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

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Rubi [A]  time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, 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 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]

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 \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*}

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Mathematica [A]  time = 0.07, 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

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fricas [A]  time = 0.41, size = 64, normalized size = 0.91 \[ -\frac {3 \, {\left (a - b\right )} \cos \left (f x + e\right )^{6} - 5 \, {\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{4} + 15 \, {\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 15 \, b}{15 \, f \cos \left (f x + e\right )} \]

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))

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giac [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),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.67, size = 92, normalized size = 1.31 \[ \frac {-\frac {a \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+b \left (\frac {\sin ^{8}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\frac {16}{5}+\sin ^{6}\left (f x +e \right )+\frac {6 \left (\sin ^{4}\left (f x +e \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (f x +e \right )\right )}{5}\right ) \cos \left (f x +e \right )\right )}{f} \]

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)))

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maxima [A]  time = 0.72, size = 62, normalized size = 0.89 \[ -\frac {3 \, {\left (a - b\right )} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a - 3 \, b\right )} \cos \left (f x + e\right ) - \frac {15 \, b}{\cos \left (f x + e\right )}}{15 \, f} \]

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

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mupad [B]  time = 12.23, size = 92, normalized size = 1.31 \[ -\frac {\frac {5\,a}{16}-\frac {35\,b}{16}+\frac {25\,a\,\cos \left (2\,e+2\,f\,x\right )}{96}-\frac {11\,a\,\cos \left (4\,e+4\,f\,x\right )}{240}+\frac {a\,\cos \left (6\,e+6\,f\,x\right )}{160}-\frac {35\,b\,\cos \left (2\,e+2\,f\,x\right )}{32}+\frac {7\,b\,\cos \left (4\,e+4\,f\,x\right )}{80}-\frac {b\,\cos \left (6\,e+6\,f\,x\right )}{160}}{f\,\cos \left (e+f\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((5*a)/16 - (35*b)/16 + (25*a*cos(2*e + 2*f*x))/96 - (11*a*cos(4*e + 4*f*x))/240 + (a*cos(6*e + 6*f*x))/160 -
(35*b*cos(2*e + 2*f*x))/32 + (7*b*cos(4*e + 4*f*x))/80 - (b*cos(6*e + 6*f*x))/160)/(f*cos(e + f*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right ) \sin ^{5}{\left (e + f x \right )}\, dx \]

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]

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

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