\(\int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx\) [95]

Optimal result

Integrand size = 23, antiderivative size = 167 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} f}+\frac {2 a^3 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^{3/2}}-\frac {4 a^2 \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {7 a \sec (e+f x) \sqrt {a+a \sin (e+f x)}}{2 f}+\frac {\sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f} \] Output:

-1/4*a^(3/2)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2) 
)*2^(1/2)/f+2/3*a^3*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^(3/2)-4*a^2*cos(f*x+e) 
/f/(a+a*sin(f*x+e))^(1/2)-7/2*a*sec(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f+1/3*se 
c(f*x+e)^3*(a+a*sin(f*x+e))^(3/2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\frac {a \sec ^3(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sqrt {a (1+\sin (e+f x))} \left (-45+6 \cos (2 (e+f x))+(3+3 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+54 \sin (e+f x)+\sin (3 (e+f x))\right )}{6 f} \] Input:

Integrate[(a + a*Sin[e + f*x])^(3/2)*Tan[e + f*x]^4,x]
 

Output:

(a*Sec[e + f*x]^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Sqrt[a*(1 + Sin[ 
e + f*x])]*(-45 + 6*Cos[2*(e + f*x)] + (3 + 3*I)*(-1)^(3/4)*ArcTanh[(1/2 + 
 I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x 
)/2])^3 + 54*Sin[e + f*x] + Sin[3*(e + f*x)]))/(6*f)
 

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3193, 3042, 3126, 3042, 3125, 4901, 2009}

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.

\(\displaystyle \int \tan ^4(e+f x) (a \sin (e+f x)+a)^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^4 (a \sin (e+f x)+a)^{3/2}dx\)

\(\Big \downarrow \) 3193

\(\displaystyle \int (\sin (e+f x) a+a)^{3/2}dx-\int \sec ^4(e+f x) (\sin (e+f x) a+a)^{3/2} \left (1-2 \sin ^2(e+f x)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (\sin (e+f x) a+a)^{3/2}dx-\int \frac {(\sin (e+f x) a+a)^{3/2} \left (1-2 \sin (e+f x)^2\right )}{\cos (e+f x)^4}dx\)

\(\Big \downarrow \) 3126

\(\displaystyle \frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\int \frac {(\sin (e+f x) a+a)^{3/2} \left (1-2 \sin (e+f x)^2\right )}{\cos (e+f x)^4}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\int \frac {(\sin (e+f x) a+a)^{3/2} \left (1-2 \sin (e+f x)^2\right )}{\cos (e+f x)^4}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\)

\(\Big \downarrow \) 3125

\(\displaystyle -\int \frac {(\sin (e+f x) a+a)^{3/2} \left (1-2 \sin (e+f x)^2\right )}{\cos (e+f x)^4}dx-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\)

\(\Big \downarrow \) 4901

\(\displaystyle -\int \left (\sec ^4(e+f x) (a (\sin (e+f x)+1))^{3/2}-2 \sec ^2(e+f x) (a (\sin (e+f x)+1))^{3/2} \tan ^2(e+f x)\right )dx-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} f}-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}+\frac {4 \sec ^3(e+f x) (a \sin (e+f x)+a)^{5/2}}{a f}-\frac {23 \sec ^3(e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f}+\frac {a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{2 f}\)

Input:

Int[(a + a*Sin[e + f*x])^(3/2)*Tan[e + f*x]^4,x]
 

Output:

-1/2*(a^(3/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f 
*x]])])/(Sqrt[2]*f) - (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) + (a*Sec[e + f*x]*Sqrt 
[a + a*Sin[e + f*x]])/(2*f) - (23*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(3/2 
))/(3*f) + (4*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(5/2))/(a*f)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 3193

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^4, 
x_Symbol] :> Int[(a + b*Sin[e + f*x])^m, x] - Int[(a + b*Sin[e + f*x])^m*(( 
1 - 2*Sin[e + f*x]^2)/Cos[e + f*x]^4), x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]
 

rule 4901

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; 
 !InertTrigFreeQ[u]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\left (1+\sin \left (f x +e \right )\right ) \left (3 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}}-8 a^{3} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-24 a^{3} \cos \left (f x +e \right )^{2}-106 \sin \left (f x +e \right ) a^{3}+102 a^{3}\right )}{12 a \left (-1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(139\)

