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

Optimal. Leaf size=98 \[ \frac{(13 a-6 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}-\frac{(11 a-18 b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{5}{16} x (a-6 b)-\frac{a \sin (e+f x) \cos ^5(e+f x)}{6 f}+\frac{b \tan (e+f x)}{f} \]

[Out]

(5*(a - 6*b)*x)/16 - ((11*a - 18*b)*Cos[e + f*x]*Sin[e + f*x])/(16*f) + ((13*a - 6*b)*Cos[e + f*x]^3*Sin[e + f
*x])/(24*f) - (a*Cos[e + f*x]^5*Sin[e + f*x])/(6*f) + (b*Tan[e + f*x])/f

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Rubi [A]  time = 0.104163, antiderivative size = 98, normalized size of antiderivative = 1., 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 = {4132, 455, 1814, 1157, 388, 203} \[ \frac{(13 a-6 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}-\frac{(11 a-18 b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{5}{16} x (a-6 b)-\frac{a \sin (e+f x) \cos ^5(e+f x)}{6 f}+\frac{b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(a - 6*b)*x)/16 - ((11*a - 18*b)*Cos[e + f*x]*Sin[e + f*x])/(16*f) + ((13*a - 6*b)*Cos[e + f*x]^3*Sin[e + f
*x])/(24*f) - (a*Cos[e + f*x]^5*Sin[e + f*x])/(6*f) + (b*Tan[e + f*x])/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 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 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 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 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 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])

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right ) \sin ^6(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6 \left (a+b+b x^2\right )}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \cos ^5(e+f x) \sin (e+f x)}{6 f}-\frac{\operatorname{Subst}\left (\int \frac{-a+6 a x^2-6 a x^4-6 b x^6}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac{(13 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{a \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac{\operatorname{Subst}\left (\int \frac{-3 (3 a-2 b)+24 (a-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-18 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{(13 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{a \cos ^5(e+f x) \sin (e+f x)}{6 f}-\frac{\operatorname{Subst}\left (\int \frac{-3 (5 a-14 b)-48 b x^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=-\frac{(11 a-18 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{(13 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{a \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac{b \tan (e+f x)}{f}+\frac{(5 (a-6 b)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 f}\\ &=\frac{5}{16} (a-6 b) x-\frac{(11 a-18 b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac{(13 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{a \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac{b \tan (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.308186, size = 78, normalized size = 0.8 \[ \frac{(96 b-45 a) \sin (2 (e+f x))+(9 a-6 b) \sin (4 (e+f x))-a \sin (6 (e+f x))+60 a e+60 a f x+192 b \tan (e+f x)-360 b e-360 b f x}{192 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*e - 360*b*e + 60*a*f*x - 360*b*f*x + (-45*a + 96*b)*Sin[2*(e + f*x)] + (9*a - 6*b)*Sin[4*(e + f*x)] - a*
Sin[6*(e + f*x)] + 192*b*Tan[e + f*x])/(192*f)

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Maple [A]  time = 0.046, size = 112, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( a \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +b \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{7}}{\cos \left ( fx+e \right ) }}+ \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) \cos \left ( fx+e \right ) -{\frac{15\,fx}{8}}-{\frac{15\,e}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)*sin(f*x+e)^6,x)

[Out]

1/f*(a*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)+b*(sin(f*x+e)^7/cos(f
*x+e)+(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)-15/8*f*x-15/8*e))

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Maxima [A]  time = 1.4814, size = 150, normalized size = 1.53 \begin{align*} \frac{15 \,{\left (f x + e\right )}{\left (a - 6 \, b\right )} + 48 \, b \tan \left (f x + e\right ) - \frac{3 \,{\left (11 \, a - 18 \, b\right )} \tan \left (f x + e\right )^{5} + 8 \,{\left (5 \, a - 12 \, b\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (5 \, a - 14 \, b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*(f*x + e)*(a - 6*b) + 48*b*tan(f*x + e) - (3*(11*a - 18*b)*tan(f*x + e)^5 + 8*(5*a - 12*b)*tan(f*x +
e)^3 + 3*(5*a - 14*b)*tan(f*x + e))/(tan(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f*x + e)^2 + 1))/f

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Fricas [A]  time = 0.502159, size = 220, normalized size = 2.24 \begin{align*} \frac{15 \,{\left (a - 6 \, b\right )} f x \cos \left (f x + e\right ) -{\left (8 \, a \cos \left (f x + e\right )^{6} - 2 \,{\left (13 \, a - 6 \, b\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (11 \, a - 18 \, 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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*(a - 6*b)*f*x*cos(f*x + e) - (8*a*cos(f*x + e)^6 - 2*(13*a - 6*b)*cos(f*x + e)^4 + 3*(11*a - 18*b)*co
s(f*x + e)^2 - 48*b)*sin(f*x + e))/(f*cos(f*x + e))

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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((a+b*sec(f*x+e)**2)*sin(f*x+e)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.2429, size = 153, normalized size = 1.56 \begin{align*} \frac{15 \,{\left (f x + e\right )}{\left (a - 6 \, b\right )} + 48 \, b \tan \left (f x + e\right ) - \frac{33 \, a \tan \left (f x + e\right )^{5} - 54 \, b \tan \left (f x + e\right )^{5} + 40 \, a \tan \left (f x + e\right )^{3} - 96 \, b \tan \left (f x + e\right )^{3} + 15 \, a \tan \left (f x + e\right ) - 42 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*(f*x + e)*(a - 6*b) + 48*b*tan(f*x + e) - (33*a*tan(f*x + e)^5 - 54*b*tan(f*x + e)^5 + 40*a*tan(f*x +
e)^3 - 96*b*tan(f*x + e)^3 + 15*a*tan(f*x + e) - 42*b*tan(f*x + e))/(tan(f*x + e)^2 + 1)^3)/f