\(\int \sqrt [3]{a+a \sin (e+f x)} \tan ^2(e+f x) \, dx\) [116]

Optimal result

Integrand size = 23, antiderivative size = 123 \[ \int \sqrt [3]{a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=-\frac {5 a \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sqrt [6]{1+\sin (e+f x)}}{3 \sqrt [6]{2} f (a+a \sin (e+f x))^{2/3}}+\frac {7 \sec (e+f x) \sqrt [3]{a+a \sin (e+f x)}}{f}-\frac {3 \sec (e+f x) (a+a \sin (e+f x))^{4/3}}{a f} \] Output:

-5/6*a*cos(f*x+e)*hypergeom([1/2, 7/6],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f* 
x+e))^(1/6)*2^(5/6)/f/(a+a*sin(f*x+e))^(2/3)+7*sec(f*x+e)*(a+a*sin(f*x+e)) 
^(1/3)/f-3*sec(f*x+e)*(a+a*sin(f*x+e))^(4/3)/a/f
 

Mathematica [A] (warning: unable to verify)

Time = 3.87 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.70 \[ \int \sqrt [3]{a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\frac {\sqrt [3]{a (1+\sin (e+f x))} \left (\frac {5 \sqrt [3]{2} \left (-2 \cos \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right )+\sqrt {\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )} \left (2 \cos \left (\frac {1}{4} (2 e+\pi +2 f x)\right )+3 \sin \left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right )\right )}{\sqrt {\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^{2/3} \sqrt [3]{\sin \left (\frac {1}{4} (2 e+\pi +2 f x)\right )}}-3 (5+\sec (e+f x)-2 \tan (e+f x))\right )}{3 f} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(1/3)*((5*2^(1/3)*(-2*Cos[(2*e + Pi + 2*f*x)/4]*Hy 
pergeometricPFQ[{-1/2, -1/6}, {5/6}, Sin[(2*e + Pi + 2*f*x)/4]^2] + Sqrt[C 
os[(2*e + Pi + 2*f*x)/4]^2]*(2*Cos[(2*e + Pi + 2*f*x)/4] + 3*Sin[(2*e + Pi 
 + 2*f*x)/4])))/(Sqrt[Cos[(2*e + Pi + 2*f*x)/4]^2]*(Cos[(e + f*x)/2] + Sin 
[(e + f*x)/2])^(2/3)*Sin[(2*e + Pi + 2*f*x)/4]^(1/3)) - 3*(5 + Sec[e + f*x 
] - 2*Tan[e + f*x])))/(3*f)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3192, 27, 3042, 3334, 3042, 3131, 3042, 3130}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3192

\(\displaystyle \frac {3 \int \frac {1}{3} \sec ^2(e+f x) \sqrt [3]{\sin (e+f x) a+a} (3 \sin (e+f x) a+4 a)dx}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec ^2(e+f x) \sqrt [3]{\sin (e+f x) a+a} (3 \sin (e+f x) a+4 a)dx}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt [3]{\sin (e+f x) a+a} (3 \sin (e+f x) a+4 a)}{\cos (e+f x)^2}dx}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {5}{3} a^2 \int \frac {1}{(\sin (e+f x) a+a)^{2/3}}dx+\frac {7 a \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f}}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{3} a^2 \int \frac {1}{(\sin (e+f x) a+a)^{2/3}}dx+\frac {7 a \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f}}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {5 a^2 (\sin (e+f x)+1)^{2/3} \int \frac {1}{(\sin (e+f x)+1)^{2/3}}dx}{3 (a \sin (e+f x)+a)^{2/3}}+\frac {7 a \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f}}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 a^2 (\sin (e+f x)+1)^{2/3} \int \frac {1}{(\sin (e+f x)+1)^{2/3}}dx}{3 (a \sin (e+f x)+a)^{2/3}}+\frac {7 a \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f}}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {7 a \sec (e+f x) \sqrt [3]{a \sin (e+f x)+a}}{f}-\frac {5 a^2 \sqrt [6]{\sin (e+f x)+1} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 \sqrt [6]{2} f (a \sin (e+f x)+a)^{2/3}}}{a}-\frac {3 \sec (e+f x) (a \sin (e+f x)+a)^{4/3}}{a f}\)

Input:

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

Output:

(-3*Sec[e + f*x]*(a + a*Sin[e + f*x])^(4/3))/(a*f) + ((-5*a^2*Cos[e + f*x] 
*Hypergeometric2F1[1/2, 7/6, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x]) 
^(1/6))/(3*2^(1/6)*f*(a + a*Sin[e + f*x])^(2/3)) + (7*a*Sec[e + f*x]*(a + 
a*Sin[e + f*x])^(1/3))/f)/a
 

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 3130

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3192

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, 
x_Symbol] :> Simp[-(a + b*Sin[e + f*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + 
Simp[1/(b*m)   Int[(a + b*Sin[e + f*x])^m*((b*(m + 1) + a*Sin[e + f*x])/Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && 
!IntegerQ[m] &&  !LtQ[m, 0]
 

rule 3334

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* 
c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) 
, x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1)))   Int[(g*Cos[e + 
f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, 
 f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((a*sin(f*x + e) + a)^(1/3)*tan(f*x + e)^2, x)
 

Sympy [F]

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

integrate((a+a*sin(f*x+e))**(1/3)*tan(f*x+e)**2,x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**(1/3)*tan(e + f*x)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

a**(1/3)*int((sin(e + f*x) + 1)**(1/3)*tan(e + f*x)**2,x)