \, a^{6} b c^{3} d + a^{7} c^{2} d^{2}\right )} x^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*((5*b^4*c^3 - 6*a*b^3*c^2*d)*x^10 + (5*a*b^3*c^3 - 6*a^2*b^2*c^2*d)*x^6)*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*((15*a*b^4*c^3 - 23*a^2*b^3*c^2*d +
4*a^3*b^2*c*d^2 + 4*a^4*b*d^3)*x^8 - 2*a^3*b^2*c^3 + 4*a^4*b*c^2*d - 2*a^5*c*d^2 + 2*(5*a^2*b^3*c^3 - 8*a^3*b^
2*c^2*d + a^4*b*c*d^2 + 2*a^5*d^3)*x^4)*sqrt(d*x^4 + c))/((a^4*b^3*c^4 - 2*a^5*b^2*c^3*d + a^6*b*c^2*d^2)*x^10
+ (a^5*b^2*c^4 - 2*a^6*b*c^3*d + a^7*c^2*d^2)*x^6), 1/24*(3*((5*b^4*c^3 - 6*a*b^3*c^2*d)*x^10 + (5*a*b^3*c^3
- 6*a^2*b^2*c^2*d)*x^6)*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*((15*a*b^4*c^3 - 23*a^2*b^3*c^2*d + 4*a^3*b^2*
c*d^2 + 4*a^4*b*d^3)*x^8 - 2*a^3*b^2*c^3 + 4*a^4*b*c^2*d - 2*a^5*c*d^2 + 2*(5*a^2*b^3*c^3 - 8*a^3*b^2*c^2*d +
a^4*b*c*d^2 + 2*a^5*d^3)*x^4)*sqrt(d*x^4 + c))/((a^4*b^3*c^4 - 2*a^5*b^2*c^3*d + a^6*b*c^2*d^2)*x^10 + (a^5*b^
2*c^4 - 2*a^6*b*c^3*d + a^7*c^2*d^2)*x^6)]

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giac [B]  time = 2.04, size = 395, normalized size = 1.90 \begin {gather*} \frac {1}{12} \, d^{\frac {7}{2}} {\left (\frac {3 \, {\left (5 \, b^{3} c - 6 \, a b^{2} 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^{3} b c d^{3} - a^{4} d^{4}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {6 \, {\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b^{3} c - 2 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a b^{2} d - b^{3} c^{2}\right )}}{{\left (a^{3} b c d^{3} - a^{4} d^{4}\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 )}} - \frac {8 \, {\left (3 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{4} b - 6 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} b c - 3 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a d + 3 \, b c^{2} + a c d\right )}}{{\left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} - c\right )}^{3} a^{3} d^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

1/12*d^(7/2)*(3*(5*b^3*c - 6*a*b^2*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^3*b*c*d^3 - a^4*d^4)*sqrt(a*b*c*d - a^2*d^2)) - 6*((sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*b^3*
c - 2*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*a*b^2*d - b^3*c^2)/((a^3*b*c*d^3 - a^4*d^4)*((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)) - 8*(
3*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^4*b - 6*(sqrt(d)*x^2 - sqrt(d*x^4 + c))^2*b*c - 3*(sqrt(d)*x^2 - sqrt(d*x^4
+ c))^2*a*d + 3*b*c^2 + a*c*d)/(((sqrt(d)*x^2 - sqrt(d*x^4 + c))^2 - c)^3*a^3*d^3))

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maple [B]  time = 0.25, size = 923, normalized size = 4.44 \begin {gather*} -\frac {5 b^{2} \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^{3}}+\frac {5 b^{2} \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^{3}}+\frac {\sqrt {-a b}\, 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^{3}}-\frac {\sqrt {-a b}\, 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^{3}}-\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^{2}}{8 \left (a d -b c \right ) \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) a^{3}}-\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^{2}}{8 \left (a d -b c \right ) \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) a^{3}}+\frac {\sqrt {d \,x^{4}+c}\, b}{a^{3} c \,x^{2}}-\frac {\sqrt {d \,x^{4}+c}\, \left (-2 d \,x^{4}+c \right )}{6 a^{2} c^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

b/a^3/x^2*(d*x^4+c)^(1/2)/c+5/8/a^3*b^2/(-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))-5/8/a^3*b^2/(-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))-1/6/a^2*(d*x^4+c)^(1/2)*(-2*d*x^4+c)/x^6/c^2-1/8/a^
3*b^2/(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^3*b*(-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^3*b^2/(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^3*b*(-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^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(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^7), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^7\,{\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^7*(a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

int(1/(x^7*(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^{7} \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**7/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Integral(1/(x**7*(a + b*x**4)**2*sqrt(c + d*x**4)), x)

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