3.1 \(\int x^3 (a+b \tan (c+d x^2)) \, dx\)

Optimal. Leaf size=73 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )}{4 d^2}+\frac{a x^4}{4}-\frac{b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{2 d}+\frac{1}{4} i b x^4 \]

[Out]

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

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Rubi [A]  time = 0.141527, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {14, 3747, 3719, 2190, 2279, 2391} \[ \frac{a x^4}{4}+\frac{i b \text{Li}_2\left (-e^{2 i \left (d x^2+c\right )}\right )}{4 d^2}-\frac{b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{2 d}+\frac{1}{4} i b x^4 \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Tan[c + d*x^2]),x]

[Out]

(a*x^4)/4 + (I/4)*b*x^4 - (b*x^2*Log[1 + E^((2*I)*(c + d*x^2))])/(2*d) + ((I/4)*b*PolyLog[2, -E^((2*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 3747

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Tan[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 3719

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

Mathematica [A]  time = 0.0417914, size = 73, normalized size = 1. \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )}{4 d^2}+\frac{a x^4}{4}-\frac{b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{2 d}+\frac{1}{4} i b x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.09, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b\tan \left ( d{x}^{2}+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a x^{4} + 2 \, b \int \frac{x^{3} \sin \left (2 \, d x^{2} + 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4 + 2*b*integrate(x^3*sin(2*d*x^2 + 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]  time = 1.63385, size = 387, normalized size = 5.3 \begin{align*} \frac{2 \, a d^{2} x^{4} - 2 \, b d x^{2} \log \left (-\frac{2 \,{\left (i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) - 2 \, b d x^{2} \log \left (-\frac{2 \,{\left (-i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) - i \, b{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1} + 1\right ) + i \, b{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1} + 1\right )}{8 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a*d^2*x^4 - 2*b*d*x^2*log(-2*(I*tan(d*x^2 + c) - 1)/(tan(d*x^2 + c)^2 + 1)) - 2*b*d*x^2*log(-2*(-I*tan(
d*x^2 + c) - 1)/(tan(d*x^2 + c)^2 + 1)) - I*b*dilog(2*(I*tan(d*x^2 + c) - 1)/(tan(d*x^2 + c)^2 + 1) + 1) + I*b
*dilog(2*(-I*tan(d*x^2 + c) - 1)/(tan(d*x^2 + c)^2 + 1) + 1))/d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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