Integrand size = 23, antiderivative size = 76 \[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}-\frac {3 a d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{4 f (d \sec (e+f x))^{4/3} \sqrt {\sin ^2(e+f x)}} \] Output:
-3*b/f/(d*sec(f*x+e))^(1/3)-3/4*a*d*hypergeom([1/2, 2/3],[5/3],cos(f*x+e)^ 2)*sin(f*x+e)/f/(d*sec(f*x+e))^(4/3)/(sin(f*x+e)^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 \left (b+a \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f \sqrt [3]{d \sec (e+f x)}} \] Input:
Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(1/3),x]
Output:
(-3*(b + a*Cot[e + f*x]*Hypergeometric2F1[-1/6, 1/2, 5/6, Sec[e + f*x]^2]* Sqrt[-Tan[e + f*x]^2]))/(f*(d*Sec[e + f*x])^(1/3))
Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3967, 3042, 4259, 3042, 3122}
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.
\(\displaystyle \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int \frac {1}{\sqrt [3]{d \sec (e+f x)}}dx-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \frac {1}{\sqrt [3]{d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle a \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3} \int \sqrt [3]{\frac {\cos (e+f x)}{d}}dx-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3} \int \sqrt [3]{\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}}dx-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle -\frac {3 a d \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right )}{4 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac {3 b}{f \sqrt [3]{d \sec (e+f x)}}\) |
Input:
Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(1/3),x]
Output:
(-3*b)/(f*(d*Sec[e + f*x])^(1/3)) - (3*a*d*Hypergeometric2F1[1/2, 2/3, 5/3 , Cos[e + f*x]^2]*Sin[e + f*x])/(4*f*(d*Sec[e + f*x])^(4/3)*Sqrt[Sin[e + f *x]^2])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
\[\int \frac {a +b \tan \left (f x +e \right )}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
Input:
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x)
Output:
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x)
\[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")
Output:
integral((d*sec(f*x + e))^(2/3)*(b*tan(f*x + e) + a)/(d*sec(f*x + e)), x)
\[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(1/3),x)
Output:
Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(1/3), x)
\[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")
Output:
integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(1/3), x)
\[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x, algorithm="giac")
Output:
integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(1/3), x)
Timed out. \[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \] Input:
int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(1/3),x)
Output:
int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(1/3), x)
\[ \int \frac {a+b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=\frac {\left (\int \frac {\tan \left (f x +e \right )}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {1}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right ) a}{d^{\frac {1}{3}}} \] Input:
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(1/3),x)
Output:
(int(tan(e + f*x)/sec(e + f*x)**(1/3),x)*b + int(1/sec(e + f*x)**(1/3),x)* a)/d**(1/3)