3.61 \(\int \frac{\sin ^4(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=238 \[ -\frac{3 \sqrt{b} \left (5 a^2+20 a b+16 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^5 f \sqrt{a+b}}+\frac{3 x \left (a^2+12 a b+16 b^2\right )}{8 a^5}-\frac{3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac{b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac{(5 a+8 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

(3*(a^2 + 12*a*b + 16*b^2)*x)/(8*a^5) - (3*Sqrt[b]*(5*a^2 + 20*a*b + 16*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(8*a^5*Sqrt[a + b]*f) - ((5*a + 8*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)
^2) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*a*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(7*a + 12*b)*Tan[e + f*x])/(8*a^
3*f*(a + b + b*Tan[e + f*x]^2)^2) - (3*b*(a + 2*b)*Tan[e + f*x])/(2*a^4*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.340288, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4132, 470, 527, 522, 203, 205} \[ -\frac{3 \sqrt{b} \left (5 a^2+20 a b+16 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^5 f \sqrt{a+b}}+\frac{3 x \left (a^2+12 a b+16 b^2\right )}{8 a^5}-\frac{3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac{b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac{(5 a+8 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a^2 + 12*a*b + 16*b^2)*x)/(8*a^5) - (3*Sqrt[b]*(5*a^2 + 20*a*b + 16*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(8*a^5*Sqrt[a + b]*f) - ((5*a + 8*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)
^2) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*a*f*(a + b + b*Tan[e + f*x]^2)^2) - (b*(7*a + 12*b)*Tan[e + f*x])/(8*a^
3*f*(a + b + b*Tan[e + f*x]^2)^2) - (3*b*(a + 2*b)*Tan[e + f*x])/(2*a^4*f*(a + b + b*Tan[e + f*x]^2))

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{a+b+(-4 a-7 b) x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=-\frac{(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{(a+b) (3 a+8 b)-5 b (5 a+8 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{8 a^2 f}\\ &=-\frac{(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{12 (a+b)^2 (a+4 b)-12 b (a+b) (7 a+12 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{32 a^3 (a+b) f}\\ &=-\frac{(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{24 (a+b)^2 \left (a^2+8 a b+8 b^2\right )-96 b (a+b)^2 (a+2 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{64 a^4 (a+b)^2 f}\\ &=-\frac{(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\left (3 \left (a^2+12 a b+16 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^5 f}-\frac{\left (3 b \left (5 a^2+20 a b+16 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^5 f}\\ &=\frac{3 \left (a^2+12 a b+16 b^2\right ) x}{8 a^5}-\frac{3 \sqrt{b} \left (5 a^2+20 a b+16 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^5 \sqrt{a+b} f}-\frac{(5 a+8 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{b (7 a+12 b) \tan (e+f x)}{8 a^3 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{3 b (a+2 b) \tan (e+f x)}{2 a^4 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 25.8581, size = 2469, normalized size = 10.37 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(((3*a^2 + 8*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr
