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

Optimal. Leaf size=177 \[ \frac {7 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{256 \sqrt {2} a^{3/2} f}+\frac {7 \cos (e+f x)}{256 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec (e+f x) (65+87 \sin (e+f x))}{192 f (a+a \sin (e+f x))^{3/2}}+\frac {a \sin (e+f x) \tan (e+f x)}{12 f (a+a \sin (e+f x))^{5/2}}+\frac {\tan ^3(e+f x)}{3 f (a+a \sin (e+f x))^{3/2}} \]

[Out]

7/256*cos(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)-1/192*sec(f*x+e)*(65+87*sin(f*x+e))/f/(a+a*sin(f*x+e))^(3/2)+7/512*a
rctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/f*2^(1/2)+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)

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Rubi [A]
time = 0.80, antiderivative size = 195, normalized size of antiderivative = 1.10, number of steps used = 20, 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, 2934} \begin {gather*} \frac {7 \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(256*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]])

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 2934

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

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))^{3/2}} \, dx &=\int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx-\int \frac {\sec ^4(e+f x) \left (1-2 \sin ^2(e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 a}-\int \left (\frac {\sec ^4(e+f x)}{(a (1+\sin (e+f x)))^{3/2}}-\frac {2 \sec ^2(e+f x) \tan ^2(e+f x)}{(a (1+\sin (e+f x)))^{3/2}}\right ) \, dx\\ &=-\frac {\cos (e+f x)}{2 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)))^{3/2}} \, dx-\frac {\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 )}{2 a f}-\int \frac {\sec ^4(e+f x)}{(a (1+\sin (e+f x)))^{3/2}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {\sec ^4(e+f x) \left (-\frac {3 a}{2}+6 a \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{3 a^2}-\frac {3 \int \frac {\sec ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 a}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a+a \sin (e+f x)}}-\frac {1}{4} \int \frac {\sec ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx-\frac {7}{8} \int \frac {\sec ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {9 \sec (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a+a \sin (e+f x)}}-\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a}-\frac {35 \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{64 a}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2}}+\frac {9 \sec (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}-\frac {45 \sec (e+f x)}{64 a f \sqrt {a+a \sin (e+f x)}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a+a \sin (e+f x)}}-\frac {15}{64} \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx-\frac {105}{128} \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {7 \cos (e+f x)}{256 f (a+a \sin (e+f x))^{3/2}}+\frac {9 \sec (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}-\frac {45 \sec (e+f x)}{64 a f \sqrt {a+a \sin (e+f x)}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a+a \sin (e+f x)}}-\frac {15 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{256 a}-\frac {105 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{512 a}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {7 \cos (e+f x)}{256 f (a+a \sin (e+f x))^{3/2}}+\frac {9 \sec (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}-\frac {45 \sec (e+f x)}{64 a f \sqrt {a+a \sin (e+f x)}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a+a \sin (e+f x)}}+\frac {15 \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 )}{128 a f}+\frac {105 \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 )}{256 a f}\\ &=\frac {7 \tanh ^{-1}\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 {9 \sec (e+f x)}{32 f (a+a \sin (e+f x))^{3/2}}-\frac {\sec ^3(e+f x)}{6 f (a+a \sin (e+f x))^{3/2}}-\frac {45 \sec (e+f x)}{64 a f \sqrt {a+a \sin (e+f x)}}+\frac {\sec ^3(e+f x)}{4 a f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.25, size = 334, normalized size = 1.89 \begin {gather*} \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} \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 )^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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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] +
Sin[(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))

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Maple [A]
time = 2.43, size = 289, normalized size = 1.63

method	result	size
default	\(-\frac {\left (-1080 a^{\frac {9}{2}}-21 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (384 a^{\frac {9}{2}}+84 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \sin \left (f x +e \right )+42 a^{\frac {9}{2}} \left (\cos ^{4}\left (f x +e \right )\right )+\left (-648 a^{\frac {9}{2}}-63 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+128 a^{\frac {9}{2}}+84 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3}}{1536 a^{\frac {11}{2}} \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(289\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 0.41, size = 294, normalized size = 1.66 \begin {gather*} \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 )}} \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))^(3/2),x, algorithm="fricas")

[Out]

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
*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*(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)

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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 {3}{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))**(3/2),x)

[Out]

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

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Giac [A]
time = 46.83, size = 231, normalized size = 1.31 \begin {gather*} \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} \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))^(3/2),x, algorithm="giac")

[Out]

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*s
qrt(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*s
qrt(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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,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))^(3/2),x)

[Out]

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

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