3.1.1 \(\int x^5 (a+b \sec (c+d x^2)) \, dx\) [1]

Optimal. Leaf size=143 \[ \frac {a x^6}{6}-\frac {i b x^4 \text {ArcTan}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b x^2 \text {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \text {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {b \text {PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b \text {PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {14, 4289, 4266, 2611, 2320, 6724} \begin {gather*} \frac {a x^6}{6}-\frac {i b x^4 \text {ArcTan}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {b \text {Li}_3\left (-i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {b \text {Li}_3\left (i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {i b x^2 \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {i b x^2 \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sec[c + d*x^2]),x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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*} \int x^5 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx &=\int \left (a x^5+b x^5 \sec \left (c+d x^2\right )\right ) \, dx\\ &=\frac {a x^6}{6}+b \int x^5 \sec \left (c+d x^2\right ) \, dx\\ &=\frac {a x^6}{6}+\frac {1}{2} b \text {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,x^2\right )\\ &=\frac {a x^6}{6}-\frac {i b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {b \text {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac {b \text {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}\\ &=\frac {a x^6}{6}-\frac {i b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {(i b) \text {Subst}\left (\int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}+\frac {(i b) \text {Subst}\left (\int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}\\ &=\frac {a x^6}{6}-\frac {i b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {b \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}\\ &=\frac {a x^6}{6}-\frac {i b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b x^2 \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {b \text {Li}_3\left (-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b \text {Li}_3\left (i e^{i \left (c+d x^2\right )}\right )}{d^3}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 146, normalized size = 1.02 \begin {gather*} \frac {a x^6}{6}-\frac {i b x^4 \text {ArcTan}\left (e^{i c+i d x^2}\right )}{d}+\frac {i b x^2 \text {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i b x^2 \text {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {b \text {PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac {b \text {PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.25, size = 0, normalized size = 0.00 \[\int x^{5} \left (a +b \sec \left (d \,x^{2}+c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (115) = 230\).
time = 3.80, size = 495, normalized size = 3.46 \begin {gather*} \frac {2 \, a d^{3} x^{6} - 6 i \, b d x^{2} {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - 6 i \, b d x^{2} {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 6 i \, b d x^{2} {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 i \, b d x^{2} {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 3 \, b c^{2} \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, b c^{2} \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 3 \, b c^{2} \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, b c^{2} \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (b d^{2} x^{4} - b c^{2}\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) - 6 \, b {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, b {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - 6 \, b {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, b {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )}{12 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \left (a + b \sec {\left (c + d x^{2} \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^5\,\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b/cos(c + d*x^2)),x)

[Out]

int(x^5*(a + b/cos(c + d*x^2)), x)

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