3.966 \(\int \sec ^6(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=73 \[ \frac{a (4 A-B) \tan ^3(c+d x)}{15 d}+\frac{a (4 A-B) \tan (c+d x)}{5 d}+\frac{(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)}{5 d} \]

[Out]

((A + B)*Sec[c + d*x]^5*(a + a*Sin[c + d*x]))/(5*d) + (a*(4*A - B)*Tan[c + d*x])/(5*d) + (a*(4*A - B)*Tan[c +
d*x]^3)/(15*d)

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Rubi [A]  time = 0.0729733, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2855, 3767} \[ \frac{a (4 A-B) \tan ^3(c+d x)}{15 d}+\frac{a (4 A-B) \tan (c+d x)}{5 d}+\frac{(A+B) \sec ^5(c+d x) (a \sin (c+d x)+a)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

((A + B)*Sec[c + d*x]^5*(a + a*Sin[c + d*x]))/(5*d) + (a*(4*A - B)*Tan[c + d*x])/(5*d) + (a*(4*A - B)*Tan[c +
d*x]^3)/(15*d)

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))}{5 d}+\frac{1}{5} (a (4 A-B)) \int \sec ^4(c+d x) \, dx\\ &=\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))}{5 d}-\frac{(a (4 A-B)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{(A+B) \sec ^5(c+d x) (a+a \sin (c+d x))}{5 d}+\frac{a (4 A-B) \tan (c+d x)}{5 d}+\frac{a (4 A-B) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [B]  time = 1.2376, size = 223, normalized size = 3.05 \[ \frac{a \sec (c) (-54 (A+B) \cos (c+d x)+18 A \sin (2 (c+d x))+9 A \sin (4 (c+d x))+128 A \sin (2 c+3 d x)-18 A \cos (3 (c+d x))+128 A \cos (c+2 d x)+64 A \cos (3 c+4 d x)+384 A \sin (d x)+18 B \sin (2 (c+d x))+9 B \sin (4 (c+d x))-32 B \sin (2 c+3 d x)-18 B \cos (3 (c+d x))-32 B \cos (c+2 d x)-16 B \cos (3 c+4 d x)+240 B \cos (c)-96 B \sin (d x))}{960 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(a*Sec[c]*(240*B*Cos[c] - 54*(A + B)*Cos[c + d*x] - 18*A*Cos[3*(c + d*x)] - 18*B*Cos[3*(c + d*x)] + 128*A*Cos[
c + 2*d*x] - 32*B*Cos[c + 2*d*x] + 64*A*Cos[3*c + 4*d*x] - 16*B*Cos[3*c + 4*d*x] + 384*A*Sin[d*x] - 96*B*Sin[d
*x] + 18*A*Sin[2*(c + d*x)] + 18*B*Sin[2*(c + d*x)] + 9*A*Sin[4*(c + d*x)] + 9*B*Sin[4*(c + d*x)] + 128*A*Sin[
2*c + 3*d*x] - 32*B*Sin[2*c + 3*d*x]))/(960*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])^3)

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Maple [A]  time = 0.099, size = 102, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{aA}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+aB \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) -aA \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) +{\frac{aB}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/5*a*A/cos(d*x+c)^5+a*B*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)-a*A*(-8/15-1/5*se
c(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/5*a*B/cos(d*x+c)^5)

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Maxima [A]  time = 1.04565, size = 116, normalized size = 1.59 \begin{align*} \frac{{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a +{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} B a + \frac{3 \, A a}{\cos \left (d x + c\right )^{5}} + \frac{3 \, B a}{\cos \left (d x + c\right )^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a + (3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*B*a
+ 3*A*a/cos(d*x + c)^5 + 3*B*a/cos(d*x + c)^5)/d

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Fricas [A]  time = 1.73992, size = 259, normalized size = 3.55 \begin{align*} -\frac{2 \,{\left (4 \, A - B\right )} a \cos \left (d x + c\right )^{4} -{\left (4 \, A - B\right )} a \cos \left (d x + c\right )^{2} -{\left (A - 4 \, B\right )} a +{\left (2 \,{\left (4 \, A - B\right )} a \cos \left (d x + c\right )^{2} +{\left (4 \, A - B\right )} a\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/15*(2*(4*A - B)*a*cos(d*x + c)^4 - (4*A - B)*a*cos(d*x + c)^2 - (A - 4*B)*a + (2*(4*A - B)*a*cos(d*x + c)^2
+ (4*A - B)*a)*sin(d*x + c))/(d*cos(d*x + c)^3*sin(d*x + c) - d*cos(d*x + c)^3)

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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(sec(d*x+c)**6*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.28708, size = 304, normalized size = 4.16 \begin{align*} -\frac{\frac{5 \,{\left (15 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 13 \, A a - 7 \, B a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}} + \frac{165 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 45 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 480 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 650 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 70 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 400 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 20 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 113 \, A a + 13 \, B a}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(5*(15*A*a*tan(1/2*d*x + 1/2*c)^2 - 9*B*a*tan(1/2*d*x + 1/2*c)^2 + 24*A*a*tan(1/2*d*x + 1/2*c) - 12*B*a
*tan(1/2*d*x + 1/2*c) + 13*A*a - 7*B*a)/(tan(1/2*d*x + 1/2*c) + 1)^3 + (165*A*a*tan(1/2*d*x + 1/2*c)^4 + 45*B*
a*tan(1/2*d*x + 1/2*c)^4 - 480*A*a*tan(1/2*d*x + 1/2*c)^3 - 60*B*a*tan(1/2*d*x + 1/2*c)^3 + 650*A*a*tan(1/2*d*
x + 1/2*c)^2 + 70*B*a*tan(1/2*d*x + 1/2*c)^2 - 400*A*a*tan(1/2*d*x + 1/2*c) - 20*B*a*tan(1/2*d*x + 1/2*c) + 11
3*A*a + 13*B*a)/(tan(1/2*d*x + 1/2*c) - 1)^5)/d