3.965 \(\int \sec ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=50 \[ \frac{a (2 A-B) \tan (c+d x)}{3 d}+\frac{(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)}{3 d} \]

[Out]

((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x]))/(3*d) + (a*(2*A - B)*Tan[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0678648, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2855, 3767, 8} \[ \frac{a (2 A-B) \tan (c+d x)}{3 d}+\frac{(A+B) \sec ^3(c+d x) (a \sin (c+d x)+a)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

((A + B)*Sec[c + d*x]^3*(a + a*Sin[c + d*x]))/(3*d) + (a*(2*A - B)*Tan[c + d*x])/(3*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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.587961, size = 97, normalized size = 1.94 \[ \frac{a \sec (c) (\sin (c+d x)+1) \sec ^3(c+d x) (-2 (A+B) \cos (c+d x)+A \sin (2 (c+d x))+4 A \cos (c+2 d x)+8 A \sin (d x)+B \sin (2 (c+d x))-2 B \cos (c+2 d x)+6 B \cos (c)-4 B \sin (d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(a*Sec[c]*Sec[c + d*x]^3*(1 + Sin[c + d*x])*(6*B*Cos[c] - 2*(A + B)*Cos[c + d*x] + 4*A*Cos[c + 2*d*x] - 2*B*Co
s[c + 2*d*x] + 8*A*Sin[d*x] - 4*B*Sin[d*x] + A*Sin[2*(c + d*x)] + B*Sin[2*(c + d*x)]))/(12*d)

________________________________________________________________________________________

Maple [A]  time = 0.084, size = 72, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{aA}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{aB \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-aA \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) +{\frac{aB}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.02611, size = 80, normalized size = 1.6 \begin{align*} \frac{B a \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + \frac{A a}{\cos \left (d x + c\right )^{3}} + \frac{B a}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.62345, size = 166, normalized size = 3.32 \begin{align*} -\frac{{\left (2 \, A - B\right )} a \cos \left (d x + c\right )^{2} +{\left (2 \, A - B\right )} a \sin \left (d x + c\right ) -{\left (A - 2 \, B\right )} a}{3 \,{\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.32178, size = 127, normalized size = 2.54 \begin{align*} -\frac{\frac{3 \,{\left (A a - B a\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{9 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, A a + B a}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(A*a - B*a)/(tan(1/2*d*x + 1/2*c) + 1) + (9*A*a*tan(1/2*d*x + 1/2*c)^2 + 3*B*a*tan(1/2*d*x + 1/2*c)^2
- 12*A*a*tan(1/2*d*x + 1/2*c) + 7*A*a + B*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d