3.964 \(\int \sec ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=29 \[ \frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)}{d}-a B x \]

[Out]

-a*B*x+(A+B)*sec(d*x+c)*(a+a*sin(d*x+c))/d

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Rubi [A]  time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2855, 8} \[ \frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)}{d}-a B x \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

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

Rubi steps

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

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Mathematica [B]  time = 0.35, size = 85, normalized size = 2.93 \[ \frac {a \left (2 (A+B) \sin \left (\frac {d x}{2}\right )+B d x \sin \left (c+\frac {d x}{2}\right )-B d x \cos \left (\frac {d x}{2}\right )\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(a*(-(B*d*x*Cos[(d*x)/2]) + 2*(A + B)*Sin[(d*x)/2] + B*d*x*Sin[c + (d*x)/2]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c
+ d*x)/2] - Sin[(c + d*x)/2]))

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fricas [B]  time = 0.52, size = 73, normalized size = 2.52 \[ -\frac {B a d x - {\left (A + B\right )} a + {\left (B a d x - {\left (A + B\right )} a\right )} \cos \left (d x + c\right ) - {\left (B a d x + {\left (A + B\right )} a\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 36, normalized size = 1.24 \[ -\frac {{\left (d x + c\right )} B a + \frac {2 \, {\left (A a + B a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-((d*x + c)*B*a + 2*(A*a + B*a)/(tan(1/2*d*x + 1/2*c) - 1))/d

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maple [A]  time = 0.46, size = 54, normalized size = 1.86 \[ \frac {\frac {a A}{\cos \left (d x +c \right )}+a B \left (\tan \left (d x +c \right )-d x -c \right )+a A \tan \left (d x +c \right )+\frac {a B}{\cos \left (d x +c \right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.43, size = 56, normalized size = 1.93 \[ -\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} B a - A a \tan \left (d x + c\right ) - \frac {A a}{\cos \left (d x + c\right )} - \frac {B a}{\cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-((d*x + c - tan(d*x + c))*B*a - A*a*tan(d*x + c) - A*a/cos(d*x + c) - B*a/cos(d*x + c))/d

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mupad [B]  time = 9.17, size = 33, normalized size = 1.14 \[ -\frac {2\,A\,a+2\,B\,a}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}-B\,a\,x \]

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)^2,x)

[Out]

- (2*A*a + 2*B*a)/(d*(tan(c/2 + (d*x)/2) - 1)) - B*a*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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