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

Optimal. Leaf size=207 \[ \frac {317 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{4096 \sqrt {2} a^{5/2} f}+\frac {317 \cos (e+f x)}{3072 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec (e+f x) (115+129 \sin (e+f x))}{384 f (a+a \sin (e+f x))^{5/2}}+\frac {317 \cos (e+f x)}{4096 a f (a+a \sin (e+f x))^{3/2}}+\frac {5 a \sin (e+f x) \tan (e+f x)}{48 f (a+a \sin (e+f x))^{7/2}}+\frac {\tan ^3(e+f x)}{3 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

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)+31
7/4096*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)+317/8192*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^
(1/2))/a^(5/2)/f*2^(1/2)+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)

________________________________________________________________________________________

Rubi [A]
time = 0.97, antiderivative size = 260, normalized size of antiderivative = 1.26, number of steps used = 23, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {2793, 2729, 2728, 212, 4486, 2760, 2766, 2956, 2938} \begin {gather*} \frac {317 \tanh ^{-1}\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 {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 {317 \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(317*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)) + (317*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*Sin[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*Se
c[e + f*x]^3)/(192*a^2*f*Sqrt[a + a*Sin[e + f*x]])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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

Rule 2729

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] + Dist[(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 2760

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

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rule 2793

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]
&& EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]

Rule 2938

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*(2*m + p
+ 1))), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2956

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

Rule 4486

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

Rubi steps

\begin {align*} \int \frac {\tan ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx &=\int \frac {1}{(a+a \sin (e+f x))^{5/2}} \, dx-\int \frac {\sec ^4(e+f x) \left (1-2 \sin ^2(e+f x)\right )}{(a+a \sin (e+f x))^{5/2}} \, dx\\ &=-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {3 \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx}{8 a}-\int \left (\frac {\sec ^4(e+f x)}{(a (1+\sin (e+f x)))^{5/2}}-\frac {2 \sec ^2(e+f x) \tan ^2(e+f x)}{(a (1+\sin (e+f x)))^{5/2}}\right ) \, dx\\ &=-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {3 \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+2 \int \frac {\sec ^2(e+f x) \tan ^2(e+f x)}{(a (1+\sin (e+f x)))^{5/2}} \, dx+\frac {3 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2}-\int \frac {\sec ^4(e+f x)}{(a (1+\sin (e+f x)))^{5/2}} \, dx\\ &=-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 f (a+a \sin (e+f x))^{5/2}}-\frac {3 \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {\sec ^4(e+f x) \left (-\frac {5 a}{2}+8 a \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}-\frac {11 \int \frac {\sec ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{16 a}-\frac {3 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{16 a^2 f}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 f (a+a \sin (e+f x))^{5/2}}-\frac {3 \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {\sec ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2}-\frac {33 \int \frac {\sec ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{64 a^2}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 f (a+a \sin (e+f x))^{5/2}}-\frac {3 \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \int \frac {\sec ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{192 a}-\frac {77 \int \frac {\sec ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{128 a}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 f (a+a \sin (e+f x))^{5/2}}-\frac {3 \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {217 \sec (e+f x)}{1536 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {35 \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{1536 a^2}-\frac {385 \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{1024 a^2}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 f (a+a \sin (e+f x))^{5/2}}-\frac {3 \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {217 \sec (e+f x)}{1536 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}-\frac {1085 \sec (e+f x)}{3072 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {35 \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx}{1024 a}-\frac {1155 \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx}{2048 a}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 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 {217 \sec (e+f x)}{1536 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}-\frac {1085 \sec (e+f x)}{3072 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {35 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{4096 a^2}-\frac {1155 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{8192 a^2}\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 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 {217 \sec (e+f x)}{1536 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}-\frac {1085 \sec (e+f x)}{3072 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {35 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{2048 a^2 f}+\frac {1155 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{4096 a^2 f}\\ &=\frac {317 \tanh ^{-1}\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 {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\sec ^3(e+f x)}{8 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 {217 \sec (e+f x)}{1536 a f (a+a \sin (e+f x))^{3/2}}+\frac {53 \sec ^3(e+f x)}{96 a f (a+a \sin (e+f x))^{3/2}}-\frac {1085 \sec (e+f x)}{3072 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {31 \sec ^3(e+f x)}{192 a^2 f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.37, size = 394, normalized size = 1.90 \begin {gather*} \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} \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1312 + (768*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 384/(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] + S
in[(e + f*x)/2])^5)/(Cos[(e + f*x)/2] - Sin[(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))

________________________________________________________________________________________

Maple [A]
time = 2.74, size = 353, normalized size = 1.71

method	result	size
default	\(-\frac {1902 a^{\frac {11}{2}} \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (-13888 a^{\frac {11}{2}}-3804 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (5632 a^{\frac {11}{2}}+7608 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \sin \left (f x +e \right )+\left (4438 a^{\frac {11}{2}}+951 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (-9920 a^{\frac {11}{2}}-7608 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2560 a^{\frac {11}{2}}+7608 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}}{24576 a^{\frac {15}{2}} \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(353\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 334, normalized size = 1.61 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*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*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 39.98, size = 186, normalized size = 0.90 \begin {gather*} -\frac {\frac {128 \, \sqrt {2} {\left (9 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{a^{\frac {5}{2}} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24576*(128*sqrt(2)*(9*sin(3/4*pi + 1/2*f*x + 1/2*e)^2 - 1)/(a^(5/2)*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)*sin(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]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________