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

Optimal. Leaf size=261 \[ \frac {b \text {Li}_2\left (-\frac {a e^{i \left (d x^2+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d^2 \sqrt {b^2-a^2}}-\frac {b \text {Li}_2\left (-\frac {a e^{i \left (d x^2+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{2 a d^2 \sqrt {b^2-a^2}}+\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{b-\sqrt {b^2-a^2}}\right )}{2 a d \sqrt {b^2-a^2}}-\frac {i b x^2 \log \left (1+\frac {a e^{i \left (c+d x^2\right )}}{\sqrt {b^2-a^2}+b}\right )}{2 a d \sqrt {b^2-a^2}}+\frac {x^4}{4 a} \]

[Out]

1/4*x^4/a+1/2*I*b*x^2*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^2*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)+1/2*b*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)-1/2*b*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)

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Rubi [A]  time = 0.54, antiderivative size = 261, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 \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*}

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Mathematica [B]  time = 1.40, 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 {Li}_2\left (\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 {Li}_2\left (\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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fricas [B]  time = 1.10, size = 1063, normalized size = 4.07 \[ \text {result too large to display} \]

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(-(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) + 2*a*b*s
qrt(-(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) - 2*a*b*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) + 2*a*b*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) - 2*(I*a*b*d*x^2 + I*a*b*c)*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) - 2*(-I*a*b*d*x^2 - I*a*b*c)*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) - 2*(-I*a*b*d*x^2 - I*a*b*c)*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) - 2*(I*a*b*d*x^2 + I*a*b*c)*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^2)

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

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(x^3/(b*sec(d*x^2 + c) + a), x)

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maple [F]  time = 0.82, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{a +b \sec \left (d \,x^{2}+c \right )}\, dx \]

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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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

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

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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