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

Optimal. Leaf size=117 \[ \frac{x \left (3 a^2+12 a b+8 b^2\right )}{8 a^3}-\frac{\sqrt{b} (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^3 f}-\frac{(5 a+4 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.168949, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{x \left (3 a^2+12 a b+8 b^2\right )}{8 a^3}-\frac{\sqrt{b} (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^3 f}-\frac{(5 a+4 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 2.13793, size = 303, normalized size = 2.59 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt{b (\cos (e)-i \sin (e))^4} \left (\sqrt{b} \sqrt{a+b} \left (a^2 \sin (4 (e+f x))-2 a^2 e+12 a^2 f x-8 a (a+b) \sin (2 (e+f x))+48 a b f x+32 b^2 f x\right )+a^2 (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )\right )+\sqrt{b} \left (34 a^2 b+3 a^3+64 a b^2+32 b^3\right ) (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )\right )}{64 a^3 \sqrt{b} f \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(Sqrt[b]*(3*a^3 + 34*a^2*b + 64*a*b^2 + 32*b^3)*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[b*(Cos[e] - I*Sin[e])^4]*(a^2*(3*a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/
Sqrt[a + b]] + Sqrt[b]*Sqrt[a + b]*(-2*a^2*e + 12*a^2*f*x + 48*a*b*f*x + 32*b^2*f*x - 8*a*(a + b)*Sin[2*(e + f
*x)] + a^2*Sin[4*(e + f*x)]))))/(64*a^3*Sqrt[b]*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e]
)^4])

________________________________________________________________________________________

Maple [B]  time = 0.097, size = 260, normalized size = 2.2 \begin{align*} -{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{2\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{5\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{3\,\tan \left ( fx+e \right ) }{8\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{\tan \left ( fx+e \right ) b}{2\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{2\,f{a}^{2}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{f{a}^{3}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) }{8\,fa}}-{\frac{b}{fa}\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}}}}-2\,{\frac{{b}^{2}}{f{a}^{2}\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) }-{\frac{{b}^{3}}{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}}}} \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),x)

[Out]

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

________________________________________________________________________________________

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.592449, size = 792, normalized size = 6.77 \begin{align*} \left [\frac{{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} f x + 2 \, \sqrt{-a b - b^{2}}{\left (a + b\right )} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) +{\left (2 \, a^{2} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}, \frac{{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} f x + 4 \, \sqrt{a b + b^{2}}{\left (a + b\right )} \arctan \left (\frac{{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt{a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) +{\left (2 \, a^{2} \cos \left (f x + e\right )^{3} -{\left (5 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, a^{3} f}\right ] \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),x, algorithm="fricas")

[Out]

[1/8*((3*a^2 + 12*a*b + 8*b^2)*f*x + 2*sqrt(-a*b - b^2)*(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*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) +
b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + (2*a^2*cos(f*x + e)^3 - (5*a^2 + 4*a*b)*cos(f*x + e
))*sin(f*x + e))/(a^3*f), 1/8*((3*a^2 + 12*a*b + 8*b^2)*f*x + 4*sqrt(a*b + b^2)*(a + b)*arctan(1/2*((a + 2*b)*
cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) + (2*a^2*cos(f*x + e)^3 - (5*a^2 + 4*a*b)*cos
(f*x + e))*sin(f*x + e))/(a^3*f)]

________________________________________________________________________________________

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),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.28923, size = 216, normalized size = 1.85 \begin{align*} \frac{\frac{{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )}{\left (f x + e\right )}}{a^{3}} - \frac{8 \,{\left (a^{2} b + 2 \, a b^{2} + 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^{3}} - \frac{5 \, a \tan \left (f x + e\right )^{3} + 4 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{2}}}{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),x, algorithm="giac")

[Out]

1/8*((3*a^2 + 12*a*b + 8*b^2)*(f*x + e)/a^3 - 8*(a^2*b + 2*a*b^2 + b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) +
arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^3) - (5*a*tan(f*x + e)^3 + 4*b*tan(f*x + e)^3 + 3*
a*tan(f*x + e) + 4*b*tan(f*x + e))/((tan(f*x + e)^2 + 1)^2*a^2))/f