\(\int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\) [111]

Optimal result

Integrand size = 23, antiderivative size = 207 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {317 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{4096 \sqrt {2} a^{5/2} f}+\frac {317 \cos (e+f x)}{3072 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec (e+f x) (115+129 \sin (e+f x))}{384 f (a+a \sin (e+f x))^{5/2}}+\frac {317 \cos (e+f x)}{4096 a f (a+a \sin (e+f x))^{3/2}}+\frac {5 a \sin (e+f x) \tan (e+f x)}{48 f (a+a \sin (e+f x))^{7/2}}+\frac {\tan ^3(e+f x)}{3 f (a+a \sin (e+f x))^{5/2}} \] Output:

317/8192*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^(5/2)/f+317/3072*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)-1/384*sec(f*x 
+e)*(115+129*sin(f*x+e))/f/(a+a*sin(f*x+e))^(5/2)+317/4096*cos(f*x+e)/a/f/ 
(a+a*sin(f*x+e))^(3/2)+5/48*a*sin(f*x+e)*tan(f*x+e)/f/(a+a*sin(f*x+e))^(7/ 
2)+1/3*tan(f*x+e)^3/f/(a+a*sin(f*x+e))^(5/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.32 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.90 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {1312+\frac {768 \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {384}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {2624 \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+2584 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-1292 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+402 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-201 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-(951+951 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+\frac {256 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {1152 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}}{12288 f (a (1+\sin (e+f x)))^{5/2}} \] Input:

Integrate[Tan[e + f*x]^4/(a + a*Sin[e + f*x])^(5/2),x]
 

Output:

(1312 + (768*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 3 
84/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (2624*Sin[(e + f*x)/2])/(Cos[ 
(e + f*x)/2] + Sin[(e + f*x)/2]) + 2584*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] 
 + Sin[(e + f*x)/2]) - 1292*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 402* 
Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 201*(Cos[(e + f 
*x)/2] + Sin[(e + f*x)/2])^4 - (951 + 951*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] 
)^5 + (256*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)/(Cos[(e + f*x)/2] - Si 
n[(e + f*x)/2])^3 - (1152*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)/(Cos[(e 
 + f*x)/2] - Sin[(e + f*x)/2]))/(12288*f*(a*(1 + Sin[e + f*x]))^(5/2))
 

Rubi [A] (verified)

Time = 2.17 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.66, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3193, 3042, 3129, 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.

\(\displaystyle \int \frac {\tan ^4(e+f x)}{(a \sin (e+f x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^4}{(a \sin (e+f x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 3193

\(\displaystyle \int \frac {1}{(\sin (e+f x) a+a)^{5/2}}dx-\int \frac {\sec ^4(e+f x) \left (1-2 \sin ^2(e+f x)\right )}{(\sin (e+f x) a+a)^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(\sin (e+f x) a+a)^{5/2}}dx-\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {3 \int \frac {1}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a}-\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {1}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a}-\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3129

\(\displaystyle -\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx+\frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {3 \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{2 a f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\int \frac {1-2 \sin (e+f x)^2}{\cos (e+f x)^4 (\sin (e+f x) a+a)^{5/2}}dx+\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 4901

\(\displaystyle -\int \left (\frac {\sec ^4(e+f x)}{(a (\sin (e+f x)+1))^{5/2}}-\frac {2 \sec ^2(e+f x) \tan ^2(e+f x)}{(a (\sin (e+f x)+1))^{5/2}}\right )dx+\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1085 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{4096 \sqrt {2} a^{5/2} f}+\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}\right )}{8 a}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {1085 \sec (e+f x)}{3072 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {1085 \cos (e+f x)}{4096 a f (a \sin (e+f x)+a)^{3/2}}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a \sin (e+f x)+a)^{3/2}}-\frac {\sec ^3(e+f x)}{8 f (a \sin (e+f x)+a)^{5/2}}+\frac {217 \sec (e+f x)}{1536 a f (a \sin (e+f x)+a)^{3/2}}\)

Input:

Int[Tan[e + f*x]^4/(a + a*Sin[e + f*x])^(5/2),x]
 

Output:

(1085*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/ 
(4096*Sqrt[2]*a^(5/2)*f) - Cos[e + f*x]/(4*f*(a + a*Sin[e + f*x])^(5/2)) - 
 Sec[e + f*x]^3/(8*f*(a + a*Sin[e + f*x])^(5/2)) + (1085*Cos[e + f*x])/(40 
96*a*f*(a + a*Sin[e + f*x])^(3/2)) + (217*Sec[e + f*x])/(1536*a*f*(a + a*S 
in[e + f*x])^(3/2)) + (53*Sec[e + f*x]^3)/(96*a*f*(a + a*Sin[e + f*x])^(3/ 
2)) - (1085*Sec[e + f*x])/(3072*a^2*f*Sqrt[a + a*Sin[e + f*x]]) - (31*Sec[ 
e + f*x]^3)/(192*a^2*f*Sqrt[a + a*Sin[e + f*x]]) + (3*(-1/2*ArcTanh[(Sqrt[ 
a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])]/(Sqrt[2]*a^(3/2)*f) - 
 Cos[e + f*x]/(2*f*(a + a*Sin[e + f*x])^(3/2))))/(8*a)
 

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. \(380\) vs. \(2(176)=352\).

