\(\int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx\) [100]

Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\frac {124 a^3 \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}+\frac {31 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}+\frac {9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f} \] Output:

124/15*a^3*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)+31/15*a^2*cos(f*x+e)*(a+a*s 
in(f*x+e))^(1/2)/f+9/5*sec(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f-2/5*sec(f*x+e)* 
(a+a*sin(f*x+e))^(7/2)/a/f
 

Mathematica [A] (verified)

Time = 6.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.51 \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\frac {a^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} (330+22 \cos (2 (e+f x))-185 \sin (e+f x)+3 \sin (3 (e+f x)))}{30 f} \] Input:

Integrate[(a + a*Sin[e + f*x])^(5/2)*Tan[e + f*x]^2,x]
 

Output:

(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*(330 + 22*Cos[2*(e + f*x)] - 
185*Sin[e + f*x] + 3*Sin[3*(e + f*x)]))/(30*f)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.11, 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, 3126, 3042, 3125}

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.

\(\displaystyle \int \tan ^2(e+f x) (a \sin (e+f x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^2 (a \sin (e+f x)+a)^{5/2}dx\)

\(\Big \downarrow \) 3192

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec ^2(e+f x) (\sin (e+f x) a+a)^{5/2} (2 \sin (e+f x) a+7 a)dx}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2} (2 \sin (e+f x) a+7 a)}{\cos (e+f x)^2}dx}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {9 a \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}-\frac {31}{2} a^2 \int (\sin (e+f x) a+a)^{3/2}dx}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {9 a \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}-\frac {31}{2} a^2 \int (\sin (e+f x) a+a)^{3/2}dx}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {\frac {9 a \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}-\frac {31}{2} a^2 \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {9 a \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}-\frac {31}{2} a^2 \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

\(\Big \downarrow \) 3125

\(\displaystyle \frac {\frac {9 a \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{f}-\frac {31}{2} a^2 \left (-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )}{5 a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}\)

Input:

Int[(a + a*Sin[e + f*x])^(5/2)*Tan[e + f*x]^2,x]
 

Output:

(-2*Sec[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(5*a*f) + ((9*a*Sec[e + f*x]* 
(a + a*Sin[e + f*x])^(5/2))/f - (31*a^2*((-8*a^2*Cos[e + f*x])/(3*f*Sqrt[a 
 + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f)))/ 
2)/(5*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 3125

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos 
[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[a^2 - b^2, 0]
 

rule 3126

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

Time = 1.94 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.57

method	result	size
default	\(-\frac {2 a^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{3}+11 \sin \left (f x +e \right )^{2}+44 \sin \left (f x +e \right )-88\right )}{15 \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(67\)

Input:

int((a+sin(f*x+e)*a)^(5/2)*tan(f*x+e)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/15*a^3*(1+sin(f*x+e))*(3*sin(f*x+e)^3+11*sin(f*x+e)^2+44*sin(f*x+e)-88) 
/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\frac {2 \, {\left (11 \, a^{2} \cos \left (f x + e\right )^{2} + 77 \, a^{2} + {\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 47 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, f \cos \left (f x + e\right )} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

2/15*(11*a^2*cos(f*x + e)^2 + 77*a^2 + (3*a^2*cos(f*x + e)^2 - 47*a^2)*sin 
(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*cos(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(5/2)*tan(f*x+e)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.62 \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=-\frac {8 \, {\left (22 \, a^{\frac {5}{2}} - \frac {22 \, a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {55 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {50 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {55 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {22 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {22 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{15 \, f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x, algorithm="maxima")
 

Output:

-8/15*(22*a^(5/2) - 22*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 55*a^(5/2 
)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 50*a^(5/2)*sin(f*x + e)^3/(cos(f*x 
 + e) + 1)^3 + 55*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 22*a^(5/2) 
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 22*a^(5/2)*sin(f*x + e)^6/(cos(f*x 
+ e) + 1)^6)/(f*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)*(sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 1)^(5/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1411 vs. \(2 (102) = 204\).

Time = 1.45 (sec) , antiderivative size = 1411, normalized size of antiderivative = 11.96 \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x, algorithm="giac")
 

Output:

-1/3840*sqrt(2)*(1080*pi*a^2*floor(1/8*(3*pi + 2*f*x + 2*e)/pi)*sgn(cos(-1 
/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^11 + 1080*pi*a^2* 
sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*ta 
n(-1/8*pi + 1/4*f*x + 1/4*e)^11 + 135*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4* 
pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^11 - 1080*a^2*arctan 
(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan 
(-1/8*pi + 1/4*f*x + 1/4*e)^11 + 3840*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2* 
e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^12 + 5400*pi*a^2*floor(1/8*(3*pi + 2*f* 
x + 2*e)/pi)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1 
/4*e)^9 + 5400*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi 
+ 1/4*f*x + 1/4*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^9 + 675*(pi - 2*f*x - 2 
*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e) 
^9 - 5400*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1 
/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^9 + 99840*a^2*sgn(cos(-1/4 
*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^10 + 10800*pi*a^2*f 
loor(1/8*(3*pi + 2*f*x + 2*e)/pi)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan( 
-1/8*pi + 1/4*f*x + 1/4*e)^7 + 10800*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 
2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^7 
 + 1350*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/ 
8*pi + 1/4*f*x + 1/4*e)^7 - 10800*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + ...
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \] Input:

int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(5/2),x)
 

Output:

int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(5/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2} \tan \left (f x +e \right )^{2}d x +\int \sqrt {\sin \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}d x +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right ) \tan \left (f x +e \right )^{2}d x \right )\right ) \] Input:

int((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^2,x)
 

Output:

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