t[a + b]])/(a + b)^(5/2) - (a*Sqrt[b]*(3*a^2 + 16*a*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + f*
x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(16384*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3) + ((a + 2*b + a
*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((-3*a*(a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2)
+ (Sqrt[b]*(3*a^3 + 14*a^2*b + 24*a*b^2 + 16*b^3 + a*(3*a^2 + 4*a*b + 4*b^2)*Cos[2*(e + f*x)])*Sin[2*(e + f*x
)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(16384*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3) - (3*(a + 2*b +
a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((2*(3*a^5 - 10*a^4*b + 80*a^3*b^2 + 480*a^2*b^3 + 640*a*b^4 + 256*b^5)*A
rcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos
[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (Sec[2*e]*(256*b^2
*(a + b)^2*(3*a^2 + 8*a*b + 8*b^2)*f*x*Cos[2*e] + 512*a*b^2*(a + b)^2*(a + 2*b)*f*x*Cos[2*f*x] + 128*a^4*b^2*f
*x*Cos[2*(e + 2*f*x)] + 256*a^3*b^3*f*x*Cos[2*(e + 2*f*x)] + 128*a^2*b^4*f*x*Cos[2*(e + 2*f*x)] + 512*a^4*b^2*
f*x*Cos[4*e + 2*f*x] + 2048*a^3*b^3*f*x*Cos[4*e + 2*f*x] + 2560*a^2*b^4*f*x*Cos[4*e + 2*f*x] + 1024*a*b^5*f*x*
Cos[4*e + 2*f*x] + 128*a^4*b^2*f*x*Cos[6*e + 4*f*x] + 256*a^3*b^3*f*x*Cos[6*e + 4*f*x] + 128*a^2*b^4*f*x*Cos[6
*e + 4*f*x] - 9*a^6*Sin[2*e] + 12*a^5*b*Sin[2*e] + 684*a^4*b^2*Sin[2*e] + 2880*a^3*b^3*Sin[2*e] + 5280*a^2*b^4
*Sin[2*e] + 4608*a*b^5*Sin[2*e] + 1536*b^6*Sin[2*e] + 9*a^6*Sin[2*f*x] - 14*a^5*b*Sin[2*f*x] - 608*a^4*b^2*Sin
[2*f*x] - 2112*a^3*b^3*Sin[2*f*x] - 2560*a^2*b^4*Sin[2*f*x] - 1024*a*b^5*Sin[2*f*x] + 3*a^6*Sin[2*(e + 2*f*x)]
- 12*a^5*b*Sin[2*(e + 2*f*x)] - 204*a^4*b^2*Sin[2*(e + 2*f*x)] - 384*a^3*b^3*Sin[2*(e + 2*f*x)] - 192*a^2*b^4
*Sin[2*(e + 2*f*x)] - 3*a^6*Sin[4*e + 2*f*x] + 10*a^5*b*Sin[4*e + 2*f*x] + 304*a^4*b^2*Sin[4*e + 2*f*x] + 1056
*a^3*b^3*Sin[4*e + 2*f*x] + 1280*a^2*b^4*Sin[4*e + 2*f*x] + 512*a*b^5*Sin[4*e + 2*f*x]))/(a + 2*b + a*Cos[2*(e
+ f*x)])^2))/(65536*a^3*b^2*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3) - ((a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e +
f*x]^6*((-6*a^2*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[
a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) +
(a*Sec[2*e]*((-9*a^4 - 16*a^3*b + 48*a^2*b^2 + 128*a*b^3 + 64*b^4)*Sin[2*f*x] + a*(-3*a^3 + 2*a^2*b + 24*a*b^2
+ 16*b^3)*Sin[2*(e + 2*f*x)] + (3*a^4 - 64*a^2*b^2 - 128*a*b^3 - 64*b^4)*Sin[4*e + 2*f*x]) + (9*a^5 + 18*a^4*
b - 64*a^3*b^2 - 256*a^2*b^3 - 320*a*b^4 - 128*b^5)*Tan[2*e])/(a^2*(a + 2*b + a*Cos[2*(e + f*x)])^2)))/(8192*b
^2*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3) + ((a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(-1536*(a + 2*b)*x
- (3*(a^6 - 8*a^5*b + 120*a^4*b^2 + 1280*a^3*b^3 + 3200*a^2*b^4 + 3072*a*b^5 + 1024*b^6)*ArcTan[(Sec[f*x]*(Co
s[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])
]*(Cos[2*e] - I*Sin[2*e]))/(b^2*(a + b)^(5/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (4*(a^4 + 32*a^3*b + 160*a^2*
b^2 + 256*a*b^3 + 128*b^4)*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e +
f*x)])^2) + (256*a*Sin[2*(e + f*x)])/f + (a*(-3*a^5 + 26*a^4*b + 736*a^3*b^2 + 2624*a^2*b^3 + 3200*a*b^4 + 128
