\(\int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx\) [633]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx=\frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}-\frac {3 a d \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{2 f (d \sec (e+f x))^{2/3} \sqrt {\sin ^2(e+f x)}} \] Output:

3*b*(d*sec(f*x+e))^(1/3)/f-3/2*a*d*hypergeom([1/3, 1/2],[4/3],cos(f*x+e)^2 
)*sin(f*x+e)/f/(d*sec(f*x+e))^(2/3)/(sin(f*x+e)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx=\frac {3 \sqrt [3]{d \sec (e+f x)} \left (b+a \cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f} \] Input:

Integrate[(d*Sec[e + f*x])^(1/3)*(a + b*Tan[e + f*x]),x]
 

Output:

(3*(d*Sec[e + f*x])^(1/3)*(b + a*Cot[e + f*x]*Hypergeometric2F1[1/6, 1/2, 
7/6, Sec[e + f*x]^2]*Sqrt[-Tan[e + f*x]^2]))/f
 

Rubi [A] (verified)

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.

\(\displaystyle \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x))dx\)

\(\Big \downarrow \) 3967

\(\displaystyle a \int \sqrt [3]{d \sec (e+f x)}dx+\frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \sqrt [3]{d \csc \left (e+f x+\frac {\pi }{2}\right )}dx+\frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}\)

\(\Big \downarrow \) 4259

\(\displaystyle a \sqrt [3]{\frac {\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)} \int \frac {1}{\sqrt [3]{\frac {\cos (e+f x)}{d}}}dx+\frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \sqrt [3]{\frac {\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)} \int \frac {1}{\sqrt [3]{\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}}}dx+\frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {3 b \sqrt [3]{d \sec (e+f x)}}{f}-\frac {3 a d \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(e+f x)\right )}{2 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{2/3}}\)

Input:

Int[(d*Sec[e + f*x])^(1/3)*(a + b*Tan[e + f*x]),x]
 

Output:

(3*b*(d*Sec[e + f*x])^(1/3))/f - (3*a*d*Hypergeometric2F1[1/3, 1/2, 4/3, C 
os[e + f*x]^2]*Sin[e + f*x])/(2*f*(d*Sec[e + f*x])^(2/3)*Sqrt[Sin[e + f*x] 
^2])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

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]
 

rule 3967

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])
 

rule 4259

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]
 

Maple [F]

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

Input:

int((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x)
 

Output:

int((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x)
 

Fricas [F]

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

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

Output:

integral((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a), x)
 

Sympy [F]

\[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx=\int \sqrt [3]{d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )\, dx \] Input:

integrate((d*sec(f*x+e))**(1/3)*(a+b*tan(f*x+e)),x)
 

Output:

Integral((d*sec(e + f*x))**(1/3)*(a + b*tan(e + f*x)), x)
 

Maxima [F]

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

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

Output:

integrate((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a), x)
 

Giac [F]

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

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

Output:

integrate((d*sec(f*x + e))^(1/3)*(b*tan(f*x + e) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{d \sec (e+f x)} (a+b \tan (e+f x)) \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right ) \,d x \] Input:

int((d/cos(e + f*x))^(1/3)*(a + b*tan(e + f*x)),x)
 

Output:

int((d/cos(e + f*x))^(1/3)*(a + b*tan(e + f*x)), x)
 

Reduce [F]

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

int((d*sec(f*x+e))^(1/3)*(a+b*tan(f*x+e)),x)
 

Output:

(d**(1/3)*(3*sec(e + f*x)**(1/3)*b + int(sec(e + f*x)**(1/3),x)*a*f))/f