3.3 \(\int x^3 (a+b \sec (c+d x^2)) \, dx\)

Optimal. Leaf size=92 \[ \frac {a x^4}{4}+\frac {i b \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {14, 4204, 4181, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {i b \text {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{2 d^2}+\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sec[c + d*x^2]),x]

[Out]

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

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4181

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 4204

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]

Rubi steps

\begin {align*} \int x^3 \left (a+b \sec \left (c+d x^2\right )\right ) \, dx &=\int \left (a x^3+b x^3 \sec \left (c+d x^2\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \sec \left (c+d x^2\right ) \, dx\\ &=\frac {a x^4}{4}+\frac {1}{2} b \operatorname {Subst}\left (\int x \sec (c+d x) \, dx,x,x^2\right )\\ &=\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {b \operatorname {Subst}\left (\int \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{2 d}+\frac {b \operatorname {Subst}\left (\int \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{2 d}\\ &=\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 d^2}\\ &=\frac {a x^4}{4}-\frac {i b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i b \text {Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{2 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 95, normalized size = 1.03 \[ \frac {a x^4}{4}+\frac {i b \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{2 d^2}-\frac {i b x^2 \tan ^{-1}\left (e^{i c+i d x^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Sec[c + d*x^2]),x]

[Out]

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

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fricas [B]  time = 0.79, size = 346, normalized size = 3.76 \[ \frac {a d^{2} x^{4} - b c \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + b c \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) - b c \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + b c \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) - i \, b {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - i \, b {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + i \, b {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + i \, b {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + {\left (b d x^{2} + b c\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (b d x^{2} + b c\right )} \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + {\left (b d x^{2} + b c\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (b d x^{2} + b c\right )} \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right )}{4 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sec(d*x^2+c)),x, algorithm="fricas")

[Out]

1/4*(a*d^2*x^4 - b*c*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + I) + b*c*log(cos(d*x^2 + c) - I*sin(d*x^2 + c) +
I) - b*c*log(-cos(d*x^2 + c) + I*sin(d*x^2 + c) + I) + b*c*log(-cos(d*x^2 + c) - I*sin(d*x^2 + c) + I) - I*b*d
ilog(I*cos(d*x^2 + c) + sin(d*x^2 + c)) - I*b*dilog(I*cos(d*x^2 + c) - sin(d*x^2 + c)) + I*b*dilog(-I*cos(d*x^
2 + c) + sin(d*x^2 + c)) + I*b*dilog(-I*cos(d*x^2 + c) - sin(d*x^2 + c)) + (b*d*x^2 + b*c)*log(I*cos(d*x^2 + c
) + sin(d*x^2 + c) + 1) - (b*d*x^2 + b*c)*log(I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1) + (b*d*x^2 + b*c)*log(-I*
cos(d*x^2 + c) + sin(d*x^2 + c) + 1) - (b*d*x^2 + b*c)*log(-I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1))/d^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x^{2} + c\right ) + a\right )} x^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sec(d*x^2+c)),x, algorithm="giac")

[Out]

integrate((b*sec(d*x^2 + c) + a)*x^3, x)

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \sec \left (d \,x^{2}+c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*sec(d*x^2+c)),x)

[Out]

int(x^3*(a+b*sec(d*x^2+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, a x^{4} + 2 \, b \int \frac {x^{3} \cos \left (2 \, d x^{2} + 2 \, c\right ) \cos \left (d x^{2} + c\right ) + x^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) \sin \left (d x^{2} + c\right ) + x^{3} \cos \left (d x^{2} + c\right )}{\cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x^{2} + 2 \, c\right ) + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sec(d*x^2+c)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 2*b*integrate((x^3*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + x^3*sin(2*d*x^2 + 2*c)*sin(d*x^2 + c) + x^3
*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/cos(c + d*x^2)),x)

[Out]

int(x^3*(a + b/cos(c + d*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a + b \sec {\left (c + d x^{2} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*sec(d*x**2+c)),x)

[Out]

Integral(x**3*(a + b*sec(c + d*x**2)), x)

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