3.6.37 \(\int \frac {x^{13}}{(a+b x^4)^2 \sqrt {c+d x^4}} \, dx\)

Optimal. Leaf size=191 \[ \frac {a^{3/2} (5 b c-4 a d) \tan ^{-1}\left (\frac {x^2 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^4}}\right )}{4 b^3 (b c-a d)^{3/2}}-\frac {(4 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{4 b^3 d^{3/2}}+\frac {x^2 \sqrt {c+d x^4} (b c-2 a d)}{4 b^2 d (b c-a d)}+\frac {a x^6 \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)} \]

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Rubi [A]  time = 0.37, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {465, 470, 582, 523, 217, 206, 377, 205} \begin {gather*} \frac {a^{3/2} (5 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 b^3 (b c-a d)^{3/2}}-\frac {(4 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{4 b^3 d^{3/2}}+\frac {x^2 \sqrt {c+d x^4} (b c-2 a d)}{4 b^2 d (b c-a d)}+\frac {a x^6 \sqrt {c+d x^4}}{4 b \left (a+b x^4\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^13/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

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

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]

Rule 582

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

Rubi steps

\begin {align*} \int \frac {x^{13}}{\left (a+b x^4\right )^2 \sqrt {c+d x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^6}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx,x,x^2\right )\\ &=\frac {a x^6 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a c-2 (b c-2 a d) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 b (b c-a d)}\\ &=\frac {(b c-2 a d) x^2 \sqrt {c+d x^4}}{4 b^2 d (b c-a d)}+\frac {a x^6 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac {\operatorname {Subst}\left (\int \frac {-2 a c (b c-2 a d)-2 (b c-a d) (b c+4 a d) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^2\right )}{8 b^2 d (b c-a d)}\\ &=\frac {(b c-2 a d) x^2 \sqrt {c+d x^4}}{4 b^2 d (b c-a d)}+\frac {a x^6 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac {\left (a^2 (5 b c-4 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 b^3 (b c-a d)}-\frac {(b c+4 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,x^2\right )}{4 b^3 d}\\ &=\frac {(b c-2 a d) x^2 \sqrt {c+d x^4}}{4 b^2 d (b c-a d)}+\frac {a x^6 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac {\left (a^2 (5 b c-4 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 b^3 (b c-a d)}-\frac {(b c+4 a d) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x^2}{\sqrt {c+d x^4}}\right )}{4 b^3 d}\\ &=\frac {(b c-2 a d) x^2 \sqrt {c+d x^4}}{4 b^2 d (b c-a d)}+\frac {a x^6 \sqrt {c+d x^4}}{4 b (b c-a d) \left (a+b x^4\right )}+\frac {a^{3/2} (5 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 b^3 (b c-a d)^{3/2}}-\frac {(b c+4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c+d x^4}}\right )}{4 b^3 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 150, normalized size = 0.79 \begin {gather*} \frac {\frac {a^{3/2} (5 b c-4 a d) \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}}+b x^2 \sqrt {c+d x^4} \left (\frac {a^2}{\left (a+b x^4\right ) (a d-b c)}+\frac {1}{d}\right )-\frac {(4 a d+b c) \log \left (\sqrt {d} \sqrt {c+d x^4}+d x^2\right )}{d^{3/2}}}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^13/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

(b*x^2*Sqrt[c + d*x^4]*(d^(-1) + a^2/((-(b*c) + a*d)*(a + b*x^4))) + (a^(3/2)*(5*b*c - 4*a*d)*ArcTan[(Sqrt[b*c
- a*d]*x^2)/(Sqrt[a]*Sqrt[c + d*x^4])])/(b*c - a*d)^(3/2) - ((b*c + 4*a*d)*Log[d*x^2 + Sqrt[d]*Sqrt[c + d*x^4
]])/d^(3/2))/(4*b^3)

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IntegrateAlgebraic [A]  time = 2.29, size = 208, normalized size = 1.09 \begin {gather*} \frac {\left (5 a^{3/2} b c-4 a^{5/2} 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 b^3 (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^4} \left (2 a^2 d x^2-a b c x^2+a b d x^6-b^2 c x^6\right )}{4 b^2 d \left (a+b x^4\right ) (b c-a d)}+\frac {(-4 a d-b c) \log \left (\sqrt {c+d x^4}+\sqrt {d} x^2\right )}{4 b^3 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^13/((a + b*x^4)^2*Sqrt[c + d*x^4]),x]

[Out]

-1/4*(Sqrt[c + d*x^4]*(-(a*b*c*x^2) + 2*a^2*d*x^2 - b^2*c*x^6 + a*b*d*x^6))/(b^2*d*(b*c - a*d)*(a + b*x^4)) +
((5*a^(3/2)*b*c - 4*a^(5/2)*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^3*(b*c - a*d)^(3/2)) + ((-(b*c) - 4*a*d)*Log[Sqrt[d]*x^2 + Sqrt[c + d*x^4]])/(4*b^3*d^(3/2))

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fricas [A]  time = 1.94, size = 1386, normalized size = 7.26

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.59, size = 337, normalized size = 1.76 \begin {gather*} -\frac {{\left (5 \, a^{2} b c \sqrt {d} - 4 \, a^{3} 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^{4} c - a b^{3} d\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {\sqrt {d x^{4} + c} x^{2}}{4 \, b^{2} d} + \frac {{\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a^{2} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2} a^{3} d^{\frac {3}{2}} - a^{2} 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^{4} c - a b^{3} d\right )}} + \frac {{\left (b c \sqrt {d} + 4 \, a d^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {d} x^{2} - \sqrt {d x^{4} + c}\right )}^{2}\right )}{8 \, b^{3} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.27, size = 953, normalized size = 4.99 \begin {gather*} -\frac {5 a^{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}}\, b^{3}}+\frac {5 a^{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}}\, b^{3}}-\frac {\sqrt {-a b}\, a^{2} 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^{4}}+\frac {\sqrt {-a b}\, a^{2} 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^{4}}+\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^{2}}{8 \left (a d -b c \right ) \left (x^{2}-\frac {\sqrt {-a b}}{b}\right ) 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^{2}}{8 \left (a d -b c \right ) \left (x^{2}+\frac {\sqrt {-a b}}{b}\right ) b^{3}}+\frac {\sqrt {d \,x^{4}+c}\, x^{2}}{4 b^{2} d}-\frac {a \ln \left (\sqrt {d}\, x^{2}+\sqrt {d \,x^{4}+c}\right )}{b^{3} \sqrt {d}}-\frac {c \ln \left (\sqrt {d}\, x^{2}+\sqrt {d \,x^{4}+c}\right )}{4 b^{2} d^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(b*x^4+a)^2/(d*x^4+c)^(1/2),x)

[Out]

1/4/b^2*x^2/d*(d*x^4+c)^(1/2)-1/4/b^2*c/d^(3/2)*ln(d^(1/2)*x^2+(d*x^4+c)^(1/2))-1/b^3*a*ln(d^(1/2)*x^2+(d*x^4+
c)^(1/2))/d^(1/2)+5/8*a^2/b^3/(-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^2/b^3/(-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^3/(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/b^4*(-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^3/(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/b^4*(-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^{13}}{{\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^13/(b*x^4+a)^2/(d*x^4+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^13/((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^{13}}{{\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^13/((a + b*x^4)^2*(c + d*x^4)^(1/2)),x)

[Out]

int(x^13/((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^{13}}{\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**13/(b*x**4+a)**2/(d*x**4+c)**(1/2),x)

[Out]

Integral(x**13/((a + b*x**4)**2*sqrt(c + d*x**4)), x)

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