\(\int (a+b \sec ^2(e+f x))^2 \sin ^4(e+f x) \, dx\) [22]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 114 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=\frac {1}{8} \left (3 a^2-24 a b+8 b^2\right ) x-\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

[Out]

1/8*(3*a^2-24*a*b+8*b^2)*x-1/8*a*(a-8*b)*cos(f*x+e)*sin(f*x+e)/f-1/4*(a^2-8*a*b+4*b^2)*tan(f*x+e)/f+1/4*a^2*si
n(f*x+e)^4*tan(f*x+e)/f+1/3*b^2*tan(f*x+e)^3/f

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4217, 474, 466, 1167, 209} \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=-\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {1}{8} x \left (3 a^2-24 a b+8 b^2\right )+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {a (a-8 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \]

[In]

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

[Out]

((3*a^2 - 24*a*b + 8*b^2)*x)/8 - (a*(a - 8*b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - ((a^2 - 8*a*b + 4*b^2)*Tan[e
+ f*x])/(4*f) + (a^2*Sin[e + f*x]^4*Tan[e + f*x])/(4*f) + (b^2*Tan[e + f*x]^3)/(3*f)

Rule 209

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

Rule 466

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[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 474

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

Rule 1167

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

Rule 4217

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 \left (a+b+b x^2\right )^2}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}-\frac {\text {Subst}\left (\int \frac {x^4 \left (5 a^2-4 (a+b)^2-4 b^2 x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = -\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {\text {Subst}\left (\int \frac {a (a-8 b)-2 a (a-8 b) x^2+8 b^2 x^4}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = -\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {\text {Subst}\left (\int \left (-2 \left (a^2-8 a b+4 b^2\right )+8 b^2 x^2+\frac {3 a^2-24 a b+8 b^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = -\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {b^2 \tan ^3(e+f x)}{3 f}+\frac {\left (3 a^2-24 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = \frac {1}{8} \left (3 a^2-24 a b+8 b^2\right ) x-\frac {a (a-8 b) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {\left (a^2-8 a b+4 b^2\right ) \tan (e+f x)}{4 f}+\frac {a^2 \sin ^4(e+f x) \tan (e+f x)}{4 f}+\frac {b^2 \tan ^3(e+f x)}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.34 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=\frac {\left (b+a \cos ^2(e+f x)\right )^2 \sec ^3(e+f x) \left (32 b^2 \sec (e) \sin (f x)+64 (3 a-2 b) b \cos ^2(e+f x) \sec (e) \sin (f x)+3 \cos ^3(e+f x) \left (4 \left (3 a^2-24 a b+8 b^2\right ) f x-8 a (a-2 b) \sin (2 (e+f x))+a^2 \sin (4 (e+f x))\right )+32 b^2 \cos (e+f x) \tan (e)\right )}{24 f (a+2 b+a \cos (2 (e+f x)))^2} \]

[In]

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

[Out]

((b + a*Cos[e + f*x]^2)^2*Sec[e + f*x]^3*(32*b^2*Sec[e]*Sin[f*x] + 64*(3*a - 2*b)*b*Cos[e + f*x]^2*Sec[e]*Sin[
f*x] + 3*Cos[e + f*x]^3*(4*(3*a^2 - 24*a*b + 8*b^2)*f*x - 8*a*(a - 2*b)*Sin[2*(e + f*x)] + a^2*Sin[4*(e + f*x)
]) + 32*b^2*Cos[e + f*x]*Tan[e]))/(24*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08

method result size
derivativedivides \(\frac {a^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+2 a b \left (\frac {\sin \left (f x +e \right )^{5}}{\cos \left (f x +e \right )}+\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )+b^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )}{f}\) \(123\)
default \(\frac {a^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+2 a b \left (\frac {\sin \left (f x +e \right )^{5}}{\cos \left (f x +e \right )}+\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )+b^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )}{f}\) \(123\)
parts \(\frac {a^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )}{f}+\frac {2 a b \left (\frac {\sin \left (f x +e \right )^{5}}{\cos \left (f x +e \right )}+\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )}{f}\) \(128\)
parallelrisch \(\frac {72 f x \left (a^{2}-8 a b +\frac {8}{3} b^{2}\right ) \cos \left (3 f x +3 e \right )+\left (-63 a^{2}+528 a b -256 b^{2}\right ) \sin \left (3 f x +3 e \right )+\left (-15 a^{2}+48 a b \right ) \sin \left (5 f x +5 e \right )+3 a^{2} \sin \left (7 f x +7 e \right )+216 f x \left (a^{2}-8 a b +\frac {8}{3} b^{2}\right ) \cos \left (f x +e \right )-45 a \left (a -\frac {32 b}{3}\right ) \sin \left (f x +e \right )}{192 f \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\) \(149\)
risch \(\frac {3 a^{2} x}{8}-3 x a b +x \,b^{2}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )} a^{2}}{64 f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a^{2}}{8 f}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a b}{4 f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a^{2}}{8 f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a b}{4 f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} a^{2}}{64 f}-\frac {4 i b \left (-3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+3 b \,{\mathrm e}^{4 i \left (f x +e \right )}-6 a \,{\mathrm e}^{2 i \left (f x +e \right )}+3 b \,{\mathrm e}^{2 i \left (f x +e \right )}-3 a +2 b \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\) \(199\)
norman \(\frac {\left (-\frac {3}{8} a^{2}+3 a b -b^{2}\right ) x +\left (-\frac {9}{8} a^{2}+9 a b -3 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+\left (-\frac {9}{8} a^{2}+9 a b -3 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}+\left (-\frac {3}{8} a^{2}+3 a b -b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (\frac {3}{8} a^{2}-3 a b +b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{12}+\left (\frac {3}{8} a^{2}-3 a b +b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{14}+\left (\frac {9}{8} a^{2}-9 a b +3 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (\frac {9}{8} a^{2}-9 a b +3 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\frac {\left (15 a^{2}+8 a b -24 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}+\frac {\left (3 a^{2}-24 a b +8 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {\left (3 a^{2}-24 a b +8 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{6 f}+\frac {\left (3 a^{2}-24 a b +8 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{6 f}+\frac {\left (3 a^{2}-24 a b +8 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{4 f}-\frac {\left (105 a^{2}-72 a b +152 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{12 f}-\frac {\left (105 a^{2}-72 a b +152 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{12 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4}}\) \(456\)

