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

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

[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)) - (Sec[e + f*x]*(65 + 87*Sin[e + f*x]))/(192*f*(a + a*Sin[e + f*x])^(3/
2)) + (a*Sin[e + f*x]*Tan[e + f*x])/(12*f*(a + a*Sin[e + f*x])^(5/2)) + Tan[e + f*x]^3/(3*f*(a + a*Sin[e + f*x
])^(3/2))

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Rubi [A]  time = 1.19927, antiderivative size = 195, normalized size of antiderivative = 1.1, 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 = {2714, 2650, 2649, 206, 4401, 2681, 2687, 2877, 2855} \[ \frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{256 \sqrt{2} a^{3/2} f}+\frac{7 \cos (e+f x)}{256 f (a \sin (e+f x)+a)^{3/2}}+\frac{\sec ^3(e+f x)}{4 a f \sqrt{a \sin (e+f x)+a}}-\frac{\sec ^3(e+f x)}{6 f (a \sin (e+f x)+a)^{3/2}}-\frac{45 \sec (e+f x)}{64 a f \sqrt{a \sin (e+f x)+a}}+\frac{9 \sec (e+f x)}{32 f (a \sin (e+f x)+a)^{3/2}} \]

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 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 2855

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]

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{\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 )}{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 \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 )}{128 a f}+\frac{105 \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 )}{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*}

Mathematica [C]  time = 0.355786, size = 334, normalized size = 1.89 \[ \frac{-\frac{192 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{32 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}-171 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+342 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-\frac{248 \sin \left (\frac{1}{2} (e+f x)\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}-\frac{32}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}+\frac{64 \sin \left (\frac{1}{2} (e+f x)\right )}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}+(-21-21 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+124}{768 f (a (\sin (e+f x)+1))^{3/2}} \]

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 = 0.619, size = 289, normalized size = 1.6 \begin{align*} -{\frac{1}{ \left ( -1536+1536\,\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( \left ( -1080\,{a}^{9/2}-21\, \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}^{3} \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( 384\,{a}^{9/2}+84\, \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}^{3} \right ) \sin \left ( fx+e \right ) +42\,{a}^{9/2} \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( -648\,{a}^{9/2}-63\, \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}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+128\,{a}^{9/2}+84\, \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}^{3} \right ){a}^{-{\frac{11}{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))^(3/2),x)

[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)/(-1+sin(f*x+e))/(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)]  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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.62036, size = 726, normalized size = 4.1 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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 [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))^(3/2),x, algorithm="giac")

[Out]

sage2