3.30 \(\int \sin ^5(e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=70 \[ -\frac{(a-b) \cos ^5(e+f x)}{5 f}+\frac{(2 a-3 b) \cos ^3(e+f x)}{3 f}-\frac{(a-3 b) \cos (e+f x)}{f}+\frac{b \sec (e+f x)}{f} \]

[Out]

-(((a - 3*b)*Cos[e + f*x])/f) + ((2*a - 3*b)*Cos[e + f*x]^3)/(3*f) - ((a - b)*Cos[e + f*x]^5)/(5*f) + (b*Sec[e
+ f*x])/f

________________________________________________________________________________________

Rubi [A]  time = 0.0620982, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3664, 448} \[ -\frac{(a-b) \cos ^5(e+f x)}{5 f}+\frac{(2 a-3 b) \cos ^3(e+f x)}{3 f}-\frac{(a-3 b) \cos (e+f x)}{f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - 3*b)*Cos[e + f*x])/f) + ((2*a - 3*b)*Cos[e + f*x]^3)/(3*f) - ((a - b)*Cos[e + f*x]^5)/(5*f) + (b*Sec[e
+ f*x])/f

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]

Rule 448

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

Rubi steps

\begin{align*} \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (b+\frac{a-b}{x^6}+\frac{-2 a+3 b}{x^4}+\frac{a-3 b}{x^2}\right ) \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{(a-3 b) \cos (e+f x)}{f}+\frac{(2 a-3 b) \cos ^3(e+f x)}{3 f}-\frac{(a-b) \cos ^5(e+f x)}{5 f}+\frac{b \sec (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0625546, size = 104, normalized size = 1.49 \[ -\frac{5 a \cos (e+f x)}{8 f}+\frac{5 a \cos (3 (e+f x))}{48 f}-\frac{a \cos (5 (e+f x))}{80 f}+\frac{19 b \cos (e+f x)}{8 f}-\frac{3 b \cos (3 (e+f x))}{16 f}+\frac{b \cos (5 (e+f x))}{80 f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*Cos[e + f*x])/(8*f) + (19*b*Cos[e + f*x])/(8*f) + (5*a*Cos[3*(e + f*x)])/(48*f) - (3*b*Cos[3*(e + f*x)])
/(16*f) - (a*Cos[5*(e + f*x)])/(80*f) + (b*Cos[5*(e + f*x)])/(80*f) + (b*Sec[e + f*x])/f

________________________________________________________________________________________

Maple [A]  time = 0.105, size = 92, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( -{\frac{\cos \left ( fx+e \right ) a}{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+b \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{8}}{\cos \left ( fx+e \right ) }}+ \left ({\frac{16}{5}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{5}} \right ) \cos \left ( fx+e \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.05209, size = 84, normalized size = 1.2 \begin{align*} -\frac{3 \,{\left (a - b\right )} \cos \left (f x + e\right )^{5} - 5 \,{\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} + 15 \,{\left (a - 3 \, b\right )} \cos \left (f x + e\right ) - \frac{15 \, b}{\cos \left (f x + e\right )}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*(a - b)*cos(f*x + e)^5 - 5*(2*a - 3*b)*cos(f*x + e)^3 + 15*(a - 3*b)*cos(f*x + e) - 15*b/cos(f*x + e)
)/f

________________________________________________________________________________________

Fricas [A]  time = 1.99045, size = 161, normalized size = 2.3 \begin{align*} -\frac{3 \,{\left (a - b\right )} \cos \left (f x + e\right )^{6} - 5 \,{\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{4} + 15 \,{\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 15 \, b}{15 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*(a - b)*cos(f*x + e)^6 - 5*(2*a - 3*b)*cos(f*x + e)^4 + 15*(a - 3*b)*cos(f*x + e)^2 - 15*b)/(f*cos(f*
x + e))

________________________________________________________________________________________

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)**5*(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError