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]

317/3072*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)

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Rubi [A]  time = 1.43, 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 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 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 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 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 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 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]

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 4401

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 \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*}

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Mathematica [C]  time = 0.54, size = 394, normalized size = 1.90 \[ \frac {-\frac {1152 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {256 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-201 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4+402 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3-1292 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2+2584 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-\frac {2624 \sin \left (\frac {1}{2} (e+f x)\right )}{\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )}-\frac {384}{\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {768 \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}+(-951-951 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )+1312}{12288 f (a (\sin (e+f x)+1))^{5/2}} \]

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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fricas [A]  time = 0.48, size = 307, normalized size = 1.48 \[ \frac {951 \, \sqrt {2} {\left (3 \, \cos \left (f x + e\right )^{5} - 4 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{5} - 4 \, \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (2219 \, \cos \left (f x + e\right )^{4} - 4960 \, \cos \left (f x + e\right )^{2} + {\left (951 \, \cos \left (f x + e\right )^{4} - 6944 \, \cos \left (f x + e\right )^{2} + 2816\right )} \sin \left (f x + e\right ) + 1280\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{49152 \, {\left (3 \, a^{3} f \cos \left (f x + e\right )^{5} - 4 \, a^{3} f \cos \left (f x + e\right )^{3} + {\left (a^{3} f \cos \left (f x + e\right )^{5} - 4 \, a^{3} f \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*2*(-1/192*(3*(
-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+21*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(
a*tan((f*x+exp(1))/2)^2+a))^4-14*a*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+39*a^2*(-s
qrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-42*sqrt(a)*a*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*
tan((f*x+exp(1))/2)^2+a))^2-7*sqrt(a)*a^2)/a^2/(-(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a)
)^2-2*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+a)^3/sign(tan((f*x+exp(1))/2)+1)+
1/24576*(-1335*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^15+20025*sqrt(a)*(-sqrt(a)*tan((
f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^14-111513*a*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(
1))/2)^2+a))^13+173379*sqrt(a)*a*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^12-26091*a^2*(
-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^11-247283*sqrt(a)*a^2*(-sqrt(a)*tan((f*x+exp(1))
/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^10+157019*a^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+
a))^9+345215*sqrt(a)*a^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^8-87061*a^4*(-sqrt(a)*
tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7-353877*sqrt(a)*a^4*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a
*tan((f*x+exp(1))/2)^2+a))^6-89371*a^5*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+95561*
sqrt(a)*a^5*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+78327*a^6*(-sqrt(a)*tan((f*x+exp(
1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+7705*a^7*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+
a))+38735*sqrt(a)*a^6*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+565*sqrt(a)*a^7)/a^2/(-
(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+2*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(
a*tan((f*x+exp(1))/2)^2+a))+a)^8/sign(tan((f*x+exp(1))/2)+1)-317/8192*atan((-sqrt(a)*tan((f*x+exp(1))/2)-sqrt(
a)+sqrt(a*tan((f*x+exp(1))/2)^2+a))/sqrt(2)/sqrt(-a))/sqrt(2)/a^2/sqrt(-a)/sign(tan((f*x+exp(1))/2)+1))

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maple [A]  time = 0.99, size = 353, normalized size = 1.71 \[ -\frac {1902 a^{\frac {11}{2}} \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (-13888 a^{\frac {11}{2}}-3804 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (5632 a^{\frac {11}{2}}+7608 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \sin \left (f x +e \right )+\left (4438 a^{\frac {11}{2}}+951 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (-9920 a^{\frac {11}{2}}-7608 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2560 a^{\frac {11}{2}}+7608 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}}{24576 a^{\frac {15}{2}} \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

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)/(
sin(f*x+e)-1)/(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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)

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