\(\int \frac {x^5}{a+b \sec (c+d x^2)} \, dx\) [16]

Optimal result

Integrand size = 18, antiderivative size = 382 \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3} \]

[Out]

1/6*x^6/a+1/2*I*b*x^4*ln(1+a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a/d/(-a^2+b^2)^(1/2)-1/2*I*b*x^4*ln(1+a*ex
p(I*(d*x^2+c))/(b+(-a^2+b^2)^(1/2)))/a/d/(-a^2+b^2)^(1/2)+b*x^2*polylog(2,-a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1
/2)))/a/d^2/(-a^2+b^2)^(1/2)-b*x^2*polylog(2,-a*exp(I*(d*x^2+c))/(b+(-a^2+b^2)^(1/2)))/a/d^2/(-a^2+b^2)^(1/2)+
I*b*polylog(3,-a*exp(I*(d*x^2+c))/(b-(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)-I*b*polylog(3,-a*exp(I*(d*x^2+c
))/(b+(-a^2+b^2)^(1/2)))/a/d^3/(-a^2+b^2)^(1/2)

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4289, 4276, 3402, 2296, 2221, 2611, 2320, 6724} \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}-\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^3 \sqrt {b^2-a^2}}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d \sqrt {b^2-a^2}}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^6}{6 a} \]

[In]

Int[x^5/(a + b*Sec[c + d*x^2]),x]

[Out]

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

Rule 2221

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

Rule 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3402

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[(c
+ d*x)^m*E^(I*Pi*(k - 1/2))*(E^(I*(e + f*x))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(2
*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4276

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 4289

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{a+b \sec (c+d x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {x^2}{a}-\frac {b x^2}{a (b+a \cos (c+d x))}\right ) \, dx,x,x^2\right ) \\ & = \frac {x^6}{6 a}-\frac {b \text {Subst}\left (\int \frac {x^2}{b+a \cos (c+d x)} \, dx,x,x^2\right )}{2 a} \\ & = \frac {x^6}{6 a}-\frac {b \text {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{a+2 b e^{i (c+d x)}+a e^{2 i (c+d x)}} \, dx,x,x^2\right )}{a} \\ & = \frac {x^6}{6 a}-\frac {b \text {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{2 b-2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {-a^2+b^2}}+\frac {b \text {Subst}\left (\int \frac {e^{i (c+d x)} x^2}{2 b+2 \sqrt {-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt {-a^2+b^2}} \\ & = \frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {(i b) \text {Subst}\left (\int x \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d}+\frac {(i b) \text {Subst}\left (\int x \log \left (1+\frac {2 a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d} \\ & = \frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-\frac {2 a e^{i (c+d x)}}{2 b-2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {b \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-\frac {2 a e^{i (c+d x)}}{2 b+2 \sqrt {-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt {-a^2+b^2} d^2} \\ & = \frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {(i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\frac {a x}{-b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {(i b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,-\frac {a x}{b+\sqrt {-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{a \sqrt {-a^2+b^2} d^3} \\ & = \frac {x^6}{6 a}+\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}-\frac {i b x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} d}+\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}-\frac {b x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2}+\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3}-\frac {i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\frac {\sqrt {-a^2+b^2} d^3 x^6+3 i b d^2 x^4 \log \left (1-\frac {a e^{i \left (c+d x^2\right )}}{-b+\sqrt {-a^2+b^2}}\right )-3 i b d^2 x^4 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )+6 b d x^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \left (c+d x^2\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 b d x^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )+6 i b \operatorname {PolyLog}\left (3,\frac {a e^{i \left (c+d x^2\right )}}{-b+\sqrt {-a^2+b^2}}\right )-6 i b \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d x^2\right )}}{b+\sqrt {-a^2+b^2}}\right )}{6 a \sqrt {-a^2+b^2} d^3} \]

[In]

Integrate[x^5/(a + b*Sec[c + d*x^2]),x]

[Out]

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

Maple [F]

\[\int \frac {x^{5}}{a +b \sec \left (d \,x^{2}+c \right )}d x\]

[In]

int(x^5/(a+b*sec(d*x^2+c)),x)

[Out]

int(x^5/(a+b*sec(d*x^2+c)),x)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1457 vs. \(2 (330) = 660\).

