3.80 \(\int \frac {\sin ^5(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=264 \[ -\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 f (a-b)^5}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 f (a-b)^5 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 f (a-b)^{11/2}}+\frac {(10 a-b) \cos ^3(e+f x)}{15 f (a-b)^4}-\frac {\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \]

[Out]

-1/5*(5*a^2+20*a*b+2*b^2)*cos(f*x+e)/(a-b)^5/f+1/15*(10*a-b)*cos(f*x+e)^3/(a-b)^4/f-1/5*cos(f*x+e)^5/(a-b)/f/(
a-b+b*sec(f*x+e)^2)^2-1/20*b*(5*a^2+4*b^2)*sec(f*x+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)^2-1/40*b*(35*a^2+40*a*b+2
4*b^2)*sec(f*x+e)/(a-b)^5/f/(a-b+b*sec(f*x+e)^2)-1/8*(15*a^2+40*a*b+8*b^2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/
2))*b^(1/2)/(a-b)^(11/2)/f

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3664, 462, 456, 1259, 1261, 205} \[ -\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 f (a-b)^5}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 f (a-b)^5 \left (a+b \sec ^2(e+f x)-b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 f (a-b)^{11/2}}+\frac {(10 a-b) \cos ^3(e+f x)}{15 f (a-b)^4}-\frac {\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-(Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(11/2)*f) - ((5*a^2
+ 20*a*b + 2*b^2)*Cos[e + f*x])/(5*(a - b)^5*f) + ((10*a - b)*Cos[e + f*x]^3)/(15*(a - b)^4*f) - Cos[e + f*x]
^5/(5*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) - (b*(5*a^2 + 4*b^2)*Sec[e + f*x])/(20*(a - b)^4*f*(a - b + b*Se
c[e + f*x]^2)^2) - (b*(35*a^2 + 40*a*b + 24*b^2)*Sec[e + f*x])/(40*(a - b)^5*f*(a - b + b*Sec[e + f*x]^2))

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]

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 462

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

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {-10 a+b+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {\frac {4 (10 a-b)}{(a-b) b}-\frac {4 \left (5 a^2+4 b^2\right ) x^2}{(a-b)^2 b}+\frac {3 \left (5 a^2+4 b^2\right ) x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{20 (a-b) f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {8 (a-b) (10 a-b) b-8 b \left (5 a^2+10 a b+3 b^2\right ) x^2+\frac {b^2 \left (35 a^2+40 a b+24 b^2\right ) x^4}{a-b}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{40 (a-b)^4 b f}\\ &=-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \left (\frac {8 (10 a-b) b}{x^4}-\frac {8 b \left (5 a^2+20 a b+2 b^2\right )}{(a-b) x^2}+\frac {5 b^2 \left (15 a^2+40 a b+8 b^2\right )}{(a-b) \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{40 (a-b)^4 b f}\\ &=-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac {(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {\left (b \left (15 a^2+40 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^5 f}\\ &=-\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{11/2} f}-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac {(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 5.81, size = 278, normalized size = 1.05 \[ \frac {\frac {(a-b) (5 (5 a+7 b) \cos (3 (e+f x))+3 (b-a) \cos (5 (e+f x)))-30 \cos (e+f x) \left (a^2 \left (-\frac {8 b^2}{((a-b) \cos (2 (e+f x))+a+b)^2}+\frac {18 b}{(a-b) \cos (2 (e+f x))+a+b}+5\right )+16 a b \left (\frac {b}{(a-b) \cos (2 (e+f x))+a+b}+2\right )+11 b^2\right )}{(a-b)^5}+\frac {30 \sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{11/2}}+\frac {30 \sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{11/2}}}{240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

((30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(11/2
) + (30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(1
1/2) + (-30*Cos[e + f*x]*(11*b^2 + 16*a*b*(2 + b/(a + b + (a - b)*Cos[2*(e + f*x)])) + a^2*(5 - (8*b^2)/(a + b
+ (a - b)*Cos[2*(e + f*x)])^2 + (18*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 7*b)*Cos[3*(e
+ f*x)] + 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b)^5)/(240*f)

________________________________________________________________________________________

fricas [B]  time = 0.78, size = 1018, normalized size = 3.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/240*(48*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 16*(10*a^4 - 31*a^3*b + 33*a^2*b^2 -
13*a*b^3 + b^4)*cos(f*x + e)^7 + 16*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 50*(1
5*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b
^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)
^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(
f*x + e)^2 + b)) + 30*(15*a^2*b^2 + 40*a*b^3 + 8*b^4)*cos(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3
+ 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 +
15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7
)*f), -1/120*(24*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 8*(10*a^4 - 31*a^3*b + 33*a^2*b^
2 - 13*a*b^3 + b^4)*cos(f*x + e)^7 + 8*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 25
*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 +
8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x +
e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(15*a^2*b^2 + 40*a*b^3 + 8*b^4)*co
s(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e
)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2
- 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f)]

________________________________________________________________________________________

giac [B]  time = 5.40, size = 885, normalized size = 3.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/120*(15*(15*a^2*b + 40*a*b^2 + 8*b^3)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*
x + e) + sqrt(a*b - b^2)))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sqrt(a*b - b^2)) + 30*(9
*a^3*b + 6*a^2*b^2 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 32*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) - 40*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2
- 54*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 24*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
48*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 16*a^2
*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 8*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 - 5*
a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x
+ e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^2) - 16*(8*a^2 + 59*a*b + 23*b^2
- 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 250*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 70*b^2*(cos(f
*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 320*a*b*(cos(f*x + e) - 1
)^2/(cos(f*x + e) + 1)^2 + 140*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 270*a*b*(cos(f*x + e) - 1)^3/(c
os(f*x + e) + 1)^3 - 90*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 45*a*b*(cos(f*x + e) - 1)^4/(cos(f*x +
e) + 1)^4 + 45*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a
*b^4 - b^5)*((cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5))/f

________________________________________________________________________________________

maple [B]  time = 0.64, size = 844, normalized size = 3.20 \[ -\frac {\left (\cos ^{5}\left (f x +e \right )\right ) a^{2}}{5 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 \left (\cos ^{5}\left (f x +e \right )\right ) a b}{5 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) b^{2}}{5 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 a^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{3 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (\cos ^{3}\left (f x +e \right )\right ) a b}{3 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (\cos ^{3}\left (f x +e \right )\right ) b^{2}}{3 f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {a^{2} \cos \left (f x +e \right )}{f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {4 a \cos \left (f x +e \right ) b}{f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\cos \left (f x +e \right ) b^{2}}{f \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {9 b \left (\cos ^{3}\left (f x +e \right )\right ) a^{3}}{8 f \left (a -b \right )^{5} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {b^{2} \left (\cos ^{3}\left (f x +e \right )\right ) a^{2}}{8 f \left (a -b \right )^{5} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {b^{3} \left (\cos ^{3}\left (f x +e \right )\right ) a}{f \left (a -b \right )^{5} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {7 b^{2} \cos \left (f x +e \right ) a^{2}}{8 f \left (a -b \right )^{5} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {b^{3} \cos \left (f x +e \right ) a}{f \left (a -b \right )^{5} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {15 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right ) a^{2}}{8 f \left (a -b \right )^{5} \sqrt {\left (a -b \right ) b}}+\frac {5 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right ) a}{f \left (a -b \right )^{5} \sqrt {\left (a -b \right ) b}}+\frac {b^{3} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \left (a -b \right )^{5} \sqrt {\left (a -b \right ) b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

________________________________________________________________________________________

mupad [B]  time = 16.33, size = 1536, normalized size = 5.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5/(a + b*tan(e + f*x)^2)^3,x)

[Out]