0*b^5)*Sec[2*e]*Sin[2*f*x] + (3*a^6 - 24*a^5*b - 920*a^4*b^2 - 4864*a^3*b^3 - 10112*a^2*b^4 - 9216*a*b^5 - 307
2*b^6)*Tan[2*e])/(b^2*(a + b)^2*f*(a + 2*b + a*Cos[2*(e + f*x)]))))/(16384*a^4*(a + b*Sec[e + f*x]^2)^3) + ((a
+ 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*(768*(7*a^2 + 32*a*b + 32*b^2)*x + (3*(a^7 - 14*a^6*b + 336*a^5*
b^2 + 5600*a^4*b^3 + 22400*a^3*b^4 + 37632*a^2*b^5 + 28672*a*b^6 + 8192*b^7)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Si
n[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] -
I*Sin[2*e]))/(b^2*(a + b)^(5/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) - (4*(a^5 + 50*a^4*b + 400*a^3*b^2 + 1120*a^
2*b^3 + 1280*a*b^4 + 512*b^5)*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e
+ f*x)])^2) - ((768*I)*a*(a + 2*b)*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]))/f + ((768*I)*a*(a + 2*b)*(Cos[2*(
e + f*x)] + I*Sin[2*(e + f*x)]))/f + (128*a^2*Sin[4*(e + f*x)])/f + (a*(3*a^6 - 44*a^5*b - 1900*a^4*b^2 - 1088
0*a^3*b^3 - 23360*a^2*b^4 - 21504*a*b^5 - 7168*b^6)*Sec[2*e]*Sin[2*f*x] + (-3*a^7 + 42*a^6*b + 2192*a^5*b^2 +
16480*a^4*b^3 + 51200*a^3*b^4 + 77824*a^2*b^5 + 57344*a*b^6 + 16384*b^7)*Tan[2*e])/(b^2*(a + b)^2*f*(a + 2*b +
a*Cos[2*(e + f*x)]))))/(32768*a^5*(a + b*Sec[e + f*x]^2)^3)

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Maple [A]  time = 0.117, size = 423, normalized size = 1.8 \begin{align*} -{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{2\,f{a}^{4} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{5\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{3\,\tan \left ( fx+e \right ) }{8\,f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{3\,\tan \left ( fx+e \right ) b}{2\,f{a}^{4} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{9\,\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{2\,f{a}^{4}}}+6\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{f{a}^{5}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) }{8\,f{a}^{3}}}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{3\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{2\,f{a}^{4} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,\tan \left ( fx+e \right ) b}{8\,{a}^{2}f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{21\,{b}^{2}\tan \left ( fx+e \right ) }{8\,f{a}^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{3\,{b}^{3}\tan \left ( fx+e \right ) }{2\,f{a}^{4} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,b}{8\,f{a}^{3}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{15\,{b}^{2}}{2\,f{a}^{4}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-6\,{\frac{{b}^{3}}{f{a}^{5}\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/f/a^4/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3*b-5/8/f/a^3/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3-3/8/f/a^3/(tan(f*x+e)^2
+1)^2*tan(f*x+e)-3/2/f/a^4/(tan(f*x+e)^2+1)^2*tan(f*x+e)*b+9/2/f/a^4*arctan(tan(f*x+e))*b+6/f/a^5*arctan(tan(f
*x+e))*b^2+3/8/f/a^3*arctan(tan(f*x+e))-7/8/f*b^2/a^3/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-3/2/f*b^3/a^4/(a+b+b
*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8*b*tan(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)^2-21/8/f*b^2/a^3/(a+b+b*tan(f*x+e)^2
)^2*tan(f*x+e)-3/2/f*b^3/a^4/(a+b+b*tan(f*x+e)^2)^2*tan(f*x+e)-15/8/f*b/a^3/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*