Time = 0.44 (sec) , antiderivative size = 1457, normalized size of antiderivative = 3.81 \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\text {Too large to display} \]

[In]

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="fricas")

[Out]

1/12*(2*(a^2 - b^2)*d^3*x^6 - 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c)
+ (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/
a^2)*dilog(-(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)
/a^2) + a)/a + 1) - 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) + (a*cos(
d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + 6*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog
(-(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)
/a + 1) + 3*I*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^2 + c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 -
b^2)/a^2) + 2*b) - 3*I*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^2 + c) + 2*a*sqrt
(-(a^2 - b^2)/a^2) + 2*b) + 3*I*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(d*x^2 + c) + 2*I*a*sin(d*x^2 + c)
+ 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 3*I*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(d*x^2 + c) - 2*I*a*sin(d
*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 6*I*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(d*x^2 + c) +
I*b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) + 6*I*a*b*sqrt(-(a^2
- b^2)/a^2)*polylog(3, -(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(
-(a^2 - b^2)/a^2))/a) + 6*I*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) + (a
*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) - 6*I*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -
(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) +
3*(-I*a*b*d^2*x^4 + I*a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) + (a*cos(d*x^
2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a) + 3*(I*a*b*d^2*x^4 - I*a*b*c^2)*sqrt(-(a^2 - b^2)/
a^2)*log((b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^
2) + a)/a) + 3*(I*a*b*d^2*x^4 - I*a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) +
(a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + a)/a) + 3*(-I*a*b*d^2*x^4 + I*a*b*c^2)*sqrt(
-(a^2 - b^2)/a^2)*log((b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) - (a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(
a^2 - b^2)/a^2) + a)/a))/((a^3 - a*b^2)*d^3)

Sympy [F]

\[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x^{5}}{a + b \sec {\left (c + d x^{2} \right )}}\, dx \]

[In]

integrate(x**5/(a+b*sec(d*x**2+c)),x)

[Out]

Integral(x**5/(a + b*sec(c + d*x**2)), x)

Maxima [F]

\[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \sec \left (d x^{2} + c\right ) + a} \,d x } \]

[In]

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="maxima")

[Out]

1/6*(x^6 - 12*a*b*integrate((a*x^5*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + 2*b*x^5*cos(d*x^2 + c)^2 + a*x^5*sin(2*
d*x^2 + 2*c)*sin(d*x^2 + c) + 2*b*x^5*sin(d*x^2 + c)^2 + a*x^5*cos(d*x^2 + c))/(a^3*cos(2*d*x^2 + 2*c)^2 + 4*a
*b^2*cos(d*x^2 + c)^2 + a^3*sin(2*d*x^2 + 2*c)^2 + 4*a^2*b*sin(2*d*x^2 + 2*c)*sin(d*x^2 + c) + 4*a*b^2*sin(d*x
^2 + c)^2 + 4*a^2*b*cos(d*x^2 + c) + a^3 + 2*(2*a^2*b*cos(d*x^2 + c) + a^3)*cos(2*d*x^2 + 2*c)), x))/a

Giac [F]

\[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int { \frac {x^{5}}{b \sec \left (d x^{2} + c\right ) + a} \,d x } \]

[In]

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="giac")

[Out]

integrate(x^5/(b*sec(d*x^2 + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{a+b \sec \left (c+d x^2\right )} \, dx=\int \frac {x^5}{a+\frac {b}{\cos \left (d\,x^2+c\right )}} \,d x \]

[In]

int(x^5/(a + b/cos(c + d*x^2)),x)

[Out]

int(x^5/(a + b/cos(c + d*x^2)), x)