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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 8

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

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 ^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 = 0.63, 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)

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 69, normalized size = 1.38 \[ -\frac {{\left (2 \, A - B\right )} a \cos \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a \sin \left (d x + c\right ) - {\left (A - 2 \, B\right )} a}{3 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

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))

________________________________________________________________________________________

giac [B]  time = 0.19, size = 94, normalized size = 1.88 \[ -\frac {\frac {3 \, {\left (A a - B a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, A a + B a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.52, size = 72, normalized size = 1.44 \[ \frac {\frac {a A}{3 \cos \left (d x +c \right )^{3}}+\frac {a B \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a B}{3 \cos \left (d x +c \right )^{3}}}{d} \]

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 = 0.32, size = 59, normalized size = 1.18 \[ \frac {B a \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + \frac {A a}{\cos \left (d x + c\right )^{3}} + \frac {B a}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 9.25, size = 107, normalized size = 2.14 \[ \frac {\frac {2\,a\,\left (\frac {3\,B}{2}+A\,\cos \left (c+d\,x\right )+B\,\cos \left (c+d\,x\right )+2\,A\,\sin \left (c+d\,x\right )-B\,\sin \left (c+d\,x\right )+A\,\cos \left (2\,c+2\,d\,x\right )-\frac {B\,\cos \left (2\,c+2\,d\,x\right )}{2}\right )}{3}-\frac {4\,a\,\cos \left (c+d\,x\right )\,\left (\frac {A}{2}+\frac {B}{2}\right )}{3}}{d\,\left (2\,\cos \left (c+d\,x\right )-\sin \left (2\,c+2\,d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sin ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

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]

a*(Integral(A*sec(c + d*x)**4, x) + Integral(A*sin(c + d*x)*sec(c + d*x)**4, x) + Integral(B*sin(c + d*x)*sec(
c + d*x)**4, x) + Integral(B*sin(c + d*x)**2*sec(c + d*x)**4, x))

________________________________________________________________________________________