3.9.10 \(\int \frac {x^9}{(a+b x^4) \sqrt {c+d x^4}} \, dx\) [810]

Optimal. Leaf size=123 \[ \frac {x^2 \sqrt {c+d x^4}}{4 b d}+\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^2}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 b^2 \sqrt {b c-a d}}-\frac {(b c+2 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{4 b^2 d^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 123, 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 = {476, 490, 537, 223, 212, 385, 211} \begin {gather*} \frac {a^{3/2} \text {ArcTan}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{2 b^2 \sqrt {b c-a d}}-\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{4 b^2 d^{3/2}}+\frac {x^2 \sqrt {c+d x^4}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 212

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

Rule 223

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 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 490

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

Rule 537

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

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Mathematica [A]
time = 0.65, size = 140, normalized size = 1.14 \begin {gather*} \frac {\sqrt {d} \left (b x^2 \sqrt {c+d x^4}+\frac {2 a^{3/2} 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 )}{\sqrt {b c-a d}}\right )-(b c+2 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^4}}{\sqrt {d} x^2}\right )}{4 b^2 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(402\) vs. \(2(99)=198\).
time = 0.38, size = 403, normalized size = 3.28

method	result	size
default	\(\frac {\frac {x^{2} \sqrt {d \,x^{4}+c}}{4 d}-\frac {c \ln \left (x^{2} \sqrt {d}+\sqrt {d \,x^{4}+c}\right )}{4 d^{\frac {3}{2}}}}{b}-\frac {a \ln \left (x^{2} \sqrt {d}+\sqrt {d \,x^{4}+c}\right )}{2 b^{2} \sqrt {d}}+\frac {a^{2} \left (-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\right )}{b^{2}}\)	\(403\)
risch	\(\frac {x^{2} \sqrt {d \,x^{4}+c}}{4 b d}-\frac {a \ln \left (x^{2} \sqrt {d}+\sqrt {d \,x^{4}+c}\right )}{2 b^{2} \sqrt {d}}-\frac {\ln \left (x^{2} \sqrt {d}+\sqrt {d \,x^{4}+c}\right ) c}{4 b \,d^{\frac {3}{2}}}-\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 b^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 b^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\)	\(408\)
elliptic	\(\frac {x^{2} \sqrt {d \,x^{4}+c}}{4 b d}-\frac {a \ln \left (x^{2} \sqrt {d}+\sqrt {d \,x^{4}+c}\right )}{2 b^{2} \sqrt {d}}-\frac {\ln \left (x^{2} \sqrt {d}+\sqrt {d \,x^{4}+c}\right ) c}{4 b \,d^{\frac {3}{2}}}-\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}-\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 d \sqrt {-a b}\, \left (x^{2}-\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x^{2}-\frac {\sqrt {-a b}}{b}}\right )}{4 b^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}+\frac {a^{2} \ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x^{2}+\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 d \sqrt {-a b}\, \left (x^{2}+\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x^{2}+\frac {\sqrt {-a b}}{b}}\right )}{4 b^{2} \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}}\)	\(408\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(x^9/((b*x^4 + a)*sqrt(d*x^4 + c)), x)

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Fricas [A]
time = 3.73, size = 739, normalized size = 6.01 \begin {gather*} \left [\frac {2 \, \sqrt {d x^{4} + c} b d x^{2} + a d^{2} \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 ) + {\left (b c + 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{4} + 2 \, \sqrt {d x^{4} + c} \sqrt {d} x^{2} - c\right )}{8 \, b^{2} d^{2}}, \frac {2 \, \sqrt {d x^{4} + c} b d x^{2} + a d^{2} \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 ) + 2 \, {\left (b c + 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{2}}{\sqrt {d x^{4} + c}}\right )}{8 \, b^{2} d^{2}}, \frac {2 \, \sqrt {d x^{4} + c} b d x^{2} - 2 \, a d^{2} \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 ) + {\left (b c + 2 \, a d\right )} \sqrt {d} \log \left (-2 \, d x^{4} + 2 \, \sqrt {d x^{4} + c} \sqrt {d} x^{2} - c\right )}{8 \, b^{2} d^{2}}, \frac {\sqrt {d x^{4} + c} b d x^{2} - a d^{2} \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 ) + {\left (b c + 2 \, a d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x^{2}}{\sqrt {d x^{4} + c}}\right )}{4 \, b^{2} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(2*sqrt(d*x^4 + c)*b*d*x^2 + a*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)) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^4 + 2*sqrt(d
*x^4 + c)*sqrt(d)*x^2 - c))/(b^2*d^2), 1/8*(2*sqrt(d*x^4 + c)*b*d*x^2 + a*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)) + 2*(b*c
+ 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c)))/(b^2*d^2), 1/8*(2*sqrt(d*x^4 + c)*b*d*x^2 - 2*a*d^2*sq
rt(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
)) + (b*c + 2*a*d)*sqrt(d)*log(-2*d*x^4 + 2*sqrt(d*x^4 + c)*sqrt(d)*x^2 - c))/(b^2*d^2), 1/4*(sqrt(d*x^4 + c)*
b*d*x^2 - a*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)) + (b*c + 2*a*d)*sqrt(-d)*arctan(sqrt(-d)*x^2/sqrt(d*x^4 + c)))/(b^2*d^2)]

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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 ) \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)/(d*x**4+c)**(1/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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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 )\,\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)*(c + d*x^4)^(1/2)),x)

[Out]

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

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