b/((a+b)*b)^(1/2))-15/2/f*b^2/a^4/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-6/f*b^3/a^5/((a+b)*b)^(
1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.796025, size = 1859, normalized size = 7.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/32*(12*(a^4 + 12*a^3*b + 16*a^2*b^2)*f*x*cos(f*x + e)^4 + 24*(a^3*b + 12*a^2*b^2 + 16*a*b^3)*f*x*cos(f*x +
e)^2 + 12*(a^2*b^2 + 12*a*b^3 + 16*b^4)*f*x + 3*((5*a^4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x + e)^4 + 5*a^2*b^2 +
20*a*b^3 + 16*b^4 + 2*(5*a^3*b + 20*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8
*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2
)*cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + 4*(2
*a^4*cos(f*x + e)^7 - (5*a^4 + 8*a^3*b)*cos(f*x + e)^5 - (19*a^3*b + 36*a^2*b^2)*cos(f*x + e)^3 - 12*(a^2*b^2
+ 2*a*b^3)*cos(f*x + e))*sin(f*x + e))/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f), 1/16*(6*
(a^4 + 12*a^3*b + 16*a^2*b^2)*f*x*cos(f*x + e)^4 + 12*(a^3*b + 12*a^2*b^2 + 16*a*b^3)*f*x*cos(f*x + e)^2 + 6*(
a^2*b^2 + 12*a*b^3 + 16*b^4)*f*x + 3*((5*a^4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x + e)^4 + 5*a^2*b^2 + 20*a*b^3 +
16*b^4 + 2*(5*a^3*b + 20*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e
)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) + 2*(2*a^4*cos(f*x + e)^7 - (5*a^4 + 8*a^3*b)*cos(f*x
+ e)^5 - (19*a^3*b + 36*a^2*b^2)*cos(f*x + e)^3 - 12*(a^2*b^2 + 2*a*b^3)*cos(f*x + e))*sin(f*x + e))/(a^7*f*co
s(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f)]

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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(sin(f*x+e)**4/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.36882, size = 439, normalized size = 1.84 \begin{align*} \frac{\frac{3 \,{\left (a^{2} + 12 \, a b + 16 \, b^{2}\right )}{\left (f x + e\right )}}{a^{5}} - \frac{3 \,{\left (5 \, a^{2} b + 20 \, a b^{2} + 16 \, b^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{\sqrt{a b + b^{2}} a^{5}} - \frac{12 \, a b^{2} \tan \left (f x + e\right )^{7} + 24 \, b^{3} \tan \left (f x + e\right )^{7} + 19 \, a^{2} b \tan \left (f x + e\right )^{5} + 72 \, a b^{2} \tan \left (f x + e\right )^{5} + 72 \, b^{3} \tan \left (f x + e\right )^{5} + 5 \, a^{3} \tan \left (f x + e\right )^{3} + 46 \, a^{2} b \tan \left (f x + e\right )^{3} + 108 \, a b^{2} \tan \left (f x + e\right )^{3} + 72 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 27 \, a^{2} b \tan \left (f x + e\right ) + 48 \, a b^{2} \tan \left (f x + e\right ) + 24 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{4}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/8*(3*(a^2 + 12*a*b + 16*b^2)*(f*x + e)/a^5 - 3*(5*a^2*b + 20*a*b^2 + 16*b^3)*(pi*floor((f*x + e)/pi + 1/2)*s
gn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^5) - (12*a*b^2*tan(f*x + e)^7 + 24*b^3*tan(
f*x + e)^7 + 19*a^2*b*tan(f*x + e)^5 + 72*a*b^2*tan(f*x + e)^5 + 72*b^3*tan(f*x + e)^5 + 5*a^3*tan(f*x + e)^3
+ 46*a^2*b*tan(f*x + e)^3 + 108*a*b^2*tan(f*x + e)^3 + 72*b^3*tan(f*x + e)^3 + 3*a^3*tan(f*x + e) + 27*a^2*b*t
an(f*x + e) + 48*a*b^2*tan(f*x + e) + 24*b^3*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan(f*x + e)^2 + 2*b*tan(f*x
+ e)^2 + a + b)^2*a^4))/f