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

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

[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) + (317*Cos[e
+ f*x])/(3072*f*(a + a*Sin[e + f*x])^(5/2)) - (Sec[e + f*x]*(115 + 129*Sin[e + f*x]))/(384*f*(a + a*Sin[e + f
*x])^(5/2)) + (317*Cos[e + f*x])/(4096*a*f*(a + a*Sin[e + f*x])^(3/2)) + (5*a*Sin[e + f*x]*Tan[e + f*x])/(48*f
*(a + a*Sin[e + f*x])^(7/2)) + Tan[e + f*x]^3/(3*f*(a + a*Sin[e + f*x])^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 1.43374, 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 = {2714, 2650, 2649, 206, 4401, 2681, 2687, 2877, 2859} \[ -\frac{31 \sec ^3(e+f x)}{192 a^2 f \sqrt{a \sin (e+f x)+a}}-\frac{1085 \sec (e+f x)}{3072 a^2 f \sqrt{a \sin (e+f x)+a}}+\frac{317 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{4096 \sqrt{2} a^{5/2} f}+\frac{317 \cos (e+f x)}{4096 a f (a \sin (e+f x)+a)^{3/2}}-\frac{\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2}}+\frac{53 \sec ^3(e+f x)}{96 a f (a \sin (e+f x)+a)^{3/2}}-\frac{\sec ^3(e+f x)}{8 f (a \sin (e+f x)+a)^{5/2}}+\frac{217 \sec (e+f x)}{1536 a f (a \sin (e+f x)+a)^{3/2}} \]

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 2714

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 2650

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 2649

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 206

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

Rule 4401

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

Rule 2681

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((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[(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 2687

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)), Int[
(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 2877

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 2859

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]

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 \operatorname{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 \operatorname{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 \operatorname{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]  time = 0.546479, size = 394, normalized size = 1.9 \[ \frac{-\frac{1152 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5}{\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{256 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}-201 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+402 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-1292 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+2584 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-\frac{2624 \sin \left (\frac{1}{2} (e+f x)\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}-\frac{384}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}+\frac{768 \sin \left (\frac{1}{2} (e+f x)\right )}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}+(-951-951 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+1312}{12288 f (a (\sin (e+f x)+1))^{5/2}} \]

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))

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Maple [A]  time = 0.843, size = 353, normalized size = 1.7 \begin{align*} -{\frac{1}{ \left ( -24576+24576\,\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) f} \left ( 1902\,{a}^{11/2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( -13888\,{a}^{11/2}-3804\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{4} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( 5632\,{a}^{11/2}+7608\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{4} \right ) \sin \left ( fx+e \right ) + \left ( 4438\,{a}^{11/2}+951\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{4} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( -9920\,{a}^{11/2}-7608\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{4} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2560\,{a}^{11/2}+7608\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{4} \right ){a}^{-{\frac{15}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

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)

[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)/(
-1+sin(f*x+e))/(1+sin(f*x+e))^3/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

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

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Fricas [A]  time = 1.90076, size = 830, normalized size = 4.01 \begin{align*} \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{align*}

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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

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]

sage2