∫ sin6(e + fx)(a + b tan2(e + fx)) dx

3.36 \(\int \sin ^6(e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=102 \[ -\frac {(a-b) \sin (e+f x) \cos ^5(e+f x)}{6 f}+\frac {(13 a-19 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}-\frac {(11 a-29 b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac {5}{16} x (a-7 b)+\frac {b \tan (e+f x)}{f} \]

[Out]

5/16*(a-7*b)*x-1/16*(11*a-29*b)*cos(f*x+e)*sin(f*x+e)/f+1/24*(13*a-19*b)*cos(f*x+e)^3*sin(f*x+e)/f-1/6*(a-b)*c

os(f*x+e)^5*sin(f*x+e)/f+b*tan(f*x+e)/f

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Rubi [A] time = 0.12, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3663, 455, 1814, 1157, 388, 203} \[ -\frac {(a-b) \sin (e+f x) \cos ^5(e+f x)}{6 f}+\frac {(13 a-19 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}-\frac {(11 a-29 b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac {5}{16} x (a-7 b)+\frac {b \tan (e+f x)}{f} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sin[e + f*x]^6*(a + b*Tan[e + f*x]^2),x]

[Out]

(5*(a - 7*b)*x)/16 - ((11*a - 29*b)*Cos[e + f*x]*Sin[e + f*x])/(16*f) + ((13*a - 19*b)*Cos[e + f*x]^3*Sin[e +

f*x])/(24*f) - ((a - b)*Cos[e + f*x]^5*Sin[e + f*x])/(6*f) + (b*Tan[e + f*x])/f

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*

(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{

a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 455

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ

uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,

x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*

ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N

eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P

olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a

*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS

um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[

{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2

)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \sin ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6 \left (a+b x^2\right )}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b) \cos ^5(e+f x) \sin (e+f x)}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {-a+b+6 (a-b) x^2-6 (a-b) x^4-6 b x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac {(13 a-19 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {(a-b) \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {-3 (3 a-5 b)+24 (a-2 b) x^2+24 b x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{24 f}\\ &=-\frac {(11 a-29 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(13 a-19 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {(a-b) \cos ^5(e+f x) \sin (e+f x)}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {-3 (5 a-19 b)-48 b x^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=-\frac {(11 a-29 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(13 a-19 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {(a-b) \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {b \tan (e+f x)}{f}+\frac {(5 (a-7 b)) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 f}\\ &=\frac {5}{16} (a-7 b) x-\frac {(11 a-29 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(13 a-19 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {(a-b) \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {b \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A] time = 0.36, size = 89, normalized size = 0.87 \[ \frac {(141 b-45 a) \sin (2 (e+f x))+3 (3 a-5 b) \sin (4 (e+f x))-a \sin (6 (e+f x))+60 a e+60 a f x+b \sin (6 (e+f x))+192 b \tan (e+f x)-420 b e-420 b f x}{192 f} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sin[e + f*x]^6*(a + b*Tan[e + f*x]^2),x]

[Out]

(60*a*e - 420*b*e + 60*a*f*x - 420*b*f*x + (-45*a + 141*b)*Sin[2*(e + f*x)] + 3*(3*a - 5*b)*Sin[4*(e + f*x)] -

a*Sin[6*(e + f*x)] + b*Sin[6*(e + f*x)] + 192*b*Tan[e + f*x])/(192*f)

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fricas [A] time = 0.60, size = 90, normalized size = 0.88 \[ \frac {15 \, {\left (a - 7 \, b\right )} f x \cos \left (f x + e\right ) - {\left (8 \, {\left (a - b\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (13 \, a - 19 \, b\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (11 \, a - 29 \, b\right )} \cos \left (f x + e\right )^{2} - 48 \, b\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6*(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

1/48*(15*(a - 7*b)*f*x*cos(f*x + e) - (8*(a - b)*cos(f*x + e)^6 - 2*(13*a - 19*b)*cos(f*x + e)^4 + 3*(11*a - 2

9*b)*cos(f*x + e)^2 - 48*b)*sin(f*x + e))/(f*cos(f*x + e))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^6*(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/