Integrand size = 16, antiderivative size = 176 \[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\frac {x}{a+i b}+\frac {3 b x^{2/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 b \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^3} \] Output:
x/(a+I*b)+3*b*x^(2/3)*ln(1+(a^2+b^2)*exp(2*I*(c+d*x^(1/3)))/(a+I*b)^2)/(a^ 2+b^2)/d-3*I*b*x^(1/3)*polylog(2,-(a^2+b^2)*exp(2*I*(c+d*x^(1/3)))/(a+I*b) ^2)/(a^2+b^2)/d^2+3/2*b*polylog(3,-(a^2+b^2)*exp(2*I*(c+d*x^(1/3)))/(a+I*b )^2)/(a^2+b^2)/d^3
Time = 0.84 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.94 \[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\frac {2 a d^3 x+2 i b d^3 x+6 b d^2 x^{2/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+6 i b d \sqrt [3]{x} \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+3 b \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{2 \left (a^2+b^2\right ) d^3} \] Input:
Integrate[(a + b*Tan[c + d*x^(1/3)])^(-1),x]
Output:
(2*a*d^3*x + (2*I)*b*d^3*x + 6*b*d^2*x^(2/3)*Log[1 + (a + I*b)/((a - I*b)* E^((2*I)*(c + d*x^(1/3))))] + (6*I)*b*d*x^(1/3)*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 3*b*PolyLog[3, (-a - I*b)/((a - I*b)* E^((2*I)*(c + d*x^(1/3))))])/(2*(a^2 + b^2)*d^3)
Time = 0.74 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4226, 3042, 4215, 2620, 3011, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx\) |
\(\Big \downarrow \) 4226 |
\(\displaystyle 3 \int \frac {x^{2/3}}{a+b \tan \left (c+d \sqrt [3]{x}\right )}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 3 \int \frac {x^{2/3}}{a+b \tan \left (c+d \sqrt [3]{x}\right )}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 4215 |
\(\displaystyle 3 \left (2 i b \int \frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} x^{2/3}}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}d\sqrt [3]{x}+\frac {x}{3 (a+i b)}\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle 3 \left (2 i b \left (\frac {i \int \sqrt [3]{x} \log \left (\frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )d\sqrt [3]{x}}{d \left (a^2+b^2\right )}-\frac {i x^{2/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x}{3 (a+i b)}\right )\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle 3 \left (2 i b \left (\frac {i \left (\frac {i \sqrt [3]{x} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {i \int \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )d\sqrt [3]{x}}{2 d}\right )}{d \left (a^2+b^2\right )}-\frac {i x^{2/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x}{3 (a+i b)}\right )\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle 3 \left (2 i b \left (\frac {i \left (\frac {i \sqrt [3]{x} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {\int \frac {\operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\sqrt [3]{x}}de^{2 i \left (c+d \sqrt [3]{x}\right )}}{4 d^2}\right )}{d \left (a^2+b^2\right )}-\frac {i x^{2/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x}{3 (a+i b)}\right )\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle 3 \left (2 i b \left (\frac {i \left (\frac {i \sqrt [3]{x} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {\operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^2}\right )}{d \left (a^2+b^2\right )}-\frac {i x^{2/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x}{3 (a+i b)}\right )\) |
Input:
Int[(a + b*Tan[c + d*x^(1/3)])^(-1),x]
Output:
3*(x/(3*(a + I*b)) + (2*I)*b*(((-1/2*I)*x^(2/3)*Log[1 + ((a^2 + b^2)*E^((2 *I)*(c + d*x^(1/3))))/(a + I*b)^2])/((a^2 + b^2)*d) + (I*(((I/2)*x^(1/3)*P olyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/d - Pol yLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)]/(4*d^2)))/ ((a^2 + b^2)*d)))
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] - Si mp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy mbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp[2*I*b In t[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Simp[2 *I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2 , 0] && IGtQ[m, 0]
Int[((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Simp[1 /n Subst[Int[x^(1/n - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ [{a, b, c, d, p}, x] && IGtQ[1/n, 0] && IntegerQ[p]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
\[\int \frac {1}{a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )}d x\]
Input:
int(1/(a+b*tan(c+d*x^(1/3))),x)
Output:
int(1/(a+b*tan(c+d*x^(1/3))),x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (147) = 294\).
