\(\int (a+b \tan (c+d \sqrt [3]{x})) \, dx\) [49]

Optimal result

Integrand size = 14, antiderivative size = 98 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3} \]

[Out]

a*x+I*b*x-3*b*x^(2/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d+3*I*b*x^(1/3)*polylog(2,-exp(2*I*(c+d*x^(1/3))))/d^2-3/2*
b*polylog(3,-exp(2*I*(c+d*x^(1/3))))/d^3

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3824, 3800, 2221, 2611, 2320, 6724} \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=a x-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i b x \]

[In]

Int[a + b*Tan[c + d*x^(1/3)],x]

[Out]

a*x + I*b*x - (3*b*x^(2/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d + ((3*I)*b*x^(1/3)*PolyLog[2, -E^((2*I)*(c +
d*x^(1/3)))])/d^2 - (3*b*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/(2*d^3)

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3800

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 3824

Int[((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*Ta
n[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[1/n, 0] && IntegerQ[p]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps \begin{align*} \text {integral}& = a x+b \int \tan \left (c+d \sqrt [3]{x}\right ) \, dx \\ & = a x+(3 b) \text {Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right ) \\ & = a x+i b x-(6 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right ) \\ & = a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {(6 b) \text {Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d} \\ & = a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {(3 i b) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2} \\ & = a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {(3 b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3} \\ & = a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3} \]

[In]

Integrate[a + b*Tan[c + d*x^(1/3)],x]

[Out]

a*x + I*b*x - (3*b*x^(2/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d + ((3*I)*b*x^(1/3)*PolyLog[2, -E^((2*I)*(c +
d*x^(1/3)))])/d^2 - (3*b*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/(2*d^3)

Maple [F]

\[\int \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]

[In]

int(a+b*tan(c+d*x^(1/3)),x)

[Out]

int(a+b*tan(c+d*x^(1/3)),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. 249 vs. \(2 (75) = 150\).

Time = 0.27 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.54 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {4 \, a d^{3} x - 6 \, b d^{2} x^{\frac {2}{3}} \log \left (-\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 \, b d^{2} x^{\frac {2}{3}} \log \left (-\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 i \, b d x^{\frac {1}{3}} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) + 6 i \, b d x^{\frac {1}{3}} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 3 \, b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right )}{4 \, d^{3}} \]

[In]

integrate(a+b*tan(c+d*x^(1/3)),x, algorithm="fricas")

[Out]

1/4*(4*a*d^3*x - 6*b*d^2*x^(2/3)*log(-2*(I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 6*b*d^2*x^(2/
3)*log(-2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 6*I*b*d*x^(1/3)*dilog(2*(I*tan(d*x^(1/3) +
c) - 1)/(tan(d*x^(1/3) + c)^2 + 1) + 1) + 6*I*b*d*x^(1/3)*dilog(2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3)
+ c)^2 + 1) + 1) - 3*b*polylog(3, (tan(d*x^(1/3) + c)^2 + 2*I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 +
1)) - 3*b*polylog(3, (tan(d*x^(1/3) + c)^2 - 2*I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)))/d^3

Sympy [F]

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

[In]

integrate(a+b*tan(c+d*x**(1/3)),x)

[Out]

Integral(a + b*tan(c + d*x**(1/3)), x)

Maxima [F]

\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { b \tan \left (d x^{\frac {1}{3}} + c\right ) + a \,d x } \]

[In]

integrate(a+b*tan(c+d*x^(1/3)),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { b \tan \left (d x^{\frac {1}{3}} + c\right ) + a \,d x } \]

[In]

integrate(a+b*tan(c+d*x^(1/3)),x, algorithm="giac")

[Out]

integrate(b*tan(d*x^(1/3) + c) + a, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right ) \,d x \]

[In]

int(a + b*tan(c + d*x^(1/3)),x)

[Out]

int(a + b*tan(c + d*x^(1/3)), x)