\(\int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [638]

Optimal result

Integrand size = 25, antiderivative size = 119 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac {3 \left (2 a^2-3 b^2\right ) d \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right ) \sin (e+f x)}{8 f (d \sec (e+f x))^{4/3} \sqrt {\sin ^2(e+f x)}}+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \] Output:

-15/2*a*b/f/(d*sec(f*x+e))^(1/3)-3/8*(2*a^2-3*b^2)*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) 
+3/2*b*(a+b*tan(f*x+e))/f/(d*sec(f*x+e))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 \left (b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{6},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+a \left (-a \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 b \sqrt {-\tan ^2(e+f x)}\right )\right )}{f \sqrt [3]{d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \] Input:

Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]
 

Output:

(-3*(b^2*Hypergeometric2F1[-1/2, -1/6, 5/6, Sec[e + f*x]^2]*Tan[e + f*x] + 
 a*(-(a*Hypergeometric2F1[-1/6, 1/2, 5/6, Sec[e + f*x]^2]*Tan[e + f*x]) + 
2*b*Sqrt[-Tan[e + f*x]^2])))/(f*(d*Sec[e + f*x])^(1/3)*Sqrt[-Tan[e + f*x]^ 
2])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3993, 27, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3993

\(\displaystyle \frac {3}{2} \int \frac {2 a^2+5 b \tan (e+f x) a-3 b^2}{3 \sqrt [3]{d \sec (e+f x)}}dx+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {2 a^2+5 b \tan (e+f x) a-3 b^2}{\sqrt [3]{d \sec (e+f x)}}dx+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {2 a^2+5 b \tan (e+f x) a-3 b^2}{\sqrt [3]{d \sec (e+f x)}}dx+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 3967

\(\displaystyle \frac {1}{2} \left (\left (2 a^2-3 b^2\right ) \int \frac {1}{\sqrt [3]{d \sec (e+f x)}}dx-\frac {15 a b}{f \sqrt [3]{d \sec (e+f x)}}\right )+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\left (2 a^2-3 b^2\right ) \int \frac {1}{\sqrt [3]{d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {15 a b}{f \sqrt [3]{d \sec (e+f x)}}\right )+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {1}{2} \left (\left (2 a^2-3 b^2\right ) \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3} \int \sqrt [3]{\frac {\cos (e+f x)}{d}}dx-\frac {15 a b}{f \sqrt [3]{d \sec (e+f x)}}\right )+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\left (2 a^2-3 b^2\right ) \left (\frac {\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3} \int \sqrt [3]{\frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{d}}dx-\frac {15 a b}{f \sqrt [3]{d \sec (e+f x)}}\right )+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {1}{2} \left (-\frac {3 d \left (2 a^2-3 b^2\right ) \sin (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(e+f x)\right )}{4 f \sqrt {\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac {15 a b}{f \sqrt [3]{d \sec (e+f x)}}\right )+\frac {3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\)

Input:

Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]
 

Output:

((-15*a*b)/(f*(d*Sec[e + f*x])^(1/3)) - (3*(2*a^2 - 3*b^2)*d*Hypergeometri 
c2F1[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]))/2 + (3*b*(a + b*Tan[e + f*x]))/(2*f*(d*Sec[e + 
f*x])^(1/3))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3993

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^2, x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m 
 + 1))), x] + Simp[1/(m + 1)   Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^2 + 
a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^ 
2 + b^2, 0] &&  !IntegerQ[m]
 

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]

\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]

Input:

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
 

Output:

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
 

Fricas [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")
 

Output:

integral((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)*(d*sec(f*x + e))^ 
(2/3)/(d*sec(f*x + e)), x)
 

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**2/(d*sec(f*x+e))**(1/3),x)
 

Output:

Integral((a + b*tan(e + f*x))**2/(d*sec(e + f*x))**(1/3), x)
 

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")
 

Output:

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(1/3), x)
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="giac")
 

Output:

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \] Input:

int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(1/3),x)
 

Output:

int((a + b*tan(e + f*x))^2/(d/cos(e + f*x))^(1/3), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\frac {\left (\int \frac {\tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right ) b^{2}+2 \left (\int \frac {\tan \left (f x +e \right )}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right ) a b +\left (\int \frac {1}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right ) a^{2}}{d^{\frac {1}{3}}} \] Input:

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
                                                                                    
                                                                                    
 

Output:

(int(tan(e + f*x)**2/sec(e + f*x)**(1/3),x)*b**2 + 2*int(tan(e + f*x)/sec( 
e + f*x)**(1/3),x)*a*b + int(1/sec(e + f*x)**(1/3),x)*a**2)/d**(1/3)