Integrand size = 14, antiderivative size = 98 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3} \] Output:
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
Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=a x+i b x-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3} \] Input:
Integrate[a + b*Tan[c + d*x^(1/3)],x]
Output:
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)
Time = 0.33 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2009}
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.
\(\displaystyle \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a x-\frac {3 b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3}+\frac {3 i b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i b x\) |
Input:
Int[a + b*Tan[c + d*x^(1/3)],x]
Output:
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)
\[\int \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]
Input:
int(a+b*tan(c+d*x^(1/3)),x)
Output:
int(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. 249 vs. \(2 (75) = 150\).
Time = 0.09 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.54 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {4 \, a d^{3} x - 6 \, b d^{2} x^{\frac {2}{3}} \log \left (-\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 \, b d^{2} x^{\frac {2}{3}} \log \left (-\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 i \, b d x^{\frac {1}{3}} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) + 6 i \, b d x^{\frac {1}{3}} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 3 \, b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right )}{4 \, d^{3}} \] Input:
integrate(a+b*tan(c+d*x^(1/3)),x, algorithm="fricas")
Output:
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
\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )\, dx \] Input:
integrate(a+b*tan(c+d*x**(1/3)),x)
Output:
Integral(a + b*tan(c + d*x**(1/3)), x)
\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { b \tan \left (d x^{\frac {1}{3}} + c\right ) + a \,d x } \] Input:
integrate(a+b*tan(c+d*x^(1/3)),x, algorithm="maxima")
Output:
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)
\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { b \tan \left (d x^{\frac {1}{3}} + c\right ) + a \,d x } \] Input:
integrate(a+b*tan(c+d*x^(1/3)),x, algorithm="giac")
Output:
integrate(b*tan(d*x^(1/3) + c) + a, x)
Timed out. \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right ) \,d x \] Input:
int(a + b*tan(c + d*x^(1/3)),x)
Output:
int(a + b*tan(c + d*x^(1/3)), x)
\[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\left (\int \tan \left (x^{\frac {1}{3}} d +c \right )d x \right ) b +a x \] Input:
int(a+b*tan(c+d*x^(1/3)),x)
Output:
int(tan(x**(1/3)*d + c),x)*b + a*x