[In]

int((a+b*sec(f*x+e)^2)^2*sin(f*x+e)^4,x,method=_RETURNVERBOSE)

[Out]

1/f*(a^2*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)+2*a*b*(sin(f*x+e)^5/cos(f*x+e)+(sin(f*x
+e)^3+3/2*sin(f*x+e))*cos(f*x+e)-3/2*f*x-3/2*e)+b^2*(1/3*tan(f*x+e)^3-tan(f*x+e)+f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=\frac {3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} f x \cos \left (f x + e\right )^{3} + {\left (6 \, a^{2} \cos \left (f x + e\right )^{6} - 3 \, {\left (5 \, a^{2} - 8 \, a b\right )} \cos \left (f x + e\right )^{4} + 16 \, {\left (3 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, b^{2}\right )} \sin \left (f x + e\right )}{24 \, f \cos \left (f x + e\right )^{3}} \]

[In]

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

[Out]

1/24*(3*(3*a^2 - 24*a*b + 8*b^2)*f*x*cos(f*x + e)^3 + (6*a^2*cos(f*x + e)^6 - 3*(5*a^2 - 8*a*b)*cos(f*x + e)^4
 + 16*(3*a*b - 2*b^2)*cos(f*x + e)^2 + 8*b^2)*sin(f*x + e))/(f*cos(f*x + e)^3)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.05 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=\frac {8 \, b^{2} \tan \left (f x + e\right )^{3} + 3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + 24 \, {\left (2 \, a b - b^{2}\right )} \tan \left (f x + e\right ) - \frac {3 \, {\left ({\left (5 \, a^{2} - 8 \, a b\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{2} - 8 \, a b\right )} \tan \left (f x + e\right )\right )}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{24 \, f} \]

[In]

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

[Out]

1/24*(8*b^2*tan(f*x + e)^3 + 3*(3*a^2 - 24*a*b + 8*b^2)*(f*x + e) + 24*(2*a*b - b^2)*tan(f*x + e) - 3*((5*a^2
- 8*a*b)*tan(f*x + e)^3 + (3*a^2 - 8*a*b)*tan(f*x + e))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1))/f

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=\frac {8 \, b^{2} \tan \left (f x + e\right )^{3} + 48 \, a b \tan \left (f x + e\right ) - 24 \, b^{2} \tan \left (f x + e\right ) + 3 \, {\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} - \frac {3 \, {\left (5 \, a^{2} \tan \left (f x + e\right )^{3} - 8 \, a b \tan \left (f x + e\right )^{3} + 3 \, a^{2} \tan \left (f x + e\right ) - 8 \, a b \tan \left (f x + e\right )\right )}}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{24 \, f} \]

[In]

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

[Out]

1/24*(8*b^2*tan(f*x + e)^3 + 48*a*b*tan(f*x + e) - 24*b^2*tan(f*x + e) + 3*(3*a^2 - 24*a*b + 8*b^2)*(f*x + e)
- 3*(5*a^2*tan(f*x + e)^3 - 8*a*b*tan(f*x + e)^3 + 3*a^2*tan(f*x + e) - 8*a*b*tan(f*x + e))/(tan(f*x + e)^2 +
1)^2)/f

Mupad [B] (verification not implemented)

Time = 18.78 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin ^4(e+f x) \, dx=x\,\left (\frac {3\,a^2}{8}-3\,a\,b+b^2\right )+\frac {\left (a\,b-\frac {5\,a^2}{8}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (a\,b-\frac {3\,a^2}{8}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+2\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (3\,b^2-2\,b\,\left (a+b\right )\right )}{f} \]

[In]

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

[Out]

x*((3*a^2)/8 - 3*a*b + b^2) + (tan(e + f*x)*(a*b - (3*a^2)/8) + tan(e + f*x)^3*(a*b - (5*a^2)/8))/(f*(2*tan(e
+ f*x)^2 + tan(e + f*x)^4 + 1)) + (b^2*tan(e + f*x)^3)/(3*f) - (tan(e + f*x)*(3*b^2 - 2*b*(a + b)))/f