(b^(1/2)*atan(((a - b)^11*(2*tan(e/2 + (f*x)/2)^2*((b^(1/2)*(40*a*b + 15*a^2 + 8*b^2)*(640*a^3*b^12 - 128*a^2*
b^13 - 240*a^14*b + 400*a^4*b^11 - 11040*a^5*b^10 + 39120*a^6*b^9 - 73344*a^7*b^8 + 84000*a^8*b^7 - 58560*a^9*
b^6 + 20640*a^10*b^5 + 1280*a^11*b^4 - 4528*a^12*b^3 + 1760*a^13*b^2))/(16*a*(a - b)^(21/2)) - (b^(1/2)*(a - 2
*b)*(40*a*b + 15*a^2 + 8*b^2)^2*(128*a^18 - 2176*a^17*b + 256*a^2*b^16 - 3968*a^3*b^15 + 28800*a^4*b^14 - 1299
20*a^5*b^13 + 407680*a^6*b^12 - 943488*a^7*b^11 + 1665664*a^8*b^10 - 2288000*a^9*b^9 + 2471040*a^10*b^8 - 2104
960*a^11*b^7 + 1409408*a^12*b^6 - 733824*a^13*b^5 + 291200*a^14*b^4 - 85120*a^15*b^3 + 17280*a^16*b^2))/(512*a
*(a - b)^(33/2))) - (b^(1/2)*(a - 2*b)*(40*a*b + 15*a^2 + 8*b^2)^2*(1920*a^17*b - 128*a^18 + 128*a^3*b^15 - 19
20*a^4*b^14 + 13440*a^5*b^13 - 58240*a^6*b^12 + 174720*a^7*b^11 - 384384*a^8*b^10 + 640640*a^9*b^9 - 823680*a^
10*b^8 + 823680*a^11*b^7 - 640640*a^12*b^6 + 384384*a^13*b^5 - 174720*a^14*b^4 + 58240*a^15*b^3 - 13440*a^16*b
^2))/(256*a*(a - b)^(33/2))))/(225*a^16*b + 64*a^2*b^15 - 1680*a^4*b^13 + 3920*a^5*b^12 + 7665*a^6*b^11 - 5077
8*a^7*b^10 + 104685*a^8*b^9 - 111960*a^9*b^8 + 57330*a^10*b^7 + 2660*a^11*b^6 - 20286*a^12*b^5 + 9240*a^13*b^4
- 35*a^14*b^3 - 1050*a^15*b^2))*(40*a*b + 15*a^2 + 8*b^2))/(8*f*(a - b)^(11/2)) - ((607*a^3*b + 64*a^4 + 274*
a^2*b^2)/(60*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^14*(128*a*b^3 + 15*a^3
*b + 24*b^4 + 85*a^2*b^2))/(2*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^12*(9
36*a*b^3 - 365*a^3*b + 64*a^4 + 936*b^4 + 1075*a^2*b^2))/(6*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2
)) + (tan(e/2 + (f*x)/2)^10*(4268*a*b^3 + 921*a^3*b - 224*a^4 + 1872*b^4 - 1545*a^2*b^2))/(6*(a - b)*(a^4 - 4*
a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^4*(6224*a*b^3 - 671*a^3*b - 128*a^4 + 1832*b^4 + 597
3*a^2*b^2))/(30*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^6*(20696*a*b^3 + 86
7*a^3*b - 448*a^4 + 6280*b^4 - 935*a^2*b^2))/(30*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e
/2 + (f*x)/2)^8*(21740*a*b^3 - 4064*a^3*b + 1312*a^4 + 12560*b^4 + 1527*a^2*b^2))/(30*(a - b)*(a^4 - 4*a^3*b -
4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2)^2*(1036*a*b^3 + 447*a^3*b + 32*a^4 + 2265*a^2*b^2))/(30*(a
- b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (b*tan(e/2 + (f*x)/2)^16*(8*a*b^2 + 40*a^2*b + 15*a^3))/(4
*(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))/(f*(a^2*tan(e/2 + (f*x)/2)^18 + tan(e/2 + (f*x)/2)^4*(2
4*a*b - 4*a^2 + 16*b^2) + tan(e/2 + (f*x)/2)^14*(24*a*b - 4*a^2 + 16*b^2) + tan(e/2 + (f*x)/2)^6*(8*a*b - 4*a^
2 + 80*b^2) + tan(e/2 + (f*x)/2)^12*(8*a*b - 4*a^2 + 80*b^2) + tan(e/2 + (f*x)/2)^8*(6*a^2 - 40*a*b + 160*b^2)
+ tan(e/2 + (f*x)/2)^10*(6*a^2 - 40*a*b + 160*b^2) + a^2 + tan(e/2 + (f*x)/2)^2*(8*a*b + a^2) + tan(e/2 + (f*
x)/2)^16*(8*a*b + a^2)))

________________________________________________________________________________________

sympy [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(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

________________________________________________________________________________________