\(\int x^3 (a+b \sec (c+d x^2))^2 \, dx\) [10]

Optimal result

Integrand size = 18, antiderivative size = 133 \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^4}{4}-\frac {2 i a b x^2 \arctan \left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac {i a b \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac {i a b \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}+\frac {b^2 x^2 \tan \left (c+d x^2\right )}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {4289, 4275, 4266, 2317, 2438, 4269, 3556} \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^4}{4}-\frac {2 i a b x^2 \arctan \left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac {i a b \operatorname {PolyLog}\left (2,-i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {i a b \operatorname {PolyLog}\left (2,i e^{i \left (d x^2+c\right )}\right )}{d^2}+\frac {b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac {b^2 x^2 \tan \left (c+d x^2\right )}{2 d} \]

[In]

Int[x^3*(a + b*Sec[c + d*x^2])^2,x]

[Out]

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

Rule 2317

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 2438

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4275

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

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

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92 \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 d^2 x^4-8 i a b d x^2 \arctan \left (e^{i \left (c+d x^2\right )}\right )+2 b^2 \log \left (\cos \left (c+d x^2\right )\right )+4 i a b \operatorname {PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )-4 i a b \operatorname {PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )+2 b^2 d x^2 \tan \left (c+d x^2\right )}{4 d^2} \]

[In]

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

[Out]

(a^2*d^2*x^4 - (8*I)*a*b*d*x^2*ArcTan[E^(I*(c + d*x^2))] + 2*b^2*Log[Cos[c + d*x^2]] + (4*I)*a*b*PolyLog[2, (-
I)*E^(I*(c + d*x^2))] - (4*I)*a*b*PolyLog[2, I*E^(I*(c + d*x^2))] + 2*b^2*d*x^2*Tan[c + d*x^2])/(4*d^2)

Maple [F]

\[\int x^{3} {\left (a +b \sec \left (d \,x^{2}+c \right )\right )}^{2}d x\]

[In]

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

[Out]

int(x^3*(a+b*sec(d*x^2+c))^2,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. 525 vs. \(2 (109) = 218\).

Time = 0.33 (sec) , antiderivative size = 525, normalized size of antiderivative = 3.95 \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\frac {a^{2} d^{2} x^{4} \cos \left (d x^{2} + c\right ) + 2 \, b^{2} d x^{2} \sin \left (d x^{2} + c\right ) - 2 i \, a b \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - 2 i \, a b \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 2 i \, a b \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 2 i \, a b \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - {\left (2 \, a b c - b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + {\left (2 \, a b c + b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 2 \, {\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 2 \, {\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 2 \, {\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 2 \, {\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) - {\left (2 \, a b c - b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) + {\left (2 \, a b c + b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right )}{4 \, d^{2} \cos \left (d x^{2} + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\int x^{3} \left (a + b \sec {\left (c + d x^{2} \right )}\right )^{2}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\int { {\left (b \sec \left (d x^{2} + c\right ) + a\right )}^{2} x^{3} \,d x } \]

[In]

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

[Out]

1/4*a^2*x^4 + 1/4*(4*b^2*d*x^2*sin(2*d*x^2 + 2*c) + 16*(a*b*d^3*cos(2*d*x^2 + 2*c)^2 + a*b*d^3*sin(2*d*x^2 + 2
*c)^2 + 2*a*b*d^3*cos(2*d*x^2 + 2*c) + a*b*d^3)*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))/(d*cos(2*d*x^2 + 2*c)^2 + d*sin(2*d*x^2 + 2*c)^2 + 2*d*cos(2*
d*x^2 + 2*c) + d), x) + (b^2*cos(2*d*x^2 + 2*c)^2 + b^2*sin(2*d*x^2 + 2*c)^2 + 2*b^2*cos(2*d*x^2 + 2*c) + b^2)
*log(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))/(d^2*cos(2*d*x^2 + 2*c)^2 + d^2*
sin(2*d*x^2 + 2*c)^2 + 2*d^2*cos(2*d*x^2 + 2*c) + d^2)

Giac [F]

\[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\int { {\left (b \sec \left (d x^{2} + c\right ) + a\right )}^{2} x^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx=\int x^3\,{\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right )}^2 \,d x \]

[In]

int(x^3*(a + b/cos(c + d*x^2))^2,x)

[Out]

int(x^3*(a + b/cos(c + d*x^2))^2, x)