\(\int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\) [107]

Optimal result

Integrand size = 23, antiderivative size = 177 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{256 \sqrt {2} a^{3/2} f}+\frac {7 \cos (e+f x)}{256 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec (e+f x) (65+87 \sin (e+f x))}{192 f (a+a \sin (e+f x))^{3/2}}+\frac {a \sin (e+f x) \tan (e+f x)}{12 f (a+a \sin (e+f x))^{5/2}}+\frac {\tan ^3(e+f x)}{3 f (a+a \sin (e+f x))^{3/2}} \] Output:

7/512*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/ 
2)/a^(3/2)/f+7/256*cos(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)-1/192*sec(f*x+e)*(6 
5+87*sin(f*x+e))/f/(a+a*sin(f*x+e))^(3/2)+1/12*a*sin(f*x+e)*tan(f*x+e)/f/( 
a+a*sin(f*x+e))^(5/2)+1/3*tan(f*x+e)^3/f/(a+a*sin(f*x+e))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.12 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.89 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {124+\frac {64 \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {32}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {248 \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+342 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-171 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(21+21 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\frac {32 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {192 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}}{768 f (a (1+\sin (e+f x)))^{3/2}} \] Input:

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

Output:

(124 + (64*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 32/ 
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (248*Sin[(e + f*x)/2])/(Cos[(e + 
 f*x)/2] + Sin[(e + f*x)/2]) + 342*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Si 
n[(e + f*x)/2]) - 171*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (21 + 21*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 + (32*(Cos[(e + f*x)/2] + Sin[(e + f*x) 
/2])^3)/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (192*(Cos[(e + f*x)/2] + 
 Sin[(e + f*x)/2])^3)/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/(768*f*(a*(1 
+ Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.75 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3193, 3042, 3129, 3042, 3128, 219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3193

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3129

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4901

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{256 \sqrt {2} a^{3/2} f}+\frac {7 \cos (e+f x)}{256 f (a \sin (e+f x)+a)^{3/2}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a \sin (e+f x)+a}}-\frac {\sec ^3(e+f x)}{6 f (a \sin (e+f x)+a)^{3/2}}-\frac {45 \sec (e+f x)}{64 a f \sqrt {a \sin (e+f x)+a}}+\frac {9 \sec (e+f x)}{32 f (a \sin (e+f x)+a)^{3/2}}\)

Input:

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

Output:

(7*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(25 
6*Sqrt[2]*a^(3/2)*f) + (7*Cos[e + f*x])/(256*f*(a + a*Sin[e + f*x])^(3/2)) 
 + (9*Sec[e + f*x])/(32*f*(a + a*Sin[e + f*x])^(3/2)) - Sec[e + f*x]^3/(6* 
f*(a + a*Sin[e + f*x])^(3/2)) - (45*Sec[e + f*x])/(64*a*f*Sqrt[a + a*Sin[e 
 + f*x]]) + Sec[e + f*x]^3/(4*a*f*Sqrt[a + a*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 3128

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

rule 3129

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

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(310\) vs. \(2(150)=300\).

Time = 0.32 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.76

method	result	size
default	\(-\frac {42 \cos \left (f x +e \right )^{4} a^{\frac {9}{2}}-1080 a^{\frac {9}{2}} \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-21 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a^{3} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}-648 a^{\frac {9}{2}} \cos \left (f x +e \right )^{2}-63 \cos \left (f x +e \right )^{2} a^{3} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}+384 \sin \left (f x +e \right ) a^{\frac {9}{2}}+84 \sin \left (f x +e \right ) \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3}+128 a^{\frac {9}{2}}+84 \sqrt {2}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}}{1536 a^{\frac {11}{2}} \left (-1+\sin \left (f x +e \right )\right ) \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(311\)

Input:

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

Output:

-1/1536/a^(11/2)*(42*cos(f*x+e)^4*a^(9/2)-1080*a^(9/2)*cos(f*x+e)^2*sin(f* 
x+e)-21*cos(f*x+e)^2*sin(f*x+e)*a^3*(a-sin(f*x+e)*a)^(3/2)*arctanh(1/2*(a- 
sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)-648*a^(9/2)*cos(f*x+e)^2-63*c 
os(f*x+e)^2*a^3*(a-sin(f*x+e)*a)^(3/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)* 
2^(1/2)/a^(1/2))*2^(1/2)+384*sin(f*x+e)*a^(9/2)+84*sin(f*x+e)*(a-sin(f*x+e 
)*a)^(3/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^3 
+128*a^(9/2)+84*2^(1/2)*(a-sin(f*x+e)*a)^(3/2)*arctanh(1/2*(a-sin(f*x+e)*a 
)^(1/2)*2^(1/2)/a^(1/2))*a^3)/(-1+sin(f*x+e))/(1+sin(f*x+e))^2/cos(f*x+e)/ 
(a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.53 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {21 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{5} - 2 \, \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (21 \, \cos \left (f x + e\right )^{4} - 324 \, \cos \left (f x + e\right )^{2} - 12 \, {\left (45 \, \cos \left (f x + e\right )^{2} - 16\right )} \sin \left (f x + e\right ) + 64\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{3072 \, {\left (a^{2} f \cos \left (f x + e\right )^{5} - 2 \, a^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, a^{2} f \cos \left (f x + e\right )^{3}\right )}} \] Input:

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

Output:

1/3072*(21*sqrt(2)*(cos(f*x + e)^5 - 2*cos(f*x + e)^3*sin(f*x + e) - 2*cos 
(f*x + e)^3)*sqrt(a)*log(-(a*cos(f*x + e)^2 + 2*sqrt(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*c 
os(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2 
)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(21*cos(f*x + e)^4 - 324*cos(f*x + 
 e)^2 - 12*(45*cos(f*x + e)^2 - 16)*sin(f*x + e) + 64)*sqrt(a*sin(f*x + e) 
 + a))/(a^2*f*cos(f*x + e)^5 - 2*a^2*f*cos(f*x + e)^3*sin(f*x + e) - 2*a^2 
*f*cos(f*x + e)^3)
 

Sympy [F]

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

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

Output:

Integral(tan(e + f*x)**4/(a*(sin(e + f*x) + 1))**(3/2), x)
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.24 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\frac {21 \, \sqrt {2} \log \left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {21 \, \sqrt {2} \log \left (-\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, \sqrt {2} {\left (21 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 312 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 507 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 240 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 16 \, \sqrt {a}\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3072 \, f} \] Input:

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

Output:

1/3072*(21*sqrt(2)*log(sin(3/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^(3/2)*sgn(cos 
(-1/4*pi + 1/2*f*x + 1/2*e))) - 21*sqrt(2)*log(-sin(3/4*pi + 1/2*f*x + 1/2 
*e) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*sqrt(2)*(21*sqr 
t(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^8 - 312*sqrt(a)*sin(3/4*pi + 1/2*f*x + 
1/2*e)^6 + 507*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^4 - 240*sqrt(a)*sin(3 
/4*pi + 1/2*f*x + 1/2*e)^2 + 16*sqrt(a))/((sin(3/4*pi + 1/2*f*x + 1/2*e)^3 
 - sin(3/4*pi + 1/2*f*x + 1/2*e))^3*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
)))/f
 

Mupad [F(-1)]

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

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

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

Output:

(sqrt(a)*int((sqrt(sin(e + f*x) + 1)*tan(e + f*x)**4)/(sin(e + f*x)**2 + 2 
*sin(e + f*x) + 1),x))/a**2