Time = 0.36 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.84

method	result	size
default	\(-\frac {1902 \cos \left (f x +e \right )^{4} \sin \left (f x +e \right ) a^{\frac {11}{2}}+4438 \cos \left (f x +e \right )^{4} a^{\frac {11}{2}}-13888 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a^{\frac {11}{2}}+951 \cos \left (f x +e \right )^{4} \sqrt {2}\, a^{4} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right )-9920 \cos \left (f x +e \right )^{2} a^{\frac {11}{2}}-3804 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) \sqrt {2}\, a^{4} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right )+5632 \sin \left (f x +e \right ) a^{\frac {11}{2}}-7608 \cos \left (f x +e \right )^{2} \sqrt {2}\, a^{4} \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right )+2560 a^{\frac {11}{2}}+7608 \sin \left (f x +e \right ) \sqrt {2}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}+7608 \sqrt {2}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}}{24576 a^{\frac {15}{2}} \left (-1+\sin \left (f x +e \right )\right ) \left (1+\sin \left (f x +e \right )\right )^{3} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(381\)

Input:

int(tan(f*x+e)^4/(a+sin(f*x+e)*a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24576/a^(15/2)*(1902*cos(f*x+e)^4*sin(f*x+e)*a^(11/2)+4438*cos(f*x+e)^4 
*a^(11/2)-13888*cos(f*x+e)^2*sin(f*x+e)*a^(11/2)+951*cos(f*x+e)^4*2^(1/2)* 
a^4*(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))-9920*cos(f*x+e)^2*a^(11/2)-3804*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*a^4*( 
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))+ 
5632*sin(f*x+e)*a^(11/2)-7608*cos(f*x+e)^2*2^(1/2)*a^4*(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))+2560*a^(11/2)+7608 
*sin(f*x+e)*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^4+7608*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^4)/(-1+sin(f*x+e))/(1+sin(f*x+e)) 
^3/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.48 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {951 \, \sqrt {2} {\left (3 \, \cos \left (f x + e\right )^{5} - 4 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{5} - 4 \, \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (2219 \, \cos \left (f x + e\right )^{4} - 4960 \, \cos \left (f x + e\right )^{2} + {\left (951 \, \cos \left (f x + e\right )^{4} - 6944 \, \cos \left (f x + e\right )^{2} + 2816\right )} \sin \left (f x + e\right ) + 1280\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{49152 \, {\left (3 \, a^{3} f \cos \left (f x + e\right )^{5} - 4 \, a^{3} f \cos \left (f x + e\right )^{3} + {\left (a^{3} f \cos \left (f x + e\right )^{5} - 4 \, a^{3} f \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate(tan(f*x+e)^4/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

1/49152*(951*sqrt(2)*(3*cos(f*x + e)^5 - 4*cos(f*x + e)^3 + (cos(f*x + e)^ 
5 - 4*cos(f*x + e)^3)*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 + 2*sqr 
t(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)*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*(2219*cos(f 
*x + e)^4 - 4960*cos(f*x + e)^2 + (951*cos(f*x + e)^4 - 6944*cos(f*x + e)^ 
2 + 2816)*sin(f*x + e) + 1280)*sqrt(a*sin(f*x + e) + a))/(3*a^3*f*cos(f*x 
+ e)^5 - 4*a^3*f*cos(f*x + e)^3 + (a^3*f*cos(f*x + e)^5 - 4*a^3*f*cos(f*x 
+ e)^3)*sin(f*x + e))
 

Sympy [F]

\[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tan(f*x+e)**4/(a+a*sin(f*x+e))**(5/2),x)
 

Output:

Integral(tan(e + f*x)**4/(a*(sin(e + f*x) + 1))**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(f*x+e)^4/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.89 \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\frac {128 \, \sqrt {2} {\left (9 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a}\right )}}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}} - \frac {\sqrt {2} {\left (201 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 1249 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 1567 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 567 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{4} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{24576 \, f} \] Input:

integrate(tan(f*x+e)^4/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")
 

Output:

-1/24576*(128*sqrt(2)*(9*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^2 - sqrt(a) 
)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi + 1/2*f*x + 1/2*e)^3 
) - sqrt(2)*(201*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^7 - 1249*sqrt(a)*si 
n(3/4*pi + 1/2*f*x + 1/2*e)^5 + 1567*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e) 
^3 - 567*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e))/((sin(3/4*pi + 1/2*f*x + 1 
/2*e)^2 - 1)^4*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int(tan(e + f*x)^4/(a + a*sin(e + f*x))^(5/2),x)
 

Output:

int(tan(e + f*x)^4/(a + a*sin(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )}{a^{3}} \] Input:

int(tan(f*x+e)^4/(a+a*sin(f*x+e))^(5/2),x)
 

Output:

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