Input:

int((a+sin(f*x+e)*a)^(3/2)*tan(f*x+e)^4,x,method=_RETURNVERBOSE)
 

Output:

1/12*(1+sin(f*x+e))/a/(-1+sin(f*x+e))*(3*a^(3/2)*2^(1/2)*arctanh(1/2*(a-si 
n(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*(a-sin(f*x+e)*a)^(3/2)-8*a^3*cos(f*x+e) 
^2*sin(f*x+e)-24*a^3*cos(f*x+e)^2-106*sin(f*x+e)*a^3+102*a^3)/cos(f*x+e)/( 
a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.38 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\frac {3 \, \sqrt {\frac {1}{2}} {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a \cos \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 2 \, {\left (12 \, a \cos \left (f x + e\right )^{2} + {\left (4 \, a \cos \left (f x + e\right )^{2} + 53 \, a\right )} \sin \left (f x + e\right ) - 51 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{12 \, {\left (f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - f \cos \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/12*(3*sqrt(1/2)*(a*cos(f*x + e)*sin(f*x + e) - a*cos(f*x + e))*sqrt(a)*l 
og(-(a*cos(f*x + e)^2 - 4*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos( 
f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*s 
in(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos 
(f*x + e) - 2)) - 2*(12*a*cos(f*x + e)^2 + (4*a*cos(f*x + e)^2 + 53*a)*sin 
(f*x + e) - 51*a)*sqrt(a*sin(f*x + e) + a))/(f*cos(f*x + e)*sin(f*x + e) - 
 f*cos(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(3/2)*tan(f*x+e)**4,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 917 vs. \(2 (142) = 284\).

Time = 0.97 (sec) , antiderivative size = 917, normalized size of antiderivative = 5.49 \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x, algorithm="giac")
 

Output:

-1/96*sqrt(2)*(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x 
 + 1/4*e)^12 - 12*a*log(2*(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 2*tan(-1/8*p 
i + 1/4*f*x + 1/4*e) + 1)/(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos( 
-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^9 + 12*a*log(2* 
(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 - 2*tan(-1/8*pi + 1/4*f*x + 1/4*e) + 1)/ 
(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
)*tan(-1/8*pi + 1/4*f*x + 1/4*e)^9 - 78*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2* 
e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^10 - 36*a*log(2*(tan(-1/8*pi + 1/4*f*x 
+ 1/4*e)^2 + 2*tan(-1/8*pi + 1/4*f*x + 1/4*e) + 1)/(tan(-1/8*pi + 1/4*f*x 
+ 1/4*e)^2 + 1))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x 
 + 1/4*e)^7 + 36*a*log(2*(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 - 2*tan(-1/8*pi 
 + 1/4*f*x + 1/4*e) + 1)/(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos(- 
1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^7 - 1089*a*sgn(c 
os(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^8 - 36*a*log 
(2*(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 2*tan(-1/8*pi + 1/4*f*x + 1/4*e) + 
1)/(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 
*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^5 + 36*a*log(2*(tan(-1/8*pi + 1/4*f*x 
+ 1/4*e)^2 - 2*tan(-1/8*pi + 1/4*f*x + 1/4*e) + 1)/(tan(-1/8*pi + 1/4*f*x 
+ 1/4*e)^2 + 1))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x 
 + 1/4*e)^5 - 996*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1...
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

int(tan(e + f*x)^4*(a + a*sin(e + f*x))^(3/2),x)
 

Output:

int(tan(e + f*x)^4*(a + a*sin(e + f*x))^(3/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int (a+a \sin (e+f x))^{3/2} \tan ^4(e+f x) \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{4}d x +\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right ) \tan \left (f x +e \right )^{4}d x \right ) \] Input:

int((a+a*sin(f*x+e))^(3/2)*tan(f*x+e)^4,x)
 

Output:

sqrt(a)*a*(int(sqrt(sin(e + f*x) + 1)*tan(e + f*x)**4,x) + int(sqrt(sin(e 
+ f*x) + 1)*sin(e + f*x)*tan(e + f*x)**4,x))