3.18 \(\int \frac{x^3}{a+b \sec (c+d x^2)} \, dx\)

Optimal. Leaf size=261 \[ \frac{b \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{2 a d^2 \sqrt{b^2-a^2}}-\frac{b \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{2 a d^2 \sqrt{b^2-a^2}}+\frac{i b x^2 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{2 a d \sqrt{b^2-a^2}}-\frac{i b x^2 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{2 a d \sqrt{b^2-a^2}}+\frac{x^4}{4 a} \]

[Out]

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

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Rubi [A]  time = 0.539928, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {4204, 4191, 3321, 2264, 2190, 2279, 2391} \[ \frac{b \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{2 a d^2 \sqrt{b^2-a^2}}-\frac{b \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{2 a d^2 \sqrt{b^2-a^2}}+\frac{i b x^2 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{2 a d \sqrt{b^2-a^2}}-\frac{i b x^2 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{2 a d \sqrt{b^2-a^2}}+\frac{x^4}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*Sec[c + d*x^2]),x]

[Out]

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

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]

Rule 4191

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 3321

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 2264

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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]

Rubi steps

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

Mathematica [B]  time = 1.33764, size = 845, normalized size = 3.24 \[ \frac{\left (b+a \cos \left (d x^2+c\right )\right ) \left (x^4-\frac{2 b \left (2 \left (d x^2+c\right ) \tanh ^{-1}\left (\frac{(a+b) \cot \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )-2 \left (c+\cos ^{-1}\left (-\frac{b}{a}\right )\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )+\left (\cos ^{-1}\left (-\frac{b}{a}\right )-2 i \tanh ^{-1}\left (\frac{(a+b) \cot \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )+2 i \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right ) \log \left (\frac{\sqrt{a^2-b^2} e^{-\frac{1}{2} i \left (d x^2+c\right )}}{\sqrt{2} \sqrt{a} \sqrt{b+a \cos \left (d x^2+c\right )}}\right )+\left (\cos ^{-1}\left (-\frac{b}{a}\right )+2 i \left (\tanh ^{-1}\left (\frac{(a+b) \cot \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )-\tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right )\right ) \log \left (\frac{\sqrt{a^2-b^2} e^{\frac{1}{2} i \left (d x^2+c\right )}}{\sqrt{2} \sqrt{a} \sqrt{b+a \cos \left (d x^2+c\right )}}\right )-\left (\cos ^{-1}\left (-\frac{b}{a}\right )-2 i \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right ) \log \left (\frac{(a+b) \left (a-b-i \sqrt{a^2-b^2}\right ) \left (i \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )+1\right )}{a \left (a+b+\sqrt{a^2-b^2} \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )-\left (\cos ^{-1}\left (-\frac{b}{a}\right )+2 i \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )}{\sqrt{a^2-b^2}}\right )\right ) \log \left (\frac{(a+b) \left (-i a+i b+\sqrt{a^2-b^2}\right ) \left (\tan \left (\frac{1}{2} \left (d x^2+c\right )\right )+i\right )}{a \left (a+b+\sqrt{a^2-b^2} \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )+i \left (\text{PolyLog}\left (2,\frac{\left (b-i \sqrt{a^2-b^2}\right ) \left (a+b-\sqrt{a^2-b^2} \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}{a \left (a+b+\sqrt{a^2-b^2} \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )-\text{PolyLog}\left (2,\frac{\left (b+i \sqrt{a^2-b^2}\right ) \left (a+b-\sqrt{a^2-b^2} \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}{a \left (a+b+\sqrt{a^2-b^2} \tan \left (\frac{1}{2} \left (d x^2+c\right )\right )\right )}\right )\right )\right )}{\sqrt{a^2-b^2} d^2}\right ) \sec \left (d x^2+c\right )}{4 a \left (a+b \sec \left (d x^2+c\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x^2])*(x^4 - (2*b*(2*(c + d*x^2)*ArcTanh[((a + b)*Cot[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] - 2*(
c + ArcCos[-(b/a)])*ArcTanh[((a - b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + (ArcCos[-(b/a)] - (2*I)*ArcTanh[((
a + b)*Cot[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] + (2*I)*ArcTanh[((a - b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Log
[Sqrt[a^2 - b^2]/(Sqrt[2]*Sqrt[a]*E^((I/2)*(c + d*x^2))*Sqrt[b + a*Cos[c + d*x^2]])] + (ArcCos[-(b/a)] + (2*I)
*(ArcTanh[((a + b)*Cot[(c + d*x^2)/2])/Sqrt[a^2 - b^2]] - ArcTanh[((a - b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]
]))*Log[(Sqrt[a^2 - b^2]*E^((I/2)*(c + d*x^2)))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cos[c + d*x^2]])] - (ArcCos[-(b/a)
] - (2*I)*ArcTanh[((a - b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(a - b - I*Sqrt[a^2 - b^2])*(1 +
I*Tan[(c + d*x^2)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))] - (ArcCos[-(b/a)] + (2*I)*ArcTanh[((
a - b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*((-I)*a + I*b + Sqrt[a^2 - b^2])*(I + Tan[(c + d*x^2
)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a + b - Sqr
t[a^2 - b^2]*Tan[(c + d*x^2)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))] - PolyLog[2, ((b + I*Sqrt[
a^2 - b^2])*(a + b - Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^2)/2]))])))
/(Sqrt[a^2 - b^2]*d^2))*Sec[c + d*x^2])/(4*a*(a + b*Sec[c + d*x^2]))

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Maple [F]  time = 0.11, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{a+b\sec \left ( d{x}^{2}+c \right ) }}\, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.51719, size = 2566, normalized size = 9.83 \begin{align*} \text{result too large to display} \end{align*}

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]

1/8*(2*(a^2 - b^2)*d^2*x^4 - 2*I*a*b*c*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) + 2*I*a*b*c*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) - 2*I*a*b*c*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos(d*x^2 + c) + 2*I*a*si
n(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) + 2*I*a*b*c*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) - 2*a*b*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*
x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a +
1) + 2*a*b*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c)
+ I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) - 2*a*b*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(
d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*(a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a
+ 1) + 2*a*b*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c
) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) - 2*(I*a*b*d*x^2 + I*a*b*c)*sqrt(-(a^2 - b^2)/a^2
)*log(1/2*(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 -
b^2)/a^2) + 2*a)/a) - 2*(-I*a*b*d*x^2 - I*a*b*c)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(d*x^2 + c) + 2*I*b*si
n(d*x^2 + c) - 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(-I*a*b*d*x^2 -
I*a*b*c)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*(a*cos(d*x^2 + c) - I*a
*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(I*a*b*d*x^2 + I*a*b*c)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(
2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) +
2*a)/a))/((a^3 - a*b^2)*d^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{a + b \sec{\left (c + d x^{2} \right )}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{b \sec \left (d x^{2} + c\right ) + a}\,{d x} \end{align*}

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(x^3/(b*sec(d*x^2 + c) + a), x)