3.6.38 \(\int \frac {x^9}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=141 \[ -\frac {\sqrt {a} (3 b c-2 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 b^2 (b c-a d)^{3/2}}+\frac {a x^2 \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{2 b^2 \sqrt {d}} \]

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Rubi [A]  time = 0.16, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {465, 470, 523, 217, 206, 377, 205} \begin {gather*} -\frac {\sqrt {a} (3 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 b^2 (b c-a d)^{3/2}}+\frac {a x^2 \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{2 b^2 \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 470

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rubi steps

\begin {align*} \int \frac {x^9}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx,x,x^2\right )\\ &=\frac {a x^2 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\operatorname {Subst}\left (\int \frac {a c-2 (b c-a d) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 b (b c-a d)}\\ &=\frac {a x^2 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,x^2\right )}{2 b^2}-\frac {(a (3 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 b^2 (b c-a d)}\\ &=\frac {a x^2 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x^2}{\sqrt {c+d x^4}}\right )}{2 b^2}-\frac {(a (3 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 b^2 (b c-a d)}\\ &=\frac {a x^2 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\sqrt {a} (3 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 b^2 (b c-a d)^{3/2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{2 b^2 \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 135, normalized size = 0.96 \begin {gather*} \frac {\frac {a b x^2 \sqrt {c+d x^4}}{\left (a+b x^4\right ) (b c-a d)}+\frac {\sqrt {a} (2 a d-3 b c) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{(b c-a d)^{3/2}}+\frac {2 \log \left (\sqrt {d} \sqrt {c+d x^4}+d x^2\right )}{\sqrt {d}}}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 1.59, size = 166, normalized size = 1.18 \begin {gather*} \frac {\left (2 a^{3/2} d-3 \sqrt {a} b 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 b^2 (b c-a d)^{3/2}}+\frac {a x^2 \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)}+\frac {\log \left (\sqrt {c+d x^4}+\sqrt {d} x^2\right )}{2 b^2 \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^9/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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fricas [A]  time = 0.84, size = 1077, normalized size = 7.64 \begin {gather*} \left [\frac {4 \, \sqrt {d x^{4} + c} a b d x^{2} + 4 \, {\left ({\left (b^{2} c - a b d\right )} x^{4} + a b c - a^{2} d\right )} \sqrt {d} \log \left (-2 \, d x^{4} - 2 \, \sqrt {d x^{4} + c} \sqrt {d} x^{2} - c\right ) + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{4} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-\frac {a}{b c - a 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^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{6} - {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right )}{16 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{4}\right )}}, \frac {4 \, \sqrt {d x^{4} + c} a b d x^{2} - 8 \, {\left ({\left (b^{2} c - a b d\right )} x^{4} + a b c - a^{2} d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{2}}{\sqrt {d x^{4} + c}}\right ) + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{4} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {-\frac {a}{b c - a 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^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{6} - {\left (a b c^{2} - a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{4} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}\right )}{16 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{4}\right )}}, \frac {2 \, \sqrt {d x^{4} + c} a b d x^{2} + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{4} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{6} + a c x^{2}\right )}}\right ) + 2 \, {\left ({\left (b^{2} c - a b d\right )} x^{4} + a b c - a^{2} d\right )} \sqrt {d} \log \left (-2 \, d x^{4} - 2 \, \sqrt {d x^{4} + c} \sqrt {d} x^{2} - c\right )}{8 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{4}\right )}}, \frac {2 \, \sqrt {d x^{4} + c} a b d x^{2} - 4 \, {\left ({\left (b^{2} c - a b d\right )} x^{4} + a b c - a^{2} d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{2}}{\sqrt {d x^{4} + c}}\right ) + {\left ({\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )} x^{4} + 3 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{4} - a c\right )} \sqrt {d x^{4} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{6} + a c x^{2}\right )}}\right )}{8 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(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*d*x^2 + 4*((b^2*c - a*b*d)*x^4 + a*b*c - a^2*d)*sqrt(d)*log(-2*d*x^4 - 2*sqrt(d*x
^4 + c)*sqrt(d)*x^2 - c) + ((3*b^2*c*d - 2*a*b*d^2)*x^4 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 2*a^2
*d^2)*x^6 - (a*b*c^2 - a^2*c*d)*x^2)*sqrt(d*x^4 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^8 + 2*a*b*x^4 + a^2)))/(a*b^
3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^4), 1/16*(4*sqrt(d*x^4 + c)*a*b*d*x^2 - 8*((b^2*c - a*b*d)*x^4 +
a*b*c - a^2*d)*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c)) + ((3*b^2*c*d - 2*a*b*d^2)*x^4 + 3*a*b*c*d - 2*a
^2*d^2)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^6 - (a*b*c^2 - a^2*c*d)*x^2)*sqrt(d*x^4 + c)*sqrt(-a/(b*c - a*d))
)/(b^2*x^8 + 2*a*b*x^4 + a^2)))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^4), 1/8*(2*sqrt(d*x^4 + c)*
a*b*d*x^2 + ((3*b^2*c*d - 2*a*b*d^2)*x^4 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*
d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a/(b*c - a*d))/(a*d*x^6 + a*c*x^2)) + 2*((b^2*c - a*b*d)*x^4 + a*b*c - a^2*
d)*sqrt(d)*log(-2*d*x^4 - 2*sqrt(d*x^4 + c)*sqrt(d)*x^2 - c))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)
*x^4), 1/8*(2*sqrt(d*x^4 + c)*a*b*d*x^2 - 4*((b^2*c - a*b*d)*x^4 + a*b*c - a^2*d)*sqrt(-d)*arctan(sqrt(-d)*x^2
/sqrt(d*x^4 + c)) + ((3*b^2*c*d - 2*a*b*d^2)*x^4 + 3*a*b*c*d - 2*a^2*d^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*
c - 2*a*d)*x^4 - a*c)*sqrt(d*x^4 + c)*sqrt(a/(b*c - a*d))/(a*d*x^6 + a*c*x^2)))/(a*b^3*c*d - a^2*b^2*d^2 + (b^
4*c*d - a*b^3*d^2)*x^4)]

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giac [B]  time = 0.54, size = 298, normalized size = 2.11 \begin {gather*} -\frac {{\left (3 \, a b c \sqrt {d} - 2 \, a^{2} d^{\frac {3}{2}}\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 )}{4 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a 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^{3} c - a b^{2} d\right )}} - \frac {\log \left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2}\right )}{4 \, b^{2} \sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.24, size = 893, normalized size = 6.33 \begin {gather*} \frac {3 a \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^{2}}-\frac {3 a \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^{2}}+\frac {\sqrt {-a b}\, a 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^{3}}-\frac {\sqrt {-a b}\, a 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^{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}}\, a}{8 \left (a d -b c \right ) \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) 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}}\, a}{8 \left (a d -b c \right ) \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) b^{2}}+\frac {\ln \left (\sqrt {d}\, x^{2}+\sqrt {d \,x^{4}+c}\right )}{2 b^{2} \sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

1/2/b^2*ln(d^(1/2)*x^2+(d*x^4+c)^(1/2))/d^(1/2)-3/8*a/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))+3/8*a/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))-1/8*a/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/b^3*(-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/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/b^3*(-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^{9}}{{\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^9/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^9/((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^9}{{\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^9/((a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

int(x^9/((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^{9}}{\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**9/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Integral(x**9/((a + b*x**4)**2*sqrt(c + d*x**4)), x)

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