Time = 0.10 (sec) , antiderivative size = 746, normalized size of antiderivative = 4.24 \[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b*tan(c+d*x^(1/3))),x, algorithm="fricas")
Output:
1/4*(4*a*d^3*x + 6*b*c^2*log(((I*a*b + b^2)*tan(d*x^(1/3) + c)^2 - a^2 + I *a*b + (I*a^2 + I*b^2)*tan(d*x^(1/3) + c))/(tan(d*x^(1/3) + c)^2 + 1)) + 6 *b*c^2*log(((I*a*b - b^2)*tan(d*x^(1/3) + c)^2 + a^2 + I*a*b + (I*a^2 + I* b^2)*tan(d*x^(1/3) + c))/(tan(d*x^(1/3) + c)^2 + 1)) + 6*I*b*d*x^(1/3)*dil og(2*((I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2) + 1) - 6*I*b*d*x^(1/3)*dilog(2*((-I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d* x^(1/3) + c)^2 + a^2 + b^2) + 1) + 6*(b*d^2*x^(2/3) - b*c^2)*log(-2*((I*a* b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan( d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2)) + 6*(b*d^2 *x^(2/3) - b*c^2)*log(-2*((-I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 + I*a* b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3 ) + c)^2 + a^2 + b^2)) + 3*b*polylog(3, ((a^2 + 2*I*a*b - b^2)*tan(d*x^(1/ 3) + c)^2 - a^2 - 2*I*a*b + b^2 - 2*(-I*a^2 + 2*a*b + I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2)) + 3*b*polylog(3, (( a^2 - 2*I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 + 2*I*a*b + b^2 - 2*(I*a^2 + 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2)))/((a^2 + b^2)*d^3)
\[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\int \frac {1}{a + b \tan {\left (c + d \sqrt [3]{x} \right )}}\, dx \] Input:
integrate(1/(a+b*tan(c+d*x**(1/3))),x)
Output:
Integral(1/(a + b*tan(c + d*x**(1/3))), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (147) = 294\).
Time = 0.22 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.53 \[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\frac {3 \, {\left (\frac {2 \, {\left (d x^{\frac {1}{3}} + c\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{2} + \frac {2 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} {\left (a - i \, b\right )} - 6 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} {\left (a - i \, b\right )} c - 6 \, {\left (i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b - 2 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c\right )} \arctan \left (\frac {2 \, a b \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - 6 \, {\left (i \, {\left (d x^{\frac {1}{3}} + c\right )} b - i \, b c\right )} {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}}{-i \, a + b}\right ) + 3 \, {\left ({\left (d x^{\frac {1}{3}} + c\right )}^{2} b - 2 \, {\left (d x^{\frac {1}{3}} + c\right )} b c\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + 3 \, b {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}}{-i \, a + b})}{a^{2} + b^{2}}}{2 \, d^{3}} \] Input:
integrate(1/(a+b*tan(c+d*x^(1/3))),x, algorithm="maxima")
Output:
1/2*(3*(2*(d*x^(1/3) + c)*a/(a^2 + b^2) + 2*b*log(b*tan(d*x^(1/3) + c) + a )/(a^2 + b^2) - b*log(tan(d*x^(1/3) + c)^2 + 1)/(a^2 + b^2))*c^2 + (2*(d*x ^(1/3) + c)^3*(a - I*b) - 6*(d*x^(1/3) + c)^2*(a - I*b)*c - 6*(I*(d*x^(1/3 ) + c)^2*b - 2*I*(d*x^(1/3) + c)*b*c)*arctan2((2*a*b*cos(2*d*x^(1/3) + 2*c ) - (a^2 - b^2)*sin(2*d*x^(1/3) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*x^(1/3 ) + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x^(1/3) + 2*c))/(a^2 + b^2)) - 6*(I*(d*x^(1/3) + c)*b - I*b*c)*dilog((I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/ (-I*a + b)) + 3*((d*x^(1/3) + c)^2*b - 2*(d*x^(1/3) + c)*b*c)*log(((a^2 + b^2)*cos(2*d*x^(1/3) + 2*c)^2 + 4*a*b*sin(2*d*x^(1/3) + 2*c) + (a^2 + b^2) *sin(2*d*x^(1/3) + 2*c)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*x^(1/3) + 2* c))/(a^2 + b^2)) + 3*b*polylog(3, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I* a + b)))/(a^2 + b^2))/d^3
\[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\int { \frac {1}{b \tan \left (d x^{\frac {1}{3}} + c\right ) + a} \,d x } \] Input:
integrate(1/(a+b*tan(c+d*x^(1/3))),x, algorithm="giac")
Output:
integrate(1/(b*tan(d*x^(1/3) + c) + a), x)
Timed out. \[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\int \frac {1}{a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )} \,d x \] Input:
int(1/(a + b*tan(c + d*x^(1/3))),x)
Output:
int(1/(a + b*tan(c + d*x^(1/3))), x)
\[ \int \frac {1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\int \frac {1}{\tan \left (x^{\frac {1}{3}} d +c \right ) b +a}d x \] Input:
int(1/(a+b*tan(c+d*x^(1/3))),x)
Output:
int(1/(tan(x**(1/3)*d + c)*b + a),x)